s
ORNL-1255
010 » ^mm mX
BASIG PERFORMANCE
CHARACTERISTICS OF THE
STEAM TURBINE-COMPRESSOR-
JET AIRCRAFT PROPULSION
CYCLE
By
Arthur P. FraasGeorge Cohen
OAK RIDGE NATIONAL LABORATORY
CENTRAL RESEARCH LIBRARYCIRCULATION SECTION
4500N ROOM 175
LIBRARY LOAN COPYDO NOT TRANSFER TO ANOTHER PERSON
If you wish someone else to see thisreport, send in name with report and
the library will arrange a loan.UCN-7969 (3 9-77)
t-fcr
OAK RIDGE NATIONAL. LABORATORYOPERATED BY
Carbide and carbon chemicals compaA DIVISION OF UNION CARBIDE AND CARBON CORPORATION
OAK RIDGE. TENNESSEE
*
4E9
Index No. ORNL-1255This document contains 37 pages.This is copy^of 152 ,Series A.
Subject Category: Reactors-Research and
Power
BASIC PERFORMANCE CHARACTERISTICS OF THE STEAMTURBINE-COMPRESSOR-JET AIRCRAFT PROPULSION CYCLE
hy
Arthur P. Fraas
and
George Cohen*
OAK RIDGE NATIONAL LABOOperated by
CARBIDE AND CARBON CHEMICALS COMPANYA DIVISION OF UNION CARBIDE AND CARBON CORPORATION
Oak Ridge, Tennessee
Contract No. W-71<-05-eng
♦Formerly with ORNL, presently employed byAEC Cincinnati Operations Office.
lKIimiIiKnIJ,ENEMY svs™s i
3 44St Q3bCH5b b
Distribution, Series
1-59- Carbide anVcarbon Chemicals Company (Y-l^jElrea)6O-69. Argonne Natmnal Laboratory70. Armed Forces Ibecial Weapons Project (S^Jdia)71-78. Atomic Energy oWaission, Washington79' Battelle Memoria^Institute80-82. Brookhaven Nation^ Laboratory83. Bureau of Ships84. Chicago Patent Group185- Chief of Naval Resean86-90. duPont Company91' H. K. Ferguson Company92-94. General Electric Company,'ItokJEdge95-98. General Electric Company, H|£]p.and99' Hanford Operations Office100-104. Idaho Operations Office105. Iowa State CollegeIO6-IO9. Knolls Atomic Power Laborajbry110-112. Los Alamos
113. Massachusetts Institute 0 Technology (Kaufmann)114-115. Mound Laboratory116. National Advisory Comm|ftee for Aero%utics117-118. New York Operations 0fjp.ce119-120. North American AviatiJE, Inc.121. Patent Branch, Washi:ra;on122. Savannah River OperaRons Office123-124. University of Califwnia Radiation Laboratory125-128. Westinghouse ElectjK: Corporation129-137. Wright Air DeveloiJent Center138-152. Technical Informqpon Service, Oak Ridge
Carbide and Carbon ChemicalsJFompany internal distribution as\follows:
1. C. E. Center
2. C. E. Larson
3« W. D. Lavers
4. A. M. Weinberg5. W. B. Humes
6. E. D. Shipley7- R. C. Briant
8. E. H. Taylor9' J. A- Swartout10. K. Z. Morgan,11. C. E. Winter12. J. A. Lane,
13- J- H. Bucl14. A. P. Frg
Issuing Office
Technical Information Department, Y-12 AreaDate Issued: >.,,.;r .
W^r UI952
K. ErgenS. ThompsonW. SavageN. LyonJ. Miller
W. Schroeder
S. BillingtonP. Blizard
E. Clifford
D. Callihan
D. Manly
L. Meem
R. Grimes
F. Poppendiek
Index Nc.
Reactoi
ORNL-1255-Research and Power
56-59.
G. F. Wi*l.icenus
G. Cohen
Gale Young''E. P. WigneiLt. Col. R. E\GreerLloyd P. Smith!Chemistry Libra!Physics LibraryBiology LibraryHealth Physics LibraryMetallurgy LibraryTraining School Libr^X-10 Central Files
ANP Reports Office
Reports Office, TID
List of Illustrations
Figure No. Title Page No.
1 Summary ofSteam Turbine-C ompressor-Jet Performance 7
2 Steam Turbine-Compressor-Jet Cycle Schematic Flow Sheet 8
3 Temperature Distribution in Condenser 10
4 Steam Turbine-Compressor-Jet Cycle Performancefor Steam Turbine Inlet Temperature and Pressure =900 F and 5000 psia, and Condenser Pressure = 660 psia 25
5 Steam Turbine-Compressor-Jet Cycle Performance forSteam Turbine Inlet Temperature and Pressure = 1200 Fand 7000 psia, and Condenser Pressure = 550 psia 26
6 Steam Turbine-Compressor-Jet Cycle Performance forSteam Turbine Inlet Temperature and Pressure = 1200°Fand 10,000 psia, and Condenser Pressure = 800 psia 27
7 Steam Turbine-Compressor-Jet Cycle Performance forSteam Turbine Inlet Temperature and Pressure = 1500 Fand 7000 psia, and Condenser Pressure = 58 psia 28
8 Steam Turbine-Compressor-Jet Cycle Performance forSteam Turbine Inlet Temperature and Pressure = 1500 Fand 10,000 psia and Condenser Pressure = 90 psia 29
9 Steam. Turbine-Compressor Jet Cycle Performance for 0Steam Turbine Inlet Temperature and Pressure = 1500 Fand 7000 psia and Condenser Pressure = 150 psia 30
10 Working Chart No. 1; Cooling Effectiveness of Condenser 31
11 Working Chart No. 2; R Factor for Condenser;Air A P/P = .05 and M = 0.9 32
12 Working Chart No. 3; R Factor for Condenser;Air A P/P = .10 and M = 0.9 33
13 Working Chart No. 4; R Factor for Condenser;Air A P/P = .20 and M = 0.9 34
14 Working Chart No. 5; R Factor for Condenser;Air Al?/? = .10 and M = 1.5 35
15 Working Chart No. 6; Compressor Air Outlet Temp. 36
16 Working Chart No. 7; Air Enthalpy Rise in Compressorand Condenser ^^^^^^^^^^^^ 37
LIST OF TABLES
Table No. Title Page No.
I Calculated Performance of a Steam-Turbine-Com
pressor-Jet Cycle For a Condenser CoolingEffectiveness of 80$, An Altitude of 45,000 ft,and a Flight Mach No. of 0.9 13
II
III
IV
Calculated Performance of A Steam-Turbine-
Compressor-Jet Cycle for a Condenser CoolingEffectiveness of 80$, an Altitude of 45,000 ft,And a Flight Mach No. of 0.9
Calculated Performance of a Steam-Turbine-
Compressor-Jet Cycle for a Condenser CoolingEffectiveness of 80$, an Altitude of 45,000 ft,and a Flight Mach No. of 1.5
Sample Calculation Sheet
14
15
22
BASIC PERFORMANCE CHARACTERISTICS OF THE STEAM
TURBINE-COMPRESSOR-JET AIRCRAFT PROPULSION CYCLE
Introduction
Recently much interest has been expressed in utilizing the supercritical
water cycle for aircraft nuclear propulsion. One of the most widely-advocated
methods of accomplishing this is to use the compressor-jet propulsion cycle.
This involves adding pressure energy to the air stream in a low compression
ratio blower, then adding heat in a radiator or condenser, and finally allow
ing the air to expand in a jet to generate an impulse for propulsion. The
supercritical steam from the reactor expands through a turbine which drives the
air blower and the feed water pump. The turbine exhaust steam is fed to the
condenser where it loses its heat of vaporization to the air stream from the
blower.
Among the advantages of this cycle is the use of water, a familiar and a
relatively non-corrosive substance, for both coolant and moderator. The dis
advantages include an inherently low specific impulse and the necessity of
developing an entirely new type of aircraft engine.
The purpose of the work covered by this report was to analyze the funda
mental cycle and discover some of the relations among the many basic parameters
involved. In particular, the condenser air inlet face area, the condenser
weight, and the impulse were evaluated per-pound of air flow in terms of the
temperature and pressure of the steam leaving the reactor and the condenser
pressure. This was accomplished for flight at 45,000 feet altitude at Mach
numbers of 0.9 and 1.5.
Reasonable values were assumed for component efficiencies and actual test
data used for the condenser performance. The results should give attainable
performance of the fundamental cycle at the two design points investigated.
Somewhat better performance can probably be obtained through a program of opti
mization of the cycle and equipment. However, the data presented here should
give performance not far from the optimum and hence should be useful for pre
liminary design studies of the system.
Summary of Results
The results of this investigation as summarized in Figure 1 indicate that,
o
for reactor outlet steam conditions ranging from 1000 to 1500 F and pressures
from 5000 to 10,000 psi, the following statements can be made:
1. The specific impulse was found to be low in all cases considered;i.e., from 15 to 20 lb/lb air/sec at a flight Mach No. = 0.9 andfrom 9 to 14 lb/lb air/sec at a flight Mach No. = 1.5.
2. Increasing reactor outlet steam temperature or pressure or condenser pressure effects some improvement in all cases and in allparameters considered; i.e., specific impulse, specific heat consumption, and specific condenser weight and frontal area.
3. Increasing steam condenser pressures above 400 psi gives relativelylittle improvement in performance.
4. Increasing the reactor steam outlet pressure from 5000 to 10,000 psigives very little improvement in performance.
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Figure 2
SCHEMATIC FLOW SHEET SHOWING THE MAJOR ITEMSOF EQUIPMENT REQUIRED FOR THE STEAM TURBINE-COMPRESSOR-JET CYCLE;
(The subscripts used in the text of this reportindicate the same points in the cycle as thecorresponding numerals in this diagram.)
Description of Cycle:
While there are many type of aircraft power plant that might be designed
around a steam generating reactor, the type considered here is one in which
the steam is generated at supercritical pressures and fed through the
turbine, condenser, and feed water pump back to the reactor. Modification
of this basic cycle can be made; e.g., a fuel oil burner could be employed
to boost the temperature of the air ahead of the jet nozzle, but such
elaborations were not considered in the cycle performance calculations
made in this study.
The proposed cycle is illustrated in Figure 2. Air compression owing to
ram takes place in the air inlet and diffuser. Further compression takes place
in the turbine-driven axial flow blower, or compressor. The air then flows to
the radiator, or condenser, where it is heated by the exhaust steam from the
turbine. The high velocity resulting from expansion of the air through the
nozzle in discharging to the atmosphere gives an impulse, or thrust.
On the steam side of the cycle, water enters the reactor from the feed pump
at supercritical pressure, i.e., above 3206 psia. The high pressure prevents
boiling in the reactor and preserves sufficient density for effective moderation.
After gaining heat energy in the reactor, the steam drives the turbine and then
transfers its heat of vaporization to the air in the condenser-radiator. The
condensed water is then raised to reactor pressure by the turbine-driven feed
water pump.
There are a multitude of parameters involved in the system. Obviously basic
are the turbine inlet conditions, T^ and P^, and the condenser pressure, Pc
(see Figure 2). No pressure drop is assumed in the condenser on the steam side,
so P^ = Pg. The condenser outlet temperature, Tg, was taken as the saturation
temperature corresponding to P^. (Actually the water would have to be cooled
below this temperature to avoid feed pump cavitation). P7 is assumed equal to
Pij.. 1c may be superheated or saturated with some moisture.
On the air side, the blower compression ratio, P2/p]_> and the air pressure
drop in the condenser, (P2-Po)/P2> must be considered. The nozzle temperature,
T3, is found to be directly related to the heat exchanger cooling effectiveness,
a widely-used heat exchanger performance parameter.
10
The cooling effectiveness,"^. , is normally defined as:
-ft _ Temperature rise in coolantInitial temperature difference
It is usually expressed as a percentage.
Steam
AT,eff
•P
<
TC
Distance from air inlet
Figure 3. Temperature Distribution in Condenser
The application of this expression is quite straight forward when there is no
change in state in a fluid passing through the heat exchanger, or when only
condensation or boiling occur. If superheated steam is fed to a condenser,
however, the peculiar temperature distribution shown in Figure 3 results. In
order to use available heat exchanger performance data, it was necessary to re
define cooling effectiveness for this special case. A reasonable approximation
was taken to be the following:
^ =T3-T2
A. T(1005&)
effective
where
=(T6 -T2) + * h^evUeat ( _ }0 * ^ h total '
The relations are such that selection of T^, P^, P5, (P2 -P3VP2 >a^d T3
identify a cycle. With values of these parameters known, the following can be
evaluated also:
Blower compression ratio
Heat exchanger cooling effectiveness = *X.
o •*• -. -i lb of thrustSpecific impulse =
lb of air per sec
Condenser airflow rate = ^ air per sec—ft air inlet face area
11
Specific condenser area = ft of air inlet face arealb of thrust
Condenser depth in direction of air flow - inches
Specific condenser weight =lb of thrust
h) Specific heat consumption = Btu/lb of thrust
The first step in the study was an attempt to select optimum values for
two of the five identifying variables. If this could be done, these two
parameters could be fixed at their optimum values and the calculations required
would be greatly reduced.
The optimization was accomplished by plotting (b) through (h) against T3
for a particular combination of T^, Pij., and P5. This was done for several
values of (P2 - P^)/P2. The results are presented in Figures 4 to 9-
These plots revealed that a pressure drop across the condenser of
AP/P = .10 seems to be the most reasonable pressure drop. They show that going
to A P/P = .20 reduces the specific area very little, but increases the specific
weight substantially. On the other hand, A P/P = .05 gives a quite severe
increase in specific area, with only a slight decrease in specific weight. The
specific weights for .05 and .10 generally converge in the region of interest.
The plots also revealed a minimum in specific condenser area. This minimum
occurs at practically the same point for each cycle. With the selection of
12
AP/P = .10, the minimum specific area is obtained at approximately 80$
cooling effectiveness. It seems reasonable to select this as the optimum value
for^ .
The identity of the cycle could now be established by T^, Pil, and Pc,
since Ap/P = .10 and ^ = 80$. Selection of >£ = 80$ is equivalent to
selection of To, because the plots reveal a linear variation between the two
parameters.
The next step was to select values for T^ and P^ and vary Pc for each
combination. The principal parameters were then plotted versus Pc in Figure 1
for five different combinations at Mach =0.9 and 1.5. The remaining parameters
are tabulated in Tables I, II, and III.
The lower limit imposed on Pc was that pressure at which the steam con
tained 10$ moisture at the turbine exit. The upper limit was either arbitrarily
taken as 1000 psi, or that pressure at which the heat from the-superheat of
the exhaust steam was 10$ of the total heat given up in the condenser by the
exhaust steam. The heat exchanger data were based on condensing coefficients
and should not be extended into the superheat range. It was felt that the
approximation for cooling effectiveness given on Page 6 should be fairly good
if limited to small amounts of steam superheat at the condenser inlet,
especially since the major temperature drop is from the metal to the air rather
than from the steam to the metal. A special 2-pass condenser would, of course,
be required with the superheated steam coursing across the air outlet end
first before entering the main part of the condenser.
Calculation Methods
The method of performing the calculations can be best explained by
carrying through the entire procedure for an illustrative cycle.
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CALCULATED PERFORMANCE OF A STEAM-TURBINE-COMPRESSOR-JET CYCLE FOR A CONDENSER COOLINGEFFECTIVENESS OF 80$, AN ALTITUDE OF 45,OOP FT., AND A FLIGHT MACH NO. OF .9
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16
Cycle B-3 will be used for this purpose. Its identifying qualities are:
M 0.9 at 45,000 ft
T4 = 1200 F
P^ = 10,000 psi
P^ = P5 = 800 psi
With the assumption of turbine adiabatic efficiency = 80$, it is possible
to specify the conditions and the energy at each point of the steam side of
the cycle.
From extrapolated steam data contained in TAB-78:
h^ = 1447 Btu/lb; S^ = I.396
Using the Mollier Diagram, the enthalpy after isentropic expansion to
condenser pressure is:
h5 » 1179 Btu/lb
A hisent
T5 "
(h4 "h5>isentropic = 268 Btu/lbApplication of the 80$ turbine efficiency gives the true enthalpy drop:
A h = (.8)(1^ -h5)igent = 214 Btu/lb
h5 m h^ -214 = 1233 Btu/lb
From Mollier Diagram at h,- = 1233, P,- = 800 psi:
o
553 F
Saturation temperature at 800 psi is 518 F, so the steam is entering the
condenser with 35 superheat. The water leaves the condenser saturated so:
Tg = 518 F
V6 "
h6 "
.0209 ft /lb
510 Btu/lb
The heat given up in the condenser is:
h^ -h6 = 1233 - 510 = 723 Btu/lb
The feed pump work to return the water to the reactor, assuming 65$
efficiency and incompressibility is:
v6 (P7 - P6)^ -h6 =
(Eff) J
= (=0209)(10,000 -8oo)(i44) = 55 Btu/lb(.65)(778)
A useful parameter is now defined:
R = Heat to condenser
Energy to blower
h5 - h6
(hi^ - h^-th-, - hg) lag - hx
723
214 - 55
(All values of enthalpy so far are per pound of water.)
Turning to the air side of the cycle, standard atmospheric conditions
at 45,000 feet are:
TQ = 393°R; P0 = 2.14 psiaThe Air Tables give the isentropic pressure ratio at Mach = 0.9-
P /P^ = .59126 (Primes represent isentropic)
P^ = 2.14/.59126 = 3.62
]?l-po= 1.48
The diffuser is assumed to achieve 85$ of the isentropic pressure recovery.
Px -P = .85(1.48) = 1.26
= 4.55
*? -H
17
The temperature is the stagnation temperature for Mach =0.9.
o
18
T, 393/.86058 = 457 R; hx = 109.2 Btu/lb of air
It is now possible to select one more condition, To, against which the
major performance parameters may be plotted. To illustrate the procedure, To
is selected as:
o
T, = 891 R
h- = 214.05 Btu/lb
The jet velocity, and therefore the thrust, can be calculated if Po is
determined. It is first necessary to determine the compression ratio of the
blower.
The energy added in the blower and condenser is:
h - hx = 214.1 - 109.2 = 104.9 Btu/lb
Since R is known from the steam side:
(112 - hj) + R(hc, - hx) = h3 - hx
ho - h, 104.9h„ - h. = -2 i = = 18.9 Btu/lb
1 + R 1 + 4.55
h2 = hx + 18.9 » 109.2 + 18.9 = 128.1 Btu/lb
From the Air Tables at this h:
= 536 R
Knowing the blower inlet conditions, and assuming an adiabatic
efficiency = 87$, it is possible to plot Tp against Pp/P-,. This has been
done in Figure 15. Entering this plot at T2 = 536 > P2/Pl is found ^°
be = 1.64.
P2 « 1.64 px = (1.64)(3.4) = 5.58 psia.
It is now necessary to select an air pressure drop in the radiator.
Three different values were used originally, but for this illustration the
optimum will be used:
PQ - Po_£ 2. = 0.10
P3 = °*9 P2 = (°-9)(5.58) = 5.02 psia
Po 5.02
P0 2.14=- 2.35
r = 0.426 = pressure ratio across jet nozzle.
The jet velocity is:
J V m
m = molecular weight
R = 1545
T3 12 gnR J n - 1/(l-r n
n - 1 V
n = 1.4 = -^cv
1 n~TT.6 v/ T_ vj (1 - r nV. = 109
The last term is tabulated in Table 25 of Keenan and Kaye Gas Tables.
V. = (109-6) ^891 (.465) = 1522 ft/secJ
The airplane velocity is:
19
V = Ma = (0.9)(49.1) V393 = 876 ft/sec
The impulse is equal to the change in momentum:
Jt - w/g (Vj -va)W = weight of air, lb
With a weight flow of one pound of air per sec, the specific impulse is:
I «VVa 1522 - 876 lb - sec
= 20.132.2 lb
20
The total heat added per pound of air is:
h, -h, = 214.1 - 109.2 = 104.9 -,, B^U .—3 1 lb of air
The specific heat consumption is:
ho - h, 104.9 Btu/secSp. Ht. Consumption = _£. ± = = 5-22
I 20.1 lb of thrust
It is necessary to calculate condenser cooling effectiveness and /T*/\ P
in order to specify the size, airflow, and weight of the condenser. Cooling
effectiveness,77 , has previously been defined as:
T " T-h = 32 (l00jt)
*•> "superheat ,(T6 -T2) + — (T5 -T6)
**"totalAhSH
All of these quantities have been found except ^_ . The^htotal
enthalpy of saturated steam at condenser pressure = 800 psi is h = 1199 Btu/lb.
AhsH h5 -hg 1233 -1199
£hT0T h5 -hf 1233 - 510
(891 - 536)(100$)
= .047
1- (978 - 536) + .047 (1013 - 978)
tf* is the ratio of the density of the air entering the condenser to the
density of air at standard sea level conditions.
/ P2 \
cr* - — —Po
R T0
519 P2 ?2= 35-3
14.69 T2 T2
£T&P = (.367)(.10)(P2) = (.0367)(5-58) = .204 psi
Entering the chart in Figure 10 at this"\_ and ^P:
, lb/secAirflow =3.42 £ Lft condenser frontal area
Condenser depth = 8.08 inches
The specific condenser area is
2
Sp. cond. area = = .0146
21
(I)(airflow) (20.1)(3.42) lb of thrust
The condenser in NACA RME7J01 is 8.75 inches deep and weighs 15.4
pounds per square foot of frontal area. Assuming that the condenser weight
is directly proportional to its volume:
15-4 . w xSp. cond. weight = (condenser depth)(Sp. area)
8.75
= (l.75)(8.08)(.0l46) =.208——^-lb of thrust
Calculations Using Graphical Working Charts
Graphical working charts have been devised to expedite the cycle cal
culations. The charts are used in conjunction with calculation sheets of
the form shown in Table IV . Use of these aids will be illustrated by once
again calculating cycle B-3.
P2 - Po AhSH± , R, , Tg and Tc are selected or calculated just
P2 Ahtotalas before. Thus lines 1, 3, 7, 8, 10, and 11 in the calculation sheet can
be filled in. A range of values of To can be selected and placed in line 2.
For this example, To = 891 R. Entering the chart in Figure 12 at To = 891
and R = 4.55, it is possible to read Po/pn = l*^ and specific impulse = 19-80.
TABLE IV
Sample Calculation Sheet
Line No Quantity Cycle No.
1 Condenser air A P — inlet air pressure=(P2-P3|/P2
B-3
.10
2
o
Condenser air outlet temperature, To, R 891
3 R factor (from Figure 12) 4-55
4 Compressor pressure ratio = P2/?i 1.64
50
Compressor air outlet temperature, T2, R 536
6o
Condenser air AT rise = To - T2 = (2)-(5)F 355
7 Ratio of steam condenser superheat A hto total A h .047
8o
Steam saturation temp, in condenser, Tg, R 978
9 T6 -T2 =(8) -(5), °F 442
10
o
Degrees of superheat = T- - (8), F 35
11 (7) x (10) 2
12 Effective temp. diff. in condenser = (9) +(11), °F 444
13 Cooling effectiveness = ~V[ = 100 x (6)/(l2) 80
14 Condenser air inlet press. = P2 = 3.4 x (4),psia 5.58
15 Condenser air inlet density ratio = <T" =35.3 x (l4)/(5) .367
16 Condenser <7~~A P = (l) x (l4) x (15),lb/ft2 .204
17 Condenser depth, in 8.08
182
Condenser air flow, lb/sec-ft inlet facearea 3.42
22
Line No.
19
20
21
22
23
TABLE IV (Cont'd)
Sample Calculation Sheet
Quantity
Specific impulse, lb/lb of air/sec
Specific Condenser area = — /.q\ rr^v -fP/lb thrust 7T8TTT19T
Specific condenser weight = I.76 x (17) x (20= lb/lb thrust
Air A h through compressor and condenser,Btu/lb (Figure l6)
Specific heat consumption = (22)/(l9),Btu/sec-lb thrust
Cycle Ho
J=l
19.8
.0148
.210
105.0
5.30
23
24
Then the chart of Figure 15 reveals that T2 = 536 R if P2 = 1.64 P1.
With the information now available, the calculation sheet can be com
pleted through line 16. Then entering the chart of Figure 10 at77 = 80$
and (TVlP = .204, the condenser depth is found to be 8.08 inches and the
airflow is 3«^2 lb/ft -sec. Combinations of calculated numbers provide
the specific condenser area = .0148 and weight = .210 for lines 20 and 21.
Chart 7 (Figure 16) gives ho - h, as a function of To . Thus ho - hn =
105.0. Dividing this by specific impulse yields specific heat consumption
= 5.30.
After learning the otpimum value for 77 , it is only necessary to guess
two or three values.of T_ until this value of 77 is obtained. The remainder
of the calculation sheet can then be filled out for this To only.
75
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PRESSURE DROP - ctaP - psiA.PFftAAS
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COMPRESSOR PRESSURE RATIO, -p2-
37