CALCULATING THE EFFICIENCY OF A TANK WITH AN INCORPORATED THERMAL CONTROL

• Document Type: CREST
• Collection: NGA Records (Formerly NIMA)
• Document Number (FOIA) /ESDN (CREST): CIA-RDP78B04747A002800110001-9
• Release Decision: RIPPUB
• Original Classification: K
• Document Page Count: 18
• Document Creation Date: December 28, 2016
• Document Release Date: May 31, 2001
• Sequence Number: 1
• Publication Date: February 1, 1965
• Content Type: REPORT

STATINTL
Ilk Approved For Release 2001/08/07 : CIA-RDP78BO4747AO02800110001-9
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 Approved For Release 2001/08/07 : CIA-RDP78B04747A002800110001-9
WITH AN INCORPORATED THERMAL CONTROL
February 1965
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REPORT 974-007
CALCULATING THE EFFICIENCY OF A TANK
WITH AN INCORPORATED THERMAL CONTROL
February 1965
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Among the major objectives of this Processor Development
Program are those to reduce equipment size, to reduce power consump-
tion, and to investigate the modular concept of processor design. In
the HTA-5 Processor, the concept of separate service units in which
all support equipment was mounted led to increased installation space
and decreased efficiency. This assignment, in conjunction with assign-
ment 974-008, investigates the possibility of eliminating the service
units and separate temperature-control equipment as a step towards
meeting these objectives.
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INTRODUCTION
1 . CONVENTIONAL TEMPERATURE CONTROL
One conventional method of regulating solution temperature in a
processing tank,is to provide a control system in which solution is
drawn from the tank,and pumped through a coil-type heat exchanger and
through a small tank containing heating units. The solution is then
returned to the processing tank through a specific recirculation system
designed to eliminate striation (Figure 1-1) .
Chilled water at a temperature of about 45?F is circulated through
the jacket of the heat exchanger to extract heat from the solution as it
passes through the coils. Should the solution already be below control
temperature, its temperature is automatically raised by electrical resist-
ance heaters as it passes through the small tank.
The temperature of the solution in the processing tank is contin-
ually monitored by a temperature probe mounted in the tank. The probe
compares this temperature against that set in a Wheatstone bridge or
other control circuit. An increase in the temperature of the solution in
the tank causes a circuit to open a valve in the chilled water line to the
heat exchanger. Conversely, a drop in the temperature of the solution
will cause a circuit to close a solenoid switch and operate the heaters
in the small Lank. This equipment is both bulky and inefficient.
2. INTEGRAL THERMAL CONTROL
To provide a built-in thermal control while permitting liquid bear-
ings to be installed at first seemed to present a technically difficult
problem. However, the concept of using the bearing pump for recircu-
lation offers one approach to the problem (Figure 2-1). If the processing
tank were divided into two sections by a heat-exchanging wall and the
bearings were mounted in this wall, a new mode of operation could be
attained. Then, solution from the bearing side of the tank could be pumped
into the pressure side of the heat exchanger wall and thence returned through
the bearings to the front section of the tank. Recirculation from one side
of the tank to the other would provide a flow of liquid around the heat
exchanger wall, with the added advantage that little energy would be lost
to the ambient environment. To further improve thermal efficiency, the
module (of which the tank would be an integral part) would be insulated to
reduce heat loss to the atmosphere.
To maintain an 88?F solution temperature with a 65?F room ambient,
a heating load of approximately 3530 BTU/hr is required. This does not
consider the heat from the bearing pump which, for the purpose of this study
is 3 horsepower, which is rated at 7635 BTU/hr. On the other hand, to
maintain a solution temperature of 68?F in a' 75?F room ambient with a
75?F solution replenishment temperature, 8720 BTU/hr would be required
as a cooling load.
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It is safe to assume that chilled water at 45?F and hot water at
160?F are available in most modern processing laboratories at flows of
1-3/4 and 3/4 gpm respectively and at 10?FTD. The tank size selected
(Figure 2-1) provides a surface area of 4.95 square feet for heat exchange.
This area can handle a 9000 BTU/hr load, which is very close to the
required 8720 BTU/hr cooling load.
If the heat exchanger is used as the back wall of the tank, then the
pulldown rate is approximately 1.4?F per hour. If the bottom surface of
the tank is used as heat exchanger, the U factors could be a little higher.
In this case, however, an area of only 3. 1 square feet and 8350 BTU/hr
would be available. The 3-horsepower pump could be used to obtain
steady state cooling, but the pulldown rate would be only 0.5?F per hour.
The following calculations assume the use of a plate-coil wall
comprised of 3/8 inch single-embossed small tubes. The pressure drop
is high, but such devices could be employed in parallel circuits to
reduce the losses . Several types of circuitry can be used for temperature
control. On the assumption that both chilled and hot water are available
in the quantity required, the system illustrated in Figure 2-2 offers a
clean compact unit with a minimum of outside controls.
Parameters:
(1) Tank Capacity: 75 gallons.
(2) Length of Film in Tank: 44-1/2 feet.
(3) Approximate Film Transport Speed: 20 fpm.
(4) Development Time Required: 2 minutes.
(5) Bearing Diameter: 2 inches.
(6) Operating Temperature: 60? to 88?F.
(7) Bearing Flow (Assumed) : 12 gpm.
(8) 13 Bearings per Tank: 156 gpm total.
(9) Replenishment Rate (Assumed): 5 to 10 gpm.
(10) Tank Insulation: 1 inch of Styrofoam.
(11) Ambient Temperature Range: 65? to 75?F at 50 percent RH.
Total Volume 38" x 32" x 14"
3
= 17024 Inches.
= 9.85 Feet (use 10)
Gallons 9.85 x 7.48
= 73.6 gallons (use 75)
Weight 73.6 x 8.345
= 615 lbs. (use 625)
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The following mathematical analysis assumes the use of
stationary sleeve-type liquid bearings. However, assignment 9 74-013 ,
which considers a type of bearing incorporating a built-in pump, would
influence the parameters given to the extent that no bearing pump would
be required. Therefore, the 3-horsepower heat input load from the pump
would not exist. Furthermore, the idea of a rear pressure compartment
would not be required. A further analysis, using the bottom and sides
of the tank, will be made when the processor module now under design
study (assignment 974-008) has advanced sufficiently.
To find heat losses during high temperature processing (88?F)
assume all sides and bottom are insulated.
A 2x 38x32 +2x 18x38 + 18x32
si 144 144 144
= 16.9 + 9.5 + 4.0 = 30.4 ft2 (inside diameter)
A = 2x 39x33 + 2x 19x39 + 19x33
so 144 144 144
= 17.85 + 10.3 + 4.35 = 32.5 ft2 (outside diameter)
32 = 0.89 ft2 (could be insulated too.)
As 4 x
144
top
A = 14 x 32 = 3, l ft2
s 144
open
surface
Let A = 33 ft2
s
As = 3. I ft2
open
U = 0.35 BTU/hr - ft2 - OF (1" Styrofoam)
Q = UA At = (0.35) (33) (8823 65) = 266 BTU/hr
sides s
.r. low
Utop= 300 BTU/hr - ft2 @ 23?TD Still air cond. (Unvented)
sens
U1
= 300 x 1.8 = 540 @ 23?TD and 300 fpm (vented)
top
sens
Assume normal B & W operation with no venting
qop = 300 x 3.1 = 930 BTU/hr
sens
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= 266 + 930 + 412 + 1920 = 3528 BTU/hr
Qwall
UA i where U = 2.0 or 6xUinsul
Qwall = 2 x 33 x 23 = 1515 BTU/hr
Q Latent = KU" As
U" = 95 BTU/hr-ft2 @ 88?F surf and 75? Room
f @ 65?F room = 1.4
KU" = 1.4x95 = 133
QLatent = 133 x 3. 1 = 412 BTU/hr
QReplen = WCp L+
Let W = 10 gph = 83.4 #/hr
C__ = 1.0 BTU/#-OF
,Ad = 88 - 65 = 23?F (assume replen supply @ room temp.)
QReplen = 83.4 x 1.0 x 23 = 1920 BTU/hr
Neglect = 0
QTotal Qwall + Qtop + Qtop + Qreplen - Qpump (Act = 7632 B/H)
Heating loss sens lat
To find max heat gain or cooling loads for low temp processing (68 0 )
Let surf areas & U-factors = approx
~pEjo~ 17
Qwall UAs -'~T = (0.35) (33) (757-68) = 81 Btu/hr
gain
Qtop = U 1 A s "T = 15 x 3. 1 x 70 = 326 Btu/hr
sens
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1 Qtot 4777 BTU/hr or 3528 35% incr w/o insul.
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974-007
Ctoo = U I IAs = 30 x 3. 1 = 93 Btu/ r
Actually 0
lat
= Wep~A:T = 83.4 x 1 x 70 = 585 Btu/hr
replen
C = 2544 x hp = 2544 x 3 = 763 Btu hr
pump =
OTotal 8717 Btu/hr
cooling
Assume hot & cold water can be supplied to HX at 160?F & 45?F respectively.
3600 0. 72 GPM
WHW. Reqd = --- -- ' 16 0 x 10 x 500
W Reqd 0720 1. 75 GPM
=
CHW 100 x 500
- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
3720
FIX LMTD`s where ATSOL C -- 156 x 500 x 1
Sn
SQL
COOLING
LMTD
Heat
670
. 11?F
6B
LMTD
Cool
180
88
13-
NOTE: To determine surf area reqd, factors
are LOAD (0), U Factor, & LMTD
O = UA LMTD
For given U
As reqd for heating = f MG
LTD
3600 53, 6
67 U
8720 484
& A reqd for cooling
s 18 U
45
Therefore cooling requirement is max for As determination.
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Surf area available in present tank design
A avail 38x32
s 144
if U = 100
then C = UA LMTD = (10 0) (8. 4 5) (18)
15, 200 Btu/hr
Cooling- --
& Q _ (100) (8. 45) (67) = 56, 500 B/hr heating
Apparently plenty of surface is avail in present design for heating;
cooling is ZX
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Now check U-Factor assumed
where _ t .06 .00052
k _ 115
4k = 1920
t
1
1 + t +
hsol Kss hwat
At 12 GPM/bearing the
bottom bearings will have
a tendency to be short
circuited from largest HX.
surface
3
A avail to bottom bearing
s flow = 33,1x42 = 0. 665 ft2
ASSUME MA~t~ v-> 9
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= B. 45 ft2
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974-007
GAM
cfs 156 x 2. 29 x 10-3 156 x 2. 29 x 144
Vsol p" 2 4 x 32/100
Vsol
U cooling
it
0.402 fps ; hsol = 400 & h sol
Assume use of single embossed, small pass platecoil
3/8" OD pipe; where acs = 0. 1 in2 000695 ft2
V = fps = cfs . 25 x 2. 29 x 10-3 25 x 2. 29
cw ft2 . 000695 . 695
Vcs = 0.82 fps P10.25 GPM ; hcw = 250
Vhw 2.46 fps @ 0. 75 GPM ; hhw = 500
r+
1 = 1
400 + . 00052 + 250
1 = 142
.00702
& Uheat . 0025 + . 00052 + . 002
200 upper
portion
.0025 + . 00032 + . 004
1
.00502
To find U-factor on wall between bearings
Ucool . 005+. 00052+. 004 . 00952 1 200 + . 00052 + 250 = 105
_ 1 1 133
Uheat - .005+.00052+.002 - .00752
cool Qlow + QHi as
00 _ 1428 o Getting
(142) (. 665)+ (105) 4 18 LMTD close
to regmnt
= L94.5+405 18
9, 000 Btu/hr cool
Below
lower
bearings
only
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& Qheat r (199) (. 665)+ (133) (4. 29) 67 " LMTD
[132+57o67= 47, 000 Btu/hr Heat
To get above cooling or heating, find flow rates !
W = 9000 0 --- = 1. 8 GPM
chw 10?x500
NOTE: Then Vchs~= . 2 8 x . 82 = 5. 9 fps
(Good)
W hw=
47, 000 = 9.4 GPM
NOTE: Vhw = 9. 8 x S. 9 = 30.8 fps (Too Hi)
%ip `~ Assuming 26 passes X2-1/2' long
AP= 7X.42 psi x (26 x 2. 5 + 50 x 3)
= 7x.42 x212= 625 psi
1. 8 GPM for cooling
yap = 7 x . 025 psi/ft = 1. 75 psi/ft x 212 = 370 psi
Way too high
This HX will need parallel flow paths and increased flow rates to maintain
steady-state velocity & h factors.
Try HX at bottom of tank (cleaning tank may be a prob.)
A avail = = 3. 1 ft-
F12c' 0 S 144
V i ~~ r_" R 1 rIjC_- S ~
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V 156x?., 29x144 = 0.92 fps
sol 4x 14x 1000
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U = 150 for cooling
c
U'lI = 200 for heating
cool (150) (3. 1) (18)= 8, 350 Btu/hr Q avail
heat (200) (3. 1) (67) = 4 1, 500 Btu/hr by A S
Add pump HP to cooling loads
Wsol 156 GPM
P = 20 psi (?)
bhp = GPM x P 156 x 20 3. 02 hp (1st try a 3HP pump is antic.)
17 14 x Neff 17 14 x . 6
= 3x2544 = 7632 Btu/hr
:. Qpump -
-------------------------------------------------------------------
What is pulldown rate w/ 9000 B/hr H.X?
wall & _ 0 ( 75? solution W/ 75? room
top gains
500 B/hr n 68? sol. W/ 75? room
N?oC)IA Tr MP.
250
8720 Reqd: not enough surface ! !
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974-007
Assume Qreplen 0 since system not in oper.
let pump 7630 B/HR
Qtotal _ 8130 B/11", P. 75?P start
gain
avail _ 8130 + mcp vT/hr = 9000
^.T/hr = 6250 1 1.4?P/hr
Pulldown from 75 --i 68? = i 4 = 5 hrs
Using 8350 Btu/hr bottom type HX
What is pulldown time?
0start _ Qpump 7630
Qend pump + Qgain 8130
L,T/hr = 8350=7630= 525 = 1. 15?F/hr no start
7
:+.T/hr = 8350-8130 =
220
625
0. 35?/hr @ end
7 = 20 hrs
.35
AT/hr actual = Avg = 6 L20 = 13 hrs
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V V V
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CONST 160? -C
RELIEF
RETURN TO
HEAT SOURCE
TH I
TEMPERATURE
BACK WALL HX
TEMPERATURE
PROBE IN SOLUTION
FIG.2-2
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TEMPERATURE (jJ ) l7 ~~ TEMPERATURE
READOUT ~T~Y' J SELECTOR
L
 STATINTL
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