ORNL-1535.txt

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CENTRAL RESEARCH LIBRARY
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ORNL-1535
This document consists of 136 pages.
Copy ;‘57 of 208 copies. Series A.

Contract No. W-7405-eng-26

THERMODYNAMIC AND HEAT TRANSFER ANALYSIS
OF THE
ATIRCRAFT REACTOR EXPERIMENT

Bernard lubarsky
NATIONAL ADVISORY COMMITTEE FOR AERONAUTICS

B. L. Greenstreet
AIRCRAFT NUCLEAR PROPULSION DIVISION
OAK RIDGE NATIONAL LABORATORY

DATE ISSUED

AUG 10 1953

 

OAK RIDGE NATIONAL LABORATORY
Operated by
CARBIDE AND CARBON CHEMICALS COMPANY
A Division of Union Carbide and Carbon Corporation
Post 0Office Box P
0ak Ridge, Tennessee

        
 

   

MARTIN MARIETTA ENERGY SYSTEMS LIBRARIES

T B0

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ORNL 1535

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FOREWORD

The Aircraft Reactor Experiment utilizes the circulating fluoride-fuel as
the primary reactor coolant. It is necessary, however, to employ an additional
coolant whose primary function is to cool the reflector and pressure shell.
Although liquid sodium will be used as the reflector coolant, the use of NakK
had been assumed during the preceding year. Consequently, aconsiderable amount
of consistent, detailed data on the performance characteristics for the reactor
system using NaK as the reflector coolant has been assembled. These data are
presented in this report. A sufficient number of calculations based upon the
use of sodium has been made to assure that the performance with sodium will not
deviate significantly from that calculated for NaK. Sodium 1s being used 1n

preference to NaK because it can be more easily sealed at the pump shaft.

 

 

 

 
 

 

 
 

CONTENTS

INTRODUCTION .
REACTOR

Pressure Pulse Resultlng from a Sudden Change 1in React1v1ty
Core Power Distribution . ..
Temperature Profile 1in a Typlcal Sectlon of the Core Lattlce .
Flow of Reflector Coolant Through Core

Heat Removal by the Beflector Coolant. .

Pressure Drop in Fuel Manifolds and Core Tubes .

Temperature Gradients in Thermal Sleeves .

Formation of Nonplugging Solids 1n Fuel Tube 1in the
Event of a Small Leak

FUEL HEAT DISPOSAL SYSTEM ..

Performance of Fuel Heat Disposal System . .

Temperature in Fuel System Because of Afterheat 1in
the Event of Complete Pump Failure .

REFLECTOR COOLANT HEAT DISPOSAL SYSTEM .

ROD AND INSTRUMENT COOLING SYSTEM.

Cooling of the Control Rods and Instruments

Cooling of the Safety Rods

Cooling of the Regulating Rod

Cooling of Fission Chambers

Helium Pressure Drops

Division of Flow of Helium . . . .o .o
Performance of Rod and Instrument Coollng System Heat Exchanger

MONITORING AND PREHEAT SYSTEM

Heat Loss Through Insulation .

Reactor Preheating . . .

Helium Leakage Through Clearance Holes 1n the Reactor
Thermal Shield

Space Cooler Performance .. . .

Temperature Patterns 1n the Monltorlng Annulus in the Event
of Heat Failure

HELIUM SUPPLY AND VENTING SYSTEM .
True Holdup of Fission Gases 1n Tanks e e e
Temperatures in the Helium Vent Lines Containing Fission Gases

Vacuum Pump Performance

 

PAGE

11
13
17
19
33
38

41

43
43

49
54

64
64
65
66
68
73
76
76

79
79
81

 

87
90

92

103
103
107
111

Vi1l
 

DUMP AND FILL SYSTEM .
Fuel Dump Tank Cooling .
Heating of Fill Tank with Centrally Located D1p Tube

OTHER INVESTIGATIONS .
Afterheat in Fission Products .o .
Temperature Difference Between Thermocouple on plpe Wall

and Bulk Fluid .

REFERENCES .

viil

 

117
117
119

122
122

123
127
Chapter 1
INTRODUCTION

The Aircraft Reactor Experiment
(ARE) is an experimental, high-tempera-
ture, circulating-fuel reactor being
constructed by the Aircraft Nuclear
Propulsion Division of the Oak Ridge
National Laboratory. The fuel is a
mixture of fused fluorides, including
uranium tetrafluoride; the moderator
1s beryllium oxide; and the structural
material 1s Inconel. Figure 1 shows
a schematic diagram of the reactor,
the heat disposal equipment, and the
other process equipment necessary to
the operation of the reactor; also
shown in Fig. 1 are the design-point
pressures, temperatures, and flows at
various polnts 1in the system. This
report contalns a summary of the more
pertinent analytical investigations
of thermodynamic and heat transfer
properties of the ARE.

Considerably more investigations
have been carried out than have been
presented; the investigations omitted
fall in one of the following categories:
1. investigations relating to systems

and items of equipment not actually

used 1n the ARE,

2. 1nvestigations relating to systems
and 1tems of equipment used in the
ARE but with different fluids than
those actually used,

3. 1investigations in which estimated
fuel properties were used that
were found to be incorrect when
additional experimental determi-
nations of fuel properties were
made,

4. routine calculations of tempera-
tures and pressures at various
points of the system that were of
interest only to the detail
designer.

For convenience, the investigations
are presented as they relate to the
following subdivisions of the ARE:

Reactor

Fuel heat disposal system

Reflector coolant heat disposal system
Control rod cooling system

Preheating and monitoring system
Helium supply and vent system

Fi1ll and dump system

the
are

The physical properties of
materials used in the reactor
given i1n Table 1.
TABLE 1.

PROPERTIES OF MATERIALS

 

 

 

MATERTAL THERMAL CONDUCTIVITY VISCOSITY SPECIFIC HEAT DENSITY REFERENCE®
[Bew/bhr*£e? (°F/fe)) (1b/hr*ft) (Beu/1b*°F) (1b/£e>)
1. Air 0.0156 at 90.3°g 4.54.x10'§ at 90, 3°F 0.2399 at 90.3°F 0.0722 at 90.3°F 1
0.0180 at 190.3°F 5.15x 10 at 190.3°F 0.2409 at 190.3°F 0.0611 at 190.3°F
2. Aluminum 116 at 64°F 0.2220 at 32°F 168.5 at 68°F 2
119 at 212°F 0.2297 at 212°F
3. Beryllium oxide 0.84 at 1100°F 0.46 at 1100°F 142 3; 68°F (Porosity 3
0.73 at 1300°F 0.48 at 1300.F o
0.68 at 1500°F 0.50 at 1500°F 1770;t 68°F (Porosity
4. Copper 222 at 64°F 0.1008 at 30°F 555.0 at 68°F 2
220 at 212°F 0.1014 at 212°F
5. Fused fluorides 1.5 30.3 at 1150°F 0.26 187
21.8 at 1325°F
16.5 at 1500°F
6. Helium 0.0885 at 100°F 0.050 at 100°F 1.24 0.0098 at IOOZF 4 and 2
0.1250 at 550°F 0.075 at 550°F 0.0054 at 550°F
0.1650 at 1200°F 0.101 at 1200°F 0.0033 at 1200°F
7. Inconel 12.4 at 1200°F 0.109 at 77 to 212°F 530 5
13.1 at 1472°F
8. Insulation
{(cf., chap. 6)
9. NaK 14,10 at 212°F 0.496 at 752°F 0.210 at 752°F 48.4 at 752°F 6
15.38 at 752°F 0.353 at 1292°F 0.209 at 1112°F 46.1 at 1022°F
0.213 at 1472°F 43.9 at 1292°F
10. Stainless steel 10.4 at 300°g 0.1178 at 212§F 489 at 32°F 7 and 8
15.7 at 1500°F 0.1519 at 752°F
11. Steel 26 at 212°F 0.1178 at 212°F 489 at 32°F 8
21 at 1112°F 0.1519 at 752°F
12. Water 0.343 at 32°F 2.43 at 68°F 0.99947 at 68°F 62.36 at 68°F 9
0.363 at 100 °F 1.59 at 104°F 0.99869 at 104 °F 62.35 at 104°F
0.393 at 200°F 1.13 at 140°F 1.00007 at 140°F 62.26 at 140°F

 

 

 

 

 

 

*The references are

given at the

end of this report,
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Yt Low . VT VAPOR TRAP — < THROTTLING INTERNAL - 1513
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g. 1. Reactor and Heat Disposal Equipment.

 
Chapter 2
REACTOR

PRESSURE PULSE RESULTING FROM A
SUDDEN CHANGE IN REACTIVITY

The self-controlling features of
the ARE are due to the expansion of
the fuel with increasing temperature.
A transient increase in reactivity
produces an attendant increase 1in
temperature that increases the volume
of the fuel. Since the reactor has a
relatively fixed volume, some of the
fuel i1s forced out of the core and
the reactivity is reduced. The entire
process occurs 1n a very short time,
and therefore 1t 1s necessary to know
what pressures will be set up in the
fuel tubes when the fuel 1is rapidly
forced out of the core. Figure 2
shows the reactor and connecting piping
to the surge tanks. The fuel can
expand out of the reactor 1n either
direction, but since the path to the
surge tank in one direction (upstream)
is considerably shorter than in the
other, i1t will be assumed that all
the excess fuel takes the shorter path
to the surge tank (this is a con-
servative simplification). Further
assumptions that have been made are:

1. The fuel 1s incompressible.

2. The pressure pulse 1is of suffi-
ciently short duration that the fuel
in the external piping remains es-
sentially fixed in temperature.

3. The Ltemperature and power
variations 1n the reactor core will be
neglected; all the reactor fuel 1s
assumed to be at the mean reactor
temperature, and the power generation
in the fuel 1s always equal to the
mean power generation.

4. The effects of the pressure
pulse on the fuel pumps will be neg-
lected. Precisely what effect this
pulse would have on the pumps 1s
unknown, but the pulse will probably
be of sufficiently short duration that
no permanent damage will be done.

Figure 3¢ 1is a schematic diagram
of the system to be analyzed. There
are two, distinct cases to consider:

the reactor in which the temperature
and the density of the fuel change
with time and the external piping 1in
which the temperature and density of
the fuel are constant., The reactor
will be considered first (Fig. 3b).
The impulse-momentum equation for any
section 1s

(1) - F dt = d(mv)
where
F = differential force on fuel volume,
1b,
t * time, sec,
m = mass in fuel volume, slugs,
v = velocityof fuel volume, ft/sec.

For a fuel volume of cross section 4,
thickness dx, and position x,

(2) m = pA Ax

where

© = mass density, slugs/fta,

A = tube cross-sectional area, ft?,

x = distance along the tube measured

from the reactor outlet, ft,

Since the fuel is incompressible,
all the incremental fuel volume being
generated between zero and x by in-
creasing temperature must pass acCcross
x (on the way to the reactor inlet and
thence to the surge tank) as rapidly
as the i1ncremental volume 1s being
generated. The rate at which incre-
mental volume is being generated
between zero and x 1is

(3) 4V BA 40
dt 7 de
where
V = volume from x = 0 to any arbi-

trary x, ft3,
coefficient of volumetric ex-

™
E

pansion, per 'F,
8 = fuel temperature, °F,

Since the rate at which the incremental
volume is being generated must be equal
to the rate at which the volume 1s
crossing x,

 

 
(4) vA = [BA df
VR TR

df

v o= v - v, = Bx—
o = B P

where v is the incremental velocity
(v - v,) 1n ft/sec and the subscript
0 refers to conditions at time zero.

The impulse-momentum equation,
Eq. 1, may be rewritten for the volume
A Dx as

dv dm

(5) AF = m— + oy —
dt dt

Differentiating Egqs. 2 and 4 gives

 

 

d d
(6) ‘—m = A Ax P
dt dt
and
dv  dv d?5
(7) — T — =
dt dt P dt? '’

and substituting Egs. 2, 4, 6, and 7
in Eq. 5 gives

 

d26
(8) -=AF = pA Ax [(Bx
dt?
d d
+ <Fx-*— + v0> A Ay —§

or

AF d%6 dp do dp
—— = A /61' o — + — + ‘Uo — .
Ox dt? dt dt dt
Dividing both sides by A and letting

Ax approach zero gives

(9) -— — = - — = Bx(p —

1 dF dP d?g
A dx dx dit?

de d
+-—E ? + v ap
dt dt 0 dt '

The

may be expressed 1in
terms of the temperature:

where P is pressure in lb/ft?,
mass density, P,

Fo
1+ 56 - 5,)
Differentiating Eq. 10 gives

(10) o =

do
P48 17

 

 

d
(11) ;fi . -
t [1+8(8 - 8,))
Substituting Egs. 10 and 11 in Eq. 9 -
gives
4?0 -

d o dt?

0 t
(12) - — = BaL———

dx 1+ 80 - 6,)

dg\?
pofi (;;)

(1 + 88 - 6,))2

 

d
voPol T

 

[1 + B8(8 - 6,)]2

Integrating Eq, 12 from x = 0 to x = L
and rearranging slightly gives

fipoL2 d?o
- P, =
2(1 -89, +80) | 42

do\? ve dB
85) 7
dt L dt

 

(13) P,

1 - 5o, + B0
where the subscript m refers to condi-
tions at x = 0 and the subscript L
refers toconditions at x = L. Equation

13 with L set equal to the length of
the reactor tubes gives the variations
with time of the pressure differential
across the reactor core,

For the external piping (Fig. 3c),
where the temperature of the fuel is
assumed constant, Eqs. 1, 2, and 5 are
valid. However, Eq. 3 1is not valid,
because no incremental volume is gener-
ated 1n the external piping. The -
external piping, however, must pass
all the incremental volume generated

 

 
  
 
 

 

   
 

 

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Im! 19098
MANIFOLD, 2 IN PARALLEL

(EQUIVALENT TO &ft OF 2-in. IPS SURGE
PLUS 22 ft OF 2-in. IPS) PUMP  _ TANK

—O{ ]

9 ft OF 1%-in-0D 1PS
2 IN PARALLEL

 
 
  

 

 

 

REACTOR
39 ft OF 1.235-in-0D x
0.0680-in-WALL IPS PUMP SURGE
6 IN PARALLEL TANK

 

 

 

(a)

N XN—— -

 

 

 

 

 

 

 

 

 

 

 

=

X=0 {c)

Fig. 3. Reactor and Connecting
Piping to Surge Tank. (a) Schematic
diagramof system. (b) Internal reactor
piping. (c¢) External piping.

in the core, and therefore Eq. 4 may
be rewritten for the external piping:
dg
dt '
BV  dé
7 o=y -yl = e— —

° A" dt
where V_ is the fuel volume in the
reactor core in ft® and the prime refers
to values 1n the external piping.
Equation 6, of course, becomes

dm

(15) =0,
dt

(14) v'A" = BV

and Eq. 7 becomes

dv' BV, d%0

4t A’ di?
Substituting Egqs. 2, 14, 15, and 16
in Eq. 5 gives

(16)

pv, d*6
(17) AF' = p'A' Dy —
A' dt?

 

Since p' is constant and equal to o,
Eq. 17 may be rewritten as

( ) 1 AFI AP: pO,BVC d29
18) - — —— - -~ - ___° ___
A" Ax Ax A' g2
or, by letting Ax approach zero,
dPl pO’BVc dza

 

 

dx A dt?
Integrating Eq. 18 from x = 0 to
x = L' gives

p BV L' d?0

(19) P -PL: = ——
n ' 2
A dt

Equation 19 with L' set equal to the
length of a runof external piping gives
the variation with time of the pressure
differential across the piping.

In addition to the pressure dif-
ferential set up by the inertia of
the fuel (Eqs. 13 and 19), there are
pressure differentials resulting from
the change in friction pressure drop
assocliated with change in velocity.
These pressure differentials may be
evaluated by methods similar to those
used above, except that the impulse-
momentum equation (Egs. 1 and 5) 1is
replaced by the customary Fanning
equation for the friction pressure

 

drop:
de Af pv? - povg
(20) — S T,
dx D 9
where
Pf = change in friction pressure drop,
1b/fe?,
f = friction factor,

D = tube diameter, ft.

Substituting Eqs. 4 and 10 for the
reactor core into Eq. 20 gives

do 2
dP, g5 |P0 Pr ot v .
(21) 2 =L = L,

dx D 1+B(9—90)
or

do\?
2_2 (&Y +
Py 2py |7 (dt>

 

 

dx D

Integrating with respect to x from x =

1+8(6 - 4,)

0 to x = L gives

d6\? d
a + 2, 24 2
<dt> LT 2t Vo

 

calculations is longer than the actual
length to account for bends, exits,
entrances, etc,, the change in friction
pressure drop in the core calculated
by using Eq. 22 must be corrected:

(23) [Py, - (P,),]

corrected

 

- —1
- L [(Pf)m— (Pf)L} 3

where Le 1s the equivalent length
for the friction pressure drop, ft.
The change in friction pressure
drop in the external piping is calcu-
lated in the following manner. Equation
20 is valid, and substituting Eq. 14 in
Eqv 20 and noting that p' = 0, glves

 

 

v do\?
dP} 2f’p0 ¢ Jt
(24) =
dx D' -
28V v d9
V. —
¢ O d
+ .
Al
Integrating from x = 0 to x = L' gives
(25) (Pf)"l - (Pf)L
dO\? do
2fp, L' |(PYe T, Ve 77
= + 2 __"
D A o Ty
10

/32L3

(22) ( 100 | 3
Since the equivalent length of
piping for friction pressure drop

- USL .
1+ 86 -4,)

 

 

and
(26) [(P})m - (P_If)L]corrected
_ eq : '
- [Py, = (Pp) I

Hence, 1f the variations of &,
df/dt, and d?9/dt* with time are known, "
the total pressure pulse can be calcu-
lated because all other quantities are
known. The variation of the tempera-
ture and its first two derivatives with
respect to time, as a result of a
0.5% step increase in reactivity, were
evaluated on an electronic analog

computer. The results are shown 1in
Fig. 4. The other constants required
for the evaluation of the pressure

pulse are:
= 0,0407 ft? (6 tubes in parallel,
1.255 in. ODby 0,060-in. wall),
8 x 10-°

I
|

T
il

per °F,

vy = -3.70 ft/sec,
Py = 5.81 slugs/ft?,
d, = 1325°F,
L = 42 f¢,
A" = 0,0233 ft? (1 pipe, 2 in, IPS),

0.0283 ft? (2 pipes in parallel,
11/2 in. IPS),
vy = =6.45, -5.31 ft/sec,
1,71 ft?,
28, 9 ft,
= 0.0929 f¢,
0.0075,
52 ft
= 0,1721, 0.1341 f¢,
0.0058, 0.0062,
= 43, 15 ft,

 
300
280
260
240
220
200
180
160
140
1380 —— 120
1370 +—

100

1360 — 80

8, RATE OF TEMPERATURE RISE (°F/sec)

1350 — 60

1340}— 40

1330 — 20

8, TEMPERATURE RISE {°F)

 

1320 — 0

| ]|6 scaLel=[8 scaLe]

o 02 04 06
TIME {sec)

0 0.050 0100 0150 0.200

 

DWG. 19300

2800

2600

2400

2200

2000

1800

1600

1400

1200

1000

800

600

400

200

-200

-400

4, RATE OF CHANGE OF TEMPERATURE RISE (°F/ seca)

-600 -

-800

-1000

-1200

-1400

-1600

-{800

0.250 0.300 0350 0400 0.450 0.500

TIME (sec)

Fig. 4. Temperature Rise, Rate of Temperature Rise, and Rate of Change of

Temperature Rise vs. Time, Ak/k = 0.005.

The final results of the evaluation
of Egqs. 13, 19, 23, and 26 are shown
in Fig. 5, where the 1incremental
pressure at the reactor outlet (position
of maximum incremental pressure) 1is
plotted against time. The maximum

incremental pressure encountered 1s

Data from ARE simulator.

about 27 psi, and it occurs about
100 msec after the introduction of the
step increase 1n reactivity.

CORE POWER DISTRIBUTION

The power distribution in the ARE
reactor was estimated theoretically,

11

 

 

 
 

30

s |

 

 

2C

 

 

TOTAL

 

 

MOMENTUM
 EXTERNAL TO
REAGTOR-

 

 

 

 

FRICTION EXTERNAL
TO REACTOR

 
  
 

 

PRESSURE INCREMENT (psi)
3

FRICTION IN REACTOR |

 

 

 

 

 

 

 

 

 

 

MOMENTUM
_5 IN REACTOR
¥
-10
N%
-20 | |
0 04 0.2 0.3 0.4 0.5
TIME (sec)
Fig. 5. Incremental Pressure at

Reactor Outlet as a Result of a Ak/k
of 0.005 vs. Time.

and the estimate was checked experi-
mentally by the physics groups of the
Laboratory, The power distributions
used 1in the calculations in this
report are presented in Figs, 6, 7, and
8. Figure 6 shows the variation of
fission heat generated in the fuel with
distance from reactor center; Fig, 7
shows the variation of neutron heating
of the moderator; Fig. 8 shows the
variation of gamma heating (average,
based on the assumption that the
reactor 1s homogenized). The maximum
of the
neutron, and gamma heating are given

in Table 2.

and average values fission,

12

 

 

t.3

DWG 12100
Pl [T |
| ! : ) L | ;

12 o ; — ——t- —
| SEE TABLE 2 FOR VALUES OF &, AND &, b
2 S LT ol AL S

4 g e e e

10 e ,,44;77:7 — 'i___‘ —— T i ‘ ._.1__i__. J“?ii
_‘\‘\\ | | ]: : ] | ' : ! -‘

 

o9l - L]

 

 

 

 

 

 

 

 

 

207} —

 

 

   

Lo b ; NG g
C 1 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 {7 18 {9
DISTANCE FROM CENTER (in)

Fig. 6. Fission Heat Generated in
Fuel as a Function of Distance from
Reactor Center.

 

Ve — — ‘

i | i i : . i : .
mEJU;—~4fi+flA+fl$\-L~w‘wiw b
o SEE TABLE 2 FOR VALUES OF A, AND A, |

   
  

 

 

 

 

 

 

 

 

 

: | | . : ] l ¢ | L ;
O12345678910111213141516171819
DISTANCE FROM CENTER {in.)
Fig. 7. Neutron Heatingof Moderator

as a Function of Distance from Reactor
Center.

 

   

 

 

10 T
o -
08— -
[ .
E e
(] : : i : i " :
£ 06— - -——1 Y < o
2 —
5 ‘ : ‘ ‘
% O 4! I T T sy e "r i P T D ‘—‘—:
K : | | i
g SEE TABLE % FOR VA:LUES OF K AND Ky et o f%:
02~ - TL e b4 . —
L w A
) + i e : {
0 2 4 & 8 10 12 14 15 18
DISTANCE FROM CENTER (in)
Fig. 8. Gamma Heating as a Function

of Distance from Reactor Center.

 
TABLE 2.

MAXIMUM AND AVERAGE VALUES OF FISSION,

NEUTRON, AND GAMMA HEATING IN THE ARE

 

 

 

AXIAL RADIAL
MAXTIMUM AVERAGE
TYPE OF HEATING HEATING HEATING MAXIMUM- MAXIMUM-
(kw/in.>) (kw/in.?) TO-AVERAGE TO-AVERAGE
’ ' RATIO, Ka RATIO, Kr
Fission (in fuel only) 2.78 1.22 1.55 1.48
Neutron (in moderator
only) 0.012 0.0052 1.55 1.66
Gamma (average for
homogenized reactor) 0.024 0,011 1,32 1.82

 

 

 

 

 

TEMPERATURE PROFILE IN A TYPICAL
SECTION OF THE CORE LATTICE

Heat is removed from the moderator
region of the reactor core by the fuel
(Fig. 9).
is transferred across the outer surface
of a column of beryllium oxide blocks,
the heat removed by the fuel flowing
in the tube through this column would
be the heat generated within the fuel,
the metal of the tube, the NaK filling
the gap between the tube and the
column of blocks, and the blocks.
For this hypothetical case, the tempera-
ture profile existing between the
center line of the fuel tube and the
outer periphery of the blocks can be
calculated.

In a case in which no heat

In the actual case, heat 1s trans-
ferred i1nto the through
the “boundary’ between the moderator
and the reflector, and through the
upper and lower surfaces of the
moderator region. Thus, the calculated
temperature profile has no meaning
other than to show the maximum tempera-
ture gradients that can exist in the

rod holes,

components of the moderator region.
For the hypothetical case, the maximum
temperature gradients will exist at
the point of maximum heat generation.

There fore, for the temperature profile

calculations, heat generation rates
corresponding to the maximum values
expected 1n the reactor

used.

core were

If the heat generated in a hollow
cylinder is transferred through the
inner surface, the temperature gradient
between the outer and inner surfaces
of the cylinder 1s given by

@1y o7 = = L e
5% | 2 re re re nri ,
where
AT = temperature difference, °F,
G = heat generation rate,
Btu/sec'fta,
k = thermal conductivity,
Btu/sec* ft? (°F/ft),
r. = inner radius, ft,
r = outer radius, ft.

o
On the other hand, the temperature
gradient necessary for the transfer
of heat across the hollow cylinder
1s

Q

-

 

(28) AT =

Q, = heat transferred per foot of
length, Btu/sec,

7 = thickness (ro - ri), ft,
A = mean area per foot of length,
fo?,

13
THERMOCOUPLE
LAYOUT

CORE ASSEMBLY

REFLECTOR CODLANT
TUBES

PRESSURE SHELL

BOTTOM HEADER

FUEL TUBES

D!! !336A

REGULATING ROD
ASSEMBLY ~_

   
 
 
  
   
 
   
  
   
    
 
   
  
   

——TUBE EXTENSICN
__——— THERMAL SHIELD AP

-

- P THERMAL SHIELD TOP

—-FUEL INLET MANIFOLD
/

SAFETY ROD
GUIDE SLEEVE

SAFETY ROD ASSEMBLY

TOP HEADER

TOP TUBE SHEET |

HEATERS

BeQ MODERATOR
AND REFLECTOR

A THERMAL SHIELD
5 ASSEMBLY
5
1
i
1

1

e 80V in. REF -

BOTTOM TUBE |
" shEET 1
_ —sTUD
-SUPPORT ASSEMBLY

T FUEL QUTLET
: MANIFOLD

- B—— e

T~ _THERMAL SHIELD

   

THERMAL SHIELD CAP BOTTOM
]
o
‘‘‘‘‘‘ <
2
M)
o
\ .
\ \ -
\SUPPORT ASSEMBLY HELIUM MANIFOLD QMG
SCALE IN INCHES
ASSEMBLY SEQUENCE
STEP 1: ASSEMBLE REACTOR ASSEMBLY TO HELIUM MANIFOLD TUBES. STEP 10: TEST PER TESTING SCHEDULE,
STEP 2: SET THERMAL SHIELD GAP AND THERMAL SHIELD BOTTOM N STEP f1. INSTALL HEATERS, CONNECT AND TEST.
PLACE AND SLIP TUBES THROUGH. STEP 11a: INSTALL THERMOCOUPLE LAYOUT.
STEP 3: INSULATE MANIFOLD TOP AND BRAZE TO TUSES. STEP 12 ASSEMBLE THERMAL SHIELD BOTTOM.
STEP 3a © INSERT SPRING AND SHOCK TUBE. STEP 13: ASSEMBLE BALANCE OF THERMAL SHIELD ASSEMBLY , THERMAL
STEP 3b: ASSEMBLE ANO ADJUST ORIFICES SHIELD TOP AND THERMAL SHIELD CAP AND WELD CLOSED.
STEP 3¢ : COMPLETE MANIFOLD, STEP 14: SEAL ALL JOINTS OF ASSEMBLY OF THERMAL SHIELD ASSEMBLY ,
STEP 4: ASSEMBLE REACTOR ASSEMBLY WITH SUPPORT ASSEMALY, THERMAL SHIELD CAP, THERMAL SHIELD BOTTOM, THERMAL SHIELD
STEP 5: ASSEMBLE STUD WITH REACTOR ASSEMBLY AND SUPPORT TOP AND THERMAL SHIELD CAP WITH INSULATING CEMENT, J-M
ASSEMBLY . NO. 400 OR FQUIVALENT.
STEP 6: ASSEMBLE THERMAL SHIELD CAP WITH HELIUM MANIFOLD, STEP 15: INSULATE HELIUM MANIFOLD WITH 2-in. LAYER J-M SPONGE FELTED
STEP 7 ASSEMBLE THERMAL SHIELD CAP LOOSELY WITH COCLANT LINE INSULATICN, SEAL ALL JOINTS WITH INSULATING CEMENT, J-M
ANNULUS OF REACTOR ASSEMBLY. NO. 400 OR EQUIVALENT.
STEP B: WELD TUBE EXTENSION TQ SAFETY ROD GUIDE SLEEVE AND STEP 16: SEE OPERATING PROCEDURE.
INSTRUMENT GUIDE SLEEVE , 6 REQ'D.
STEP 9: CONNECT ALL PIPING,

14

Fig. 9. Vertical Cross Section of Reactor.
 

when heat 1s both trans-

There fore,
ferred across and generated within a
hollow cylinder,

Qt G 1

 

(29) AT =

e
o
+ r? 1n —
0 r.
1
The temperature at any radius in a

hollow cylinder
internal heat generation and the heat

in which there 1is

is transferred through the 1inner
surface 1s given by
G |1
(30) T = — |= (r? - r?)
2k |2 :
9 r
+ r® ln —| + T, |
o r. l
t
where
T = temperature at any radius r, °F,
r = any radius within the cylinder,

ft,
Ti-=?temperature at r, °F,
The heat gernerated per foot of length
of the cylinder can be calculated from

(31) Q= Gr(r? - r2)

r

where Q 1s the heat generated per foot
of length, Btu/sec.

The calculations of temperatures
and temperature differences were made
from the following data for beryllium
oxide, NaK, and the metal tube.

Beryllium Oxtde Block. The beryllium
oxi1de block 1s hexagonal in cross
section (3.73 in. across flats). For
simplicity in calculation, the block
was assumed to be a circular cylinder
of equivalent cross-sectional area,
with an outside radius of 1.960 1in,
and an inside radius of 0,630 inch,
The temperature difference between the
outer and inner surfaces is due entirely

tointernal heat generation. Therefore,

G |1 T
S B - 2 2 2
AT = 5; 9 (ri - ro) tor: ln‘:f .

i

 

NaK. The outer and inner radii of
the NaK annulus, measured from the
center of the cylinder, are 0.630 and
0,618 in,, respectively.,
NaK 1s nearly stagnant, the tempera-
ture gradient through the NaK 1is

Since the

Q ¢ |1 r,
AT = —— +— | = (r2-r2)+r? In
k 2k | 2 ' ° ° r.

L

 

T m

where . is the heat generated 1in the
beryllium oxide.

Metal Tube. The outside radius of
the metal tube 1s 0.618 in., and the
inside radius is 0.558 1inch., The

expression for the temperature gradient
in the tube 1s the same as that for
NaK, except that Qt is the heat gener-

ated in the beryllium oxide plus the
heat generated in the Nak,

The amount of heat generation 1in
the fuel 1s proportional to the length
of time the fuel resides in the reactor
core. Thus high-velocity fuel will
have less heat generated within 1t
than fuel flowing at a low velocity
and, correspondingly, 1ts temperature
will be lower. Upon looking at the
velocity profile of fuel flowing in a
fuel tube, 1t 1s seen that the maximum
velocity 1s at the center and that the
minimum velocity is at the tube wall.
From this, 1t that a
temperature gradientwill exist between
the outer surface and the center of
the fuel. This temperature gradient
can be determined, and the expression
(r, - T_,.)/Gr?, in which T  is the
temperature of the outer surface of
the fuel, °F, and T ., 1s the tempera-
ture of the fuel at the center of the
tube, °F, has plotted as a
function of Reynolds’ and Prandtl’s
(10)

can be seen

been

moduli by Poppendiek and Palmer,
The fuel also has a temperature gradient
between the center and outer surface,
which 1s given by

k
= — 0. 0.
b= 0.023 = NS-® N34

15

 
where

h = coefficient of heat transfer by
convection, Btu/sec*ft?«°F,

A = area of the inside surface of
the metal tube foot of
length, ftz,

D = hydraulic diameter of the fuel

per

tube, ft,
Ny = Reynolds’ modulus,
Np, = Prandtl’s modulus,
Q, = heat generated in the beryllium

oxide plus the heat generated
in the NaK plus the heat gener-
ated in the metal tube,

From this it is seen that the total

temperature gradient in the fuel 1is
t
AT = — + AT (resulting from internal
hA heat generation).
For the fuel, the physical charac-

teristics are:

Radius, ry 0.558 in.
Temperature, T 1170°F
Viscosity, # 8.27 x 10~3
1lb/sec* ft
Density, £ 187 1b/t?
1400

TUBE WALL

1360

0.26 Beu/1b*°F
4,17 x 10°*
Btu/sec'ft2
(°F/fe)

The following are the maximum
expected heat generation rates for
the various components:

Specific heat, €p

Thermal conductivity, k

COMPONENT ¢
(Btu/seceft”)
BeO 62.21
NaK 41,47
Metal tube 41, 47
Fuel 4838
The results of the calculations of
temperature differences for the
various components are:
COMPONENT AT (°F)
Fuel 81.08
Metal tube 22,97
NaK 3.08
BeQ 108,00
Total 215.13
Thus the maximum temperature 1is
1170 + 215 = 1385°F. A plot of

distance from fuel
tube center is shown on Fig. 10,

temperature vs,

DWG, 19103

 

1320
©
o
Ld
g
g 1280
jaed
L
a
=
58]
—

1240

1200

1160

0 1.0 20
RADIUS (in.)
Fig. 10, Temperature Profile in Core Lattice.

16

 
FLOW OF REFLECTOR COOLANT THROUGH CORE

In the ARE, the pressure shell, the
moderator, and the thimbles of the
control rod and instrument holes are
cooled by the circulation of NaK, The
NaK circuit (reflector coolant cir-
cuit) is shown in Fig. 1.

The coolant enters the pressure
shell at the bottom, flows through
the lattice, and leaves at the top
(Fig. 9). The flow through the core
lattice 1s divided among the annulus
between the outer periphery of the
reflector and moderator can and the
pressure shell, the tubes leading
through the reflector blocks, and the
annuli around the rod and instrument
holes. For adequate cooling of the
moderator and pressure shell, it 1is
necessary to have only a minimum amount
flowing in the
around the rod and instrument holes.
The additional heat that must be
removed from the thimbles will be
carried away by the helium that cools
the rods and instruments. To obtain
this minimum flow, orifice plates
(Fig. 11) were placed in the annuli.

of coolant annuli

There will be a pressure drop in
the fluid when it enters the pressure
shell because of expansion, and when
it leaves there will be a further
pressure drop because of contraction.
In the spaces between the tube sheets
and the pressure shell above and below
lattice, which will act as
plenum chambers, the coolant velocity
will be effectively Thus the
pressure drop in these regions will

the core
ZeT O.
be very small. The pressure drop in
the coolant flowing through the core
lattice will be due to contraction,
friction, and expansion. Since the
pressure drop of the fluid inside the
pressure shell 1is due almost entirely
to that i1n the core lattice, the flow
through the various paths will be
directly proportional to the respective
resistances of the paths,

The loss in pressure by contraction
is given by

 

(32) AP = k

where k_ is a constant given by McAdams

(cf., p. 122 of ref. 11); the expansion
pressure drop 1s given by
plvy - vy)?
(33) AP
e 2g

and the friction pressure drop for
turbulent flow 1n either a tube or an
annulus 1s given by

 

(34) o, = apt 2
f 7 p 2
The equation for the friction factor 1is
0.046
(35) f=—
0.2
NR

The meanings and units of the symbols

32 through 35 are:

used 1n Egs.

AP = pressure drop, lb/ft?,
f = friction factor, dimensionless,
L = length, ft,
D = hydraulic diameter, ft,
o = density, 1lb/ft3,
v = average velocity, ft/sec,
g = gravitational acceleration,
32.2 ft/sec?,
kc = constant, dimensionless,
v, = average linear velocity upstream,
ft/sec,
v, = average linear velocity down-
stream, ft/sec,
Np. = Reynolds’ number.

The subscripts have the

meanings:

following

¢ = contraction,
e = expansion,
f = friction,

All the annuli around the control
rod and instrument holes have the same
dimensions: the outside diameter, D ,
is 3.652 in, and the inside diameter,
Di’ 1s 3.000 i1nches. There are two
instrument holes, three safety rods,
and one regulating rod; thus there are
six holes and six corresponding annuli,
Where the NaK enters at the bottom of
the annuli (Fig. 9), there are orifice
plates (Fig. 11). These orifice plates
reduce the outside diameters of the
annuli to 3.140 in,; the length of
each orifice plate 1s 1.00 in,; and
the length of the remainder of each
annulus 1s 35.25 inches., The annulus

17

 

 
81

 

UNGLASSIFIED

e @ DWG.E-A-1-24
3930, N %D e
| J

  
 

-~ 35640 in. 0.060 in. PARTS LIST
INCONEL TUBE GOIL A
INCONEL TUBE CCIL B
INCONEL. INSTRUMENT TUBE SUB-ASSEMBLY
INCONEL CAN
BeO REFLECTOR EDGE BLOCK
BeO REFLECTOR BLCCK
BeO CORE BLOCK
INCONEL REFLECTOR COOLANT TUBES
INGONEL. CORE SLEEVE
INCONEL SAFETY ROD GUIDE SLEEVE
INCONEL TOP TUBE SHEET
INCONEL BOTTOM TUBE SHEET
INCONEL SUPPORT STUDS
INCONEL SPACER
INCONEL ORIFICE PLATE

 

 

 

 

 

__/,//®

   

 

 

=

Aoy
S

- -=-2.035in.

 

 

 

 

ey

 

 

 

 

 

l IR /{ e e o e e -
{% NN NN R e NS (®
' 4

O @O~ OO s WM -

4

 

",

.
s

"
|
|

N

 

 

o

 

 

 

 

 

 

 

 

 

 

ASSEMBLY SEQUENGE

 

STEP {: WELD ITEMS 8,9 AND 15 TO ITEM 2,
9 AND 15 WELDED TOGETHER TO ITEM 12

STEP 2: ASSEMBLE ITEM 13 WITH ITEM {2

STEP 3: ASSEMBLE ITEMS 1 AND 2 WITH ITEM 12,
THEN ITEM 14 WITH ITEMS 1,2 AND 8

STEP 4: ASSEMBLE ITEM 7 WITH ITEMS { AND 2
STEP 5° ASSEMBLE ITEM 6 WITH ITEM 8
STEP &° ASSEMBLE ITEM 5

 

 

 

 

 

 

 

 

 

 

. ’i ;///’5,//,;”' N

 

 

 

 

 

 

 

 

e > STEP 7° ASSEMBLE AND WELD ITEM 4 TO ITEM 12
() 7 S NN
% SOENSRONNNNN 4 STEP 8: WELD {TEMS 8 AND 9 TO ITEM 1
§§§g§§\s STEP 9: WELD ITEM 4 TO ITEM {f
;/,7;/,///

 

 

 

 

 

 

 

 

Fig. 11. Core and Reflector Assembly Elevation Section.
between the pressure shell and the
outer periphery of the moderator can
has an outside diameter of 48.57 in.:
the gap width 1s 1/16 in.; and the
path length through this annulus is
36.25 inches.

In the reflector region,
79 tubes, one through each column of
beryllium oxide blocks. Each of these
has an inside diameter of 0.49 in.,
and 1s 36.25 in. long.

The coolant is carried to and from
the pressure shell by 2 1/2-1n.
schedule-40 pipe. Since the NakK
flowing through the reactor pressure
shell and core lattice is almost
isothermal,

there are

the physical properties
were considered to be constant. The
values of density and viscosity used

were, respectively,
p = 44.5 1b/ft3
and
pmo= 0.36 1b/hr: ft .

By using Egqs. 32 and 33, the inlet
and outlet losses of the pressure
shell were calculated to be

AP (inlet) = 156.3 1b/ft?
and
AP (outlet) = 78,15 1b/ft? ,
From Eqs. 32 through 35 and the

assumption that the distribution of
flow through the core lattice depends
only on the relative resistance of each

path, the distribution of flow was
found, by trial and error, to be:
FLOW RATE
PATH
T (ft3/sec)
Large annulus 0.108
Tubes through reflector 0.272
Six annuli 0.120

The pressure drop through the
lattice, which 1s the loss in pressure
experienced by the fluid flowing

through one path, was found to be

AP = 19.49 1b/ft? .

If the pressure drop i1in the spaces
above and below the core lattice 1is
neglected, the total pressure drop 1is

AP = 253,9 1b/ft?
= 1.76 lb/in. 2% .

 

HEAT REMOVAL BY THE REFLECTOR COQLANT

A cross section of the reactor 1is
shown in Fig, 9;
in Fig., 12. Molten NaK is circulated
through the reactor to remove the heat
generated i1n the reflector and to cool
the pressure shell and the walls of
rod and instrument holes. In passing
through the reactor, heat is removed

a plan view 1s shown

from the following sources in the

reactor by the NakK:

1. heat generated in the reflector,

2. heat generated in the pressure
shell,

3. heat generated in the Nak,

4, heat transferred from the reactor
core

a. through the serpentine elbows,
through the tube sheets,

c¢. through the “boundary’ between
the core and the reflector,

d. through the walls of the rod
and i1nstrument holes,

The NaK i1in the interstices of the
moderator blocks 1n both the reactor
core and reflector is almost stagnant
and, for the purposes of the following
calculations, is assumed to be entirely
stagnant, Jt 1s assumed that the
electric heaters on the pressure shell
add sufficient heat to the system to
Just balance the heat loss to the
environment; therefore it 1s also
assumed that there 1s no net heat flow

across the outside of the pressure
shell.

Heat Generated in the Reflector.
The following values for gamma and
neutron heating of the reflector were
obtained from the section on “Core
Power Distribution.”

Gamma Heat Generation:

Peak gamma heating (at center of
core), Btu/sec.in.3

0.023

Ratio of maximum gamma heating at
a radius of 18 in. to peak gamma
heating 0.20

Maximum gamma heating at a radius

of 24 1in. 0.0

Ratio of maximum gamma heating at
any radius to average gamma
heating at the same radius 1.32

19
DWG.E-A-1A

 
  
   
 
 

THERMAL SHIELD ASSEMBLY

PRESSURE SHELL
TOP HEAD ASSEMBLY

........

 

FUEL OQUTLET
MANIFOLD

 

    

SAFETY ROD
EXTENSION GUIDE

INSTRUMENT TUBE
EXTENSION GUIDE

Fig.

Neutron Heat Generation:
Peak neutron heating in beryllium
oxide (at center of core),
Btu/sec-in.

Ratio of maximum neutron heating
at a radius of 18 in. to peak
neutron heating

Neutron heating of the reflector
at a radius of 22 1/2 in.

Ratio of maximum neutron heating
at any radius to average neutron
heating at the same radius

20

12.

 

  
  

INGHES
6 9

   
 

2

 

Plan View of Reactor,.

Neutron heating aqccurs only in the
beryllium oxide, which constitutes

about 78.5% of the reactor.

0.011 It is assumed that within the
reflector the gamma and the neutron
heat generation vary linearly with

0.30 radial distance. That this assumption
1s fairly valid may be observed from

0.0 Figs. 7 and 8. The following equations
for the average gamma and neutron heat
generation at any radius within the

1.55 reflector can be written:
 

(36)

g, =0.00460 -0.000767 (R - 18) ,
- 18 <R < 24,

g, =0.00260 -0.000575 (R - 18) ,
18 < R < 22.5,

(37)

(38) q, = 0, R > 22.5 ,
where
q = heat generation, Btu/sec-in.?3,
R = radius, 1inches.
The total heat generation, Q, 1is
24
(39) Q, =.L8 [0.00460
- 0.000767 (R - 18)] 27R dR
= 40,7 Btu/sec ,
22.5
40 = 0.00210
(40) @, =f "I
- 0.000467 (R - 18)] 27R dR
= 9,5 Btu/sec ,
(41) Q = Qy + Qn = 50,2 Btu/sec .

The heat generated in the reflector
is removed by the NaK through various
interfaces; the temperature limitations
of the beryllium oxide and associated
structure are sufficiently far above
the NaK temperature that temperature
patterns in the reflector need not be
calculated.

Heat Generated in the Pressure Shell.
The heat generated in the Inconel
pressure shell 1s due to gamma attenu-
ation and fast-neutron scattering. The
actual variation of heat generation
with position in the top head is shown

in Fig., 13. The curve may be approxi-
mated by the exponential relationship
(42) q'= 13,200 ¢-9-20=
where
9" = heat generation, Btu/hr-ft?,
x = distance from inside surface, ft.

The heat generation in the bottom head
and barrel is actually less than this,
it 1s assumed
that the curve may be used. The total
heat generation, Q, in a flat plate
of thickness 7 is given by

but to be conservative,

(43) Q = 13,200 e~ %2°% gy
0

= 1450 (1 - ¢~%207) |

where 7 1s in feet.

The

structed of 2-in.-thick plate reinforced
to 4 1in,
locations 1in the heads.
43 for thicknesses, T,

1/3 ft gives

Eq.

and

respectively,

If the barrel 1s treated as a flat
plate,
thick material is 72.5 ft?,
total area of the 4-1in.-thick material

is 9.5

generation is 95,300 Btu/hr,
Btu/sec.
but because Fig,
conservative value 1s desirable.
Temperature of Pressure Shell.
Inasmuch as the allowable design stress

Inconel pressure shell 1s con-
in thickness at critical
Evaluation of

of 1/6 and

Q = 1135 Btu/hr- ft?2

Q = 1380 Rtu/hr-ft? ,

the total area of the 2-1n.-
and the
ft?2., Therefore the total heat
or 26.5
This value 1s probably high,
13 1s approximate, a

 

in the pressure shell is a function of

the pressure shell temperature,

the

temperature pattern in the pressure
shell head (the most critical portion

of the

pressure shell from the stand-

point of both stress and temperature)
must be calculated,.

The

as a

flat plate
internal heat generation.

pressure shell will be treated
in which there 1is
One side of

the plate is insulated and the other

side

(44)

is cooled by circulating Nak.

 

The differential equation, then, 1is:
d26 q'
dxz- k-

where

x =

Btu/hr- ftd =

heat generation,

13,200 ¢~ 720

T - T ,

x Nak

temperature in pressure shell at
position x,

of NakK,

thermal conductivity of Inconel
= 11 Btu/hr* ft? (°F/ft),
distance from 1nside surface,

ft.

temperature

The boundary conditions are:

df

— =0 , at x = T ,
dx

dé h

— =—26&, at x = 0 ,
dx k

21

o

 
HEAT (watts /cm3)

22

DWG. 19301

 

5
* TOTAL HEATING SHCULD BE INCREASED
APPROXIMATELY 10% TO ALLOW FOR
FAST-NEUTRON SCATTERING
2
10!
TOTAL HEATING ¥
5
2.5-Mev
GAMMA (CORE)
2
1072
5 7-Mev GAMMA (CORE
CAPTURE GAMMAS
2 CAPTURE PLUS INDUCED ACTIVITY
FROM GAMMAS IN SHELL
1073
0 2 4 6 8 10 12

THICKNESS OF SHELL {cm)

Fig. 13. Heat Generated in Pressure Shell at 3-Megawatt Power Level.

 
 

where h 1is the heat transfer coef-
ficient of NaK in Btu/hr-ft?-°F. The
heat transfer coefficient, h, may be
calculated by the following formula:

kNaK

(45) h (5.8 + 0.02 Pe®?®),

 

where
knag = thermal conductivity of Nak,

D = NaK passage equivalent diameter,
Pe = Peclet’s number for NakK.

To assure a conservative value for the
heat trans fer coefficient, the hy-
draulic diameter of the NaK passage
without the tube elbows was used as
the equivalent diameter of the NaK
passage. The diameter and other
values are

D = 0.667 ft,
kyag = 16.0 Btu/hr-fe? (°F/ft),
Pe = 148, using a velocity of 0.1

ft/sec (cf., chap. 2, “Flow of
Reflector Coolant Through
Core ),
Therefore
h = 166 Btu/hr-ft2-°F,
The differential equations and
boundary conditions thus become

 

d?6
(46) = 1200 =920
dx ?
and
dé
(47) — =0 , at x = T ,
dx
dé
(48) — = 15.16 , at x = 0 ,
dx

The solution 1is, of course, obvious:

dé
(49) -E— = 130.5 e %% + cy

x
where, from Ea. 47,

c, = -130.5 e” 9207

and therefore

d6

d_ = 130.5 (6'9.201 _ 6‘9.207') ,

x
(50) 6 = —14,19 ¢79-20=
- 130.5 x e~ 9%+-207 4 c,
c, = 8.64 (1 - e %) + 14.19

6 = 14.19 (1 — ¢™9:20%)
- 130.5 x e-9-207
+ 8.64 (1 — e~ 2-20%)

The temperature at x = 7, which 1s,
of course, the maximum temperature
difference, can be determined:

(51) 6&_ = 22.83 (1 — e %207)

- 130.5 7 e~ 9207

For 7 = 2 in., B = 13.2°F; for 7 = 4
in., 8 = 19.8°F.

As a check on the calculation,

6 _, = 8.64 (1 - e79-207)
= 6,78 , for 7 = 1/6 ft ,
= 8.23 , for 7 = 1/3 ft ,
he oo = 166 0,
= 1130 , for 7 = 1/6 ft ,

1/3 ft

These values check very closely the
heat generated per square foot in the

1370 , tor T

two different sections, The pressure
shell, therefore, will be about 5 to
20°F hotter than the NaK.

lleat Generated in the NaK. The
calculation of heat generation 1in the
NaK in the reactor circuit is rather
complex, and only a gross estimate of
the heat generation is available. The
maximum heat generation in the NaK
will probably be of the order of 10
Rtu/sec- ft®, with the average heat
generation being of the order of 3
Btu/sec* ft®. Since there is approxi-
mately 10 ft® of NaK in the pressure
shell, the total heat generation in
the NaK will be of the order of 30
Btu/sec.

The total heat added to the re-
flector coolant circuit by heat
generation in the reflector, pressure

shell, and NaK 1s
Q = 50.2 + 26.5 + 30 = 106.7 Btu/sec

This heat addition 1s independent of
NaK temperature.

Heat Transferred from the Reactor
Core. Heat is transferred to the NakK
in the reflector coolant circuit from
the reactor core, The boundaries

23

 

 

 
 

across which fhe heat 1s transferred
are the serpentine elbows, the tube
the boundary between the core
and the walls of
the rod and instrument holes.
Serpentine Elbows. The fuel tubes
that
(Fig.
9); there are six tubes in parallel,
There are therefore sixty 180-deg bends
or elbows
pressure shell,
30 below 1t. In addition, between the
core and the pressure shell,
s1x straight sections at the inlet of
the tubes and six straight sections at
the outlet. The length of the 180-deg
elbows is 8.5 in., for the elbows above

sheets,
and the reflector,

are wound as serpentine coils
pass through the core 11 times

between the core and the
30 above the core and

there are

the core and 6.5 in. for elbows below
the core; the straight sections are
4 i1in, long above the core and 3 in.
long below 1t. The tubes have an
outside diameter of 1,235 in. and
0.060-in, walls. The total length of
the bends and the internal and external
areas of the bends are

6.5 8.5) 4 3>
L = 30— + =2 ) 4+ g2+ =2
19 12 12 192

 

= 41.0 ft ,
1.235
A . = 41,0 x 7 x = 13,3 ft? |
ex 12
1,235 - 0.120
A, =410 x 7 x —m — —
1n 12

= 12,0 ft?

The fuel-side heat transfer coef-
ficient can be calculated by first
calculating the heat transfer coef-
ficient for a straight pipe and cor-
recting for the effect of the bend:

! Dl
(52) hf=hf<l+3.55->,

H
where
hf = fuel-side heat transfer coef-
ficient, Btu/sec*°F,
h} = heat transfer coefficient in a
straight section, Btu/sec*ft?-°F,
D. = tube inside diameter,

Dy = bend diameter,

h, = h'(1 + 3 L.115 = 2.04 h'
f_f 05 _o4.'

 

3.75 f

24

(53) k% = 0.027-/i> Re Pr}“" fi)“’“
\D f Hs/f
where
Ref = Reynolds’ number for fuel
= 10,600,
Prf = Prandtl’s number for fuel
= 3.78,
0.14
<fi> = 0.96,
Moo f
0.004409,

TN
o |~
\_/
“—h

I

Therefore

h} = 0,303 Btu/sec-ft?2-°F,
h 0.618 Btu/sec*ft?*°F,
(hA)f= 7.42 Btu/sec*°F,

),

The value of the NaK-side heat transfer
coefficient 1s somewhat in doubt be-

0,135,

cause very little work has been done
on the heat transfer coefficient of
ligquid metals flowing across round
tubes, For an approximation, the
formula for flow inside a round tube
will be used, with the outside diameter
of the tube being considered as the
equivalent diameter:

.
I 0.8
(54) k. = <D>N K <7 + 0,025 PeNaK> ,

where
PeN“(= Peclet’s number for NaK
= 22,90,
<k> 0.0432
D Nak
Therefore
hNflK = 0.316,
(hA)NaK = 4,20 Btu/sec*°F,
1
<-> = 0.238.
h4 NakK
The tube wall resistance is
DO
ln —
1 D
(55) — == 0,130 ,
hA 2k L

‘metal

where D, is the outside diameter; and

 
 

the over-all resistance 1is

oo () (x) « (&)
)

metal

0.503 .

For the calculation of the effective
temperature difference between the fuel
and the NaK 1t 1s assumed that the NaK
temperature 1s constant and 1s the
same as the mean temperature of the

This simplifying
to some 1lnaccuracles

NaK 1in the core.
assumption leads
in the calculation of the heat trans-
ferred to the NaK; however, in view of
the extremely complex flow of the NaK
through the reactor, it i1s doubtful
whether any approximation would be
the variation in
the fuel temperature from 1150 to
1500°F is due primarily to heating in
the core, the arithmetic-average
temperature difference between the NaK
and the fuel should be used.

The arithmetic-average elbow temper-
ature 1s about 1330°F, The effective
temperature difference and the heat
the NaK through the
are in the

more valid. Since

rejected to
serpentine elbows given
following tabulation for various mean

temperatures:

MEAN NaK AVERAGE TEMPERATURE HEAT REJECTED

TEMPERATURE DI FFERENCE TO NaK
(°F) (°F) (Btu/sec)
1300 30 60
1200 130 258
1075 255 507
1000 330 656
800 530 1054

Tube Sheets. The tube sheets are the
metal plates at the upper and lower
boundaries of the core through which
the fuel tubes pass (Fig., 9). Figure
14 shows a cross section of a typical
fuel tube and 1ts associated moderator
blocks and tube sheet section., Heat 1s
generated 1n the moderator and trans-
ferred to the fuel and the NaK; 1in
addition, heat flows from the fuel to
the NaK through the moderator blocks.
The differential equation and hboundary
the two-dimensional

conditions for

 

 

 

 

 

 

 

 

 

 

 

 

  

 

 

 

 

 

 

  

 

system shown 1in Fig. 14 are given 1in
the following:
%68 9% q'
(57) A
Ox? dy 2 k
where
— 0
g =T - Tk °F»
T = temperature at any point (x,y),
OF,
Ty.k temperature of the NaK, °F,
g’ = heat generation in moderator,
Ptu/sec- ft3,
k = thermal conductivity of moder-
ator, Btu/sec*ft? (°F/ft).
OWG 19104
TUBE SHEET
NaK FLOW /
. | :fié::::::j
x |
g |
M
x |
2 |
q f 1
s |l
D i
o i
= ‘ % i
|
o @
> X
e L TR e N
REACTOR CENTER LINE
Fig. 14. Cross Section of Moderator

Block Showing Tube Sheet and Fuel Tube.

The heat generation in the moderator
block 1s assumed to be constant in the
x direction and to the y
direction 1in accordance with the axial
moderator heating curve found in the

The

vary 1in

section on power distribution.

25

 

 
 

boundary conditions are

 

a6
(58) — =0 , at x = 0 ,
Ox
o6
(59) — =0 , at y = 0 ,
dy
h
a8 f
(60) 5;'= T (9 T Thak = TS> ,
at x = a ,
h
o NakK
61 — = - g =
(61) 5 . , at y b

No closed or series solution for
this boundary value problem could be
readily found; consequently, solutions
to specific numerical problems were
undertaken by the relaxation method.
Figure 15 shows the temperature pattern
across the tube sheet face of atypical
moderator block, as found by this
method, for several NaK temperatures.
The heat loss from the section of the
tube sheet adjacent to one fuel tube
and moderator hblock and the total heat
loss from both tube sheets, which 1is
132 times the loss from one of the
sections, are tabulated in the follow-

Nak TEMPERATURE HEAT LOSS PER pap,; peaT LoOSS

a SECTION

("F) (Btu/sec) {Btu/sec)
900 1.86 246
1075 1.15 150
1250 0.25 33

Boundary Between Core and Reflector.
The fuel tubes at the periphery of the
core (near the reflector) are at a
considerably higher temperature than
the reflector, and they transfer heat
into the reflector through the beryl-
lium oxide blocks., An estimate of the
heat transferred into the NaK circuit
by this method has been made. The
system of fuel tubes, NaK tubes, and
beryllium oxide blocks was simplified
and considered as a hollow beryllium
oxide cylinder with internal heat
generation in which the inner surface
1s washed by fuel and the outer sur-
face 1s washed by NaK. The actual
heat transfer coefficients of the fuel
and the NaK in the tubes were calcu-
lated and, from these coefficients,
hypothetical heat transfer coef-
ficients on the sides of the cylinder
were computed so that

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ing: (62) R*A* = hA
Dd! 15105
l—~e—TUBE (¢ AT O} ~=— ~TUBE (4. AT 3.75)%
mafl°”
1300
\ /
e 1250
“ 1200 /
w \
o
2
b ,—’///////
x 1075 R
S 100 A\ /
w
- //
|
J
t000
\\\\\‘h“‘- 900 _—___—____,,,///
--———-_—
900
0 05 { 1.5 2.0 2.5 30 3.5
RADIAL DISTANCE FROM CENTER OF FUEL TUBE (in)
Fig. 15. Temperature Pattern Across Tube Sheet Face.
26
 

h* = actual heat transfer coef-
ficient, Btu/sec* ft?:°F,
A* = actual heat transfer area, ft?,
h = hypothetical heat transfer
coefficient, Btu/sec* ftQ'OF,
A = hypothetical heat transfer area,
ft2,

The mean radius of the peripheral
fuel tube circle 1s about 15 1in, and
there are 29 tubes; the mean radius of
the inner circumference of the NakK
tubes 1s about 18 1/2 in. and there
are 35 tubes; and the reactor core 1s
36 in., 1n height. Therefore the actual
heat transfer areas are

(63) A

* 2
Af 25.8 ft* |
* _ 2 .
ANaK = 11.86 ft° ;

and the hypothetical heat transfer
areas are

il

7DLN

(64) A = 2nrL
- 2
Af = 23.6 ft* ,
- 2
ANaK = 29,1 ft© ,
where
D = tube 1nside diameter, ft,

tube length, ft,
number of tubes,

L
N
ro=

= average radius of tube caircle,
fe.

The heat transfer coefficients 1in the
tubes are found as follows:

1. For the fuel tuhles

. kf w 0.14
(65) hy=10.027 — Re}-B pr}x’3 <—~>
D ’usf
where

k

thermal conductivity of fuel,
Btu/sec* ft? (°F/ft),
Ref = Reynolds’ number for fuel,

Prf

f

I}

Prandtl’s number for fuel,

H . : :
<~—> ratio of viscosity of fuel at
B /g bulk temperature to viscosity
: at surface temperature,

For a fuel velocity of 3.7 ft/sec and
a Df of 0.09285 ft, the values of the

terms in Eq. 65 are

o
H
1

2,98 ,

 

hy = 0.373
2. For the NaK tubes
* kNaK
(66) hy, , = b, (7.0+0.025 Pey: &),

where Pe \ is Peclet’'s number for NakK.
For a NaK velocity of 3.0 ft/sec and a

DNaK of 0.03582 ft, the values of the

terms 1n Egq. 66 are

PeNaK = 236.4

%*

hy.o = 1.093

The hypothetical heat transfer coef-
ficients are therefore

h*A*
hf =%f = 0.4-08
f
and . .
b _ hNaKANaK - 0. 44
NaK A = 0.446
NakK

The differential equation for steady-
state heat conduction with internal
heat generation 1s

!

 

(67) vip = - -
BeO
where
T = temperature at any polint in
the BeO, °F,
g’ = internal heat generation,

Btu/sec* ft3,
kpeog = thermal conductivity of BeQO,
Btu/sec- ft? (°F/ft).
Although g’ varies with radial distance
and kBeo varies with temperature, both
will be assumed constant and equal to
their average values., This simplifying
assumption undoubtedly results 1in some
error and the following solutions,
accordingly, are only estimates:

qg' = 7.74 Rtu/sec* ft?

?

27

 
 

and
ko.o = 0.0054 Bru/sec*ft? (°F/ft) .
The solution of the differential
equation for the i1nfinite cylinder
case 1S

dT “
(68) —_ = =717 + — ,

dr r

(69) T = -358.5 r? + ¢

+
, Inor c,

where r 1s the radial distance from
the center of the core in feet and €,

 

 

and ¢, are arbitrary constants, The
boundary conditions are
h
dT
(70) — = - f (1460 - T) ,
dr BeO
at r = 1.25 ft
(the mean fuel temperature in the
peripheral! tubes 1s 1460°F),
(11) 2L - nex (T - T,..)
o T ‘NaK !
dr kBeo
at r = 1.5417 ft
(TNaK 1s the mean NaK temperature).

The values of ¢
by using the

, and ¢, are obtained
boundary conditions:

(72) c, = 4.3823 TNaK ~ 5008.,53
and
(73) c, = 3072.71 - 0.93146 T
Nak
(the solution requires that many

significant figures be carried). The
temperature differences at the ReO-to-
fuel and BeO-to-NaK interfaces and the

heat removed from the fuel and added
to the NaK are given in Table 3. It
1s obvious from the data in Table 3

that heat 1s transferred from the fuel
to the NaK and that all the heat
generated 1n the beryllium oxide goes
into the NaK. A simple check of the
numerical work, by the method of

superposition, 1s therefore possible:

(74) Q5.0 q'W(rfiaK—r;)L
= 59.5 Btu/sec,
fp.0 = temperature difference
across the beryllium
28

oxide resulting from
internal heat generation

only = 61.1°F,

PR

|-

—
n

0.1039 |
f
1
(——) = 0.0771 |,
hA NakK
1
<——> = 2.060 ,
hA'BeO
1
<——> = 2.241
hA'f-NaK

The fuel-to-NaK temperature differences
for the various NaK temperatures are
given in Table 4. As can be seen, the
results quite well with the
results given in Table 3.

A portion of the heat transferred

agree

to the NaK comes from heat generated
in the beryllium oxidein the reflector,
as previously discussed. It 1s
therefore necessary to subtract from
the heat transferred to the NaK an
amount equal to

2
Q:?_m[(}f_-__s_)
12

18. 0 \?
- — X 3 = 4.65 Btu/sec
12

The net heat added to the NaK 1s then

TNaK (QNaK)net
800 322
1000 233
10675 199
1200 143
1300 99
Walls of the Rod and Instrument

Holes.
through the reactor core and pressure
shell to admit control rods and
nuclear instruments (Fig. 16). In
each hole there are three concentric
tubes (Fig. 17); NaK flows through

the passage between the outer

There are six vertical holes

two

 
TABLE 3.

TEMPERATURE DIFFERENCES AND HEAT TRANSFERRED AT THE

Be0-TO-FUEL AND BeO-TO-NaK INTERFACES

 

 

 

TEMPERATURE DIFFERENCE (°F) 5
Tyag (°P) Qs ) QNaK( ) Opeo = Unax * ¢ ()

BeO- to-Fuel BeO-to-NaK e a f

800 ~27.77 25.20 -267.4 326.7 59.3

1000 ~-18.49 18.31 ~178.0 237.4 59.4

1075 ~15.01 15.73 ~144.5 203.9 59. 4

1200 -9.21 11.42 ~88.7 148.0 59.3

1300 -4.57 7.97 -44.0 103.3 59.3

 

 

 

 

 

 

(G)Qf = heat transferred to the fuel, Btu/sec,
(b)QNaK = heat transferred to the NaK, Btu/sec.

(C)QBeO = heat generated in BeO, Btu/sec.

 

 

TABLE 4. TEMPERATURE DIFFERENCES AND HEAT TRANSFERRED AT THE
FUEL-TO-NaK INTERFACE

TEMPERATURE DIFFERENCE _ -
1§BK FUEL-TO-NaK, 6, \ o gf“Na§ = "Beo T ) ek Ppnak = Fpeo) | Ovax = 0p * Opeo
(°F) (°F) - ("F) (Btu/sec) (Btu/sec)
800 660 598.9 -267.2 326.7
1000 460 398.9 -178.0 237.5
1075 385 323.9 -144.5 204.0
1260 260 198.9 —88.7 148. 2
1300 160 98.9 ~44.1 103.6

 

 

 

 

 

tubes and removes the heat that would
otherwise be transmitted to the hole
from the reactor; the passage between
the inner two tubes 1s packed with
insulation; helium flows inside the
inner tube and cools the rod or
instrument and the inner tube wall.
Heat 1s therefore added to the NaK
from the reactor and removed from the
NaK by the helium. Since the two
instrument holes are located in the
reflector and the heat addition to
the NaK from the reflector has already
been accounted for in the calculation
of heat generated 1n the reflector,
only the heat added to the NaK in the
four control rod holes remains to be

calculated. The calculation of the

 

heat removed from the NaK by the
found

and Instrument

helium in the six holes can be

‘‘Rod

”

in chap. 35,
Cooling System.
Each rod hole through the reactor
1s surrounded by six beryllium oxide
blocks, each of which contains a fuel
tube (Fig. 16). The mean fuel tempera-
ture 1n the fuel tubes 1s about
1200°F. As a simplification (similar
to that used 1n the calculation of
heat transfer across the boundary of
the core and reflector), the holes
and associated structure were considered
as a hollow beryllium oxide cylinder
with internal heat generation and with
the 1nner surface washed by NaK and
the outer surface washed by fuel.

29

 
   
   
   
   
  
   
 
 
 

TUBE COIL B
{INCONEL)

 

TUBE COIL A
(INCONEL)
0 o
O o
o
O o
— 5 o
o o
o
O O
© o

 

 

 

Fig. 16.

There is some overlapping of fuel
tubes in the sense that some of the
tubes are members of the tube sets
encircling more than one rod hole.
This effect was neglected i1inasmuch as
the heat added to the NaK in the rod
holes smaller 1n
magnitude than the heat added elsewhere
and, a rough estimate of the
heat additionis all that 1s necessary.

1s considerably
hence,

The actual heat transfer coefficient
in the fuel tubes was calculated and,
from this coefficient, a hypothetical
heat transfer coefficient on the side
of the cylinder was computed so that:

(75) h*A* = hA

3

30

 

DWG. 15847

—INSTRUMENT TUBE
(INCONEL)

——_> REF_ECTOR EDGE BLOCK
{ BERYLLIUM OXIDE)

\\\\\\ - REFLECTOR BLOCK

{BERYLLIUM OXIDE)

  

_.—-CORE BLOCK
{BERYLLIUM CXIDE)

REFLECTOR COOLANT
TUBES {INCONEL)

— GAN
(INCONEL )

——— CORE SLEEVE
(INCONEL)

 

SAFETY ROD GUIDE
SLEEVE ({INCONEL)

]

SGALE IN INGHES

Plan View of Reactor Core Lattice,

where

h* = actual heat transfer coefficient,
Btu/sec'ftz'oF,

A* = actual heat transfer area,

ft?,

h = hypothetical hecat transfer
coefficient, Btu/sec'ftz'oF,

A = hypothetical heat
area, fe 2,

The inside radius of the outer
NaK-containing tube 1s 1.826 1in.;
the mean radius of the fuel tube
circle i1s 3.75 in.; and the height
of the reactor core 1s 36 1inches.
The heat transfer coefficients and

transfer
DWG. 12106

;// 72

  
 
    

 

 

_

DIATOMACEOUS EARTH

Fig. 17. Cross Section of Helium
Passage and Sleeve,

 

 

areas are therefore:
1. For the NaK annulus

 

(76) ANnK = AaaK = 7TDNaKL !
(77) hNaK = h;aK
“NaK 0.8
) (5.8 + 0.020 Pel:2)
Nak
where
D = equivalent diameter, ft,
L = length of reactor core, ft,
k = thermal conductivity, Btu/sec-ft?
(°F/f¢e),
Pe = Pelcet’s number.

For a NaK velocity of 0.81 ft/sec,
an Ay.x = Af.x of 2.87 ft?, and a
Dg.x of 0.0543 ft, the values of the
terms in Egs. 76 and 77 are

PeNaK = 97,4 ,
hy.x = hy . x © 0.530 Btu/sec* ft?* °F.

2. For the fuel tubes

(78) A? = WDfLNf ;
(79) Af = 2erL ,

(80) h;

kf [ 0.14
= 0.027 — Re%® pr /3 (——) ,
Df f f

N = number of tubes,
r = mean radius, ft,
Re = Reynolds’ number,
Pr = Prandtl’s number,

(fi;> = ratio of viscosity of fuel at
f bulk temperature to viscosity
at surface temperature.

For a fuel velocity of 3.7 ft/sec,
an A* of 5.25 ft?, an Af of 5.90 ft?,
and a Df of 0.09285 ft, the values of
the terms in Eqs. 78, 79, and 80 are

Re = 7970 ,
Pr = 5.02 ,

h% = 0.294 Btu/sec* ft2-"F
h, = 0.262 Btu/sec* ft?+°F

The differential equation for
steady-state heat conduction with
internal heat generation 1s

i

 

(81) V2T= - 1 ’
kBeO
where
g' = internal heat generation,

Btu/sec* ft?,
= thermal conductivity of BeO,
Btu/sec* ft? (°F/ft).

Although q' varieswith radial distance

kBeO

and ky,, varies with temperature,
both will be assumed constant and
equal to their average values. This
is done for simplicity and undoubtedly
results in solutions that are only
approximate. The values are

g' = 39.8 Btu/sec*ft?
and

k.o = 0.0058 Btu/sec:ft? (°F/ft) .

The solution of the differential

31

 

 

 
 

equation for, the infinite cylinder
case 1s

dT ¢y
(82) — = 3431 r +—,
r r
(83) T = 17155 r* + ¢, In r + ¢, ,
where
T = temperature, °F,
r = radial distance from center of

core, ft,
¢; and ¢, = arbitrary constants.

The boundary conditions are

 

 

dT hyak
(84) — = (T - Ty K) ,
dr kBeO )
at r = 0,15217 ft,
-h
dT
(85) -7 (1 - 1200)
dr kg.o
at r = 0,13250 ft (the mean fuel temper-

ature 1s approximately 1200°F),
Solving for ¢; and ¢, by using the

boundary condition gives

(86) c, = 1573.87 - 1.1596 T,  ,
(87) c, = 3110.40 - 1.2666 T,
(the solution requires that many

significant figures be carried).

The temperature differences at the
BeO-to-NaK and BeO-to-fuel interfaces
and the heat added to the NaK and to

the fuel are given in Table 5. The

values given in Table 5 are for one
of the four rod holes. A check of
some of the numerical work, by the
method of superposition, is possible:

QBeO

The temperature difference from
the fuel to the NaK across the beryllium
oxide because of internal heat gener-
ation only is 131,7°F when all the
generated heat enters the NaK, and
-88.8°F when all the generated heat
enters the fuel,

q'W(ri—rfiaK)L =27.9 Btu/sec

The resistance to heat transfer
between the NaK and the fuel 1is

1
<__> - 0.6574 |
. hA N a K
1
<—A> = 0. 6475 ’
h f
1
(-) = 6.5818 |
hA BeO
1
(——) = 7,887
hA f-Nak

The fuel-to-NaK temperature differences
for the various NaK temperatures are
given 1n Table 6.
results

As can be seen, the
agree qulte well with the
results given 1in Table 5.

 

 

 

 

 

 

 

 

 

TABLE 5, TEMPERATURE DIFFERENCES AND HEAT TRANSFERRED AT THE
BeO0-TO-NaK AND BeO-TO-FUEL INTERFACES
TEMPERATURE DIFFERENCE (°F)
TNK (OF) QN K(G) Q (b) QBO=QNK __Q (C)
& BeO-to-NaK BeO-to-Fuel a f © a f
800 40.74 -22.01 61,97 ~33.,97 28,00
1000 24,06 -5.57 36. 60 —8.60 28,00
1075 17.80 0.60 27.08 0,93 28.01
1200 7.38 10, 87 11,23 16,78 28.01
1300 ~0, 96 19,09 ~1, 46 29,47 28.01
(a) -
Quag = heat transferred to NaK, Btu/sec.
(b)Qf = heat transferred to the fuel, Btu/sec.

(

C)QBeO = heat generated in BeO, B:u/sec.

32

 
 

TABLE 6. TEMPERATURE DIFFERENCES AND HEAT TRANSFERRED AT THE
FUEL-TO0-NaK INTERFACE

 

 

Tyak ef-NaK GENERATED ef,NaK = %Be0 Qf Onak
(°F) (°F) HEAT ENTERS (°F) (Btu/sec) (Btu/sec)
800 400 NaK 268.3 -34.0 62,0
1000 200 NaK 68, 3 -8.7 36.6
1075 100 NaK and fuel

1200 0 NaK and fuel

1300 ~100 Fuel -11.3 +29.3 -1.4

 

 

 

 

 

The heat removed from the NaK by
the helium flowing through the rod
holes is calculated in chap. 5, ‘Rod
and Instrument Cooling System,” for
the various rods and instruments.

Summary. A plot of the heat added
to the NaK circuit vs, the mean Nak
temperatures for the various sources
of heat addition is shown in Fig. 18,
A plot of the net heat that must be
removed from the NaK in the reflector
coolant heat exchangers vs. the mean
NaK temperature is shown in Fig. 19,
By cross-plotting these two figures,
the mean NaK temperature 1s obtained;
the cross-plot is shown in Fig, 19,
The mean NaK temperature 1is about
1172°F, and at that temperature,
640 Btu/sec (675 kw) 1s removed from
the reactor by the reflector cooling
system. The minimum and maximum NakK
temperatures are about 1105 and 1235°F,
respectively (Fig. 1). The maximum
pressure-shell temperature 1s about
20°F greater than the maximum NakK
temperature (see previous work in
this section) and is therefore about

1255°F,

PRESSURE DROP IN FUEL MANIFOLDS
AND CORE TUBES

The fuel flows from the heat
exchangers through the surge tanks
and pumps and into the inlet manifold
of the reactor, whereit i1s distributed
into the six parallel passes through
the core lattice. Upon leaving the
core lattice, the fuel enters the
outlet manifold. From the outlet
mani fold, the fuel is returned to the

 

heat exchangers. For a fuller under-
standing of the circuit, see Figs. 1
and 9, The problem is to determine
the pressure drop (or drops) of the

Da! 19107

700

 

 

600

 

 

500

 

400

 

300

 

 

 

 

200——

 

HEAT ADDED TO NaK IN REACTOR (Btu/sec)

 

 

 

100

    

 

 

 

 

 

 

 

 

SHELL, AND NaK
O [ TRANSFERRED TO HELIUM IN -
ROD-COOLING SYSTEM
-100
900 1000 1100 1200 1300

MEAN NaK TEMPERATURE (°F)

Fig. 18, Heat Removed by NakK.

33

 
DWG. 19108

 

1400 r

1200 \

HEAT ADDED TO NeK IN REACTOR

1000 \
\ !
800

 

 

 

 

 

HEAT TRANSFERRED (Btu/sec)

 

 

 

 

 

 

 

 

 

600
\
HEAT REMOVED FROM
NaK IN HEAT EXCHANGERS
400
200 |
0 I
900 1000 1100 1200 1300 1400
MEAN NgK TEMPERATURE {°F)
Fig., 19. Heat Added to and Removed

from NaK vs. Temperature.

fuel from the time 1t enters the inlet
mani fold until 1t leaves the outlet
mani fold.

The pressure drop of the fuel in
flowing through the i1nlet manifold,
core lattice, and outlet manifold 1is
due to factors:
curvature of the tubes or pipes,
contraction losses, expansion losses,
and losses 1n the pipe fittings and
square elbows. The formula for loss
1in pressure because of friction 1is

five friction,

L 2
op, = 45 =2

(88) ,
D 2g

 

where f 1s the friction factor given

by f= 0.046/Ny? for turbulent flow. (")

 

The curvature loss 1s given more
simply by
2
v
(89) 6P, = k2
2g

In this expression, k_ is a constant
that 1s taken from the curves of Cox,

34

Glen, and Germano (cf., p. 191 and
193 of ref. 12). The expression for
contraction loss 1s very similar to
that for curvature;

pvy
2g

where k 1s a constant that 1s dependent

AP = k

X

(90)

 

1

upon the ratio of the smaller area to
the larger. Values for k are found
in the work of McAdams (cf., p. 122 of
ref. 11). The loss caused by expansion
1s determined from
(v, = v,)?

(91) AP = p

2g
a fitting or
square elbow 1s calculated by using
the frictiondrop equation and replacing
L by L,; that is,

The pressure drop 1n

L 2
AP, = 4f S EY

(92) .
F
D 2g

 

 

The L_,'s for square elbows and various
fittings are given in ref. 13; they
depend upon the tube or pipe size.
The symbols used i1in the equations
above have the following meanings and
units:

AP = pressure drop, lb/ft?,

f = friction factor, dimensionless,

L = length, ft,

D = hydraulic diameter, ft,

£ = density, lb/fta,

v = average velocity given by fe?
of fluid per second/ft? of
cross section, ft/sec,

g = gravitational acceleration,
32.2 ft/sec?,

k = constant, dimensionless,

kc = constant, dimensionless,

v, = average linear velocity upstream,
ft/sec,

v, = average linear velocity down-
stream, ft/sec,

L_ = equivalent length of straight
pipe, ft,

Np, = Reynolds’ number.

The subscripts have
meanings:

the following

f = friction,
¢ = curvature,
X = contraction,

 

 
Y © expansion,
e = equivalent,
F = fitting.

In the pressure drop calculations,
it was assumed that the flow of fuel
would be equally distributed between
the six passes through the core.
Although this 1s a reasonable as-
sumption, 1t 1S not qulte true, as
will be demonstrated by the calculated
data.

The same value of density was used
throughout the calculation, but the
viscosities corresponded to the mean
temperatures in the inlet manifold,
core lattice, and outlet manifold.
The properties used were:

Density, p 187 1b/ft>
Viscosity, p
Inlet manifold 12,5 cp
Core lattice 9.0 ¢cp
Outlet manifold 6.8 cp

The flow rate of the fuel was considered
to be 0.150 cfs. Figure 20 shows
sketches of the inlet and outlet
manifolds. In this figure, the
outlets of the 1nlet manifold are
numbered 1 to 6 and the inlets of the
outlet manifold are numbered corre-
spondingly. Thus outlet 1 of the
inlet manifold 1s connected by a tube
to inlet 1 of the outlet manifold,
etc., and the fluid paths from the
entrance tee of the 1nlet manifold
to the exit tee of the outlet manifold
are given designations such as “l1-1.7
The geometries of the two manifolds
are shown in Figs. 21 and 22. The
these manifolds
were calculated by using Eqs. 88
through 92.

The tubes carrying the fuel through
the core lattice are often referred
to as “serpentine” tubes because each
tube has 10 bends and 11 straight
lengths. For each tube the equivalent
straight length 1s

L = 42,20 f¢.

pressure drops 1n

The pressure drop for each path through
the system being considered was easily
calculated by using the equal flow
rate assumption and Eqs. 88 and 89.

O

The results are tabulated in Table 7.
Of course, as can be seen, the flow
will not be distributed equally among
the six parallel paths through the
core lattice. In fact, the flow rate
may vary as much as 2.5%.  Thus a
maximum temperature difference of the
order of 9°F may exist in the fuel

exit temperatures from the various

DW!. i9109

tubes.

 

 

 

INLET MANIFOLD

 

 

OUTLET MANIFOLD

Fig. 20. Fuel Manifolds.

35

 
 

9¢

INCONEL CONCENTRIC \
REDUCER 1'% x 1-in. IPS~

INCONEL WELDING COLLAR

 

 

INCONEL LEG
{-in. IPS SCH 40

    
   
  

 

 

 

INCONEL LEG —~_ ;
1Y% -in. IPS SCH 40 .

| T AL
2
|

x

 

 

 

 

 
  

 

 

 

 

o N SECTION A-A

 

INCONEL STRAIGHT TEE
2x2x2-in. IPS——____|

 

 

 

DWG. E-A-1-1-3 B

—INCONEL LEG
1-in. IPS SCH 40

1.049-in.D REAM TO

1.053-in. AFTER WELDING

 

 

   

 

 

 

 

   

\ ey 2
|
INCONEL CASING TUBE

 

3l-in. 0D x Yg-in. WALL

INCONEL. LEG 7Y, - R

2-in IPS SCH 40

S

  
   

INCONEL OQUTLET LINE
2-in. IPS 5CH 40

™

&:s
<

  
 

"o ——INCONEL CONCENTRIC
" REDUCER 2 x 1%~ in. IPS

  
  
  

 

INCONEL LEG
1‘/2-1‘n. IPS 5CH 40

 

Fig. 21. Fuel Outlet Manifold.

O

 

24 in.

D500

 

 
OWG D-A-1-1-1-3A

 
  
 
 
  

LINLET LINE 1% IPS SCH 40 PIPE

ECCENTRIC REDUCER
12 IPS = 1 IPS—— -~

_REDUCER 2IPS - 1% IPS

 

 

 

 

 

 

 

 

 

 

 

 

INLET LEG o
1 IPS SCH 40 PIPE- - L
- TEE 2x2x2
e
1‘//}
«~—INLET LINE 2 IPS SCH 40 PIPE
“
COVER REMOVED
iN THIS VIEW——
- \‘\\\‘ o f ;
- i} /.‘
.. 59 /
e /
- - /.r/
//
.
. \ i
|
‘ /
— 21 H‘
COVER. BODY

 

 

 

 

 

 

ST

 

T— : f
| | f | \ > \\
| ! i 5 MINLET LINE \ INLET LINE
1 | i ‘ N TEE
: ‘ | 1 e
! } | h
: 1 ‘ \SLEEVE
NOTES:
- et ALL MATERIAL 1S INCONEL.

 

 

 

 

 

 

 

 

 

| |
i |
| | ALL DIMENSIONS ARE IN INCHES.
\
|
|

Fig. 22. Fuel Inlet Manifold.

37

 
 

TABLE 7.

 

PRESSURE DROP IN FUEL MANIFOLDS AND CORE TUBES

 

 

 

 

 

 

 

 

 

 

PRESSURE DROPS (lb/ftz) TOTAL PRESSURE DROP
1-1 627.17 613.0 172.5 1413 9.81
2-2 605.7 613.0 155.7 1374 9.54
3-3 627.7 613.0 155.7 1396 9.70
4-4 627.17 613.0 172.5 1413 9.81
5-5 605.7 613.0 121.90 1340 9.30
6-6 627.17 613.0 121.0 1362 9. 46
TEMPERATURE GRADIENTS IN THERMAL where

SLEEVES

The fuel tubes pass through the
pressure shell heads in the manner
shown schematically in Fig. 23a. The
sleeve connecting the pressure shell
and fuel tube 1s at the pressure shell
temperature at one end and at the
fuel tube temperature at the other
end. At the fuel tube outlet, the
fuel tubes are at 1500°F; the pressure
shell temperature at this point 1is
difficult to determine, but it 1is
probably not less than 1150°F, The
sleeve gains heat from the fuel tube
through the NaK between the tube and
sleeve; the NaK i1s assumed to be
stagnant. Heat lost by the sleeve
to the surrounding helium atmosphere
may be neglected because the helium
is at about 1200°F, The problem
was simplified by considering the
sleeve, the fuel tube, and the NakK
between the sleeve and the fuel tube
as a single cylindrical tube. The
temperature was assumed to be constant
across any cross section of this
cylindrical tube, and an
thermal

average

conductivity for the tube

was used. (The conductivities of NaK

and Inconel are nearly equal.) The

differential equation governing the

flowof heat in the sleeveis (Fig. 23b)
d*T

2

 

(93) ~ka = hc(1500 - T)

dx

38

k = thermal conductivity of the

composite tube, Btu/hr*ft?
(°F/fc),
a = cross-sectional area of the

composite tube, ftz,

BWG. 90

 

 

 

 

 

 

 

 

 

 

 

 

 

 

13, in. e 1 %, ~in-IPS SLEEVE
PRESSURE
SHELL
FUEL
TUBE —=— {
A--‘ 125N

| .
1

PRESSURE
SHELL

 

 

/COMPOS!TE TUBE

—h

(&)

FUEL

Fig. 23. Thermal Sleeve.

 

 
’s s
T = temperature, °F,

x = distance from pressure shell, ft,
h = fuel heat transfer coefficient,
Btu/hr+ ft?: °F,
¢ = circumference of inside of fuel
tube, ft.
The boundary conditions are
(94) T = 1150°F, at x = 0 ,
(95) T = 1500°F, at x = 0.1146 .

The fuel heat transfer coefficient
maybe evaluated by using the following
formula:

k 0,14
(96) h=0.027-—ae°-spr‘/3<i> ,
D v
3
where

k = thermal conductivity of the
fuel, Btu/hr-ft? (°F/ft),

D = fuel tube inner diameter, ft,
Re = Reynolds’ number for the fuel,
Pr = Prandtl’s number for the fuel,
(L

il

ratio of fuel viscosity at bulk
Fs temperature to fuel viscosity
at surface temperature,

For this case,

k = 1.5 Beu/hr-ft? (°F/ft),
D= 0,0929 f¢,
Re = 14,100,
Pr = 2,85,
0.965

ST
LE
\___/o
o
F-Y
i

h = 1250 Btu/hr-ft?:°F,
In Eq. 93,
k = 13.7 Btu/hr*ft? (°F/ft),
a = 0.0129 fg2,

c = 0,422 ft;
therefore

2T
6
- 2980 T = -4.48 x 10
dx?

The solution of Eq.

(97)

 

97 1is
(98) T=c, e~®H%r+ 1500 ,
The first boundary condition, Eq. 94,

gives

(99) ¢, * ¢, = =350,
The second boundary condition, Eq. 95,

b
654 I+Cz

gives

(100) 520 ¢, + 0.001923 ¢, = 0 .

Therefore

c, = 0,001294
and

c, = - 350
Thus Eq. 98 becomes

(101) T = 0.001294 54-6=x
- 350 e=%%-6x 4+ 1500

and the first derivative of T with
respect to x 1s

54,64«

dT
(102) — = 0.07065 e

dx
+ 19,110 e
Plots of Eqs. 10l and 102 are shown

in Fig. 24, The maximum temperature
gradient is about 1600°F per inch.

-54. 6!

Temperature in Fuel Tube Elbows as
a Result of Afterheat of Residual
Fuel. There 1s a possibility of fuel
remaining 1in the U bends of the
reactor fuel tubes after dumping.
This residual fuel will rise in
temperature because of internal heat
generation, and it 1s desirable to
know the maximum temperature that
will be attained. Since the activity
of the fuel decreases with time after
shutdown, an equilibrium will be
reached when the heat generation rate
1s the same as the rate of the heat
Once this equilibrium has been
the fuel temperature will

loss.
reached,
decrease.

The total heat generation rate of
the fuel is given 1in another portion
of this report. From this, the heat
generation rate of the fuel in a
U bend 1is

volume of fuel in U bendx

 

total fuel volume

where

Q = heat generation rate, Btu/sec
Qf = total heat generation rate of

fuel, Btu/sec.

The temperature rise in the U bend
will be greatest when the reflector
coolant system 1s filled with helium

(Fig. 9). Under this condition,
heat 1s removed from the tube by free
convection and radiation. The free

39

 
DWG. 19111

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1600 2000

1500 1600
£
T ,///// 5
£ 1400 1200 ~
w =
o =2
> / L
é 3
i S
= w
~ 1300 800 %
\\ c
@
w
a
\ s
L
-

1200/ : 400

\\
1100 0
0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

DISTANCE FROM PRESSURE SHELL (in.)

Fig. 24. Temperature and Temperature Gradient in Thermal Sleeve vs. Distance

from Pressure Shell.

convection loss 1s
(104) q, = hAG

where

q, = heat transferred, Btu/sec,

h

free convection heat transfer
coefficient, Btu/sec*ft?+°F,
A= surface area of tube, ft?,
¢ = temperature difference between
the surface of the tube and the
helium, °F.

The radiation loss 1s given by
0.173
3600

4 4
« <Ttube> (Tsink>
“1'\7100 / "~ \Too ’

€ = radiation emissivity, di-
mensionless,

absolute temperature of the
surface of the tube, °R,

(105) g4 =

 

 

where

tube

40

T,;,x = absolute temperature of the
sink (surrounding material),
°R.

The total heat loss 1s
(106) q = g9, 1t qp

Jakob has given Nusselt’s number
(cf., p. 525, Fig. 25-1, of ref. 14)
as a function of the product of
Grashof’'s and Prandtl’s numbers.
The value of Nusselt’s number for the
existing helium conditions was taken
from Jakob’'s work, and the free-
convection heat transfer coefficient
was determined.

The temperature change of the
fuel and the metal of the tube per
unit time 1s
(107) pr =279

chp

where

AT = temperature change, °F/sec,

 

 
chp = summation of heat capacities

of fuel and metal of tube,
Btu/°F,

w = weight of material, lb,

¢, = specific heat of material,
Btu/lb- °F.

The data required to make the
calculations are listed 1n the follow-
ing tabulation; only the portion of
the tube that 1s below the tube sheet
1s considered:

1. For the metal tube,
Area, A

Volume of metal

0.226 ft?

1.86 in.°
0.307 1b/in.>
0.11 Btu/1b*°F

Density, p
Specific heat, ¢,

2. For the fuel,

Volume of fuel
Density, p

4.74 % 1073 f£¢3
187 1b/f¢?

Specific heat, c, 0.26 Btu/1b*°F

Total fuel volume
of entire system

7.75 fi°

The temperatures of the fuel at
various times after shutdown were
calculated by numerical methods for
sink temperatures of 1200 and 1300°F,
The calculated data for the two sink

temperatures are given in Fig. 25.

 

 

 

 

 

 

 

 

 

 

 

1600 DWG 19112
1500
/z§\<rsoo¢ SINK

e \i\
W 1200°F SINK- — |
2 1400
<L
(4]
i
0o,
=
Ll
=
-
wJ
=
T 1300

1200

| 10 102 103 104 10°
TIME (sec)
Fig. 25. Temperature of Fuel at

various Times after Shutdown with Fuel
Remaining in U Bend.

FORMATION OF NONPLUGGING SOLIDS IN
FUEL TUBE IN THE EVENT OF A
SMALL LEAK

In the event of a small leak 1n a
fuel tube, 1t 1s possible that the
NaK entering the fuel tube will
precipitate solids that will adhere
to the walls of the fuel tube. This
deposit can build up to appreciable
thickness before causingany measurable
change in the flow through the fuel
tube. The temperatures which might
existin such a deposit were calculated
by treating the solid deposit as a
flat plate, with internal heat gener-
ation, cooled on one side by the
flowing fuel. The differential
equation and boundary conditions
governing the heat flow are

 

 

d%6 !
(108) = .4
dx? ko
dt
(109) =0, at x =0,
dx
d6
(110) -k — = h8 , at x =T,
x
where
€ = temperature difference between

a point 1n the solid and the
flowing fuel, °F,

x = distance from tube wall, ft,
q' = internal heat generation,
Beu/hr- fe?,
k = thermal conductivity of fuel,
Btu/hr* ft? (°F/fv),
7 = thickness of solid deposit, ft,
h = heat transfer coefficient from
solid to flowing fuel, Btu/hr- ft?
(°F/fu).
The heat transfer coefficient 1s
given by

k L 0,14
(111) h-= 0.027-5-Re°-8 Pr1/3<——) ,

ey,
where
D = equivalent diameter, ft,
Re = Reynolds’ number,
Pr = Prandtl's number,
M . fatio of viscosity at bulk

Hs temper ature to viscosity at
surface temperature.

41

 

 

 
Two cases will be considered; a
tube near the 1nlet where the fuel
temperature is about 1170°F and the
internal heat generation is l.64 x 107
Btu/hr*ft® on the reactor center line,
and a tube near the outlet where
the fuel temperature is about 1490°F
and the i1nternal heat generation 1s
about 0,85 X 107 Btu/hr*ft?® on the

reactor center line:
Fuel temperature, °F 1170 1490

Internal heat generation,

Btu/hr* ft> 1.64 x 107  0.85 x 107
Equivalent diameter, ft  0.0929 0. 0929
Reynolds’ number 7700 14, 100
Prandtl's number 5.20 2.85

u 0,14
— 0.97 0.97
p's

Heat transfer coef-

ficient, Btu/hr*ft2*°F 945 1260

The solution of the differential
equation, Eq. 108, 1s
df q'x
(112) —_= - t o, ,
dx k

 

2
q x
2k

The boundary conditions, Egs. 109
and 110, give

(113) g = -

 

+ c,x + €,

 

(114) c, = 0,
q.rT2 q"T
(115) c, = + —,
2k h

and Eq. 113 becomes

42

! !

 

-
(116) 6= (72 - 27 + 12
2k h
The maximum temperature in the
solid occurs at x = 0; therefore
qr7_2 q"T
= + = + — +
(117) T, = Oha* T, ot T Ty
where
T ., ° maximum temperature 1in the
solad, °F,
Tf= temperature of the flowing
o
fuel, °F.

A tabulation of values of some of the
terms of Eq. 117 for values of 7 of
0.1 and 0.2 in. is presented below:

 

Tf, °F 1170 1490
T, in 0.1 0.2 0.1 0.2
2
9'7’ o
, °F 380 1418 197 787
2k
4'7'
—\, °F 145 289 56 112
h
T , °F 1694 2878 1743 2388
max

The values for the maximum temper-
atures of the solid indicate that
unless the solids found when NaK and
fuel are mixed have a lower melting
point than that of Inconel there 1is
danger that a small leak might result
1in the formation of a solid deposit
which would reach a high enough
temperature to melt the Inconel tube

wall.
Chapter 3
FUEL HEAT DISPOSAL SYSTEM

PERFORMANCE OF FUEL HEAT DISPOSAL
SYSTEM

Heat Transfer. The heat added to
the fuel 1n the reactor 1s disposed
of by circulating the fuel to a heat
exchanger where 1t gives up its heat
to helium. The hot helium 1s then
passed through a heat exchanger where
it gives up 1ts heat to water. The
fuel and helium recirculate; the
water 1s dumped. There are actually
four fuel-to-helium heat exchangers
and four helium-to-water exchangers.
These exchangers are arranged (Fig. 1)
so that all four fuel-to-helium
exchangers are 1n parallel on the
fuel side and all four helium-to-water
exchangers are in parallel on the
water side. The helium circuits are
arranged so that there are two parallel
circuits and each circuit contains
two fuel-to-helium and two helium-to-
water exchangers in series (the
exchangers alternate, of course).
Figures 26 and 27 show the details
of the two types of heat exchangers.
The performance of the fuel heat
disposal loop over a range of operating
conditions 1s calculated below.

The fuel heat disposal
obeys the following relations:

system

(1) Q = Wf ¢ (Tf - Tf) ,

(2) Q = wHe CHe (Tl;e - THe) !

(3) Q=W ¢ (T -T) ,

(4) Q = hf-He Af-He 6f-He !

(5) Q - hHe-w He s w He-w !
where

Q = heat transferred, Btu/sec,

W = weight flow, 1b/sec,

T' = maximum temperature, _F,

T = minimum temperature, F,

¢ = specific heat, Btu/lb-*°F,

h = heat transfer coefficient,

Btu/sec‘ft2'°F,
A = heat transfer area, ftz,
§ = effective log mean temperature

difference, °F,

and the subscripts refer to

f = fuel,
He = helium,
w = water,
f-He = fuel-to-helium heat exchanger,

He-w = helium-to-waterheat exchanger.

The heat disposal systemis operated
with the following restraints and
conditions:

1. The fuel flow 1s held fixed at
28.2 1lb/sec.

2. The water outlet temperature
1s thermostatically controlled and
held fixed at 135°F.

3. The water inlet temperature
will vary with the time of year, but
for this analysis 1t has been assumed
to be fixed at 70°F.

4, The mean temperature of the
fuel 1s fixed because the reactor 1is
sel f-controlling to a constant mean
temperature. The design mean tempera-
ture is 1325°F. Calculations were
also carried out for mean temperatures
of 1225 and 1425°F ¢to give some
indication of the effect of a change
in mean fuel temperature.

The helium weight flow 1s the
independent variable; the flow 1s
controlled by variable speed hydraulic
motors which drive the two helium
blowers. The dependent variables are

Q, W, T}, Tf, T;e, and Ty _. Thus

there are six unknowns and si1x equa-

ti1ons: the five equations listed
above and the one i1nherent i1n the
fourth condition, namely:

T + T

f f
(6) ———= a constant .

2
Before these equations can be

solved, the heat transfer coefficients

and areas must be calculated for the

two exchangers. The over-all hA’s are

(1) (hA), .

 

1 1 1 -1
= + +
(hA) (hA) g, (hA) J

f-He

43
vy

 

  
  

————2f1 3in —— e e e in. = g S

- 2Y-in. SCH 160 INGONEL
s PIPE FOR MANIFOLDS
- e o

MATERIAL AND MATERIAL SPEGIFICATIONS

 

 

 

 

DESGRIPTION MATERIAL SPECIFICATION QUALITY
B SHELL, TUBE SUPPORT PLATES | STAINLESS STEEL | ASME-5A-240 | GR 5 TYPE 304
f MANIFOLD , WELD ELL AND NICKEL , CHROME

 

PIPE AND IRON ASTM*B-16? INGONEL

SHELL SUPPORTS, MANIFOLD
SUPPORTS, FLANGE BARS AND | STAINLESS STEEL |COMMERGIAL | TYPE 304
BAGKING BARS

 

 

 

6 in—
. 16 in_44444““ -

-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

   
 
   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

: TUBES N FOME | ASTM-B-167 | INGONEL
| FINS _ | sTANLESS STEEL | TYPE 304
- BOLTS STAINLESS STEEL | ASME-SA-193 | GRADE B-B
”””””””””” 'DESIGN DATA
SEGTION G- { | DESIGN PRESSURE SHELL SIDE 6 in WATER
2. DESIGN PRESSURE TUBE SIDE 25 psi
< 3 DESIGN TEMPERATURE SHELL SIDE 1500 °F
2 WY e @ 4. DESIGN TEMPERATURE TUBE SIDE 1500 °F
~ 5. HYDROSTATIC TEST PRESSURE TUBE SIDE 500 psi
. -3~in. TAPPED HOLES o ’
Y, in, E 38-g-in | et | . _
2% 1 FOR GAP BOLTS f% 20 7 in.
77_:: “A;"'""'fff'fff” i E fi
o] Lo o o o o Q o ,
Lo A D e e e - o
T | j T T A
€ | o |- ——— i + 1 TUBE VENT AND i Yy |0 o
‘F\:\l - / - T T T 7 DRA!N,? STRAIGHT A \/ “ i
™ e e . LENGTHS " " S
oL = N (U ‘ l
¥ 5] T [ q’ |
T HET —_— Pl e | | : ;/,fmj'—l%;_f I-in. OD NO. 12 BWG
o = - - : T N M ; (‘: | | . INGONEL TUBES, 43
c R } ) G | o~ | ||+ STRAIGHT LENGTHS,
| u ] - 1 || | 2=inOD STAINLESS
o |- - - \ (J O | | | STEEL FINS 0.024-in
I i _ i ! ! [ | | THICK .
"I [ | I \ A e P! !
TP — — SUPPORT PLATE TO | ! ; A =
| b—— 1. ! -
=—lerr— — = BE GUT IN TWO | M | |
A e S SEGTIONS AND | | 0 1 N
o | o o - WELDED TOGETHER | | ! T |
. AFTER ASSEMBLY—. | | A% St ||
S s — i 0O 1.
. . - F
o= (T I
0 - ! - 1
L — —-- £ ! ‘
1‘ O‘/- f/\/;_—“ — e = O }
c | H} — ——————— — w L |
o R | !
- 4 Y _ P |
o l fiv[} - S - @ : 1 }
L 5 L
i T — T T — 1 -
© Lo o Q\ ' o < 2 e . /
L N . e ey 6 _
I 54 ~in. OD ¥ 0065 WALL L] %8 * 17pmin. BAR
DRAIN TUBE SECTION E-E

Fig. 26. Fuel-to-Helium Heat Exchanger.
  
 
 

2 ft—a'gin-

 

 
  

T

i =

1 o

| =z

w

i -

: ¢ =Z

5

. ="

£ rc\]>

&L =

" g

w [

| e

- ul
=
<

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

       
    
  

L 3 ft=2 in——————e

UNCLASSIFIED

DWG. A-3-2-214

 

"~ Y- in DIA. RODS

“Yg-in. THICK FULL FACE ASBESTOS
CLOTH GASKET.

5
Y

 

0 0 C 0 0.0.0 0 O

 

 

 

 

 

 

 

 

 

“Yg-in. THICK FULL DURABLA GASKET Yp-in
WIDE AT PARTITION.

 

 

 

 

 

 

 

 

|
r
|
|

e 2 H=B Y i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e e e — 3ft—2'% in. e ‘
' 3, . P DESIGN DATA
*"—'[ =— 29 in. i =
% T - - T SHELL | TUBES
} © ! O; DESIGN PRESSURE 6-in Hy0 | 50 psi
b = o s, DESIGN TEMPERATURE | 1S00°F | 340°F
= —_
3 o
}
l ] ! ; ° FEET
] L | ; ;
o i f T T
0O 0 & 0 0 0 0 0 0 O 1 1 0 !
HALF VIEW CF STAT. HEAD WITH COVER REMOVED
MATERIAL AND MATERIAL SPECIFICATIONS
DESCRIPTION MATERIAL SPEC. QUALITY
SHELL FLANGE AT TUBE SHEET, GAS INLET FLANGE, ASME Sa—
SHELL AND SHELL COVER PLATE, BLANKING-OFF PLATE, |STAINLESS STEEL | 40 GRS TYPE S-304
TUBE SUPPORT BAR, BACKING-UP BAR.
STAT. HEAD PLATES, STAT. HEAD FLANGES, GAS GUTLET ASME
FLANGE, STAT. HEAD PARTITIONS, STAT. HEAD TIE PLATES,| FIREBOX STEEL Ao GRADE C
STAT. HEAD COVER PLATE, TUBE SHEET.
3-in,, 150 Ib. WELDING NECK FLANGE. FORGE STEEL ASME SA-{BI GRADE 1
3-in SCH. 40 PIPE. SEAMLESS STEEL | ASME SA-106 GRADE A
TUBES, TUBE BENDS. SEAMLESS STEEL | ASME SA-179 MIN WALL
FINS COPPER COMMERCIAL
STAT. HEAD GASKETS. DURABLA COMMERCIAL
SHELL GASKET SMEARED BOTH SIDES WITH PLASTISEAL. | ASBESTOS CLOTH | COMMERCIAL
BOLTING FOR TUBE SHEET AND SHELL FLANGE. STAINLESS STEEL | ASME SA-193 B8
NUTS AT TUBE SHEET AND STAT. HEAD COVER PLATE. BETH OIL QUENCH | ASME SA-194 CLASS 2H
BOLTING FOR STAT. HEAD COVER PLATE. ALLOY STEEL

 

 

 

 

Fig. 27.

Helium-to-Water Heat Exchanger.

ro 4L

45

 
and

(8) (hA),, _

w

 

1 1 1 ]!
= + +
{ (hA),,  (hA)_ (hA) } ’

He= w

where the subscript m refers to the
metal of the tube wall.

The heat transfer coefficients of
the fuel are calculated from the
curve of ref. 11 (p. 193). The heat

transfer coefficient of the water 1is

k ; 6,14
(9) h =0.027 —* Re%:8 ppi/3( L
w D v w /-Ls

 

w v

(15))

(Sieder-Tate relation where

k = thermal conductivity, Btu/sec* ft?
(°F/ft),
D = equivalent diameter, ft,
Re = Reynolds' number,

Pr = Prandtl’s number,
L = viscosity at bulk temperature,

1b/hr* ft,

p, = viscosityat surface temperature,
1b/hr- ft.

The liquid-side areas are:

(10) Af = WDfo_He

and

(11) A, = 7D Ly,

where L 1s the total effective length
of the heat exchanger tubes 1n feet.
The resistance of the metal 1s

DO
In 'B“
f/f-He

 

 

 

1
(12) { J =
()] o 2k ae Ly,
and
Do
In| —
Df He.w
(13) |——— = ,
{(hA).J 27T(km)l'le-w LHe w
He - w -

where D is tube outside diameter 1in
feet.

The heat transfer coefficients on
the gas side are taken from a corre-

46

lation by Kern of data by Jameson
and Tate and Cartinhour (cf., p. 555
of ref. 8). Kern’s correlation curve
may be approximated by the following
equation:

(14) h

He
k 0,14
He Ret723 p 173 [ K
D eHe I‘He s
He /J": He

= (0. 092

 

where DHe for the fin and tube con-
figuration 1s defined as

 

2(A0+Af)
(15) Dy, = —— . ,
€ m(projected perimeter)
where
A = tube outside bare area, ft2,

0
A, = fin area, ft?,
and the projectedperimeter 1s the sum
of all the external distances 1in the
plan view of the finned tubes (ft).
The mass velocity used i1n the
Reynolds’ number is computed from the
free flow area 1in a single bank of
tubes at right angles to the gas flow.
The helium—-side heat transfer area 1s

(16) Ag, = A, + an

f ’

where i 1s the fin efficiency.(16)

Table 8 lists the pertinent infor-
mation obtained from the heat exchanger
designs (Figs. 26 and 27) and Egs. 7
through 16. Values are given for
three different flow rates of helium
and of water; curves were plotted
from these values and used 1n the
solution of Egs. 1 through 6.

The data of Table 8 were used to
solve the six simultaneous equations
(Eqs. 1 through 6} for various values
of the independent variable - helium
weight flow. (Inasmuch as the water
heat transfer coefficientis a function
of the water flow rate, 1t was necessary
to assume a water flow rate for each
helium flow rate and solve the equations
by i1teration.) The results of the
solution of Egqs. 1 through 6 are
shown in Figs. 28 and 29, where the
heat transferred, maximum fuel temper-
ature, and maximum and minimum helium

 
TABLE 8. HEAT EXCHANGER DATA

 

 

 

 

 

FUEL-TO-HELIUM HELIUM-TO- WATER
HEAT EXCHANGER HEAT EXCHANGER
. Tube outside diameter, in. 1.0 0,625
Tube wall thickness, in, 0.109 0,049
Tube material Inconel Steel
- Fin outside diameter, in. 2.0 0.75
Number of fins per inch 7 16
Fin thickness, in. 0.024 0,014
| Fin material Stainless steel Copper
| Tube transverse pitch, in. 2,75 2,25
Tube longitudinal pitch, in, 2.43 1,47
Number of tubes per exchanger 43 360
Number of tubes transverse to flow
(per exchanger) 9 12
Exchanger height, in. 24,75 27.0
Exchanger width (also active tube
length), in. 26,25 27.5
Total active tube length per exchanger,
ft 93.65 825
Liquid flow area per exchanger, fe? 0.01006 (First bank 0.0545
group*), 0,00671
(Second bank group*)
. Liquid flow rate per exchanger, lb/sec 7.05 10,61, 7.08, 3,54
Liquid velocity, ft/sec 3.74, 5.61 3.11, 2,08, 1.04
Reynolds’ number 5890, 8840 19,000, 12,700, 6,340
. Liquid heat_transfer coefficient,
Btu/sec* ft?*°F 0.24, 0,38 0.27, 0.20, 0.11
Liquid heat Eransfer area per exs-
changer, ft 11.15, 8,02 114
(hA) || 414 PeT exchanger, Btu/sec*°F 2.676, 3.048 30.8, 22.8, 12.5
Total 5.724
(1/hA) |; .uiqa Per exchanger, sec*°F/Btu | 0.175 0.0325, 0.0439, 0,080
(1/hA). per exchanger, sec*°F/Btu 0.137 0.00456
Helium-side equivalent diameter (DHe)’
in. 1, 447 0.670
Helium free flow ratio 0.575 0.71
Helium flow area per exchanger, fe2 2,60 3.66
| Helium flow rate per exchanger, 1b/sec 1.5, 1.0, 0.5 1.5, 1.0, 0.5
Helium Reynolds’ number 4050, 2630, 1290 1330, 860, 420
Helium heat transfer coefficient, 0.0083, 0,0061, 0.0080, 0.0059, 0,0035
Bru/sec*ft?+°F 0.0036
Fin efficiency 0.37, 0.43, 0.55 1.0, 1.0, 1.0
| Helium heat transfer area,** 2 115, 131, 162 402, 402, 402
| (hd),, per exchanger, Btu/sec*°F 0.955, 0.800, 0.580 3.22, 2.37, 1.40
i (1/hA),_ per exchanger, sec* F/Bru 1,05, 1.25, 1.72 0.310, 0,422, 0.714

 

*The liquid sideof the fuel-to-helium heat exchanger is divided into two groups of tubes. The first
bank group has three tubes in parallel and is connected in series with the second bank group, which has
two tubes in parallel.

. 44 % ) ea’ varies with helium flow rate because the fin efficiency varies with helium flow rate {(cf.,
Eq.

47

 

 
 

 

700 — ‘ E—— OWG 19113

MEAN FUEL TEMPERATURE (°F |
| | | T

   

 

1600 - e
1500

1400

 

1300

MAXIMUM FUEL TEMPERATURE (°F)

 

1200—

  

3000

 

 
     

 

 

 

 

 

 

T
=
2 L
D 2000F-- - I
gy
i |
L
w) : :
S ' ' ]
g 1000— e b
T o - L
O 2000 4000 6000 8000 10000

HELIUM VOLUMETRIC FLOW (cfm)

Heat Transferred and
Helium

Fig. 28.
Maximum Fuel Temperature vs,
Yolumetric Flow.

temperatures are plotted against
helium volumetric flow for the three
different values of mean fuel tempera-
ture investigated.

At design point,
by the reflector coolant
cooling circuits. Therefore,
1s to be removed by the fuel.

Film Temperature. Because the fuel
hasa high melting point (about 970°F),
the minimum fuel film temperature must
be calculated to determine the situation
with respect to freezing of the fuel
film. The minimum fuel film tempera-
ture at any point may be calculated
from the relation

700 kw 1s removed
and rod

2300 kw

Q

(17) T =T -,
fmin fbulk (hA)f
where
Tf = minimum fuel film temper-
min ature at the point 1in
question, °F,
48

DWG. 19114
900

BOO

700

600
< |
o QUTLET OF FUEL-TO-HELIUM EXCHANGER
..D_ 500 b
< o>
& 5
[
a :
= Pa
i bog
= 400 - O
2 5
% - | MEAN FUEL
@ S | TEMPERATURE
T @ {°F)
300 o
L
g
o

200

j
160 INLET OF FUEL=TO-HELIUM E

 

o 2000 4000 6000 8OO0
HELIUM VOLUMETRIC FLOW {cfm)

10,000

12,000

Fig. 29. Maximum and Minimum Helium
Temperature vs. Helium Volumetric Flow.

Tf = bulk fuel temperature at the
bulk . . : OF
point in question, .

The other symbols mean the same as
they did previously, except that the
fuel heat transfer coefficient, hf'
must be evaluated locally, that is,
at the point 1n question. Several
critical points were investigated 1in
the fuel-to-helium exchanger, and a
plot of the lowest of the temperatures
vs. helium volumetric flow (and various
mean fuel temperatures) is shown 1in
Fig. 30a.

Pressure Drop. The pressure drops
in the fuel and in the helium have
been calculated for the exchangers.
The fuel pressure drop was calculated
by using the customary Fanning equa-
tion¢!!) for the straight sections,
and the curves presented by Cox and
Germano¢1%) for the bends, exists, and
entrances. Since the fuel flow 1is
constant and the properties do not
vary substantially i1n the range of
mean fuel temperatures considered, the

 
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

DWG. 19415
1500 - ,
| | ]
: | |
— i
S 1400 xr——-—-————------ e ey |
o |
@
2 |
g
a
g 1300 N ™ i | I
= co
3 N . MEAN FUEL
1200 S TEMPERATURE (°F }|
> \‘\
s l =~ 1425
Z 100 \ 1 |
= \\\\\J o] s
(G’) \ |
1000 I T~ 1225 |
=T
e
Q)
i
[&)
2 Sr—1
% o =
o O|
5T @
wE ol /
o8 ‘ p A
” s | &
T3 | |
oz FUEL-TO- HEL UM EXCHATEEB//’
%04 | |
oo
o // HEL {UM - TO- WATER
! EXCHANGER
(5) |
C | ‘ | i |

 

 

 

 

 

0 2000 4000 6000 8000 10,000 12,000
HELIUM VOLUMETRIC FLOW (cfm)

Fig. 30. Minimum Fuel Film Tempera-
ture and Helium Pressure Drop vs.
Helium Volumetric Flow.

fuel pressure drop 1s constant and
equal to about 17 psi1.

The pressure drop in the helium
was calculated by the method given
by Gunter and Shaw: ¢17)

H, = viscosity at surface tempera-
tures, lb/hr-ft,
D;. = equivalent diameter (cf.,

Eq. 19), ft.
The equivalent diameter in the
Gunter-Shaw correlation 1s

, 4 X net free volume
(19) DH = ,
© A, + A
f ©

where Af and A have the same meanings
as previously.

The values of Dy,  are 0.0438 ft for
the fuel-to-helium exchangers and
0.1365 ft for the helium-to-water
exchangers. The friction factor, f,
which is a function of Reynolds'
number, 1s evaluated from the Gunter-
Shaw correlation. The helium pressure
drop 1s plotted against helium volu-
metric flow for both exchangers 1in
Fig. 30b. The helium pressure drop 1s
not greatly affected by the mean fuel
temperature, and thus the curves of
Fig. 30b are valid for the range of
mean fuel temperature considered.

The pressure drops in the fuel and
in the helium for the remainder of
the fuel and helium circuits have been
calculated by using the Fanning
equation and the Cox and Germano data.
Figure 1 shows the pressure at various
points in the fuel circuit. The fuel
pressure drop external to the heat
exchangers is 40 psi; the total fuel
pressure drop is 57 psi (Fig. 2 shows
the fuel piping). Figure 31 shows the

 

 

 

(18) AP = 5,305 x 10°1'°

;. <

AP = pressure drop, in. of H,0,

f = friction factor,

Gy, = helium unit weight flow,
1b/sec* ft2,
LP = length of helium flow path, ft,
s = average helium specific gravity,
S, = transverse pitch, ft,
S, = longitudinal pitch, ft,
@ = viscosity at bulk temperature,
1b/hrft,

2 ! > .
fGHeLp DHe 0.1 Sl o
" 0.14 \ S, 3: ’
p’s He

 

helium ducting for the fuel circuit.
The pressure drop external to the
heat exchangers and the total pressure
drop are plotted against helium volu-
metric flow in Fig. 32.

TEMPERATURE IN FUEL SYSTEM BECAUSE
OF AFTERHEAT IN THE EVENT OF
COMPLETE PUMP FAILURE

If there were a complete power
failure and all the fuel remained 1n

49

 
50

 

ELBOW

 

 

 

 

 

 

 

 

 

 

 

 

He TO H,0
HEAT EXCHANGER-— =

 

 

 

 

 

UNCLASSIFIED
DWG. D-A-3-2A

ELBOW

 

 

 

HEAT BARRIER

 

 

 

 

 

 

 

HEAT BARRIER

 

 

 

 

 

 

 

te——He TO Hy0
HEAT EXCHANGER

 

 

 

 

 

 

 

i /FUEL TO He
! HEAT EXCHANGER
L 7 /
/
/ /1/ ( TRANSITION PIECE

ELBOW

 

Fig

 

 

 

 

 

 

PLAN

ELEVATION

31. Helium Loop for Fue

—— HELIUM BLOWER

 

 

-———HYDRAULIC MOTOR

L

 

 

 

 

 

 

 

 

 

 

1 Circuit.
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

DWG. 19116
2.5 i /
o
oJ
T
£ 20 |
0 /1
O '
;
:
£ 15 -
’ Y, > |
2 T =
@ | : |
: / 5l
o 1.0 S ;
: gl
T g‘
: oD
/ 2
05 5 ql
i [0
i | l
| ; NGERS
EXTERNL\L 10 HE AT EXCHA | -

 

0 2000 4000 6000 8000 10000
HELIUM VOLUMETRIC FLOW (cfm)

Fig. 32. Helium Pressure Drop vs.
Helium Volumetric Flow.

the system, 1t would be desirable to
know the maximum temperature the fuel
would attain (Fig. 1). The tempera-
ature rise 1n the fuel will be due to
internal heat generation. However,
since the heat generationrate decreases
with time after shutdown, at some
temperature an equilibrium will be
reached between the heat generated and
the heat lost from the system. The
maximum fuel temperature will be
attained at equilibrium.

reached,

As soon as
the maximum has been the
temperature will decrease.

The total heat generation rate of

the fuel 1s given as a function of

time after shutdown in chap. 9. Heat
is lost from the components of the
fuel system by free convection to
the helium in the pits and by thermal
radiation to the equipment and walls.
This heat loss is given in chap. 6.

The change in temperature of the
fuel and metal of each fuel system
component per unit time 1s

Qf "Ql

 

(20) AT =
2we
P
and
(21) ©
_ volume of fuel 1n component < 0
total fuel volume £
where
AT = temperature change, °F/sec,

total heat generation rate 1in
the fuel, Btu/sec,

heat generation rate in the fuel
in the component, Btu/sec,

Q, = heat loss from the component,
Btu/sec,
Zwfl;= heat capacity of component
plus fuel, Btu/°F,
w = weight of material, 1b,
c, = specific heat of material,

Btu/1b- °F,

The most critical pointsin the fuel
system under the condition postulated
are those with the smallest surface-
to-volume ratio. The components in
this categoryare the reactor (Fig. 9),
the surge tanks (Fig. 33), and the
2-in. pipe lines.

Table 9 gives the data needed to
calculate the temperature rise 1in
these components.

 

 

TABLE 9. DATA FOR CALCULATION OF TEMPERATURE RISE OF VARIOUS COMPONENTS
2we FUEL VOLUME TEMPERATURE
COMPONENT
(Btu/sec) (£.9) AT FAILURE (°F)
Reactor 4166 1. 96 1400
Surge tank 48. 6 0. 83 1150
2-in. pilpe
(per foot of length) 1.52 0.023 1500

 

 

 

 

51

 
 

S

 

 

 

 

   

NOTES:
ALL DIMENSIONS ARE IN {NCHES
ALL MATERIAL S INCONEL

WELDING CAP
6% oD x % THiCK

HANDHOLE
6% 0D x Y% walLL

WELDING CAP
10Y% 0D x 1 THICK

  
 
   
   
   

-7

SPARK PLUG PROBES
#4mm SPECIAL -~ -

UNCLASSIFIED
DWG E-A-3-1-33A

 

LIQUID-LEVEL WELL
3PS x 18% LONG—={

 

TANK SHELL
10%2 0D x 24 LONG —

 

 

 

AR £ 4!

 

 

 

 

BAFFLE PLATE B

 

 

 

 

 

LIQUID-LEVEL
STATIC LEG
3% IPS x 26 LONG—-~—|

    
   

 
 

LIQUID INLET AND LIQUID QUTLET
ROTOMETER 2% SCH 40

3 1PS — ———fi ] LR 90° ELL-

 

 

WELDING CAP
1050 OD % ¥ THICK

 

 

%_-, - 5 [ - ,._;_.- e 8 . - - __—:‘ e 9 . .
‘,__.....__ - I 26 . S
Fig. 33. Fuel Surge Tank.

 

 

 

 

 

 

 

 

       
   
  

 

—-WELDING CAP
“Jo'% 0D x 1 THICK

 

 

 

 

 

 

 

 

 
s

For the metal

c, = 0.11 Btu/lb-°F,
o = 0.307 1b/in. 3,

. and for the fuel
c, = 0.26 Btu/lb*°F,
o = 187 1b/ft3,

The total volume of fuel 1n the system
was taken to be 7.75 ft3, and the sink
temperature was takenas 130°F or 590°R.

The maximum temperature to be
expected 1in each component was calcu-
lated numerically by using Eqs. 20
and 21, The following results were
obtained:

E
COMPONENT MAXIMUM TEMPERATUR

(°F)
Reactor 1400
Surge tank 1900
2-in. pipe 1520

Curves of temperature vs. time after
shutdown are shown in Fig. 34.

G.

"7

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1900 e
1800 (
1700 ]
SURGE TANK#
1600
L
& 2-in. PIPE
o
< 500
& %
a
p-
LJ
-
REACTOR
14009 <> 1/
1300
1200 —
¢
1100
f 10 102 10> 1ot 10°
TIME {sec)
Fig. 34. Component Temperature vs.

Time after

Shutdown.

53

 

 
 

Chapter 4
REFLECTOR COOLANT HEAT DISPOSAL SYSTEM

The heat added to the reflector
coolant (NaK) in the reactor is dis-
posed of by circulating the NaK to a
heat exchanger where it gives up its
heat to helium., The hot helium is then
passed through a heat exchanger where
1t gives up its heat to water. The
NaK and helium recirculate: the water
1s dumped. There are two NaK-to-helium
and two helium-to-water
1) so that
the liquid sides of each pair of ex-
changers are in parallel. The helium
two para]_le]
circuilts, and each circult contains one

exchangers
exchangers arranged (Fig.

circuit consists of

NaKk-to-helium and one helium-to-water

heat exchanger in series. Figures 35
and 36 show the details of the two
tvpes of heat exchangers. The per-

formance of the NaK heat disposal loop
over a range of operating conditions
1s calculated below.

The NaK heat disposal syster obeys
the following relations:

(1) @ = Fyak CNak (7\Iax - Ty.x) >
(2) Q = Wfle Che (THE - THe) )
(3) Q=W,c, (T, - T,)
(4) Q - hNaK-He ANaK-He rNal‘{-He ’
(5) Q - hHe-w AHe-w 6He-w '
where
) = heat transferred, Btu/sec,
W = weight flow, lb/sec,
T' = maximum temperature, °F,
T = minimum temperature, °F,
h = heat transfer coefficient,

Rtu/sec* ft? (°F/ft),

= heat transfer area, ftz,

= effective log mean temperature
difference, °F,

and the subscripts refer to

DA
|

NaK = Nak,
He = helium,
w = water,
NaK-He = NaK-to-heliumheat exchanger,
He-w = helium-to-water heat ex-
changer,

54

The heat disposal system 1s operated
with the following restraints and
conditilons:

1. The NaK flow
23 lb/sec.

2. The helium volumetric flow per
exchanger 1s held fixed at 2000 cfm at
This fixes

1s held fixed at

blower inlet temperature.
Wy, once Ty  is known, _

3. The water outlet temperature is
thermostatically controlled and 1is

held fixed at 100°F.

4. The water inlet temperature
will vary with the time of the year,
but for this analysis it has been

assumed to be fixed at 70°F.

The heat transferred is chosen as
the independent variable in the system,
by the heat
This
I

NaK'

since 1t 1s determined
added to the NaK in the reactor.
leaves as dependent variables T
TNaK’ Tée, Ty, and W _. There are
therefore five unknowns and five
equations (Egs. 1 through 5). Before

can be solved, the
heat transfer coefficients and areas
must be calculated for the
exchangers.

these equations

two heat

The over-all hd4's are:

(6) (h), . .
o
(hA)NaK (hA)He (hA)m NaK-He
and
(1) (ha),
e, o
@, G, G,

where m refers to the metal of the

tube wall,

The liquid-side heat transfer coef-
ficients are:
kNaK 0.8
(8) hyak = (7.0 + 0,025 Pe”* %)

Nak

 

 
S¢S

EWG. A-3-3-21A

         

 

 

 

 

 

 

 

 

 

 

 

 

17
——I 4l SUPPORT PLATES MATERIAL AND MATERIAL SPECIFICATIONS

———————————————— PART MATERIAL SPECS. QUALITY
SHELL, TUBE SUPPORT PLATES. STEEL ASTM A-283 | GRADE C
MANIFOLD, WELD ELL, AND REDUCER. | NICKEL , CHROME, IRON | ASTM B-167 | {NCONEL
SHELL SUPPORTS, FLANGE BARS ~
BACKING BARS AND AIR STOPS. STEEL ASTM A-283 | GRADE C
TUBES NICKEL, CHROME, IRON | ASTM B-167 | INCONEL
MANIFOLD CAP AND TRANSITION. NICKEL, CHROME, IRON | ASTM B-167 | INCONEL
FINS STAINLESS STEEL TYPE 304
MANIFOLD SUPPORTS STAINLESS STEEL TYPE 304

INCHES
2 0 4 8 12 16

VENT TUBE

NOTE. ALL DIMENSIONS IN INCHES.

SECTION A-A

c
‘1 25%, 39Y% __J‘

 

16Y,

 

 

 

 

21,
16 Yy

 

 

 

 

 

 

 

 

 

 

 

 

SECTION C~-C

ELEVATION

Fig. 35. NaK-to-Helium Heat Exchanger.

 

 

 
9¢g

 

WG I'l---!-lt“.;lflu

 

lg-in TH'E. FULL FACE ASBESTOS

 

 

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e o MATERIAL AND MATERIAL SPECIFICATIONS
Yo % 1-in BAR WELDED TO . —- DESCRIPTION MATERIAL | SPECFICATION | OQUALITY
_ BLANKING-OFF PLATE - | = SELL, SHELL OOVER B, |
T—@—' b : BLANKING-OFF &, TUBE
| ) f SUPPORT BAR, BACKING-UP
- i BARS FOR SHELL FLANGE, | swEEL ASTM - A-283 | GRADE C
: SHELL REINFOR(ING BARS,
| = SHELL SUPPORT R, ‘
, WA BAFFLE BOX .
| STAT. HD FLANGES, SUPP'T [
| R'S, STAT HD P ; 5 |
| - STAT WD TIE RS, STAT, HO. e DO |ASTM-5-285 | GRADE C
1 %% -in BARS WELDED TO — COVER R, TUBE SHEET STAT | X
-in WELL | | O R'S i
APIERET TLAtET—= | | - B50-1b WELDING NECK FLANGE, FORGE STEEL |ASTM-A-181 | GRADE |
L [~ .’ . < 1'5-in 5CH 4D PPE | SEAMLESS STEEL| ASTM-A-106 | GRADE A
= | .| H-'_ — o T
= | I = | i i B e TUBES | SEAMLESS STEEL| ASTM-A-1T9 | umy. wall
: ' E | | : | | s FING | COPPER COMME RCIAL
¥ 1 | | ! (11 ' STAT HD GASMETS DURSBL A COMMERCIAL
\ — | | | g oa. -4 !;L‘s-rn THE
~in BLANKING-OFF PLATE- , e—————y TAT -
iy : | ‘ J ‘ i 1 j“r SHELL GASKET &F TUBE SHEET ASRESTOR Lo msm%
| | | 1 ye TAP BOLTS AT BLANKING-OFF BLLOY STEEL COMMERTIAL
| ‘ . | | | :s,l_l_T BAFFLE : f
| | . BOLTING AT STAT HD AND | asTh- | 125000 TS
| L ' T SRET ALLOY STEEL | ASTM-8-93 B7| \65000 e
i | TiE AOCS, SPACERS AND NUTS STEEL | COMMERCIAL |
i . = T
i MUTS AT TUBE SHEET AND BETH OIL A
| | bl _Ir::- CNVER i APE UENCH | ASTM-a-134 |cuass 21
e | | | | i t
| !
T
& 4 Ii ] 1)
Ya-in TH'N FuLL F{E‘E ASBESTOS fil iy ba-it. DIA VENT HOLE
ke, g — 8N~ 0Tg In = _l—in THK DURABLA GASKEY
- —t
r | DESIGN (ATA
SHELL TUBES
| CESISN PRESSURE 6 o WATER  SCips
CESIGN TEMPERATURE 000°F 340°F
< | HYDROSTATIC TEST PRESSURE 75 psi
2 DESIGN TEMPERATURE AT TUBE SHEET  BSO"F
e MANXIMUM METAL TEMPERATURE  SHELL SI0E SO0™F

 

 

 

2

 

 

SOE FLEVATION

Zf I%-in o Yy-in X 2-in STEFENMG BAR

e ! Ya-in Doa DR HOLE
S x1% ~in BAR ALL AROUND DPENING BOTH SIDES ¥5g-in TH'K DURABLA GASKET

t% -0 IS0HD F5 WELDING NECK FLANGE-

Fig. 36. Helium-to-Water Heat Exchanger.
(Lyon equation(%)) and

k

0.1

(9) h, = 0.027 — Re®8 Prls3 (—“) !
p, * ®

Hs

w w

(Sieder-Tate equation(!3)), where

k = thermal conductivity, Btu/sec* ft?
(°F/ft),
D = equivalent diameter, ft,
Pe = Peclet’s number,
Re = Reynolds’ number,
Pr = Prandtl’s number,
[ = viscosity at bulk temperature,
lb/hr- ft,
ft, = viscosity at surface tempera-
ture, 1b/hr-ft,
The liquid-side areas are
(10) Avex = ™yak Inax-ne
and
(11) A, =7D, Ly .y s

where L 1s the total effective length
of the heat exchanger tubes in feet.

The resistance of the metal 1s

 

 

 

 

DO
In
1 DNaK NaK-He
(12)  |—— =
(hA)m NaK-He QW(km) NaK-He LNaK-He
and
DO
In D
1 He -
(13) [____ ] v w ’
(hA)m He-w 27T(km)He-w LHe-w

 

where DO is the tube diameter 1in feet.

The heat transfer coefficients on
the gas side are calculated in a manner
identical to that used for the fuel
heat disposal system calculations (cf.,
Egs. 14, 15, and 16 of chap. 3).

Table 10 contains alist of pertinent
information obtained from heat ex-
changer designs (Figs. 35 and 36) and
the heat transfer relations discussed
above. Values are given for four
different flow rates of helium and four
different flow rates of water; curves
were plotted from these values and used
in the solution of Egs. 1 through 5.

The data of Table 10 were used to
solve the five simultaneous equations
(Egs. 1 through 5) for various values

of the independent variable - heat

Inasmuch as the helium
flow 1s constant (see
the helium weight flow

transferred.
volumetric

restraint 2),
cannot be precisely determined until
the helium temperature at the blower
(Ty,) is known. It was necessary
therefore to assume a helium weight
flow for each power and to solve the
equations by i1teration. The results
of the solution of Egqs. 1 through 5
are shown in Figs., 37 and 38, in which
the minimum and maximum NaK and helium
temperatures are plotted against heat
transferred. The mean NaK temperature
is also shown 1in Fig. 37.

The pressure drops in the NaK and
in the helium have been calculated for
the exchangers. The NaK pressure drop
was calculated with the use of the
customary Fanning equation‘!!) for the

 

 

 

 

 

 

 

 

 

 

 

  

 

 

2000 | i |
i
| i /
1800 i -
| [
1600 e
f 5 MAXIMUM
! ‘ ? MEAN ~
| MINIMUM
1400
T | //
o ‘ !
W 1200 . ey
0 i
= :
<[
m '
L
o
= 1
& 1000 i
x !
o
g

 

800!

 

 

 

600

 

 

 

 

 

 

 

0 200 400 600 800 1000

HEAT TRANSFERRED (kw)
Fig. 37. NaK Temperatures vs. Heat

Transferred in Heat Exchangers.

57

 
TABLE 10.

HEAT EXCHANGER DATA

 

NaK-TO-HELIUM
HEAT EXCHANGER

HELTUM-TO-WATER
HEAT EXCHANGER

 

Tube outer diameter, in.

Tube wall thickness, in.

Tube material

Fin outer diameter, 1in.
Number of fins per inch

Fin thickness, in.

Fin material

Tube transverse pitch, ft
Tube longitudinal pitch, ft
Number of tubes per exchanger

Number of tubes transverse to flow
(per exchanger)

Exchanger height, in.

Exchanfier width (also active tube
length}, in.

Total active tube length, ft

Liquid flow area per exchanger, £t 2

Liquid flow rate per exchanger, 1b/sec

Liquid velocity, ft/sec

Peclet’s number

Reynolds’' number

Liquid heat transfer coefficient,
Btu/sec: ft2' °F

Liquidheattransfer.areaperexchanger,ft2
(hA)liquid per exchanger, Btu/sec'°F
(l/hA)liquid per exchanger, sec- F/Btu

(l/hA)! per exchanger, sec'oF/Btu

Helium-side equivalent diameter (DHe)’ in.

Helium free flow ratio

Helium flow area per exchanger, ft2

Helium flow rate per exchanger, 1lb/sec

Helium Reynolds’' number

Helium heat transfer coefficient;
Btu/sec- ft2-°F

Fin efficiency

Helium heat transfer area,* fr 2

(hA)He per exchanger, Btu/sec'°F
(1/hA)y, per exchanger, sec’ °F/Btu

 

1.0

0.109

Inconel

2.0

7

0.024
Stainless steel
2.75

2.43

75

13.75

16.25
101.6
0.0503
11.5
4. 97
790

0.830

20.8
17.3

0.058

0.126

1. 447

0.575

0.893

0.33, 0.22, 0.11
2,230, 1,470, 720

0.0063, 0.0046, 0.0028
0.43, 0.50, 0.60

142, 162, 190

0.889, 0.743, 0.526
1.12, 1.35, 1.90

 

0.625
0.049
Steel
0.875
12
0.010
Copper
2.80
1.684
180

6
16.8

17.0

255.1

0.0273

16.67, 10.0, 6.67, 3.33
9.89, 5.40, 3.91, 1.96

49,300, 29,600, 19,700, 9,860

¢.610, 0.406, 0.290, 0.168
35.2

21.5, 10.2, 5.92
0.047, 0.070, 0.098, 0.169
0.015

0.722

0.768

1.524

0.33, 0.22, 0.11

650, 430, 210

14.3,

0.0051, 0.0038, 0.0023
1.0, 1.0, 1.0

187, 187, 187

0.957, 0.709, 0.424
1.04, 1.41, 2.36

 

"
“Area” varies with helium flow rate because the fin efficiency varies with helium flow rate.

58

 
 

JR—
DWG. 19119
1800

 

 

1800 /

1400

1200 /

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=
wl
r
e 1000
& /
w
a
2 MAXIMUM
-
< 800
2
|
L
I
600 /
400
/
P—
200 flfifl@ia
-——/
0
0 200 400 600 800 1000
HEAT TRANSFERRED (kw)
Fig. 38. Helium Temperatures vs,

Heat Transferred in Heat Exchangers.

straight sections and the curves pre-
sented by Cox and Germano{1%) for the
bends, exits, and entrances. Since the
NaK flow 1s relatively constant, the
NaK pressure drop is approximately
constant and is equal to about 1 psi.
The helium pressure drop is calculated
in a manner identical to that used for
calculating the helium pressure drop
in the fuel system., The helium pressure
drop 1s plotted against heat trans-
ferred in Fig. 39.

DWG. 19120

 

 

 

 

o

HELIUM PRESSURE DROP {in HEO)
Q
\\ N

 

 

 

o
@

 

o
o

 

 

 

 

 

 

 

 

 

 

 

 

0.4
0.2
HE LIUM -TO-WATER EXCHANG ER
HELIUM - e ="
EXTERNAL CIRCUIT
0 l [
0 200 400 600 800 1000 1200

HEAT TRANSFERRED (kw)

Fig. 39. Helium Pressure Drop vs.
Heat Transferred.

The pressure drops in the remainder
of the NaK and the helium circuits
have been calculated with the use of
the Fanning equation and the Cox and
Germano data. Figure 40 shows the NaK
piping; Fig. 41 shows the helium duct-
ing., The pressures at various points
in the fuel circuit are shown in Fig.
1. The NaK pressure drop, exclusive
of the heat exchangers, is 16 psi; the
total NaK pressure drop is 17 psi. The
helium pressure drop exclusive of the
heat exchanger and the total helium
pressure drop are plotted against heat
transferred in Fig, 39,

59

 

 
 

 

UNCLASSIFIED
DWG, A-3-3-5{A

 

 

 

ELEV. B40 ft-6in ‘ [——!

i i . . - - REFLECTOR COOLANT
HEAT EXCHANGER LOOP

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 
 
  
   

 

 

 

 

 

 

 

 

 

 

 

T -
/////////// T M////////////i///////////%/f i
ELEV 832 119 Fg i /////// 0 |
9 R #/M/////////Y/////////////W//////// ’J’// v i
ELEV. 832 f1-9%g in. i %
¢ OF TEE . -
7 Wl
i W
7 =
7 21
| a=|
fi 23
c Z L > |
o ] ; - ouwl £
; % l-ly-f//’mm/mx//mxmmmm%{fi N o
£ R / ; i
- J % /"'////l/#//l//f////lfllllffllfllflé'rz % — \! -
REACTOR % | F L 7 ©
% . % ‘i ! E M ! / //[/[IlllllllllllIlllflfllfllllllfff‘y,,llf I
i ! )
| = | FLOWMETER .| [ N :
g T ‘ g ! g [l § :
a 52
_ i NaK PURIFICATION SYSTE 7 !. g .fi 7 &L .
e e T, ‘ A\ e wl £
. . _ELEV. 842 f1-3i L B % ?’ Hn % fiz oL ©
W é ///Illfllfllfll Illl/fl%fllfllllf’/f/j RTITE) J -“--I"/””/% L /’//,¢ St |
e L -
/ REFLECTOR COOLANT ‘ é/ y z e — “’ 8 E
/ ’;f / : e ;
f ////////// Z ‘ ;2 z % ONITORING He BLOWER é
1 : ! e ' P O / '/
’ ) ' o i L
i " ) | | /
/ o .’fi;m mfga/é___,_%-”,,u / '////////////
' & & cievain- o [\ Qy  d =T
- / / Loy 1“ %W i tr 1
BLDG. ¢ / / . ,,,, / U ! ELEV 841 -3 | -
| / ~ a1} - in. BLDG. ¢
: ! é T T ///////////‘Agwf/f////mfmmmmmf,,,é ‘ [ Ay,
: A 4 % P
DUMP VA ) dl | ELEV.838 fi-6in. - ? %
: : ‘ / ; z
é Vo )
/

 

i
I
[R—

 

S TTTSS

e

= -

L 710 :

- ) A - ELEV. 832 ft-9in, fimflm#flfilmmmwmfl// !

. ) . 7 Y ¢ OF TEE i
SR Ry D , :

 

 

MONITORING SYSTEM

 

 

\\\\!\Q\\\\\\\\\w

\“\\\.
[

 

 

 

 

 

 

 
 

 

 

 

 

 

 

eSS

   

 

 

 

 

]
NaK PURIFICIA_lON SYSTEM

==t
< |

 

 

| | I\I |

MONITORING He BLOWER -

 

WWW

 

|

 

VR

 

 

 

 

 

 

-

 

 

 

 

 

 

 

 

 

REFLECTCOR COOLANT PIPING ==

 

///I/I/I/I//II//M’I LR R L ///‘
—

 

 

 

 

 

 

 

\\\\\\\\\,\\K\\\\\\‘K\\\\\\‘\\\‘%\\\\k\\\\\‘\‘%\\\\\\&\\

 

 

 

 

 

 

 

 

 

 

 

 

 

T LR : : FLOWMETER :
A \\\\ 3 ‘ Q / : R Z
\\\\\ \\\ \\ \ \.: T & E S h
NN | = = ol
N - 3
W \ \ A . ) . \\ i >

AN : el e J// ///// e —— // ~— @

< e /////// HEAT EXCHANGER INLET
C.I: - ; ELEV. B39 ft-0in.
«

 

HEAT EXCHANGER OUTLET
ELEV. 839 ft -1 in.

REFLECTCR COOLANT HEAT EXCHANGER

 

 

 

 

 

 

 

 

 

   

.——5 ft-0in. — -t 9 ft-0in. ]

 

 

 

 

[2ft-2in.

9ft-2in.
EDGE OF PLATFORM

 

 

Fig. 40. Reflector Coolant Piping.

 
      
  

UNCLASSIFIED
DWG D-A-3-4-51A
rem —— 14 ft-0%in REF. e ‘—— e

10 f1-10%in. —

 

NO. 20-SM200 LAMSON BLOWER

VICKERS HYDRAULIC MOTOR

TRANSITION PIECE, 10-iniD TO 20'gx16% in

90° ELBOW, 20Ygxi6% in

DUCT, 20% x16% x 1134-in. LONG

DUCT, 20% x16% in x3 f1-0-in. LONG

TRANSITION PIECE, 10-inID TO 20%x 6% in.
GRISCOM-RUSSELL He TO H,0 HEAT EXCHANGER
GRISCOM-RUSSELL NaK TO He HEAT EXCHANGER
ORNL HEAT BARRIER

11 ASBESTOS GASKETS, REQUIRED DIA X Ygin. THICK
NOTE:

ITEMS NO 3,4,5 6, AND 7 ARE SAE NO. 10-20.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

O ® N O D DL -

 

 

 

 

 

 

 

 

 

—7ft-3%in ——
o

 

 

 

 

 
  
 

 

 

 

 

-8 in.r 3f1-QYgin |~
| :
I

 

 

 

 

 

 

 

 

 

 

-2 ft-gin

 

ELEVATION END VIEW

Fig. 41. Reflector Coolant Heat Exchanger Loop.

 
 

Chapter 5
ROD AND INSTRUMENT COOLING SYSTEM

COOLING OF THE CONTROL RODS
AND INSTRUMENTS

There are six vertical holes in the
reactor of the ARE (Fig. 16) into which
a regulating rod, three safety rods,
and two chambers

fission can be

lowered, when required. Helium 1s
blown through the passages

between the rods and i1instruments and

annular

the reactor to cool the various surfaces
In the case of the rods,
this cooling 1s desirable because the

present,

vanes on the rods slide 1n

contact with the sleeve 1n the reactor;

guide

considerable uncertainty exists as to
what would happen 1f this sliding con-
to be made at the high
temperatures 1in the reactor., The
require cooling because
the electrical

tact were

instruments
insulators used 1n
their construction emit false signals
at temperatures above about 700°F,
The rod and instrument cooling circuit
each of 1000
cfm capacity, which blow helium through

in the

consists of two blowers,

the annular passages reactor

holes, and a heat exchanger which
removes from the helium the heat
picked up 1n the reactor holes.
Although the blowers have a total

capacity of 2000 c¢fm, they are operated
so that the helium flow 1s 1000 cfm,

Figure 1 shows a diagram of the rod
and instrument cooling system. Figure
42 shows the dimensions of the helium
passages through the reactor. Figure

17 shows a cross section of the helium
passages and details of the sleeve 1n
the reactor. This sleeve,
is actually triple-walled. The

as may be
seen,
helium passes between the 1inner wall
and the rod or instrument 1n question,
The space between the 1nner wall and
the middle wall 1s packed with diato-
maceous earth, which serves as thermal
insulation between the reactor and the
NaK from the
reflector coolant flows
between the middle and outer wall and

rod or instrument hole.
circuuil

provides some cooling because 1t 1s

64

colder than the surrounding sections
the 1000
may be ap-

of the reactor core. Since
cfm of helium available
portioned differently to the various
holes, the cooling of the
holes was investigated for a range of
helium flows. Also, at the time of

the calculations, the mean NaK tempera-

various

ture was unknown and, consequently,

a range of mean NaK temperatures was

investigated. The cooling of the
various rods and instruments will be
considered separately and 1in the
following order: safety rods, regu-

The
division of the total flow among the
different holes will be discussed last.

lating rod, fission chambers.

DWG. 19421

PLENUM CHAMBER

fee— 7 -

2%, 0Dx0.043 WALL

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  
 
 
 

 

 

 

 

 

 

BELLOWS 3Y%g ro@ _S A P
\ i N
‘ o
297 ODx0.035 e —2.4500x0.035 WALL
WAL~ N
‘ &<
L &
ROD SLEEVE—= = INSTRUMENT SLEEVE
e |
! oot
et
i | RS TOP OF PRESSURE
F___- sefl H'@ L sHELL
b ‘ | * &\
3.7500x0.049 | |1 1 ot TOP OF CORE
WALL——L—-- ; ‘ 1 f LATTICE
I ‘ -
3.000Dx0.085 {}=——3.75 0D x 0.049 WALL
WALL = ;
o 1} 1.75 0D x 0.035 WALL
o L
2.46 0Dx 0.035 oo 300 0D x0.065 WALL
WAL L ‘ |
: BOTTOM OF
DIATOMACEQUS | = 1 ,/ CORE LATTICE
EARTH-T—L : T {
| ©
I 1o i \\BOTTOM OF
3
2 9700x0035 | 258 1 PRESSURE SHELL
WALL: 2y 2.97 0D x 0.035 WALL
PLENUM CHAMBER
i
VARIABLE ORIFICE
DIMENSIONS ARE IN INCHES

 

Fig. 42, Rod and Instrument Sleeves.

 
COQ LING OF THE SAFETY RODS

During normal reactor operation,
the safety rods are not in the reactor
but are withdrawn into sleeves above
the reactor. The safety rods themselves
are 1n a cool area, and the only cooling
problem, then, 1s that of cooling the
sleeve i1in the reactor. The equations
governing this process are quite
straightforward and are listed below:

(1) Q= We (T, - Ty,
(2) Q = hAG + Q

’

 

(3) RA = | ——— 4 1
(hA) .o (hd)_

1 1 1 -1
— o+ — oy ,
(hA),  (hA)_ (hA),.
where
) = heat transferred to helium,
Btu/sec,
W = weight flow of helium, 1lb/sec,
¢, = specific heat of helium,
Btu/1b-°F,
T' = maximum temperature, F,
T = minimum temperature, °F,
h = heat transfer coefficient,
Btu/sec* ft? (°F/ft),
A = heat transfer area, ft?,
f = effective log mean temperature

difference between the NaK and
the helium,

heat generated in the sleeve by
nuclear radiation, Btu/sec,

and the subscripts

Q, =

He = he lium,

NaK = Nak,
m = middle metal wallin sleeve,
I = insulation (diatomaceous earth),
n = 1inner metal wall.

The various terms in Eq. 3 may be

evaluated as follows:

 

kNaK \
(4) hy . = (5.8 + 0.020 Pe®-#) |
NaK
where
k = thermal conductivity, Btu/sec* ft?
(°F/ft),
D = equivalent diameter (four times

hydraulic radius), ft,
Pe = Peclet’'s number:

(5) A = d' h

%-:b Nak m '
where
d' = tube outside diameter, ft,

L = length of tube =
pressure shell, ft;

hei1ght of

 

 

 

 

 

 

1 In 2=
(6) = -,
(hA)m 27Tk L
where
d = tube inside diameter, ft:
dm
In o
1 n
(7) = —
(hA)I 277k (L
d,
In ~
1 n
(8) = ;
(hA) 21k L
kHe
(9) hy = 0.023 ~D—Re°-3 Pelet

for Re 2 2100

k}[e DHe 1/3
e = 1.86 — Re!/3 Pr1/3<——> ,
Dy L

’

-
I

for Re < 2100

(10) Ay, = 7d L

Equations 1 through 10 and the
dimensions shown 1in Figs. 42 and 9
were used to calculate the exit tempera-
ture of the helium and the heat trans-
ferred to the helium; an inlet helium
temperature of 150°F was assumed. The
properties of helium were eva luated
at the mean helium temperature. The
range of helium weight flows considered
was from 0,005 to 0,100 1b/sec, and
the calculations were made for mean
NaK temperatures of 1000, 1150, and
1300°F, The of the com-
putations are shown in Figs, 43, 44,

and 45,

resul ts

of the

sleeve wall can be calculated from the

The maximum temperature

65

 

 

 
 

 

n DWG.
] 1400 7

 

 

 

 

 

 

 

 

 

 

 

 

1000 |
g ! |
|
9004~ ot 5.0 1200 ' e
i /QTOTAL ’(T/
i : /////’ i T T ™o, To1aL
‘ i 1
eoo% | / } 1000 '/ 5
i / : _ §
‘ W 2
i ; ! = =
7ooé i | | 40 4 800 J , B
! = —
| 1 2 | 2
! \ ‘ : « o
: ; E \\</M£\XI'\AUM WALL TEMPERATURE =
| - 5 600 ‘ —3 °

 
  
 
 
 
 
 

MAXIMUM WALL TEMPERATURE \

30 T >
N |

400 ) ‘ —
\ - \\QTT TEMPERATURE

\ 200 -

=0

t
!
t
i
|

 

 

 

 

TEMPERATURE (°F)}

Q, TOTAL (Btu/sec)

 

 

 

 

 

 

|
| | 0
80 100

; 120 0 i
HELIUM QUTLET TEMPERATURE o 20 40 60

; HELIUM FLOW (lb/sec X 103)

 

 

 

Fig. 44. Safety Rod Cooling at a
| ; Nak Temperature of 1150°F.

 

 

 

 

; 1.0
0 0020 0.040 0.060 0.080 0100

 

HELIUM FLOW RATE {ib/sec)

 

 

 

 

 

 

      

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 43. Safety Rod Cooling at a 1800y . ‘O
NaK Temperature of 1000°F. | |
1400} #70
following equation:
1200 6.0
. Q
(11) (T ) = The .
n max e (hA) .
He 1000 —50 3
The helium heat transfer coefficient, g _ _ [%
h, in Eq. 11 is evaluated by means of < 800 et e e 40 T
Eq. 9, with the helium properties § \KMAXIMUMWALLTE‘MPERATURE B
corresponding to the maximum helium M eoo N | _ J—fl——aog;
temperature being used. The maximum \
wall temperature was evaluated by ‘. 50
. 400 — 2
using Eq. 11, and the results are ‘ N HELlUMOUTLETTl:MPERATURE |
shown in Figs. 43, 44, and 45. | \‘< ! | 5
— o]
200 o
COOLING OF THE REGULATING ROD | |
During normal operation, the o} i o __io
position of the regulating rod may O 0020 0040 0060 0080 0100 0120
vary from a position completely out HELIUM FLOW {ib/sec)
of the reactor to a position where
the midpoint of the stainless steel Fig. 45. Safety Rod Cooling at a
slug is coincident with the center line NaK Temperature of 1300°F.

66

 
of the reactor. When the regulating
rod is completely out of the reactor,
the situation is exactly the same as
that analyzed in the previous section
on the safety rods, the triple-walled
sleeve being i1dentical for the two
types of rods. When the regulating
rod 1s i1n the reactor, the situation
1s more complex; the case of the
regulating rod all the way into the
reactor, which is the extreme condition,

HELIUM~,

 

 

 

 

will be analyzed. The regulating rod
hole will be divided into three
sections in the reactor (Fig. 46).
The top section of the rod hole is
the section that contains the portion
of the regulating rod which 1is in the
reactor but contains no appreciable
amount of neutron and gamma absorber.
The middle section of the rod hole is
the section that contains the portion
of the regulating rod which contains

DWG. 19125

CONTROL ROD
|-

TOP OF PRESSURE SHELL

 

TOP OF LATTICGE

 

 

 

 

 

 

48.50
35.25

| INSULATION

¢

 

 

 

‘/NCIK GAP

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 46.

 

 

 

BOTTOM OF LATTICE

 

 

 

 

 

BOTTOM OF PRESSURE SHELL

 

Regulating Rod.

67
the heavy stainless steel slug. The
bottom section of the rod hole 1s the
section that contains no regulating
rod. The governing the
cooling process in the three sections
of the rod hole are

(12) Q= We (T, =~ Ty), ,

e 1

;i =1,2,3,

the symbols have the
meanings as given previously and the

equations

where same

subscript 1 refers to the sections of
YR .
the regulating rod hole:

1 = top section,
2 = middle section,
3 = bottom section;
(13) Q, = (hA6 + Q. + q), ,
where
g = the heat which must be removed from
the rod, Btu/sec;
(14) q; = (g, * qg), ,
where
q, = heat generated in the rod by
nuclear radiation, Btu/sec,
qp = heat transferred from the sleeve
to the rod by thermal radiation,
Btu/sec;

it may be noted that (qr)i=1 = 0 and
qi'—'3 = 0:

1 . 1
(hA) g x  (hA)

 

(15) (hA), =

 

 

1 1 1 -1
+ — + ,
(hA), (h4) (hA) . |

where the various terms are evaluated
exactly as the corresponding terms 1in

Eq. 3;

!’

 

(16) (0,), = il -
7T 3600 1 a1
— o+ [EN -
€ (A;)(g )
(qR)i=3 = 0 ’
where
a = surface area of control rod, fe?,
t' = maximum rod temperature, °F abs,
t = minimum rod temperature, °F abs,
€ = emi1ssivity, |

68

It may be noted that the arith-
metical-average temperature 1is used
in Eq. 16 to evaluate the heat trans-
ferred by thermal radiation. This 1is
only an approximation, but a check
that 1t 1s
greatly 1in error. Equations 12
through 16 and the dimensions shown
in Figs. 42, 46, and 47 were used to

calculate for the various sections the

has i1ndicated not too

exit temperature of the helium and the
heat transferred to the helium; an
inlet temperature to the rod hole
of 150°F was assumed. The prop-
erties of helium were evaluated
at the mean temperatures, and the values
of Qr and q_ are shown in Fig. 48,
The range of helium weight flows
considered was 0.015 to 0.025 1b/sec;
of 1000,

as before.

average NaK temperatures
1150, and 1300 were used,

The results of the calculations are
in Figs. 49, 50, 51, and 52,
which show the helium temperature
and the heat transferred to the helium
at the end of sections 2 and 3 plotted
against helium flow for the various

shown

NaK temperatures.

The maximum temperature of the
sleeve and the rod can be calculated
from the following equations:

Q
(17) (T = A
max = Ty )ies [(hA)He}

and

Q
(18 ) o= (Th) . —
e T T )i [(hA)HeJ,=

the
coefficients

 

helium heat transfer
are evaluated locally.

{(Tn + T;>4 (t - tij}
200 200 /1,

The results of the evaluations of
Eqs. 17 and 18 are
Figs. 49, 50, and 51.

COOLING OF FISSION CHAMBERS

The

the reactor core during startup of the

Again,

 

 

also shown 1n

fission chambers will be 1n
69

T T . .
&Efi'&ffis 2TEEL . . ENQ'QDLES’ISLU%EELSLUGS STA'E‘[SEfiOSRTEEL 1% -in. 00 x 0.065-in. WALL x 170 in.
ING ROD NOSE STAINLESS STEEL LOWER DRIVE TUBE
TACK WELDAL WELD

 

5 1 - | —
(1 “ (

 

 

 

 

 

 

 

 

 

 

UNCLASSIFIED
DWG. D-A-2-5A

 

 

 

 

 

 

 

 

 

 

 

 

 

   

 

 

 

 

 

RN /.. /v
TACK WELD
69.031 in.
68.969 in.
/:‘28312' DIA. 0.577 in.
' ' 0.547 in.
SNUG FIT WITH LOWER MAKE FROM 1%-in. OD x Q.312-in. WALL
DRIVE TUBE
\ /15° 3-in. DIA. HOLE~ | [ ]
B 7 7 77 6V 777 TR
. _ ] 1265-in. o \ V _ ) ) ) .
( 1235-in. = ( c
s S S 774 Z 77 Py
0125-in. MAX—={{=— L
WELD AND
0.765~in. GRIND FLUSH 1140-in,
0735-in. 1410-in.

 

 

  

19g-in. OD x Yg~in. WALL x 69-in.
STAINLESS STEEL UPPER DRIVE TUBE

Fig. 47. Regulating Rod Assembly.

 

 
......

  

DWG. 19126

6.6 watts/linear in.

DIATOMACEQOUS EARTH

 

 

SLEEVE
IN CORE REGION

  

56.2 watts/linear in.

4.0 watts/linear in.

CONTROL ROD

Fig. 48. Cross Sections of Sleeve,
Nuclear Heat Generation Rates.

reactor, but later they will be removed
from the core the
fission

entirely. For
following analysis, the
chambers are considered as being in
the core, which

Each

into

is the most extreme

condition, instrument hole 1s
divided in much the
same fashion as the regulating rod
hole was divided.

of the hole t he

the fission chamber;

sections

The top section
portion which
the

1s
contains

70

   

295 watts /linear in.

r———~5.0 watts /linear in.

 

4.0 watts /linear in.

 

SAFETY ROD

Control Rod, and Regulating Rod Showing

lower portion contains nothing (Fig.

53). to be used

The equations in
calculating the cooling requirements
are
- ’ - -
(19) Q, = ch (Tge ~Tgedyr =1, 2, _
where the subscript i refers to
1 = top section .
and
2 = bottom section;

 

 
 

 

 

 

  
 

 

 

 

 

 

 

 

 

 

DWG. 19127
600
MAXIMUM
WALL TEMPERATURE
- 500
MAXIMUM ROD
S TEMPERATURE
) 400
o HELIUM TEMPERATURE
w
T 300
2
<1
o
w
a
S 200
’_
100
0
0015 0.020 0.025
HELIUM FLOW (Ib/sec)
Fig. 49. Regulating Rod Cooling at

a NaK Temperature of 1000°F.

. WG. 19128

 

 

 
   

 

 

 

 

 

 

 

 

 

 

 

700
600 T\
\ MAXIMUM WALL
TEMPERATURE
MAXIMUM ROD
500 o=z TEMPERATURE
&
~ 400
g 7T
[+
E MAXIMUM HELIUM TEMPERATURE
&
o 300
=
w
’—
200
100
0 ;
0.015 0020 0025
HELIUM FLOW (Ib/sec)
Fig. 50. Regulating Rod Cooling at

a NaK Temperature of 1150°F.

 

 

o

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

700
MAXIMUM WALL
TEMPERATURE
eoo\\\*
(\ \
[
- 400
2 a
E MAXIMUM RCD YEMPERATURE
z
& 300
>
— MAXIMUM HELIUM TEMPERATURE —
200
iCO
0
0015 0020 0025
HELIUM FLOW (Ib/sec)
Fig. 51. Regulating Rod Cooling at

a NaK Temperature of 1300°F.

850

800

7.50

700

@, TOTAL (Btu/sec)

650

 

 

 

 

 

 

 

 

 

 

1000°F _ ‘

 

 

 

 

 

 

 

0.026

6.00 - - i —r
O// ; f
550 | | l | '
0014 0.018 0.022
HELIUM FLOW (lb/sec)
Fig. 52. Regulating Rod Cooling at

Various NaK Temperatures,

71

 
el

DWG. 19131

| INSTRUMENT

HELIUM

TOP OF PRESSURE SHELL

 

 

 

. : : TOP OF LATTICE

 

 

 

| INSULATION

 

 

 

35.25

 

-t

 

 

NaK GAP

e

 

 

 

 

 

 

 

| BOTTOM OF LATTICE

V - ' BOTTOM OF PRESSURE SHELL

 

 

 

 

 

 

 

 

 

 

 

Fig. 53. Instrument Hole.

72

 
(20) Q. = (hdO + qp), ,

where the symbols have the same meanings
as before;

Lo,
(hA)y.x  (hA)_

 

(21)  (h4), - [

 

1 1 1 -1
+ + ,
(hA)I (hA)n (hA)He}_
where the various terms are also
evaluated the same as in Eq. 3, except
that the heat transfer coefficient 1in
an annulus is given by

 

k 0.14 /D 0.
He He
(22) h = 1.02 —— Re®-*5 <3i> ( )
D M l

He

where
i = dynamic viscosity evaluated at
the mean temperature of helium,
lb/sec* ft,

. = dynamic viscosity evaluated at
inner wall temperature,
lb/sec" ft,

l = length of section, ft,

df = outside diameter of fission

chamber, ft,

Grashof number;

&
—
i

 

dimensions of the sleeve are shown 1in
Fig. 42, The calculations are based
on the same NaK temperatures as used
before, and the results are given 1in
Fig. 54.

Figure 55 shows the heat removed
and the maximum temperature of the
helium vs, helium flow rate for a NakK
temperature of 1150°F,

HELIUM PRESSURE DROPS

The helium pressure drops in the
safety rod, regulating rod, and
instrument sleeves are calculated by

 

0.8

4 d
( ") Gr°:%5 | for Re < 2100 ,
dy

 

 

using various helium flows and a Nak
temperature of 1170°F. The equations
governing the pressure drops are

 

 

1 pv?
24 AP, = Af —
pv’
(25) " --ix. APC = X ,
2
g

 

r 4 4
(23) _0.173 2 (Tn + T, (tf * tf)
" 73600 1 200 200 ’

where
a, = surface area of fission chamber,
fe?,
ty = minimum fission chamber tempera-
ture, °F abs,
t} = maximum fission chamber tempera-
ture, °F abs,
€, = emissivity of fission chamber,
€ = emissivity of tube.

The cooling requirement calculations
for the fission chambers are made with
the assumption that the bottom of
each instrument (excluding nose) 1is
at the center line of the core. Also,
the maximum temperature of the instru-
ment is taken to be 600°F, and the
inlet temperature of helium 1is taken
to be 150°F, The outside diameter of
the instrument is 1.25 in., and the

 

 

2
(26) ap = Pr = Vo)
e 2g
L, ov?
27 AP = - .
(27) , Af D 2

The symbols have the following meanings
and units:
AP = pressure drop, lb/ft?,

f = friction factor,

l = length of section, ft,
l, = equivalent length, ft,

D = hydraulic diameter, ft,

v = average velocity, ft/sec,

g = gravitational acceleration,

32.2 ft/sec?,

constant, dimensionless,

S
I

73

 
 

v, * average linear velocity up-
stream, ft/sec,

v, = average linear velocity down-
stream, ft/sec.

The meanings of the subscripts are the
following:

f = friction,

c = contraction,

e = expansion,

y = elbow,

The friction factor, f, is given by
(28) £ = lé‘, for Re £ 2100 ,

Re
0.125
f=0,0014 + —M8Mm — ,
(Re )?-32

for Re > 2100 ,

 

However, for an annulus
6
(29) f = , for Re £ 2100 .
Re
Values for X are given by McAdams

(cf., p. 122 of ref. 11), and [, values

are taken from ref. 13,

DWG 19132
2.6 T 1.6

 

 

 

24

   
 

M
mn

 

™
O
O

 

1.C00

 

o
®
@,TOTAL (Biu/sec)

 

REQUIRED FLOW RATE (ib/sec X10%)

 

 

 

 

 

086
0.4
1.2 0.2
|
00 \ o
{000 100 1200 {300 1400
NoK TEMPERATURE {°F)
Fig, 54. Instrument Cooling.

74

 

Each sleeve is equipped with a
variable orifice where 1t enters the
the pressure

bottom chamber., However,

drops calculated here do not include

the pressure losses across these
orifices. The dimensions of the
sleeves are given in Fig. 42,

The lower end of the safety rod

will be 1 in, above the top of the
core lattice when the safety rod is
in the “full out' position. The
pressure drops were calculated for
the rod in this position with helium
flows varying from 0,005 to 0,090
Ib/sec. The safety rod dimensions
are given in Fig., 56, and the results
of the calculations are given in Fig,
37,

The regulating-rod helium pressure
drops were calculated for the rod 1in
the “ full in”and’ full out” positions.
The *“full out” position for this rod
was taken to be the same as that for

DWG. 15133
1.80 800

 

 

 

1.70 700

 

{60 600

 

 

/

 

 

 

 

 

 

 

 

 

 

 

 

.. 150 500 ©
o o
@ ) ‘ L
3 | : 5
o i | 5
4 1.agQ / ' -1400 (5
= \x\<:,,MAMMUM HELIUM i
& TEMPERATURE =
= L
Q' |
1.30 /// ‘*~‘5=~¢300
120 4 ' 1200
110 100
| |
|
.00 f o
0 0.0020 0.0060 0.0100

HELIUM FLOW {Ib/sec)

Fig. 55. Instrument Cooling with
Bottom of Instrument at Center Line of
Core and at a NaK Temperature of
1150°F.

 
 

5 BOTTOM INSERT NQOSE
\ //‘

\\

\\

. \
A

N

/ \
/
RO SLUG WITH SF’ACERS—/ SLUG WITHOUT SPACERS

ALTERNATE, 9 EACH, 18 TOTAL
NOTE:
INSERTS, NOSE, BRAIDED HOSE, TAIL
AND COLLAR ARE STAINLESS STEEL

Fig. 56.

DWG. 19134

 

300

280

 

260

 

240

 

 

 

200 /

180 /

160
140

120 /
/

80 /

- /
, /

20

 

 

 

 

PRESSURE DROP [lb/ft¢)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0 0.030 0.060 0.090
HELIUM FLOW (fb/sec)

Fig. 57.
Rod Sleeve,

Pressure Drop in Safety

the safety rod. Figure 58 gives the

results drop com-

from 0.015

of the pressure
putations for flow rates
to 0.025 1b/sec.

The helium pressure drops for the
instrument hole were also calculated

AM. METAL BRAIDED HOSEfj

Safety

UNCLASSIFIED
DWG D-A-2-4A

TAIL--5 TOP INSET —
/ /

/ /

     
  
  

s

o .
/ 0156 in _. F s

Z 00949 in Y
CONNECTING ROD

COLLARV'\

   

 

 

 

 

 

 

 

 

 

 

Rod Assembly.
135
55 —
" /
40 3
Q
&
-
=
o
o
O
x
o
)
x
o
¥ 30
w
o
a )
g st
o
20
e~
10
0.015 0.020 0.025

HELIUM FLOW (Ib/sec)

Fig. 58. Pressure Drop inRegulating
Rod Sleeve.

for the instrument in the ‘“full in” and
in the ““full out’” positions. Here,

the bottom of the instrument (excluding

75

 
above the top of the

instrument

nose) 1is 15 1in.

pressure shell when the 18
in the “full out” position. The flow
rates used 1n these calculations

ranged from 0.0015 to 0,0100 1b/sec.

The results of the calculations

are
given 1n Fig. 59,
DIVISION OF FLOW OF HELIUM
The flow of the helium to the

various holes 1s apportioned so that

the minimum flow to the instrument
in the ““full is 0,005
lb/sec, and the minimum flow to the
regulating rod in the ““full out”
0.034 1b/sec. The flow
rates per hole for all combinations of
reguTgyfinglwxlandinstrumentpositions,
with the safety rod out of the
are given 1in Table 11,

C L
1in’’ position

position 1s

core,

PERFORMANCE OF ROD AND INSTRUMENT
COOLING SYSTEM HEAT EXCHANGER

The control rods and nuclear instru-
in the of the ARE are

cooled by helium, which gives

ments core
up 1ts
heat 1in a helium-to-water exchanger
(Fig, 60), governing

the performance of this heat exchanger

The equations

are
(29) Q = Wfle CPHe (The - THe ) ’
(30) Q=4H c (T; - T)
(31) (} = hAS |
where

@ = heat transferred, Btu/sec,

W = weight flow, lb/sec,

c, = specific heat, Btu/lb:°F,

T' = maximum temperature, _F,

T = minimum temperature, °F,

h = over-all heat transfer coef-
ficient, Btu/sec*ft?+°F
A = heat transfer area, ftz,
f = effective log mean temperature
difference,
The heat exchanger operates under the

following restraints and conditions:
1. The helium flow is 1000 cfm at

blower temperature.

DWG. 19136
]

 

18.0:

  
 
  
   

 

 

|
|
|
16.0F—— e tmmed
|

% ‘ INSTRUMENT IN-
: \

 

I

140 e

 

 

10.0 -

 

PRESSURE DROP (Ib/f1°)

@
@]
4*,]7 IR
|
. __i.__. .

20

 

INSTRUMENT QUT
|
0.0060
HELIUM FLOW (Ib/sec)

 

©0.0100

Fig.
Sleeve,

59. Pressure Drop in Instrument

 

 

 

 

 

TABLE 11, HELIUM FLOW RATES IN ROD AND INSTRUMENT HOLES
POSITION HELIUM FLOW (lb/sec)
R lati Rod Inst t ;
ceo avine 7o peramen Safety Rod Regulating Instrument
Qut In Out In Rod
x X | 0.035 0.034 0.005
X X 0,037 0.026 0.005
X X 0.037 0,026 0.005
X X 0.035 0.034 0,005

 

 

 

 

 

 

 

76

 
 

bt 3ft 5in.

—-— 4*/4 in,

 

‘ 2 fr 8%in. }

  

T

 

\ri/

 

 

 

 

 

Z3—in. P.T. INLET

 

 

 

 

fl ‘
‘ /32
Yyin P, 5’//—2-m.ar.ouTLET E-m.RFINLETj\ GASKET
[
| 1 'Z
4, in. T a'hin. | T -—-///
- * - i . OD___LL -

UNCLASSIFIED
DWG. 3-5-21A

——

4‘/4 N —-]

-in,-THK.

   

HEX. HEAD CAP SCREWS

BONNET

 

& in.

 

 

Fig. 60. Helium-to-Water Heat Exchanger.

2. The water outlet temperature 1is 1 1 1 -1
thermostatically controlled and is (32) h4d = |—— t ;
held fixed at 100°F., (hi)y, (W), (h4),

0.14 1/3
Bue /H 1/3 1/3 Pye
- (33) hy, = 1.86 — Rey.” Pry, A
DHe ’u's He ’
. 3. The water inlet temperature (McAdams, p. 190 of ref. 11), where

| will vary with the season of the year,
| but has been assumed to be 70°F for
| this calculation.

The rod cooling system heat ex-
changer of three standard
shell-and-tube heat exchangers arranged
in parallel. The helium flows in the
tubes, and the water flows
the outside of the tubes. The follow-

consists

AaCross

ing are pertinent physical dimensions
of the exchanger; all values are for
one of three units:

The heat transfer coefficient may
the

Shell inside diameter, in. 6 1/8
Tube length, ft 3
1 Tube outside diameter, in. 0,375
| Tube inside diameter, in. 0.319
Number of tubes 116
Surface area per tube (inside), fr? 0.251
Surface area per tube {outside), fi2 0.295
Surface area (inside), fi? 29.1
. Surface area (outside), fr? 34.2
Baffle spacing, in. 2
Tube pitch (triangular pattern), in. 29/64

be evaluated from
equlations:

following

the calculation 1ndicates that the
helium flow is laminar;

 

 

 

(34) h =— [— Re?: %% Pr!/3
Y DW ’LLS w v Y
(Kern, p. 137 of ref. 8), where
GWDW
Re = ,
Fy
4 x f
b = ree ared (free area

¢ wetted perimeter

measured in a plane perpen-
dicular to the tubes),

 

 

dCB
G = :
p
1 27k, L
(35) ) e
ha m DHe
In
DW
where
kR = thermalconductivity,Btu/sec'ft2

(°F/ft),

77

 
D = equivalent diameter, ft,

@ = viscosity at bulk temperature,
Ib/hr-ft,

#. = viscosity at surface temperature,
1b/hr-ft,

Re = Reynolds’ number,
Pr = Prandtl’s number,
L = tube length, ft,
G = unit weight flow, lb/hr-«ft?,
d = shell inside diameter, ft,
¢ = tube clearance, ft,
B = distance between baffles, ft,
p = tube pitch, ft,
and the subscripts
He = helium,
w = water,
The helium and water heat transfer
areas are tabulated above: 29,1 and
34,2 ft? per one unit of three,
respectively,

By using Egs. 29 through 35 and
noting that the large number of
baffles on the shell side makes the
effective log mean temperature dif-
ference almost equal to the log mean
temperature difference for counterflow,
the performance of the rod cooling
system heat exchanger can be calcu-
lated. Figure 61 shows the result of
such a calculation: the helium inlet
and outlet temperatures are plotted
against the heat transferred. Some
of the pertinent information not
shown 1n Fig. 60 1s tabulated in
Table 12,

The heat removed from the core in
the rod cooling circuit, calculated
in the first section of this chapter,
is about 25 Btu/sec. From Fig. 61,
it may be seen that the minimum and
maximum temperatures in the rod cooling
heat exchanger will be about 110 and
240°F, respectively., The helium will
actually pick up a considerable amount
of heat between the heat exchanger and
the 1nlet to the reactor and, con-
sequently, the 1nlet and outlet
temperatures of the heat exchangers
wi1ll be greater than those listed
above. The inlet temperature to the
reactor will be about 150°F,

78

 

650 o
|

   

 

550

4505 -

DWG 19137
- |

 

 

%onu-

250 —

HELIUM INLET TEMPERATURE (°F)

150 —-

HELIUM OUTLET TEMPERATURE (°F)

O 20 40 60
HEAT TRANSFERRED (Btu/sec)

 

 

 

200 e

80 100 120

Fig. 61. Helium Inlet and Qutlet
Temperature vs. Heat Transferred.

TABLE 12, HEAT EXCHANGER DATA FOR
HELIUM-TO-WATER HEAT EXCHANGER

 

 

Helium flow area (total),
f12

Helium velocity, ft/sec
Helium Reynolds' number

Helium heat transfer coef-
ficient, Btu/sec*ft“*°F

(hA)He, Btu/sec* F
Water weight flow, lb/sec

Water heat transfer cogf«
ficient, Btu/sec*ft?*°F

(hA)w, Btu/sec® F
h4, Btu/sec'OF

0.246
65
950

0.0039
0.344
1.0, 2.0, 3.0

0.13, 0.19, 0.24
13.5, 19.8, 24.6
3.01, 2.99, 2.98

 

 
Chapter 6
MONITORING AND PREHEAT SYSTEM

HEAT LOSS THROUGH INSULATION

The heat loss to the environment of
the various pileces of equipment in the
three ARE pits is complicated by the
fact that the pits are filled with
helium. The helium tends
the air in the pores of the insulation
and thus to 1ncrease the thermal
conductivity of the insulation. The
for the thermal
conductivity of porous 1insulating
materials was derived by Maxwell and

to replace

following formula

 

 

reported by Jakob (cf., p. 85 of ref,
14):
akg
1 - {1 ~-——1%
kS
(].) k = ks ’
1 + (a - 1)b
where
3k
a = ,
2kt k
s g
V V
p-—8& -y °* -1 _£2
V. Vg V. + Vg P
k = thermal conductivity, Btu/hr-ft?
(°F/ft),
V = volume, ft3,
o = density, lb/ft?,
and the subscripts
s = solad,
g = gas,

The ARE uses three types of insu-
lation; diatomaceous earth and Superex
for high-temperature insulation, and
felted Asbestos-Sponge for lower-
temperature insulation. The properties
of these insulations are given 1n
Table 13. By using the data of Table
13 and mean thermal conductivities of
0.033 and 0.021 Btu/hr-ft? (°F/ft) for
the air in the high- and low-temperature
the thermal
can be

insulation, respectively,
conductivity of the solid, &k,
determined from Egq. 1. Since k_  1s
known, k can be calculated for the
case with helium in the pores by using

0.15 and 0,11 Btu/hr-ft? (°F/f¢),

respectively, for the high- and low-
temperature mean thermal conductivities
of helium. The results of the calcu-
lations are tabulated in Table 14,

The
the ARE piping and equipment 1s a com-

insulation on the outside of

bination of Superex and Asbestos-

Sponge, and diatomaceous earth 1s used
for i1nsulation i1n the control rod
cooling loop. The heat loss to the

environment through the combination
Superex and Asbestos-Sponge insulation
may be calculated from the following
equations:

 

 

1. For insulating cylindrical
shapes (1.e., piping),
2k O
(2) _ LTRs Us ,
DS
In
DI
s
2mk 6’(1-3
(3) —
¢ D
In—"
DI
a=3s

 

 

 

4y . 0113 (Ts )4”<T6 )4
¢ 3500 © 100, 100.

+ h AT, - T,)

e »

Q = heat loss per lineal foot,
Btu/sec*lineal ft,

k = thermal conductivity, Btu/sec"* ft?

(°F/ft),
& = temperature difference, °F,
D = outside diameter, ft,
D' = inside diameter, ft,

€ = emissivity for radiant energy,

A = outside surface area per lineal
foot, ft?/lineal ft,

T, = temperature of outside surface,
OB,

Te = temperature of environment, °R,

h. = free convection heat transfer

coefficient, Btu/sec*ft?-°F,
and the subscripts
s = Superex,
a-s = Ashestos-Sponge.

79
 

 

 

TABLE 13. PROPERTIES OF VARIOUS INSULATING MATERIALS
DIATOMACEOUS SUPEREX FELTED
EARTH ASBESTOS- SPONGE

Primary solid constituent 510, 510, MgS10, ’
Density, o, lb/ft? 18 24 30
Density of solid, p_,

1b/fe’ 145 145 205
Mean thermal conductivity

(air in pores), k,

Btu/hr* ft? (°F/ft) 0,060 0.070 0.045

 

 

 

 

TABLE 14. THERMAL CONDUCTIVITIES OF VARIOUS

INSULATING MATERIALS

 

 

 

 

 

 

 

THERMAL CONDUCTIVITY [Btu/hr+ft? (°F/ft)]
MATERIAL
Selid, ko, E (helium in pores)
Diatomaceous earth 0.34 0.17
Superex 0,34 0.18
Felted Asbestos-Sponge 0.25 0.13
The free convection heat transfer the pipe, annulus, heaters, and 1insu-

coefficient 1s determined from the
data of Nusselt,(18) McAdams,
King,('%) who correlated Nusselt's
number with Grashof’s and Prandtl’s
numbers (as given by Jakob¢'*’).

2. For

and

insulating plane shapes,

 

 

 

 

Eas. 2, 3, and 4 become
kg
(5) q = —;— 693 ,
s
(6) _ ka’s 6
q = ¢ s
aq= S
4 4
0.173 T, T,
(7} gq = € _
3600 100 100
t+th (T -T7) ,
c S e
where
q = heat loss per unit area, Btu/sec-ft?,
t = thickness, ft.

The heat loss per lineal foot of
the piping was
Egs. 2, 3,

calculated by using
and 4, The dimensions of

80

lation of the fuel and NaK piping are

given in the following:

Mo B
Pipe

Annulus
Heater
Superex

Asbestos~
Sponge

Table 15 gives
lineal foot for a range
outside wall temperatures
emlssivity, €,
temperature,

Equations 3,
calculate the heat loss
foot of a flat surface for
insulation thicknesses
outside wall temperatures.
shows the heat loss

1325,

of 1150,

~ W W et

OUTSIDE DIAMETER (in.)

11/2 2 2 1/2
315 1.900  2.375  2.875
.0 3.0 3.5 4.0
94 3.94 4.5 5,25
.94 7.94 10 10.75
.94 9.94  12.5 13,25

the heat loss per
of heater
for an
of .90 and an ambient

T,, of 590°R (130°F).

e
6, and 7 were used to
per square
various
and heater

Figure 62
for temperatures

and 1500°F for combined

insulation thicknesses of from 2 to 6

 

 
TABLE 13. HEAT IDSSES FOR VARIOUS TEMPERATURES OF THE HEATER OUTSIDE WALL

 

 

 

 

 

 

 

 

 

TEMPERATURE OF HEATER NOMINAL PIPE DIAMETER (in.)
OUTSIDE WALL 1 to 1Y% 2 2
(°F) HEAT LOSS (Btu/sec*lineal ft)
. 1150 0.23 0.21 0.23
1325 0,27 0.25 0.27
1500 0.31 0.29 0.31
in.: th thick f th S
in e 1C ness O e uperex Dm

layer was assumed to be twice the

 

. 0.30

thickness of the Asbestos-Sponge layer.

The emissivity and ambient temperature

were the same as those used above, \\
0.25

 

The values calculated above for
heat loss through the insulation were
used to roughly estimate the heat loss \\

 

 

 

 

 

power output) of the various items of e 1500
equipment in the three ARE pits (Fig. 3 E:Ezzfg
. 63). The results are given in Table 16. 2 015 \\<‘
. u)
The total heat loss (kw) in each g
pit 1s: - \\\\
8 0.10 AN
- Reactor pit 30 r - N \\\\\\\\\\\\
Heat exchanger pit 135 \\
Dump tank pit 75 T — ]
— 0.05
Total 240
REACTOR PREHEATING o

The reactor of the ARE, with its
large mass of beryllium oxide, has the
largest heat capacity of any piece of
equipment i1in the ARE. The reactor must
be raised to a high temperature before
the system can be filled with fuel
carrier, since the fuel carrier has a
melting point of approximately 950°F,
Figures 9 and 12 show elevation and
plan views of the reactor; Fig. 9 shows
the reactor andits thermal insulation,
which is attached to a metal shell
that 1s separate from the reactor,
The heat 1is supplied by electrical
strip heaters strapped to the reactor
itself, The procedure for preheating
the reactor will be the following:
1. turn the heaters on at a power of

10 kw;
2. 1ncrease the heater power as the
pressure shell temperature 1in-

 

 

 

 

 

 

 

 

! 2 3 4 5 6
INSULATION THICKNESS (in))

Fig. 62. Heat Loss Through Insulation.

creases so that the heater power
is equal to 10 kw plus the heat
loss through the insulation;

3. when the pressure shell tempera-
ture reaches 1200°F, gradually
reduce the heater power to hold
the pressure shell temperature at
1200°F,

When no further changes in heater

power are required to hold the pressure

shell at 1200°F, equilibrium will have
been attained and the preheat process
completed.

Steps 1 and 2 in the preheat process
imply that a constant amount of heat
1s being added to the reactor, namely,
10 kw. This method of preheating will

81

 

 
TABLE 16.

HEAT LOSSES IN THE

ITEMS IN THE ARE PITS

 

 

 

 

 

 

HEAT

ITEM LOCATION LOSS

(kw)

Reactor and reactor thermal insulation Reactor pit 15
Fuel system heat exchangers Heat exchanger pit 5
Fuel system surge tanks Heat exchanger pit 4
Fuel pumps Heat exchanger pit 2
Reflector coolant system heat exchangers Heat exchanger pit 6
Reflector coolant surge tanks Heat exchanger pit 4
Reflector coolant pumps Heat exchanger pit 2
Fill and flush tanks Dump tank pit 25
Fuel dump tank Dump tank pit 8
Fuel surge tank vapor traps Heat exchanger pit 2
Fuel dump tank vapor trap Dump tank pit 2
NaK purification system Heat exchanger pit 10
Valves Heat exchanger pit 15
Dump tank pit 12

Piping Reactor pit 15
Heat exchanger pit 85

Dump tank pit 30

result i1n a quasi-stationary tempera- shells, 3 ft long and 3.75 in. thick,

ture state being set up in the reactor
after some initial period. A quasi-
stationary temperature means that the
temperature difference between any two
points remalns constant even though
the entire temperature level isrising.

The preheating process can therefore

be broken up into three periods:

1. the initial period before the
quasl-stationary state is obtained,

2. the period when the reactor is in
the quasi-stationary temperature
state,

3. the period after the pressure shell
has reached 1200°F, during which
the heater power is being gradually
reduced and the quasi-stationary
state no longer exists,

The following assumptions were made
for the calculation of the reactor
preheating:

1. The reactor will be represented

by six concentric beryllium oxide

82

separated by helium shells 0,050 1in,
wide.

2. The pressure shell will be con-
structed of Inconel and will be 48 1in.
in inside diameter and 2 in. thick.

3. There will be a 1/16 in. helium
the shell
outer beryllium oxide cylinder.

4. The heat loss from the ends of
the reactor will be considered, but
longitudinal heat flow within the
reactor will be neglected,

gap between pressure and

The following physical properties,
which represent mean values between

100 and 1200°F, were used in the
analysis:

BeQ INCONEL. ~ HELIUM
Thermal cogductivity,

Btu/hr* ft° (°F/ft) 10 12 0.13
Density, lb/ft> 184 518 0.0055
Specific heat,

Btu/ 1b*°F 0.38 0.15 1,24

 

 
 

 

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Composite Plan View o

63.

Fig.

83

 

 
The temperature difference between
any radial station in the reactor and
the outside of the pressure shell when

the reactor is in the quasi-stationary
temperature state can be calculated

from the following equation:
dé Q

(8) — = ,
dt (WCP)BEO'TL (WCP)IHC+(WCP)He

where

d@

 

= rate of temperature rise,
dt °F/hr, of any point in the
reactor (in the quasi-stationary
temperature state),
Q = heater input to the reactor,
Btu/hr,
W = weight of material in the
reactor, lb,
Cp = specific heat, Btu/lb*°F,
and the subscripts
BeO = beryllium oxide,
Inc = Inconel,
He = helium,

The weight of the beryllium oxide
in the reactor 1is approximately 5800
Ib; the Inconel in the pressure shell
weighs about 7900 1b; the helium weight
can be neglected., From Eg. 8,
do 34,100

dt 5800 x 0.38 + 7900 x 0.15
- 10.1°F/hr .

 

The temperature difference across
any of the cylindrical shells may be
approximated in the following manner:

r

 

nt+1
q, In —
(9) ANO = ,
n 21k L
n-1 d6
(10) q, = 34,100 - -’;1 (ch)i"é? ,
i =
where
A = radial temperature differential

across a shell (in the quasi-
stationary temperature state),
OF’
= heat entering shell, Btu/hr,
outside radius of shell, ft,
k = thermal conductivity of shell,
Btu/hr ft? (°F/ft),
L = length of shell, ft,

~
il

and the subscript n refers to the
number of the shells, counting inward
from the outside. (For a more accurate
estimate, the fact that all the g, 1n
FEg. 10 1s not transferred across shell
n should be considered. The above
approach 1s a conservative simplifi-
cation, however, because i1t leads to
larger values of A .)

The results of the solution of Egs.
9 and 10 are shown in Fig. 64, which
gives the difference in temperature
(in the quasi-stationary state) be-
tween the outside of the pressure
shell and any radial position. The
difference between the outside of the
pressure shell and the center of the
reactor is 162°F.

The time required to reach the
quasi-stationary temperature state may
be readily calculated. The reactor 1is
originally at some uniform temperature
04+ The quasi-stationary state is
reached while the temperature of the
center of the reactor remains fixed at
€, and the outer shells heat up to a

0
temperature of

N
1
(11) ¢ =6+ E AG . +— A
ny 0 ) 1 2 n
1=nt1
where
6, = temperature of the nth shell at

! the beginning of the quasi-

stationary state,
N = total number of shells.
The time required to reach the quasi-
statlonary temperature state 1is,

DWG. 19133

INCONEL PRESSURE SHELL

—~=~ == SMOOTHED CURVE f;/
TRUE RA s

HELIUM GAPS~

OUTSIDE OF PRESSURE SHELL
AND RADIAL POSITION (°F)

TEMPERATURE DIFFERENCE BETWEEN

 

Q 4 B 12 16 20 24 28
RADIUS {in.)

Fig. 64. Temperature Profile of

Reactor Core and Pressure Shell (Quasi-
Stationary State).

85

 

 
therefore,

N

L

i=1
1 Q !

the time required to reach

(We )y (8, ~84)

t

(12) t

i}

 

where t  1s
the quasi-stationary temperature state
in hours. This time was calculated
from Eq. 12 to be about 10.0 hours.
Thus, after about 10 hr, the tempera-
ture of the outside of the pressure
shell will be at about 162°F above
ambient temperature, while the center
of the core will still be at about
ambient temperature. The actual
ambient temperature in the pits will
probably vary during the preheating
period; most of the calculations of
this report assume an ambient tempera-
ture of 130°F in the pit. It was
assumed for these calculations that
the reactor and pit were both, at the

 

until

the reactor pressure shell outer

temperature reaches 1200°F, During this

period,
quasi-stationary state,

the temperatures are 1in the
and every

temperature in the reactor is raised .
uniformly at the rate of 10.1°F/hr:

(13)

where

quasi-
period,

1200 - 292
10.1

t, is the time at the end of the
stationary state. During this

t, — t, 89.9 hr, -

the electrical heater power

must be continually increased as the

pressure shell temperature

increases

to malntailn a constant rate of heat

f1low

power

into the reactor.

The heater
must therefore be equal to 10 kw

plus the heat loss through the 1insu-

lation.
insulation may
function of reactor pressure

outer

The heat loss through the
calculated as a

shell

temperature from the following

be

equations:

Ay

 

A

1 < 1
Ay \ €,

G- _

 

 

(14) S hA (T, - T, ) il
q 171 1 = THe 3600
— 4
“1
0.173
(15) g = h,A, Ty, - T,) ¢ 3600
1
4+
€1
kf
(16) g =— A, (T, - T;) ,
71
0.173
(17) q = h,A, (T, - Ta) + —ggaafesAs
start, at a temperature of 130°F, The

time required to raise the mean temper-
ature of the reactor from 70 to 130°F
is about 6 hours. It may therefore be
considered that t 1s really about
16 hr and that at the end of the 16-hr
period the reactor pressure shell
outer temperature will be about 292°F
and the center-line temperature will
be about 130°F,

The next period of the preheating
process 1s the heating of the reactor

86

T

where

q."—'

e~y

() - (%

T

a

 

]

heat loss through
Btu/sec,

heat transfer
Btu/sec* ft?:°F,
heat transfer area,
temperature, R,
thermalconductivity,Btu/sec‘ft2
(°F/ft),
thickness,

insulation,
coefficient,

ft2,

ft,
emissivity,
and the subscripts

l] = outside of reactor pressure
shell,
2 = inside of thermal insulation,
3 = outside of thermal insulation,
I = insulation,
He = helium between pressure shell

and thermal insulation,
a = ambient,

Equations 14 through 17 and the
following data were used to calculate
the heat loss through the insulation
for various pressure shell tempera-
tures:

1. € = 1.0,

2. The data of King(!?) were used
to evaluate h, ,

3. The dimensions of thermal insu-
lation were:

Outside diameter 76.25 in.

Outside length 80.5 1in.

Thickness 6 in. (4 in. of Superex, 2
in. of Asbestos-Sponge})

4. The thermal conductivity value
for the insulation, as found in the
previous section on “Heat Loss Through
Insulation,” was used.

The results of the calculation are
shown 1n Fig. 65, where the heat loss
through the insulation 1s plotted
against the reactor pressure shell
outer temperature. Temperatures at
other points 1n the reactor thermal
insulation are shown by the dotted line.

The last portion of the preheating
process, that is, when the heater power
is reduced to maintain the pressure
shell outer temperature at 1200°F, 1is
evaluated by a graphical method de-
veloped by Schmidt.(2%) Asa simplifi-
cation, each helium shell 1s divided
between adjacent beryllium oxide
shells, and the Inconel pressure shell
is replaced by an equivalent thickness
of beryllium oxide. The reactor 1s
then treated as a homogeneous cylinder
with a thermal conductivity equal to
the average thermal conductivity of
the composite beryllium oxide and
helium shells. A plot of the tempera-
tures in the reactor for the first 8
hr of the last portion of the pre-
heating period is shown in Fig. 66.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1200

1000

800
c
L
w
o
2
L 600
o
w
a
=
w
-

400 /

200 / OF INSULATION __ e |

[ | OUTER SIDE OF INSULZ
e e
AMBIENT TEMPERATURE
0
0 4 8 12 16

HEAT LOSS THROUGH INSULATION (kw)

Fig. 65. Temperature of Outer Side
of Pressure Shell vs. Heat Loss Through
Insulation.

After 8 hr, the temperature of the
core center line is about 1125°F,

A summary of the temperatures and
powers in the entire preheating period
is shown in Fig. 67.

The maximum temperature gradient 1in
the pressure shell at any time 1is
approximately 6°F/inch,

HELIUM LEAKAGE THROUGH CLEARANCE HOLES
IN THE REACTOR THERMAL SHIELD

The tubes which contain the control
rods and fission chambers pierce the
reactor thermal insulation at both the
top and bottom, and there are six such
tubes. Figure 68 shows a sketch of a
typical tube and the passage through
the reactor thermal shield. The helium
inside the reactor thermal shield,
being at a higher temperature than the
room helium, will flow through the
clearance holes in the reactor thermal
insulation and consequently will add

87

 

 
Dwg. 19441

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0 1200
= 7 = 1200°F
= SURFACE
o 20 t 1180
a AB=0.804hr /
o
s
= 40 10A8 =~ Bhr 7 1160
S 9Af /////
.62 _8A8 N /
= 60 T 1140
/// 2’
ST gop 046 — / nzo ¥
LéJ S—f A ~ " ___-———/;//_/// : %
m g 100—'__'——'—_—— ey _—// / 1100 Ej
- =" = 7 7 o
g é j:: ah ——////// =
5 =z — 5A8 % r —-///// w
x © 420 e anb - T | 4 1080
w [
2A0 —— /—/ //
w 140 —— 1060
o a6 L] <
2 L 17 _INITIAL TEMPERATURE |
< — DISTRIBUTION {APPROXIMATE)
x 160 = 1040
a |
5 |
— 180 1020
1.24 2 3 4 5 6 7 8 9 10 15 20 30
RADIAL DISTANCE FROM CORE CENTER (in.)
Fig. 66, Temperature vs. Time Relationship for Reactor Core and Pressure

Shell.

to the room cooling requirements and
the reactor electric heater require-
ments.

It was assumed that the flow through
the bottom clearance holes was negli-
gible and that the flow through the
upper clearance holes was that due to
the difference in head between the

helium in the clearance holes and
room helium. The head available for
flow 1s, then,
(18) P=h('oa_pc):
where

P = pressure head, 1b/ft?

P, = density of ambient helium,
1b/ft3
P, = density of helium in clearance
hole, 1b/ft3,
h = height of clearance hole, ft.
To be conservative, it was assumed that

the temperature of the helium in the

88

1200°F;
temperature was taken as about 130°F,

Thus

clearance hole was the room

 

6.125
P = T x (0.0092 - 0.00335)
= 0,00299 1b/ft?
The loss in pressure head in a gas
flowing through an annulus in laminar
flow is
12.GL
(19) p = Fr
yig p
where
f = viscosity, lb/hr-ft,
G = flow, 1lb/hr- ft?
L = length = height of clearance
hole, ft,
y = annulus gap, ft,
g, = gravitational constant, ft/hr?
© = gas density, lh/ft?3

 
DWG. 19142

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1200 26
—
/ -
1000 / ry 22
800 N 18
™
° -
W 3
g St
[0
Y 600 14 W
s =z
w O
a o
s
w
[t
400 10
200 \ 6
N
0 2
0 20 40 60 80 100 120 140

TIME (hr)

Fig. 67. Summary of Temperature and Power in Entire Preheating Period.

 

An evaluation of Eg. 19 gives

 

 

6.125
12 x 0,102 x G x T
P = = 4,24 x 10°% G 1b/ft? ,
0.39\? )
—EE— X 32,2 x (3600)° x (0.00335)

 

The flow, G, 1s then evaluated by
equating the available pressure head 7 3.75\2 2.97\2
to the loss 1n head because of flow: A= 6 X— || — - |-
4 12 12
G = 7.05 1b/hr-ft? ,
= 0.1715 fe?

Since there are six 1dentical . )
Therefore the weight flow 1is

clearance holes, the total flow area

1s W =0.1715 x 7,05 = 1,21 1b/hr ,

89
(<

 

2.97in.

 

 

 

 

_— 33/4 T —

Fig. 68. Rod Sleeve Through Thermal
Shield.

and the heat loss 1s

Q=MWe (T, -T,) =1.21x1.24 (1200 -130)

= 1605 Btu/hr
= 0.446 Btu/sec ,

where
T = temperature in clearance hole, °F,

T, =ambient temperature, °F,

The heat loss through the clearance
holes 1s therefore approximately 0,45
Btu/sec.

SPACE COOLER PERFORMANCE
The bulk of the equipment of the ARE

is contained 1n three sealed pits,

 

B4 -

 

14 —— —= — -14 -—

OUTSIDE : QUTSIDE

  

—— e — i o L e L

 

bmee e B Y e 27l — 18% -3l —-
NOTE: DIMENSIONS IN INCHES

Fig. 69.

90

", 6-DIA HOLES FOR
CEILING SUSPENSION

which are kept cool by the use of
space coolers. A typical space cooler
is shown in Fig. 69. Fach space cooler
contains two helium blowers in parallel
(total capacity 7200 cfm) and a water
cooling coi1l. The entire unit was
supplied by the Trane Co. The cooling
coil is a Trane Co., series 93, type S
coil, with a 12- by 48-in. face and
four rows of tubes longitudinal to the
gas flow., The heat transfer coefficient
of such a coil, operating with air on
one side and water on the other, 1s
given 1n ref. 21, for various water
and air velocities. As used 1in the
ARE, these co1ls wi1ll have helium
instead of air on the gas side.

Since 1t was necessary to convert
the performance data from those for
alr-to-water operation to those for
helium-to-water operation, the follow-
ing equation, from which values can be
obtained that are approximately the
same as the actual performance data
given 1in ref. 21, was used:

1 0.00374 | 0.189

 

(20) — = ,
hA ,0-8 L0 6
w a
where

h = over-all heat transfer coef-
ficient, Btu/hr:ft?-°F,

A = heat transfer area per row per
square foot of coil face area,
fo?,

v _ water velocity, ft/sec,
- air velocity, ft/min.

To use Eq. 20, an equivalent air

UNCLASSIFIED
DWG A-3-5-224A

————— e 3 - - - -

. "Mg-DIA HOLES FOR
/ CEILING SUSPENSION
/ 4% -P TAP RETURN

3t
376" ./ NEAR END
L NEARE!

ey

Dy

 

 

 

 

 

 

 

 

; BRERS
| | 25
| o =
i : oxr ™
: | »x O o
! | et
: i ol
= i
A S S—— Y .
/u’ g |
ALEMITE FITTING // —~ 9%~
re <y ‘/2-P TAP SUPPLY
BOTTOM PANEL REMOVABLE 7 NEAR END.

Space Cooler.
 

F:__________________________——————————————————————————————————————————————————————__________________,44444ffi

velocity, that is, the air velocity
which would give the same heat transfer
coefficient as the actual helium
velocity, had to be calculated. Since
the fin efficiency of the coil fins 1s
very close to 100% at all times (sheet
fins of aluminum, 0.012 in. thick,
with the tubes on a triangular pitch
of approximately 1.2 in.), 1t may be
assumed that the gas-side heat transfer
coefficient is a function of the
Reynolds’ number to the 0.6 power.
Since the Prandtl’s numbers for aair
and for helium are about the same at
this temperature and since the geometry
of the co1l is unchanged, the equivalent
air velocity can be calculated from
the following equation:

 

 

 

 

 

 

0.6 0.6 ,,0.6
ha kava pa #He
(21) = 1,0 =
0,6 0.6 0.6
He KieVHe PHe Ha
or ; 1.667
Ua _ He pHe #a
UHe ka pa /J“He
where
k = thermal conductivity,

© = density,

f = viscosity,

h = heat transfer coefficient,
v, = equivalent air velocity,

Vo = helium velocity.

Since the equivalent air velocity
(calculated above) and the water flow
are known, the over-all hA can be
calculated from Eq. 20.

The equations governing the operation

of the space coolers are

(22) Q= Wy, (Ty, - Ty.)

He

(23) Q= W,e (T, -T,),

(24) Q = h46 ,

where

Q = heat transferred, Btu/sec,

£ = log mean temperature difference,
OF’

W = weight flow, 1lb/sec,

c, = specific heat, Btu/1b*°F,

T' = maximum temperature (also equals
ambient temperature), °F,

T = minimum temperature, °F,

and the subscript w refers to water.
The system operates under the following
restraints and conditions:

1. The helium flow is 7200 cfm at
blower temperature.

2. The outlet water temperature 1s
thermostatically controlled to be 100°F.

3. The inlet water temperature
varies with the season of the year but
is assumed to be 70°F for this calcu-
lation,

Equations 20 through 24 and the
restraints and condition listed above
can be used to evaluate the performance
of the space coolers. Figure 70 shows
the variation of heat transferred vs.
ambient temperature for one of the
space coolers. (The heat exchanger
pit contains four such units; the
other pits each contain two units,)
The difference between the maximum
helium temperature (ambient) and the
minimum helium temperature 1is small;
that 1s, 1t varies from about 0.2 deg
at a heat load of 25 Btu/sec to about
1.6 deg at a heat load of 200 Btu/sec.

The heat rejected to the ambient
atmosphere in the three ARE pits from
the various items of equipment 1in the
pits, calculated in the previous
section on “Heat Loss Through Insu-

lation,” is tabulated in the following:
Reactor pit 30 kw
Heat exchanger pit 135
Dump tank pit 75
Total 240 kw

The three pits are interconnected by
passages of relatively small flow area.
Therefore the pits will not operate at
the same temperature, nor will they
operate at the temperatures which
would exist 1f the pits were com-
pletely sealed from one another. The
pit operating temperatures were calcu-
lated for the two extreme cases (pits
at the same temperature; pits completely
independent) by using the data of
Fig. 70, which gives the space cooler
performance, It was assumed that the
heat loss from the pits to the outside
through the concrete walls could be
neglected; this 1s a conservative
assumption. Table 17 gives the heat
losses 1n the pits.

91

 

 
 

UNCLASSIFIED
DWG.19144

 

250

 

200 //

/
/

100 ///

L
S

 

 

HEAT TRANSFERRED (Bfu/sec/unit}

 

 

 

 

 

 

 

 

O
{00 120 140 160 180 200
AMBIENT TEMPERATURE (°F)
Fig. 70. Space Cooler Performance.

The pits will therefore operate at a
temperature of about 110 to 120°F, Tt
should be noted that all the heat loss
calculations in the section on “Heat
Loss Through Insulation’’ were made with
an assumed ambient temperature of 130°F
Therefore the heat losses listed in
Table 17 are slightly low. The dif-
ference is quite small, however, inas-
much as the surface temperature of most
of the equipment losing heat is from

1150 to 1500°F,

TEMPERATURE PATTERNS IN THE MONITORING
ANNULUS IN THE EVENT OF HEAT FAILURE

The fuel and reflector coolant
systems of the ARE are entirely enclosed
in ametallic envelope (Fig. 71). This
envelope consists of concentric tubing

around all the pipes, and conveniently
shaped containers around the other
such as valves and
pumps . circulated in the
space between the metallic envelope
and the main systems to

items of equipment,
Helium 1s
serve two
purposes, The helium 1s continuously
sampled by instruments which can detect
the presence of small quantities of
fuel or NaK vapor and thus give warning
the system
“monitoring system'’),

1f a leak develops 1n
(hence the name,
The flowing helium also acts as a
distributor of heat in the following
Most of the electric heaters
of the main system are on the outside
of the Should a heater
failure occur, the flowing helium, by
transporting heat from the heated to
the unheated areas, will tend to make
uniform the heat distributed to the
systems and, consequently, to reduce
the temperature gradients which would
otherwise occur,

manner.,

envelope,

The locations where the heat distri-
bution function 1s particularly
critical are in the fuel dump-and-fill
lines. As may be noted from the
position of the valves in Fig. 2, the
fuel dump-and-fill lines may stand full
of stagnant fuel during the operation
of the reactor., Consequently, in the
event of a heater failure, it 1is
possible for the fuel carrier to freeze
in these lines and make dumping of the
reactor impossible, However, a helium
bleed line will be connected to the
fuel-carrier line to prevent drainage.
The temperature patterns that would
occur 1in these fuel-carrier lines and
their envelopes 1f there
particularly bad heater

were a
failure were

 

 

 

TABLE 17. HEAT LOSSES IN THE PITS
TOTAL HEAT NUMBER OF HEAT LOSS AMBIENT
LOSS SPACE PER COOLER TEMPERATURE

(Btu/sec) COOLERS (Btu/sec) (°F)
All pits 228 8 28.4 118
Reactor pit 28.5 2 14,3 111
Heat exchanger pit 128.0 4 32.0 120
Dump tank pit 71.1 2 35.6 121

 

 

 

 

 

92
 

  
  
 

 

LEGEND ROD COOLING SYSTEM ugfiéAhsgllgEgA
H — HELIUM A 04
W— WATER & o3 ,_1L*‘— i
IS— INERT SALTS 602 G_L - I | eos
F — FUEL i
602 — L INE NUMBER e ‘:
LD—LEAK DETECTOR § Neoi—_ 6017 A b
@ — SNIFFER POINT : o AH)
-——- INSULATED LINES TEMPORARILY INSTALLED HW How ] eos LJ“{,“M o
DURING PRETESTING PERIOD WHEN ROOF T N e |
SLABS ARE OFF. T -

 

 
   

SURGE TANK

 

 

 

 

 

TANK 4 TANK 5 TANK & TANK 3 TANK 2 QUMP TANK

TANK §
Nak NaK NaK Nak FUEL CARRIER FUEL CARRIER

Fig. 71. Monitoring System Flow Sheet.

€6

 

 
Y6

1 Mo ——

  
 
 

COOLEY ELECTRIC CO. UNIT
HEATERS, 650 WATTS, 120V
CLAM SHELL TYPE, 8 in. LONG

  
    
 

i oy
'k R
= TC OTHER
HEATING UNITS

5 COOLEY ELECTRIC CO UNIT
i HEATERS, 325 WATTS 120 v
CiAM SHELL TYPE, 4in LONG

 

—
b
. e
Lz L L2
y éx LT e
' B
/ A 2 WIRE, 3 UNIT. USE ONLY WHEN
/7 \‘\ NECESSARY FOR 3 UNITS
u)//f/ \ L3 Lt L‘Z L3
L by i J e 3y Ty Uy T J

A C
3WIRE, 9 UNIT DELTA 3 WIRE, 6 UNIT. USE ONLY WHEN NECESSARY FOR 6 UNITS

WIRING DIAGRAMS A, B, AND C

Fig. 72. Electrical Wiring Diagram for

 

DWG D-13296A

 

 

   

KEY PLAN

TN NO. 16 GAGE.
/ N STAINLESS STEEL WIRE
{

DETAIL E
HEATING UNIT HELD IN
PLACE BY NO. 16 GAGE
STAINLESS STEEL WIRE

 

N ~HEATING UNIT
{ N
\\\ \x‘\\
L .

\\\ \\

T - AN

\\\ F’

DETAIL D \\é/\zg’i

CLAM SHELL HEATING UNIT S Sop
\"th “
Q\\\'\C co o
e, :
.
PN
PORCELAIN INSULATOR . e -
BUSHING ‘

Monitoring Annulus Heaters.

 
investigated for several helium
velocities 1in the monitoring system.
For the investigation, 1t was assumed
that the lines were filled with the
fuel carrier. Figure 2 shows the pipe
line to the fuel dump tank; Fig. 72
shows a wiring diagram of a typical
set of electric heaters, which are in
the form ofhalf cylinders and surround
the lines shown in Fig. 2. Tt may be
noted from Fig. 72 that three sections
of electric heater, each consisting of
one half cylinder 8 1in. long, are
connected 1n series and that the
failure of any one heater results 1in
the failure of the other two. The
three heaters are separated so that
they are not adjacent to one another.
For the type of heater failure con-
sidered i1n the following analysais, 1t
1s assumed that two sets of heaters
fail and that these are exactly opposite
each other on the pipe line,

The system, as set up for analysas,
1s shown in Fig. 73e. For simplicity,
the heat produced by the electric
heaters 1s assumed to be generated
directly in the wall of the concentric
tube which serves as the metallic
envelope. The pipe line 1s divided
into seven sections:

1. Sections 1 and 7 are the un-
affected areas of the pipe lines up-
stream and downstream of the area
where the heater failures occurred.

2. Sections 2, 4, and 6 are the
areas where the heaters on both sides
of the pipe failed.

3. Sections 3 and 5 are the areas
where the heaters (which separate the
failed heaters) are still operative.

In the steady-state condition, the
fuel pipe temperature will be equal to
the temperature of the flowing helium.
Therefore only the heat transfer be-
tween the helium, themetallic envelope,
and the outside environment needs to
be considered. The heat balance 1n a
given section of pipe line 1in the
steady-state condition 1s given by

(Fig. 73b)
(25)  Q, +Qu * Qyo_p = QU * Q,_,
and

(26) Qe = Qe t Qeoun -

Dl! 19145

INSULATION
METAL TUBE
gsECTIONSG) @, 8. 9.6 ., 0 /7 {

N 3

 

 

4

 

 

 

 

 

 

 

_¢\ 7 STAGNANT FUEL { !
FLOWING HELIUM PIPE WALL
{a)
x x+dx
INSULATION | |
? o |
) METAL TUBE @pm —— Q0 |
FLOWING HELIUM @He™™ , — Q'
OHe—m ’
) PIPE WALL | | /
1

 

l I
(&)

Fig. 73. Sections Showing Fuel Pipe
and Annulus Walls.

where Q and Q' are heat transferred in
Btu/sec and the subscripts
m = along the metal of the envelope,
He = in the flowing helium,
He-m = from the helium to the metal of
the envelope,

m-e = from the metal of the envelope
(through the insulation) to the
environment,

g = internally generated.
The various terms 1n Eqs. 25 and 26 are
expressed quite readily as follows:
From the common equation for one-
dimensional heat conduction,

< d 28
(27)  Q -0Q' =k.a — " dx ,
n mn m m
dx?
where
Qm = difference between metal temper-
ature and ambilient temperature,
OF,
k = thermal conductivity, Btu/sec* ft?
(°F/ft),
a = cross-sectional area, ft?,
x = distance from origin, ft (the

left end of each section, except
section 1, 1s considered as the
origin of that particular section;
the right end of section 1 1is
used as the origin of that
section).

85

 
The heat gained by the helium
(28) vQHe - QHe - wHeCpHe daHe ’
where

g = di1 fference

He helium

between
temperature and ambient temper-
ature, °F,

W = weight flow, lb/sec,

c, = specific heat, Btu/lb-°F.

The heat transferred from the helium
to the metal

(29) QHe-m - hHe—mCm dx(fifle - Qm) ’
where
h = heat transfer coefficient,

Btu/sec* ft2-°F,
C = circumference of inside wall of
metallic envelope, ft,

The heat transferred from the metal
through the insulation to the environ-
ment 1s a complicated function which
was evaluated for various metal temper-
atures in the section on ‘“Heat Loss
Through Insulation,” Egqs. 2, 3, and 4,
and is shown in Fig, 74. The curve in
Fig. 74 can be approximated from about
500 to 1500°F by a straight line for

which the formula is
(30) Qn_e = b0, + d ,

where b and d are constants,

The internal heat generation in the
metal 1s equal to the heat supplied by
the electric heaters in those sections

H

in which the heaters are still operative,

and 1t 1s equal to zero in those

 

 

 

   

 

 

 

040
030 —- o R R e -;;) .
2 e
= | |
3 !
&
—
o
w
2 010 — I
Ing .
Approximation Formula
{0=2.29x107%9-00237}
0 - ’/ - - - ——_— S —— _—
0 400 800 1200 1600
METAL TEMPERATURE (°F}
Fig. 74. Heat Loss from Insulated
2-in, Pipe vs. Wall Temperature of the

Pipe (Sink Temperature, 130°F).

96

sections 1in which the heaters have
failed. The heat supplied by the
heaters which are still operative can
be varied by the operator. At design
conditions, the heaters will be
operating at about 5/9 capacity; if
indicate that
failures in the
line, the adjacent heaters which were
stil]l operative could be run at full

thermocouple readings
there were heater

For the following analysis,
the more dangerous situation

power.,
however,
will be considered, namely, that the
heater failures go unnoticed and the
heaters which are stil]l operative
remain at their normal power:

(31)  Q, = gq'a, dx = P dx ,

for sections 1, 3, 5, and 7, and

Q = q'am dx = 0

g
for sections 2, 4, and 6, where
g' = unit internal heat generation,
Btu/sec* ft?,
P = normal power input to the

electric heaters per lineal foot,
Btu/sec*lineal ft,

Substitution of Egs. 27 through 31 in
Egs. 25 and 26 gives

 

 

d?6,
(32) k a + q'a
nom 2 m
dx
* hHe-mCm (QHE - Qm) = bem - d
and
d@He
(33) W

HECPHe dx
= *hHe-mcm (EHe - @m) .

Solving Eq. 33 for 6  and differ-

entlating twice gives

 

 

 

 

WHECPHe dgHe
(34) 6, =6, t——"
hHe—mCm dx
and
5) d%, d%e,, ) uepy, d%6,,
dx2 dx2 hHe—mCm dx3

Substitution of Eqs. 34 and 35 in Egq.
32 gives
d36 h cC d?2f

 

(36) He + He~-m ™ m He
dx3 WHeCpHe dx?
hHe-an t b dof hHe-mCm b o
kna, dx kmamWHeCpHe

hHe-mCm (ld ,
= —_— . — - q ,
km%eCpHe o

 

which may be solved quite readily., It
is first necessary to evaluate the
constant in Eq. 36 for the conditions
of the problem. Three different helium
velocities were considered: 1, 3, and
10 ft/sec. For these three different
velocities, the values of the constants
of Eq. 36are tabulated in the following:

Helium velocity, ft/sec 1 3 10
hHe-m’ Btu/sec* ft2-°F 3.03 x 10-4 4.97 x 10-% 8.54 x 10-4
nr It 0.899
Wy.,» lb/sec 1.03 x 10-* 3.09 x 10°* 10.30 x 10°*
k., Btu/sec-ft? (°F/ft) 0.00333
a., fr 2 0.00487
b 2.29 x 10-*
d -0.0237
qg', Btu/sec*ft’ 45.5 (sections 1,
3, 5, and 7)
0 (sections 2,
4, and 6)
hHe-mCm
v 2.13 1.17 0.601
c
He pHe
hHe—mCm t b
_— 30.9 41.6 61.5
kmam
He-mcm b
30.1 16.5 8.48
E a W ¢
m m He PHe
hHe-mCm (d ’>
N
kaHecpHe a,
For sections 1, 3, 5, and 7 -32,300 -17,600 -9090
For sections 2, 4, and 6 -3120 -1700 -879

The helium heat transfer coefficient, hH ]

formula: ¢(1%)

H

(37) He-m D He He

He

k
h = 1.02 — Rel: 4% Pr°~5(

.+ may be evaluated from the following

0.14 /D 0.4 /D \o.8

) <_Hi> <_?.> Grl-05
He !

He LHe' Di

97

 
 

where
D = equivalent diameter, ft,
D, = outside diameter of helium
annulus, ft,
D. = inside diameter of helium

annulus, ft,

Re = Reynolds’ number,

Pr = Prandtl’s number,

e : . :

— =ratio of viscosity at bulk
Fs temperature to viscoslty at sur-

face temperature,
L = length of passage, ft,
Gr = Grashof’s number,

Having evaluated the constants 1n
Eq. 36, the solution of the equation
is straightforward. Once an expression
for @He 1s found, Qm
from Eq. 34,

can be evaluated
The solutions for helium

velocities of 1, 3, and 10 ft/sec
follow:
1. For a helium velocity of 1 ft/sec,
(38) Gy, = ny g0 10% 4 n, e 8- 29%
+n 6-0.9411 + K
3
and

(39) & =3.39n, e>1%% _1,95n, ¢ 6:29%
+0.559 e=0.941x + | |

where

n = arbitrary constant,

K = 1070 for sections 1, 3, 5, and 7
= 104 for sections 2, 4, and 6.

2. For a helium velocity of 3 ft/sec,

(40) €, =n

6.10x
He 4 € *

-6.87x
nse

+n6 6'0.3921 +K
and

(41) & = 6.23 n, eb.10x __ 4 9 n o~ 6.87x
t0.664n, e 0-992% + g |

3. For a helium velocity of 10 ft/sec,

3 _ 7.62x -8.08x
(42) Gy, = n; e tng e

+ n9 6'0.1371 + K
and

(43) & = 13.7n_ e7-62% _ 19.4 n -8.08x
m 7 8

+ 0,772 ng e 0. 1372 4 g |

There are nine arbitrary constants
in Egqs. 38 through 43. These constants
have a different value in each of the

98

seven sections; there are therefore,
altogether, 63 arbitrary constants.
These may be evaluated by means of the
following boundary conditions, which
apply to all three helium velocities:

(44) lim #

X, — ~®
1

He

is bounded. Since each section has a
different origin (see definition of x,

following Eq. 27}, the subscript on x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

refers to the number of the section
being considered. Thus
(45) 6 -9 ,
He xl=0 He x2=0
(46) g = & ,
n 11=0 " x,=0
d@m d@m
(47) = ,
dx xl=0 dx x2=0
(48) 0 = g ,
He x,=2/3 He %, 4y=0
r =2,3,4,5,6 ,
(49) ¢ = g
_ H - !
" xr—2/3 € Ir+1_0
r = 2’ 3’ 4, 5’ 6 r
d@m dé
* X, =2/3 * T4y =0
r = 2’ 3) 4, 5! 6 )
(51) lim QHe
17—'(1)

1s bounded.
Equations 44
the nine arbitrary con-
, to n, for each of the
seven sections. The results of the

evaluation are given in Table 18,

through 51 were used
to evaluate
stants, n

The values of the arbitrary con-
stants and Eqs. 38 through 43 were used
to calculate the temperatures of the
helium and the metallic envelope for
all posttions along the pipe line for
the three helium velocities. The
in Figs. 75, 76, and
77. The minimum helium temperature

results are shown

 

 
1000

900

800

700

TEMPERATURE (°F)

600

500

100

1000

900

800

TEMPERATURE (°F)

700

600

500

Fig. 75.

DWG. 19147

 

SECTION 4

SECTION 5

 

 

T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

\ / )
L\\\c HELIUM L//}r’,éf,/*r' \\r

\\\\O<:::,—WALL

o 94,4
2 4 6 8 2 4 8
DISTANCE ALONG PIPE LINE (in))
SECTION & SECTION 7
’/0/_/()
y
>\ /
‘k\“x HELIUM Y//;f/,
\ /
—WALL

2 49 6 8 2 4 8

DISTANCE ALONG PIPE LINE (in)

Temperature Patterns of Helium and Wall of Monitoring Annulus.
Helium velocity of 1 ft/sec.

99

 
TEMPERATURE (°F}

TEMPERATURE (°F)

He l

100

DWG. 19148

 

1100

1000
HELIUM
900¢ ~ 1//}

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

i

800 | /
L WALL

700

| SECTION 4 SEGTION 5
600 E

2 4 6 8 2 4 6 8

DISTANGE ALONG PIPE LINE (in)
1000
900(*‘~a\\\1
HELIUM )
< /O/(/O/(

 

 

 

WALL

 

700 %

~

SECTION 6 SECTION 7

 

 

 

 

 

 

 

 

 

600 |
2 4 6 8 2 4 6 8

DISTANGE ALONG PIPE LINE (in.})

 

Fig. 76. Temperature Patterns of Helium and Wall of Monitoring Annulus.
ium velocity of 3 ft/sec.

 
101

1300

12004

1100

t000

TEMPERATURE (°F)

900

800 -

700

HELIUM E
1 I S
; ‘: |
i | | |
e i S— S e e : — — —

| SECTION 1 ‘ SECTION 2 SECTION 3 SECTION 4 1

|
| | : ,

: 1 : 1 i j ’ ‘
-8 -6 -4 -2 0 2 4 6 8 2 4 8 4 6 8
DISTANCE ALONG PIPE LINE (in)

1100 [
HELIUM |
|
__ 1000 —
5
¢
& LwALL
'—.-..
2 900 e k}\( ~ _ -
Ll ; |
a | |
S ! i _o-"('/{
L i i
’_—
800 1 . .
| | |
SECTION 5 SECTION 6 SECTION 7
700 ’
2 a 6 8 2 4 6 8 4 8
DISTANCE ALONG PIPE LINE (in)
77. Temperature Patterns of Helium and Wall of Monitoring Annulus. Helium velocity of 10 ft/sec.

Fig.

DWG 19149

 

 

l

i
i

 

 

e ]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 
 

 

TABLE 18,

VALUES OF THE ARBITRARY CONSTANTS

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[ SECTION NUMBER
1 2 3 4 5 6 1
n, -80.6 2.69 -2.72 2.88 ~2.70 2.38 0
n, 0 -76.1 75.5 ~76.6 15.6 =75.7 89.5
ng, 0 960 -447 722 =575 654 -637
ng -30.5 0.476 -0.459 0.467 -0.461 0.473 0
ng } 0 -27.6 27.7 -27.5 28.0 -27.6 27.9
neg | 0 965 -222 796 -354 695 -431
n. ~8.64 0.0554 -0.0537 0.0546 -0.0538 0,0548 0
ng 0 -8.20 8.19 ~7.95 8.05 -7.96 8.04
ngy 0 968 ~84.2 891 ~-155 826 -213
and the minimummetallic envelope temper- G 19150
ature have been plotted against helium 1000
velocity in Fig, 78. Since the temper-
ature of the fuel 1in the pipé line 1is
equal to the helium temperature and 900
since the fuel melts 1n the vicinity o
ot 950°F, 1t can be seen from Fig, 78 o
that the fuel will remain molten, 1in E 800
this particular case, 1f the helium &
velocity exceeds about 8 ft/sec. %
S 700 |-
2
=
z
=
600 |-
500 |
o 2 4 6 8 o
HELIUM VELOCITY { ft/sec)
Fig. 78. Minimum Wall and Helium

102

 

Temperatures vs.
Monitoring Annulus.

Helium Flow in
 

Chapter 7
HELIUM SUPPLY AND VENTING SYSTEM

TRUE HOLDUP OF FISSION GASES IN TANKS

The fission gases removed along
with helium through the various gas
vents 1n the fuel circuit are held up
in the pipes and tanks of the off-gas
system (Fig, 79) until the activity
in the fission gases is reduced suf-
ficiently to permit their being
discharged into the stack. The holdup
time of the gases in pipe lines or
similar shapes of large length-to-
diameter ratio 1s readily calculated
as the pipe volume divided by the gas
volume flow rate. All the gas is held
up substantially the same amount of
time. In large tanks or similar shapes
of small length-to-diameter ratio,
the average gas holdup time is the
same as 1n a pipe line of similar
volume. However, because of the
mixing of the gas in the tank, not
all the gas 1s held up for substantially
equal periods; that 1s, some of the
gas leaving the holdup tank is held
up for much less than average time and
some for much more than average. The
true activity of the gases leaving the
holdup tank (Fig. 80) of the off-gas
system has been calculated and
compared with the activity which would
be present 1f all gases were held up
for equal periods.

The mixing of the gas 1n the tank
1s due to convective and diffusive
forces and occurs fairly rapidly
compared with the average holdup time
in the tank. Thereasonably conservative
assumption was made that the gas
entering the tank mixes instantaneously
with the gas i1n the tank. The follow-
ing definitions were made for the
calculations:

N = flow rate of gas,

V = tank volume, ft3,

C(t)

cfm,

concentration of gases of
age t, that is, C(t) dt equals
the ratio of the volume of gas
with age from t to t + dt
compared with the total tank
volume, V (age is used here to

mean time measured from the
time the gases enter the off-
gas system),

t = time,

Then,

min,

(1) C(t + dt) dt
N
= C(t) dt - C(¢t) dt'jfdt ,

C(t + dt) - C(t)

 

(2) P = - C(t)fif’
(3) dCCt) ey N

dt y '
(4) C(t) = K e~ (N V)t

where K 1s an arbitrary constant.

When t = t,, where t, 1s the age

of the youngest gases in the tank,
N
(5) C(t) dt =-—dt
v
and
N
(6) Clty) =—.
V
Substituting Eq. 6 in Eq. 4 and
solving for K gives
(7) _fi1= K e-(N/V)t0
v
(8) K = W/
v
N
(9)  C(1) =——em W/ V) Ctato)
V

The energy, E(t), in the fission
gases xenon and krypton (bromine and
iodine removed previously) at time
t 1s plotted against t in Figs. 81,
82, and 83 for the separate xenon and
krypton 1sotopes and for the total
of these isotopes. For the determi-
nation of E(t), 1t was considered that
there had been 200 hr of reactor
operation prior to time zero and that
the fission gases had subsequently
decayed time t.

103

 
 

Y01

=

- 36in —L- 4ft-3in.

 

 
 
   
               
      
 
   

 

      

 

 

    
 
 
 

MATERIAL AND MATERIAL SPECIFICATIONS UNCLASSIFIED
aTy NAME [ SIZE | MATERIAL DWG E-A-3-8-524
P2 KINNEY VAC PUMP NO Y26166 = 1'% hp
L2 DRUMS 55 gal 55
1 VAPOR TRAP, NoK " 55 gal S5 T
1 VAPOR TRAP, SUMP 55 gaf 55 . b
2 SURGE TANK VAPOR TRAP 55 gal 55 i
13 FLEXIBLE HOSE | %inx3f1-0in LG SS P
6 RESEARCH CONT. CO VALVE 4 in 1IPS 5% !
ASREQ | MANIFOLD, VAC 2in IPS SCH 40 | WS i fft-6in
AS REQ | ARLINE FROM 1O cfm BLOWER | 2in. tPS SCH 40 | WS
6 SWING CHECK VALVES | 2iniPs ©58 ) =
g VALVE ; G b 1
t i OIL CATCH BASIN LoGin IPS x 24in $5 4 16 4o )
10 | CHECK VALVE 3 ss 1832 - goa = df-dnm 5
5 | SOLENOID VALVES | 1in. IPS-SCREW  BRASS L - =
ASREQ | VAC PUMP LINES | tin IPS SCH 40 ot N T
1 FULFLO FILTER : ol f g9
2 RADIATION MONITORS AND SHIELD , b !
3 PACKLESS VALVE | ‘4in BAR STOCK : 58 £
ASREQ | VAC LINE ' Yin IPS SCH 40 WS o
2 MASON NEILAN VALVE 1 :/4 in {TYPE 108) 55 Ll
.6 SOLENOID VALVES DY 1PS 55 Lo
o DUMP TANK VAPOR TRAP FUEL SURGE TANK NOf 5
; 2 SWING CHECK VALVE o1l in 1PS 55 !
b MONITORING SYSTEM HEATERS
1 RELIEF VALVE ¥, in 1PS | BRASS 37

TANK NO 3 FILL AND FLUSH TANKS 4 i
. (SAME TYPE VALVES AND B
1805~ ~ HOSE ON EACH TANK) Jc
N v ah . -
6.~ 35 2t 2 FUEL ENRICHING
- o SYSTEM TFANKS

 

 
  
 
  
  
 

   
  

TEST PIT
— 2 ft-in

   

L8075 Laoa
4ft-0in- == 3ft 4in = - 1Z2in

   

    
  
   

 
  

S eri -
TANK NO 4 TANK NO.S  TANK NO 6 L8o2

  
   
    
  
  
 

DUMP TANK .~ FUEL SURGE TANK NO. 2

G6ft-0in

  

A

  
  
  
     
 
  

   

tft-6in —

  
  
 

[
e

‘.14_1- "_:‘l’ _= ! ~
38 10

STACK

19 in

1Q ¢fm BLOWER

Fig, 79. 0ff-Gas Disposal System,
 

1-in. NPS TO DIMENSIONS SHOWN
MATERIAL STAINLESS STEEL

XB/Q—in. EXISTING PIPE COUPLING
11 ggfi%f?zzzfigfiz/ DN
__/—___/

 

 

 

 

 

 

e

l I

 

 

 

 

I

L

] LTI

o
et

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1D

 

 

UNCLASSIFIED
DWG. C-A-3-8-37A

N

 

 

 

Fig. 80. 0ff-Gas Disposal System Drum.

If R 1is the ratio of the true
activity of the gases leaving the
holdup tank to the activity which
would be present i1f all the gases
were held up equally,

 

‘s N /v
f —en (N/V)Ct-t0) E(¢)dt
ty V

(10) R =

 

E( +V)
t ————
o N

where t, 1s the age of the oldest
gases 1n the tank. The total energy
in all the fission gases developed
may be used since the final answer
desired 1s a ratio.

It 1s 1mportant to note that for
Eq. 10 1t is assumed that all the

gases 1n the tank contain equal amounts
of fission gases. This is obviously
not true of the gas originally in the
tank., Thus, at the beginning of
operation, the value of R is actually
less than 1s predicted by Eq. 10. The
highest possible values of R will
occur after long periods of operation,
and Eq. 10 will therefore be evaluated
with t, sufficiently large for the
contribution to the total activity of
the gases of age greater than t, to be
neglected.

The curve of E(t) vs. t shown in
Fig. 83 for the total energy of all
xenon and krypton i1sotopes was broken
up into a series of M segments between
t, and t_, and an equation of exponential
form was fitted to each segment.

|
105

e

 
UNCLASSIFIED
3 DWG. 19302

 

W=0496 X £ X PXF (/) WHERE
W=WATTS
£ =ENERGY OF RADIATION, Mev
£ =REACTOR POWER, MEGAWATTS
F(r)=ACTIVITY FUNCTION

i

NQOTE.: ! :
ALL ISOTOPES CONSIDERED AS
SATURATED AFTER 200 hr OF
OPERATION EXCEPT Xe'33

M

o
[

SUBSCRIPTS:
{1=153-m HALF LIFE
(2)=9.2-h HALF LIFE

o

xe'2266% OF
SATURATION)

xe135

(1)

™

o

 

_

 

wn

 

 

ACTIVITY FUNCTION, £ {#) (DISINTEGRATIONS /10,000 FISSIONS)

M

f 10 100 1000 10,000 100,000
172 hr thr fahr  24br 4 days 40 days

TIME AFTER REACTOR SHUTDOWN (min)
Fig. 81. Integrated Xenon Activity, F(t), vs. Time after Reactor Shutdown.

Thus, for the nth segment,

(11) E (t) = A e bnt, t

and then

o
¥ f‘n le-(N/V)(t_to) A e“bnt gt
- t ) n

 

(12) R -2 | ,
E + v
t —
° N
where ty = tp,
1 . N
t
(13) R = ———————”17—-223 An-TTftn 1 em LN/ tb, L et/ vy ey gy ,
E{t, +—) =2 n-
0 N n=1
N
¥ A — .
(14) R = _ 2 "V {e-[(N/V)+bn]t+(N/V)to J ,
E + v N + b thoi
t o = - na
o Ty =1 v "
_...IY.._G(N/V)tO N

A
R =__l/___—v Z _N—" [ e Lewsvy+enle, _ - [(N/V)+bn]tn_1]

0 N n=1 7_*_ bn

106

 
n

o
r

th

N

o

ACTIVITY FUNCTION, £ (/) (DISINTEGRATIONS /10,000 FISSIONS)
o,

172 hr {hr
TIME AFTER REACTOR SHUTDOWN (min)

Fig. 82.

The data and results of the calcu-
lations for the particular problem of
the first holdup tank in the off-gas
system are given in Table 19,

As a check on whether the intervals
selected were small enough, the
interval between t;, and t, was divided
into two parts. Recalculation of the
data for the smaller intervals is
given in Table 20,

The valueof 295 checks very closely
with the value of 296 calculated for
the original intervals, and thus the
intervals selected were probably small
enough.

The final answer, then, is that the
actual activity of the gases coming
out of the tank 1s 23% greater than
the activity which would be present

 

UNCLASSIFIED
DWG. 19303

    

W=0.496 X £ X P X F(#) WHERE
W=WATTS  — .
£= ENERGY OF RADIATION, Mev
P = REACTOR POWER IN MEGAWATTS
- F{#) =ACTIVITY FUNCTION

NOTE
k85,1 =0.0017 SATURATED
F() AT 200hr =0.04
SUBSCRIPTS:
()= 4.4-h HALF LIFE
(2)=~10-y HALF LIFE

100 1000 10,000

{12hr 24 hr

Integrated Krypton Activity, F(t), vs. Time after Reactor Shutdown.

1f all the gases were held up for
equal periods.

TEMPERATURES IN THE HELIUM VENT LINES
CONTAINING FISSION GASES

The off-gas system carries gases
from the surge tanks, the dump tank,
and the pits to the exhaust stack
(Fig. 79). At times, a portion of
these gases will be radioactive fission
products. The gas flow rate is
generally quite low, and the question
arises as to what temperatures will
exlist on the off-gas system with
essentially stagnant radioactive
gases 1n the system. The total energy
ofall the fission gases produced after
200 hr of reactor operation at 3
megawatts is about 14,8 Btu/sec for a
delay time of 1 1/2 minutes.(??) The

107
TABLE

19. DATA AND RESULTS
FIRST HOLDUP TANK IN

OF CALCULATIONS FOR THE
THE OFF-GAS SYSTEM

 

 

 

 

 

Data:
t, = holdup time in off-gas system before reaching tanks
= 6 hr = 360 min
N = gas flow
= 1/2 ¢fh = 1/120 cfm
V = tank volume
= 7.5 fe?
vy 900 mi
- = min
N v
E(to ' z—v> - E(1260) = 350
A
Vo ) e-[(N/V)+bn]tn
y N
V n
t, (min) E(t) A b
n n
Lo 1e ||
- e n n-—- ]
t, = 360 800
t, = 1,000 | 419 | 1152 | 1.011 x 10°° 296
t, = 2,000 | 233 752 | 0.587 x 107° 106
t, = 4,000 113 480 | 0.362 x 10°° 25.7
t, = 7,000 56 288 | 0.234 x 10°° 1.5
ts = 12,000 25 173 | 0.161 x 10°° negligible
t, = 20,000 10 99 | 0.115 x 1073 negligible
Y- 4292
429,92
R = = 1.23
350

 

 

 

 

energy is divided among the fission
gases as follows:

Btu/sec
Xenon 2,2
Krypton 2.0
Bromine 1,4
Iodine 9.2

This energy may be distributed over
the gas volume 1n the surge tanks,
which 1s about 1.3 £t3, or 1t may come
out of the fuel i1n the dump tank, 1in
which case 1t will be distributed over

The

a greater gas diluent volume.

108

maximum energy release in the off-gas
which occurs 1n the release
of gases from the surge tanks, 1is
therefore about 11.4 Btu/sec*ft”.

The fission gases on leaving the
surge tanks enter a fuel vapor trap
(Fig. 84) and then a 2-in. IPS pipe.
The trap and the pipe are the critical

system,

points 1n the off-gas system, since
they have the greatest volume-to-
surface ratio and since the fission
gases are youngest at these points.
The energy dissipated per square foot

of outside surface by radiation and

 

 
 

1000

ENERGY (watts)

UNCLASSIFIED
10,000 OWG. 19304
5

NOTE:
) Xe'33 % 0.66 SATURATED
Ke®® 5, » 0.0017 SATURATED
100
\

! 10 100 1000 10,000 100,000
TIME AFTER REACTOR SHUTDOWN (min)

Fig. 83. Energy Emission of Xenon and Krypton after Shutdown from Operation
of 200 hr at a Power of 3 Megawatts.

TABLE 20, RESULTS OF CALCULATION FOR A SMALLER TIME INTERVAL

 

 

 

 

 

 

 

 

 

 

N (N/V)e, A {—[(N/V)"'bn]tn
—_— -_———t e
v ° N
V n
t (min) E(t) A b
" n n -[(N/v)+qn]:n_4
- e
ty = 360 800
t] = 600 600 1233 1.20 x 10°° 159
t, = 1000 419 1023 0.90 x 10-3 136
Y = 295
convection may be written 1n the where
’ familiar form Q = heat dissipated, Btu/sec-ft?,
T A\ T\ € = emlssivity,
. (15) o 0.173 ( s ) < e ) T, = surface temperature, °R,
S — € —— — .
3600 100 100 Te = environment temperature, 'R,
hc = free convection heat transfer
* hc(T, - Te) ! coefficient, Btu/sec*ft?*°F,

109

 
UNCLASSIFIED

OWG. D-A-3-8-73A
He {NLET { '/, -in. SCHEDULE 4G PIPE — F {

 

Z.

~
’-—* 2-in. SCHEDULE 40 PIPE CAP

INLET ANNULUS {2-in. SCHEDULE 40 PIPE }

— 10in. -—1‘ 5
/“1 ————————— ==
’d —[=TO VENT

 

o

s

5
]

o S \Wl//é

 

 

 

 

<

 

 

FROM SURGE |___ __________ pay
TANK ? &

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

|
\
I ‘ -
PAN LY " J ' N
VAPOR INLET ( {-in. SCHEDULE 40 PIPE) /] # | A \\
/ | J !Zfi \ VENT{ % -in. SCHEDULE 40 PIPE )
} \ ™ Y,~in. SCHEDULE 40 ELBOW
6-in. SCHEDULE 40 PIPE CAP o
.4 ‘F l ‘-4! S § /
4-in. SCHEDULE 40 PIPE CAP J |
|
MAX. LIQUID LEVEL ON BACK SURGE \\; L
I ]
MAX. LiQUID LEVEL DURING |
GAS DISCHARGE ———— | 7 o
N=UE == V=
NNER SHELL ( N = E II —— 4=
! 4-in. - — — -
SCHEOULE 40 PIPE ) B o T
] N
|| /
FILLLEVEL———Y P4 o |
| f
|
OUTER SHELL {6-in. | c
SCHEDULE 40 PIPE) —————————==N | ©
L/ | -
7 17
N | } v
N | 2 Y
N (4 ‘+ ARAAANANN N
| : A
V-NOTCHES TO BE APPROX.
Ya=in. WIDE x Y4-in, DEEP —— " [N B
/
Y
6-in. SCHEDULE 40 PIPE CAP —— A

 

 

 

 

 

 

 

- ———— 6,625 in.———-l

ALL MATERIAL TYPE-316 STAINLESS STEEL

Fig. 84. Surge Tank Vapor Trap.

110
The free convection heat transfer
coefficient i1s determined from the
data of Nusselt,(ls) McAdams, and
W. J. King, ¢!'?) who correlated Nusselt’s
number with Grashof’s and Prandtl’s
numbers (as given by Jakob(14)),

The pertinent physical dimensions
and unit energies of the fuel vapor
traps and the 2-in,
listed in the following:

IPS pipe are

FUEL VAPOR

2-in
TRAP ;
(por trap) 1PS PIPE
Gas volume, ft’ 0.13 0.0233
{per ft)
Surface, fr2 2.6 0.622
(per ft)
Equivalent outside 0.55 0.198
diameter, ft
Total energy, Btu/sec 1.48 0.265
(per ft)
Energy per ft2of sur- 0.246 0.427
face, Btu/sec‘ft2
For proper operation, the fuel

vapor traps must be maintained at
1500°F, The traps are
covered with electric heaters and
insulated. The heater power 1is
controlled by thermocouples which
measure the trap temperature. The
heat input required to hold a trap
at 1500°F is about 0.71 Btu/sec
(cf., chap. 6, section on ‘‘Heat Loss
Through Insulation” ). The energy
loss through the insulation would more
than be balanced by the energy in the
and the heaters could

therefore

fission gases,

 

be shut off as the fuel vapor trap

began to heat up slowly. The heat
capacity of the vapor trap, including
contained liquid, is about 13.2 Btu/°F.
A tabulation of the energy dissipations
and of the temperatures of the vapor
trap at various times 1s given 1in
Table 21. The vapor trap temperature
would be increased by only about 40°F
at the end of 20 min, and the heaters
would have to be turned on again
as the fission gases further decayed.

The 2-in. IPS pipeis neither heated
nor insulated, and therefore Eq. 15
may be used to calculate the pipe
outside temperature (the inside temper-
ature 1s not greatly different). For
an emissivity of 0.5, the pipe outside
temperature necessary to dissipate

0.265 Btu/sec*lineal ft is about 380°F,

VACUUM PUMP PERFORMANCE

There are two vacuum pumps in the
off-gas system (Figs. 79 and 85).
They serve many purposes, but their
most severe requirement is that they
must reduce the pressure in the fuel
system sufficiently to permit the
boiling off of NaK which cannot be
completely drained from this system
(NaK is circulated in the fuel system
for precleaning and hot testing),
Figures 86 and 87 give the experi-
mentally determined performance of the

two vacuum pumps. Figure 86 shows

the variation of pressure with time
12, 3-f¢?°

rate of

in a pump-down test of a
tank; Fig. 87 the
pressure rise due to leakage after the

shows

 

 

 

 

TABLE 21. ENERGY DISSIPATIONS AND TEMPERATURES
OF THE VAPOR TRAP AT VARIOUS TIMES
HEAT IN HEAT LOSS HEAT RATE OF
TIME FISSION THROUGH INTO TEMPERATURE TRAP TEMPERATURE
{min) GASES INSUIATION TRAP RISE IN TRAP (°F)
(Btu/sec) (Btu/sec) (Btu/sec) (°F/min)
1.5* 1.48 0.71 0.77 3.49 1500
10 0.91 0.74 0.17 0.77 1530
20 0.75 0,75 0.0 0 1538

 

 

 

 

 

 

*The gas is assumed

to enter the trap after a delay of 1.5 minutes.

111
cll

UNCLASSIFIED
-——SQOLENCID VALVE DWG 20225

     
  

 

 

 

 

 

 

 

 

 

 

 

 

 

   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

NOTE: Y%, THREE WAY COCK o)
ALL DIMENSIONS FOR ALTERNATE OIL 5 T
ARE IN INCHES FEED LINE 1 %6 ,
7% ] \ - PLAN OF BASE 11080-29-N 16 Y5
|
I o FOUR HOLES FOR
Y, BOLTS - l
O
Y, PT R | ! 33 Ve
Vo PT. FILLE 7 J_z/% P.T. DISCHARGE
|
25 | | —] |
12 x18
SEPARATOR TANK
| 1 1
. Y4 PT. WATER ‘
PT. SUCTION
‘ P OUTLETS Va
Ya PT. WATER
37% INLET
M gta
| dl Y%, PT. DRAIN
1l L VALVE
i @ O 1
T il - - 18‘/2 15373
B8,
T TT TT
th t4 +h ‘3
Lo 1y 1 1) 37
tHH || by 1!
11 11 L

 

 

Fig. 85. Kinney Vacuum Pump.

 
 

DWG. 131 5!

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1250
1000
750
Pump 8
500
PUMP A
250
o ' 2 3 4
PRESSURE {(mm Hg)
400 \
300 \ \\\
S \ \
@
&
LIEJ \
E 200 ———
100
o 10 20 30 40 50
PRESSURE {(mm Hg)
300
200 /

4

 

100 e
)/

 

 

 

 

 

 

 

 

 

 

 

P A
PUM KINNEY VACUUM PUMP
0
0 5 10 15 20 25 30 35
VACUUM {in)}
Fig. 86. Vacuum Pump Performance.

pumps are shut off. From these curves,
curves of the variation of rate of
change of pressure with pressure were
drawn, as shown in Fig, 88,

If a leak develops 1in the systenm,
it may be desirable to remove all the
NaK from the system. In order to boil
off NaK at 1200°F, a pressure of
about 100 mm Hg or less must be
maintained. The following calculations
were made to determine the maximum

size of leak which might exist and
still permit the vacuum pumps to
maintain a pressure of 100 mm Hg.

The process of evacuating gas from
a tank or contalning system is 1so-

thermal, and therefore
(16) PV = constant = K
or
K
17 P =—,
(17) v

113
here
P = pressure, mm Hg,
V = volume, fts,
K = a constant.
!56.15152
4
<
ifiig/
3
g ,/////,PUMPA
£
o
o ¥
[9p] .
T 7
?
|
0 |
0 90 120 150 180
TIME (min)
Fig. 87. Leakage Test.
0.00 0.0t
1000
500
200
&
I
£
E
&
T 400
w)
w)
V4]
ot
o
50
20
10
0.1 0.2 0.5 1
Fig. 88. Vacuum Pump Performance.
114

 

 

 

 

 

 

 

 

 

 

 

 

 

Differentiating with respect to time,

t, gives

(18)

K dV

V2 dt

and dividing both sides by P gives

(19)

However, K

(20)

 

dP
dt _ K dV
P Pv2dt-
= PV, and therefore
dP dV
dt _ dt
p v

Since the volume of the tank tested
was 12.3 ft?, Eq. 20 becomes

AP/AF (mm Hg/sec)

 

2

AP/AF (mm Hg/sec)

Volume of test system,

!!G. 19163

PRESSURE (mm Hg)

12,3 i@,

 
 

dP dV

dt dt dV 12.3 | 4P
(21) — or | — —_
P 12.3 dt P dt

 

The values of volumetric flow, dV/dt,
as calculated from the pump-down test
results, are given in Table 22, The
values are for pump 1nlet conditions;
that 1s, the pressure 1s equal to the
system pressure and the temperature 1is
approximately 560°R, since the gas is
cooled before 1t reaches the vacuum
pumps.

The vacuum pumps are of the rotary
positive displacement type, and 1t
would therefore be expected that the
volumetric flow would decrease slightly
with decreasing pressure, since the
leakage past the rotor increases. It
1s apparent from the calculated values
of volumetric flow, dV/dt, that the
curves of Fig. 88 are probably somewhat
inaccurate., The 1naccuracy 1s not
entirely unexpected because the
information for Fig. 88 was obtained
by measuring the slopes of the curves
of Fig. 86, and the measurements were
not highly accurate. The accuracy 1s
of the orderof t10%, however, which 1s
sufficient for the purpose of the

problem.
The valuesin Table 22 for volumetric
flow are for pumping air. The leakage

past the rotor when pumping helium may
be somewhat higher than when pumping
air, but 1t is probably accurate
enough to assume that the volumetric
flows are the same for both helium and
air.

The volumetric flow of helium
through a leak can be approximated by
considering a hypothetical “leak” in

TABLE 22. CALCULATED RESULTS OF

the form of a round orifice:

dV
(22) (———) = CAv ,
dt CE
or1 i1ce

where

dV
(;?) = orifice volumetric flow,
orifice cfs,
C= orificecoefficient =0,60,
A= qrifice area, fe?,
v = gas velocity 1norifice,
ft/sec.
The critical pressure ratio for a
perfect gas 1s given by

2 Y/ (ry-1)
23 = :
(23) e (5 1> :

thus with », = ratio of specific
heats = 1,67,

 

It

0.487

e

The throat pressure for sonic velocity
is therefore

(24) P, = 0.487 x 760 = 370 mm Hg.

The throat temperature 1is

(25) Tth = Tnmb (rC)(y‘l)/y!
where
T ., = ambient temperature in the
helium monitoring annulus =
1660 °R,
T,, = throat temperature = 1660

(0.487)(Y-1)/7 = 1235°R.

For system pressures less than
370 mm Hg, the volumetric flow through
the orifice will be constant. The
velocity in the throat will always be

THE PUMP-DOWN TEST OF A 12.3-ft® TANK

 

 

 

 

P, SYSTEM dP/dt (mm Hg/sec) dV/dt (cfs)

(mm He) Pump A Pump B Pump A Pump B Pumps A + B
300 5,95 5.55 0.244 0,277 0,571
150 2.56 2,28 0.210 0,187 0.397
100 1.63 1,41 0.200 0.173 0.373

50 0.86 0.70 0.211 0.172 0.383

 

 

 

 

 

 

115

 
sonic velocity, which 1s given by and thus

(26) v = \Jg)/RT , v = 5060 ft/sec

where

g = 32.2 ft/sec?, Therefore

v = 1.67, 4V

R = gas constant = 386, (27) (d—t> = 0.60 A X 5060 = 3040 A .
T = temperature = 1235°R, orifice

116

 
 

Chapter 8
DUMP AND FILL SYSTEM

FUEL DUMP TANK COOLING

The heat generated in the fuel
after shutdown of the reactor can
easily be taken up by the heat capacity
of the fuel and the fuel system without
overheating. However, once the fuel
is in the hot fuel dump tank, 1t is
desirable to keep the temperature of
this tank below a reasonable level.
Therefore heat is removed by allowing
helium to pass through the 91 coolant
tubes of the tank (Fig. 89).

It 1s postulated that the maximum
heat removal rate necessary is 50
Btu/sec. The problem of cooling this
tank does not lend i1tself readily to
analytical calculation. However, the
cooling was analyzedby free-convection
calculations and by chimney-effect
methods., Both methods are outlined
below.

The equation for free convection
heat transfer is

(1) Q. = hAs |
where
¢, = heat removed by convection,
Btu/sec,
h = free-convection heat transfer

coefficient taken at mean helium
temperature, Btu/sec:ft?-°F,

A = 91 times surface area of inside
of a coolant tube, ft?,
f = temperature difference between

the 1nside surface of tube and
mean helium temperature, °F,

The value of the free-convection
heat transfer coefficient for each
helium temperature was calculated from
the Nusselt’s number. The Nusselt’s
number 1s given as a function of the
product of Grashof’s and Prandtl’s
numbers on p. 525 of ref. 14. It
1s realized that the length-to-diameter
ratio of the coolant tube 1s too
great for use of these correlations;
however, they were theonly correlations
available, and an order of magnitude
estimate could be gained by using
them,

The i1inside diameter of each coolant
tube 1s 2.067 in. and the length is
3.1 feet. The temperature of the
inside surface was taken to be 1325°F
over the entire length of the tube.

Mean temperature valueswere assumed
for the helium 1n the coolant tube,
and from these and Eq. 1, a curve was
obtained of heat removed vs. mean
helium temperature. The mean tempera-
ature corresponding to a heat removal
rate of 50 Btu/sec is 1025°F. This
corresponds to an outlet temperature of
1920°F, since the inlet temperature
was assumed to be 135°F. A lower
outlet temperature would mean a greater
heat removal rate. Thus, apparently
the dump tank will be adequately
cool ed.

The chimney-effect method makes use
of the fact that a buoyant force acts
on the air in the coolant tube because
of a density difference. This buoyant
force 1s

(2) Fe V(o —p)

where

"y
1

buoyant force, lb,

V = volume, fts,
Powp, = density of helium at room
temperature, 1b/fe3,

P, = density of helium at mean
temperature of helium in tube,
1b/ft?.

From Eq. 2, the pressure acting on the

column of helium 1is

(3) AP:L:

A

where

AP = difference in upward and downward
pressures, lb/ft?,

A = cross-sectional area of coolant

tube, ftl,

The pressure can also be expressed by

AP = 4 <L>p'“
(4) P" f E 2g 1

 

117

 
 

NOTE !
ALL MATERIAL 1S

346 STAINLESS STEEL.

 

 

’ UNCLASSIFIED
| DWG D-A-3-1-31-1A

3-in.R, 5 PIPES

6%-in. R, 12 PIPES

10%-in. R, 18 PIPES
14%-in.R, 24 PIPES

18Y%-in. R, 30 PIPES

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

89 COOLING PIPES,
2~in. SCHEDULE NO. _
40x36-in LONG —
\\_4_1// DOUBLE-VEE Ju.NT NELDED FROM
BOTH SIDES Wil COMPLETE
JOINT PENETRATION
PLAN
3ft-5in.
I
ELBOW, 2-in. SCHEDULE | tin END PLATE
NO. 40 90° SR ELL - i
. \m 41-in. DIA -4'm-r
J 1_/ i LA Ll A i
HiH ) 3in.
4 1 |
a /
1! i VENT, 1-in. SCHEDULE NO.
] N 40x4%-in LONG
! : {
14 1 ']
] 1 i
/
]|} ! —~=——FILL PIPE, 2-in. ————'%-in. TANK WALL PLATE,
) 11 SCHEDULE NO. 40 128.8 x 34 in.
36in. o i 1 f 35-in. LONG
34 in. i) ' : : ; f
N 1 i 1 I i N
N : 4 : H
N 1 1 oo ™
N 4 H i N
N ' i ' /] I | ;:
N ' 1 q i1 )
N | 0 1 1] D
N H . 1 N
N 1 nln ' N
" H 4 u 4
; H | ; N
] 1| 1 d //<' N
| ! H r N
4 H H o
; L e 457 t-inEND PLATE [
/ ] E a/ i ’ & 41-in.DlA7 N
] t .
i { . A4 Z /.IV Ll ////////5
ELEVATION

118

Fig. 89. Fuel Dump Tank.

 
16
f "N
R
= friction factor for laminar
flow, dimensionless,
L = length of coolant tube, ft,
D = equivalent diameter of coolant
tube, ft,
v = helium velocity, ft/sec,
g = gravitational acceleration,

32.2 ft/sec2,
Np = Reynolds’ number, dimensionless,

The velocity of helium through the
tube 1s, from Eq. 4,

1/2
(5) v = ( 2eD ) :
4fLp,

The amount of heat removed per tube
can be expressed by

 

(6) Q = we, AT
and
(7) Q= hd 0 .

In Eqs. 6 and 7,
v = vAp

k[ u 0,14 /4 \V3
h =186 — — <__mL
D\u, 7Tk L ’

@ = amount of heat removed, Btu/sec,

and

w = weight flow rate, 1b/sec,

¢ = specific heat, Btu/lb*°F,

AT = difference between inlet and
outlet temperatureof helium, °F,

h = heat transfer coefficient,
Btu/sec'ftz'oF,

A, = surface area of inside of tube,
ft?,
¢ = log mean temperature difference,

k = thermal conductivity of helium,
Btu/sec*ft?+°F,

(o = viscosity of helium at mean
temperature, lb/sec-ft,

t, = viscosity of helium at surface

temperature of tube, lb/sec’ft.

Equations 6 and 7 were solved by

assuming various helium outlet temper-

atures, and the values thus obtained
for each equation were plotted against
helium outlet temperature. From these
curves, the helium outlet temperature
was found to be 805°F, the heat removal
rate per tube i1s 0.59 Btu/sec, and
the total heat removed is 91 x 0. 59
= 53.7 Btu/sec.

HEATING OF FILL TANK WITH CENTRALLY
LOCATED DIP TUBE

Upon heating a flush-and-fill tank
containing helium, the temperature
of the dip tube will lagthe temperature
of the helium and metal walls of the
tank during the heating process
(Fig. 90). The temperature of the
helium will follow very closely the
temperature of the tank walls,

Heat is transmitted to the dip tube
by free convection of the helium and
by thermal radiation from the wall,
The difference in temperature between
the dip tube and the wall has been
determined at various wall temperatures
for certain heating rates.

The net heat input to the tank,
excluding loss, is

Q=qT+qH+qD:

where
dt At
qr = (WCP)T E;—g (ch)fjg;"
and
dt At
qy = (WCP)HE";'}-.‘ (WCP)HA—T-

The term dt/d7 can be replaced by
At/AT without much error if ATis
small.

The value of g, is given by

 

qp, = hADQ
0,173 4 4
. Ae(i_i_
3600 P 100 100

The emissivity 1is

 

119

 
UNCLASSIFIED
DWG. D-A-3-1-32A

 

3 LUGS EQUALLY SPACED

 

 

6-in. SCHEDULE 40 PIPE —————

Z
Z

 

 

 

 

 

 

1/2 in.

 

 

 

 

: NY
=
—
210 SF —— s O

 

2 ft—0in ——— - k-—ain.-—w-—

<
N c
N ™ b
—2 ft—& in DIA ———— x
i c
N ©
175 in. NOM.~s] |e— !
r o
N
x
N 1
A X
N X
X

 

 

 

 

 

 

 

JE— l—t—— 2 IN. SF

 

Y

 

ELEVATION

 

MATERIAL—INCONEL

Fig. 90. Fill Tank.

120

 
In these equations, the symbols
have the following meanings and units:
Q = total net heat input, Btu/sec,
g = heat input of component, Btu/sec,

W = weight of material, lb,

c, = specific heat of material,
Btu/1b* °F,
At = temperature change, °F,

AT = time interval, sec,
h = free-convection heat transfer
coefficient, Btu/sec*ft?:°F,
A, = outside surface area of dip

tube, ft?,

f = temperature difference between
helium and dip tube, °F,

€ = radiation emissivity, dimension-
less,

T = absolute temperature, °R,

o
f

diameter, ft,
and the subscripts
= tank,

helium,

oo
i

dip tube.

The free-convection heat transfer
coefficient can be calculated from the
Nusselt’s number. A curve of Nusselt’s
number vs. the product of Grashof’s
and Prandtl’s numbers is given on
p. 525 of ref. 14,

The heat input to the tank wall
and the helium 1s

qT+qH=Q"qD'
Therefore
Q"qD

A'T= ;
(WCP)H At + (WCP)T At

 

the change in temperature of the dip
tube 1s

q DA’T

D T
(WCP)D

and the temperature at any time 1is

(19) t, = t; t At

where t; 1s the temperature at the
beginning of the time interval (°F).

The emissivity of the tank wall
was taken to be the same as that of
the dip tube. The value used in
making the calculation was

€, =€, = 0,50

T D

The dip tube was taken to be a 2-in.,
schedule 40 pipe that extended to the
bottom of the tank.

In calculating the temperature
differences between the wall and the
dip tube, net input heat rates of 50,
12, and 8 Btu/sec were used. The
results of these calculations are
shown in Fig. 91.

 

»
@
o

400 N

SN
N\
S -

/- @=12 Btussec
160 A\\\

 

 

 

|5
™
<

N
N~
o

 

 

 

 

 

 

 

 

 

 

 

 

80 / T~ \0\
T — “\(\{
/ @=8 Btu/sec —
0 | B
O 200 400 e00 800 1000 1200

DIFFERENCE IN TEMPERATURE BETWEEN TANK WALL AND DIP TUBE (°F)

TANK WALL TEMPERATURE {°F)

Fig. 91. Results of Fill Tank Pre-
heat Calculations.

121
Chap

ter 9

OTHER INVESTIGATIONS

AFTERHEAT IN FISSION PRODUCTS

In order to analyze many of the
problems which arise 1n the ARE,
1t 1s necessary to know the decay
energy of the fission products at
various times after reactor shutdown.
Reference 23 gives the average dis-
integration energy of the fission
products as a function of time after
fission. This curve (ref. 23, Fig. 6a),
reproduced as Fig. 92 of this report,
cannot be used directly, however,
because the fissions 1n the reactor
occur over a considerable period of
time and the fission products, conse-
quently, have a wide range of ages.
The decay energy of the fission products

may be expressed as

(1)

where

Q

q(t)

fl

Q=Kf” q(t) P dt ,
t

energy in the fission products,
Btu/sec,

the average disintegration
energy of the fission products
as a function of time after
fission (Fig. 92), Mev/sec* fis-
sion,

fission rate 1n the reactor,
fission/sec,

a constant for
Mev/sec to Btu/sec =
X 10.16'

converting

1.517

DdG. 19155

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10— T T
oo o j | | | | | |
| /
| -
1 .:}\\\//35«(”+1"m=(39d“-2+u.?d"“)1o‘b
|
1 ; I = 3273 (/" %+ 20436 7%
387+ (71 ={3.8-0861/)7 ' g
\ ! | !
! i i
ot : \’7 3BT (1) =0708 X #7078 o
=
o
o
@
w fO—Z
a
ul
o
)—
o \ 380+ T (1) =6.425 ¢y~ 19
W o0? N
w : ’ N \
=
S BB(HFF(f)THEORETCAL(chef25,p1328)//;;§\\\\\\
= i N
p! -4 i | ~
g 10 L . ) N\
LpLJ | ; \\
=z | 5 M
- | >
= T ; ;
= i 5 [ ' ,
+ & () = ENERGY OF BETA PARTICLES IN Mev per second per fission -1.0
; & L3+ T 7 =1.00 "
= 10—6 I' (F) = ENERGY OF GAMMA RADIATION IN Mev per second per fission ‘/\'
@ ! =TIME AFTER FISSION |
: Ve 47TER FSSION e W
i | ~
N
\\
107 - - > —
N
N
| ~
| N
-B ‘ \ \\
10 :
\\
: ~
| N
i N
! | \\
0L L | S
4 -3 2 —1 3 9
10 10 1o 10 | 10 10 e 0 16° 1o 10 10 10
TIME AFTER SHUTDOWN (sec)
Fig. 92. Disintegration Energy per Fission vs. Time after Shutdown.

122

 

 
t = age of fission product group
under consideration, sec,
t; = age of youngest fission

products {(also equals time
since reactor shutdown),

t, = age of oldest fission products
(also equals time since reactor
startup), sec.

sec,

For these calculations, 1t was
considered that the power had been
constant and equal to 2850 kw, which
equals 2700 Btu/sec or 1.78 x 101!°
Mev/sec. If it is assumed that there
is 190 Mev of energy released per
fission, then the fission rate, P, 1is
9.37 x 10'® fission/sec. Substitution
of these values for P and K in Eq. 1
gives

(2) Q=14.22 {2 q(t) dr .
t)

In order to integrate Eq. 2, it 1is
necessary to find a function to fit
the curve of g(t) as given in Fig. 92,
This was done by dividing the curve
into three parts and fitting each of
these parts with a different power

function. The resulting equations are
(3) g(t) = 0,708 ¢0%-72

(4) q(t) = 6.43 ¢~ 116

(5) gq(t) = 3.27 ¢=1-2 4+ 95, 4 ;=14
(Eq. 5 is the equation recommended in

ref. 23 for old fission products).

Equations 3, 4, and 5, were used for
integrating Eq. 2 for various values
of time after reactor shutdown (1t,)
and for 100 hr of reactor operation
(t, = t, + 3.60 x 10°). The results

are tabulated in the following:

TIME AFTER DECAY ENERGY OF THE
REACTOR SHUTDOWN FISSION PRODUCTS
(sec) (Btu/sec)
10° 207
103 119
10* 58.8
10° 17.7
3.16 x 10° 4.1
10° 1.4
107 0.1

These values of decay energy include
the energy carried by neutrinos. The
energy of the neutrinos is essentially
unabsorbable and, consequently, the
actual absorbable decay energy of the
fission products 1is less than that
tabulated above. A rough estimate of
the absorbable energy can be obtained
by takingone half thetotal energy, ¢ %%’
with the results shown in Fig. 93,
where the absorbable decay energy in
the fission products is plotted as a
function of the time after reactor
shutdown for the conditions of this
case (2850 kw, 100hr, 190 Mev/fission).

TEMPERATURE DIFFERENCE BETWEEN
THERMOCOUPLE ON PIPE WALL
AND BULK FLUID

The bulk temperaturesof the fluids
in the fuel and NaK loops of the
ARE are measured by thermocouples
welded to the outside of the pipe
walls 1n which the fluid flows. The

 

1< t< 100,
100 € t < 3.16 x 10°

t > 3.16 x 10°

’

 

difference between the measured temper-
atureand the actual fluid bulk temper-
ature has been calculated for the
most serious case, that is, the fuel
circuit (Fig. 94).

For the configuration shown, the
heat loss per lineal foot, as given

in chap. 6, 1is:

TEMPERATURE OF OUTSIDE
OF PIPE WALL (°F)

HEAT LOSS
(Btu/sec*lineal ft)

0,23
0,31

1150
1500

The temperature drop across the fuel
film 1s given by
). O
hA

123
120 :

 

 

o
o

@
o

 

 

o
o

 

DISINTEGRATION ENERGY (Btu/sec)
o
<

 

20

 

 

 

i
102 103 104 10° 108 el 108
TIME AFTER SHUTDOWN (sec)

 

Fig. 93. Disintegration Energy vs.
Time after Shutdown.

and the area i1s given by

X 1. 049

T = 0,275 ft?/lineal ft,

A=17

where

€ = temperature difference across

fuel film, °F,

Q =heat loss per
Btu/sec*lineal ft,

lineal foot,

h = fuel heat transfer coefficient,
Btu/sec'ftz'oF,

A = heat transfer area per lineal
foot, ft?/lineal ft,

D = pipe diameter, ft (1l-in., IPS,

schedule 40 pipe has an inside
diameter of 1.049 in. and an
outside diameter of 1. 315 in.).
Theheat transfer coefficientis givenby

K 0,14
h = 0.027 —Re0® Pr"3<ii> :
D ©,
where
k= fuel thermal conductivity,

Btu/sec*ft? (°F/ft),
Re = Reynolds’' number,
Pr = Prandtl's number,

#4 = viscosity at bulk temperature,
1b/hr-ft,

4, = viscosltyat surface temperature,
1b/hre ft.

124

For fuel with the properties listed 1in
chap. 1, and a velocity of 6.26 ft/sec,

Re, so0p = 11,700

and

= 22,300 .

Re 1500 °F

If it 1s assumed that (#/#3)0.14 ~ 1,
h = 0.918 Btu/sec*ft’*°F

1150°F )
hisoor = 1.245 Btu/sec*ft?* °F,
(hA) | 150°F = 0.253 Btu/sec*°F,
(hA) 5900°F = 0.342 Btu/sec* °F,
B1150° = 0.91°F,
61500°p = 0.91°F.

The temperature drop across the pipe
wall 1s '

 

Q 1n D,
g =—
= 21k
n
where
€ = temperature difference across

pipe wall, °F,
D, = pipe ft,
thermal conductivity of metal,
Btu/sec- ft? (°F/ft),
= 2.72°F,
3.67°F,

outside diameter,

&
n,1150°F
5 =
n,1500°F

Since the fuel film temperature
drop is small and the Reynolds’
numbers are high, the difference
between the fuel bulk temperature
(the so-called ‘mixing cup” tempera-
ture) and the fuel maximum temperature
will be very small. The total differ-
ence between the fuel bulk temperature
and the outside pipe wall temperature
1s
(6 +6_)
(6 +6.)

"

= 0.91 + 2.72 = 3.63°F,

4. 58 °F .

1150°F

1500°f = 0.91 + 3.67

At other thermocouple locations in
the fuel circuit, the temperature
drop 1n the fuel film 1s less than the
drop calculated above because the above
conditions i1nclude the minimum i1nsu-

lation thickness and the minimum
Reynolds’ number encountered in the
system. The temperature drop in the

filmin the NaK circuit 1s, of course,

very small. The temperature drops
 

el

MUST BE GAS TIGHT FOR {1 psi PRESSURE

 

]—~'/4-in. SLOT IN 3-in,

LENGTH ONLY

TYPE-307 STAINLESS STEEL FLANGE

MH 42347-3 2W INSULATION
20-GAGE OVAL x 3-in. LONG

FILLET WELD ALL AROUND

 

 

 

 

- — SAUEREISEN
- @I PrTITn AR
06 « #
O

 

 

 

————6 in + Yyin.

 

THERMQCOUPLE CONNECTOR /

— L'/,G in.

 

 

{THERMO-ELECTRIC €O.)

 

 

 

 

 

MH 42347-3 2W INSULATION
20-GAGE OVAL x f1-in. LONG

‘ PORCELAIN INSULATOR (FISH-SPINE)
FOR 20-GAGE THERMOCOUPLE WIRE

TYPE A,

TYPICAL INSTALLATION AT SLEEVE

 

 

 

TYPICAL INSTALLATION AT ELBOW PIECE

12 in.; TYPE B, 8in.;

 

 

 

TYPE C, 3in.

PORCELAIN INSULATOR (FISH-SPINE)
FOR 20-GAGE THERMOCOUPLE WIRE

%_

, /WEL{D GAS-TIGHT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

", 4
. - 7% =
i < e "
/. =
HELIARC WELD’
= 3
- - - —__jWELDED

Fig.

94.

 

 

ALREADY

|

MH 42347-3 2w INSULATION
20-GAGE OVAL x 1-in. LONG

WELD
GAS TIGHT

   

—

70185 0D x 0135+

1D INCONEL

WELD COVER A

Y, -in. LONG

CHROMEL-ALUMEL
THERMOCCUPLE

 

 

 

 

 

 

)

Thermocouple Assembly.

 

 

UNCLASSIFIED
DWG. A-6—0-25A

 
  
   

17
()
Pl

     

P

  
   
 
   
 
   

T

 

T =]
L

 

 

 

 

HELIARC

 
through the pipe wall at various other
thermocouple locations are given 1in
Table 23. The maximum temperature
differential between the outside of a
pipe and the bulk of the fluid 1is
approximately 4. 6°F. It is important
to note that 1f the flow rates in
either the fuel or NaK circuits are
reduced,
during operation,

for some unforseen reason,
the temperature

differential between the outside of the
pipe wall and the bulk of the fluid
will This i1ncrease 1in
temperature differential, since 1t
occurs only within the fluid, should
not become serious so long as the
flow 1s turbulent; in the laminar
reglion, the temperature
differential can become great enough
to render the thermocouple unreliable.

increase.

however,

 

 

TABLE 23. TEMPERATURE DROPS THROUGH THE PIPE WALL AT VARIOUS
THERMOCOUPLE LOCATIONS
NOMINAL PIPE PIPE WALL HEAT LOSS o (°F)
DIAMETER (in.) TEMPERATURE (°F) (Btu/sec*lineal ft) "
11/2 1150 0.23 1.98
1500 0.31 2. 67
2 1150 0.21 1.53
1500 0.29 2.11
2 1/2 1150 0.23 73
1500 0.31 2.33

 

 

 

 

126

 
  
       

iMd& . . RER

1. J. H. Keenan and J. Kaye, Gas
Tables, Wiley, New York, 1948,

2, C, D. Hodgman, Handbook of
Chemistry and Phystics, 31°%' ed.,
Chemical Rubber Publishing Co., 1949,

3. M, C, Udy and F. W. Boulger,
The Properties of Beryllium Oxide,
BMI-T-18 (Dec. 15, 1949),

4, NBS-NACA Tables of Thermal
Properties of Gases, U. S, Department
of Commerce, National Bureau of
Standards:

H., W. Wooley, Table 6.10, “Helium

(Ideal Gas State),” July 1950;

R. L. Nuttall, Table 6.39, “Helium
(Coefficient of Viscosity),"” December
1950;

R. L. Nuttall, Table 6.42, “Helium
(Thermal Conductivity),’” December
1950,

5. The Engineering Properties of
Inconel, International Nickel Co. Tech.
Bull, T-7.

6. R. N. Lyon (Editor-in-Chief),
Liquid Metals Handbook, 2d ed., NAVEXOS
P-733 (rev.), June 1952,

7. Stainless Steel for Elevated
Temperature Service, U.S, Steel Corp.

8. D. Q. Kern, Process Heat Trans-
fer, McGraw-Hill, New York, 1950.

9, J.H. Perry, Chemical Engineers’
Handbook , 3d ed., McGraw-Hill, New York,
1950,

10. H. F. Poppendiek and L. D,
Palmer, Forced ConvectionHeat Transfer
in Pipes with Volume Heat Sources
Within the Fluid, ORNL-1395 (Dec. 2
1952).

’

)L i, 8
g v

 

o

PO +nt hiath

 

o .jfif¥#*$i?§5? ) "fihfifiw

Heade

11, W, H., McAdams, Heat Trans-
mission, 2d ed. (rev.), McGraw-Hill,
New York, 1942,

12. G, N, Cox and F, J. Germano,
Fluid Mechanics, Van Nostrand, New
York, 1941,

13, Flow of Fluids Through Valves,
Fittings, and Pipe, Crane Co. Tech.
Paper No. 409, May 1942,

14. M, Jakob, Heat Transfer, Vol,
I, Wiley, New York, 1949,

15, E. N, Sieder and G. E, Tate,
Ind. Eng. Chem. 28, 1429-1435 (1936).

16, K. A, Gardner, Trans. Am. Soc.
Mech. Engrs. 67, 621-631 (1945),

17. A. Y., Gunter and W. A. Shaw,
Trans. Am. Soc. Mech. Engrs. 67, 643-
660 (1945), |

18, W, Nusselt, Z. Ver. deut. Ing.
73, 1475 (1929),

19. W, J, King, Mech. Eng. 54,
347 (1932).

20, E., Schmidt, Foppls Festschrift,
p. 179, Springer, Berlin, 1924; and
Z. Ver. deut. Ing. 75, 969 (1931),

21, The Trane Co. Bulletin D-S 365,
September 1951,

22, H. L. F. Enlund, personal
communication, 1952,

23. K. Way and E. P. Wigner, Phys.
Rev. 73, 1318-1330 (1948).

24, S. L. Jameson, Trans. Am. Soc.
Mech, Engrs. 67, 633-642 (1945),

25. Johns-Manville Products General
Catalog, DS Series 2.

26. M. W. Zemansky, Heat and
Thermodynamics, 3d ed., McGraw-Hill,
New York, 1951,

127