1. A six-cylinder, two-stroke marine diesel engine operates at a piston
speed of 1200 rpm. The 5 in. x 5.6 in. engine has an 18:1
compression ratio. If the air intake is at 14.8 psia and 82° F,
determine:
(A) the displacement volume, ft3;
(B) the clearance;
(C) the ideal air inlet volumetric flow rate, ft3/min; and
(D) the mass flow rate for a volumetric efficiency of 85%, lb/min.

p1 = 14.8 psia;
5
t1 = 82° F
( A) compression ratio 
2
  5.6 
ft
ft  6  =0.3818 ft 3
4  12   12 
1
1
1
(B) rV = 1+ ; c =
=
= 0.0588
c
rV -1 17
(A) VD =
•
(C) V D =





14.8 lbf in2 144 in2 ft 2
P
=
=0.07372 lb 3
ft
RT  53.34 ft-lbf lbm°R   542°R 




1.4
V 
 13.2 
(B) p2  p1  1   15 
  252 psia
V
 1.76 
 2
252 144  1.76
pV
T2  2 2 
 1198 R
R
53.3
q
352
T3  S  T2 
 1198  3256 R
cV
0.171
0.3188 ft 3 stroke 1200 rev min 
VDN
=
n
1 rev intake stroke 
= 458.16 = 458.2 ft 3 min
(D)  =
k
V1 13.2

 7.5
V2 1.76
p3 
RT3 53.3 3256 

 685 psia
V3
1.76 144 

k 1
V 
m  V  V D   0.85 458.2 ft 3 min 0.07372 lb ft 3  28.71 lb/min
 1 
(C ) T4  T3  3 
 3256 
  1454 R
 7.5 
 V4 
T 
 1454 
p4  p1  4   15 
  40.8 psia
2. A diesel engine (in the above EXAMPLE) delivers 200 bhp at a piston
 535 
 T1 
speed of 1200 rpm. If the indicated power is 250 ihp find:
(D) qR  cV T4  T1    0.1711454  535   157 Btu lb
(A) engine torque, ft-lbf;
(B) brake mean effective pressure, psi;
(C) indicated mean effective pressure, psi;
(D) mechanical efficiency, %; and (e) friction power, fhp.


wk  net 
(E )
J
(g) t 
200 
(h) W k 
2  T  1200 
PLANc
33,000  n rev power stroke 
200 33,000  1 
2
PLANc
33,000  n
 250 ihp  33,000 ft-lbf hp-min 1 rev power stroke 
2
 5 12 ft   4   5.6 12 ft  1200 rev min   6 
IMEP = 16,078 lbf ft 2 = 112 lbf in2
BHP
200
89 psi
=
=
= 0.794 = 79%
IHP
250
112 psi
(E) FHP = IHP - BHP = 250 - 200 = 50 fhp
(D) em =
3. At the beginning of the compression stroke and ideal Otto cycle has
an air pressure of 15 psia, a temperature of 75° F and a specific
volume of 13.2 ft3/lb. At the end of compression the specific volume is
1.76 ft3/lb. The heat supplied to the cycle is 352 Btu/lb. Calculate the
following:
A. the compression ratio
B. the highest temperature and pressure of the cycle
C. the temperature and pressure at the end of expansion of the air
D. the heat rejected, Btu/lb
E. the net work of the cycle, Btu/lb
F. the thermal efficiency of the cycle, %
G. the horsepower developed by an ideal engine operating on this
cycle using 0.5 pound of air per second.

p1 = 15 psia;
t1 = 75° F;
qS = 352 Btu/lb
M wk  60
J  42.42 

1
7.5
0.4
 55.3%
0.5 195  60
42.42
 138 hp;
(A) sketch the p-v diagrams for the cycle and then calculate:
(B) the temperatures and pressures at the end of compression, at
the end of addition of heat and at the end of the expansion
process
(C) the heat rejected, Btu/lb
(D) the net work, Btu/lb
(E) the thermal efficiency, %
(F) the horsepower developed by an ideal engine operating on the
cycle and using 0.5 pound per second of air
(G) the thermal efficiency of a cycle having the same initial conditions
and compression ratio, but with a constant pressure heat addition
of 500 Btu/lb.
P = IMEP
=
1
4. At the beginning of compression an ideal Diesel cycle using air has a
pressure of 15 psia, a temperature of 75° F and a specific volume of
13.2 ft3/lb. For a compression ratio of 15 and a heat addition of 352
Btu/lb,
 5      5.6 
 12   4   12  1200   6 

 

BMEP  12,862 lbf ft 2 = 89 lbf in2
(C) IHP =
k 1
1 hp = 42.42 Btu min
P = BMEP

1
 rV 


33,000
T= 875.35 in-lb = 875 ft-lbf
(B) BHP 
 qS  qR  352  157  195 Btu lb
qS  qR 195

 55.4%
qS
352
or t  1 
2 TN
(A) BHP =
33,000
0.4
3
v1 = 13.2 ft /lb
v2 = 1.76 ft3/lb













p1 = 15 psia;
t1 = 75° F;
qS = 352 Btu/lb
v4 = v1 = 13.2 ft3/lb
v1/v2 = 15.0
v1 13.2

 0.88 ft 3 lb
rV
15.0
(B) v2 
1.4
v 
p2  p1  1 
 v2 
 15 15 
1.4
 664.7 psia
664.7 144  0.88
p2v2

 1580 R
R
53.3
p3  p2  664.7 psia
T2 
T3  T2 
qS
352
 1580 
 1580  1467  3047 R
cp
0.24
RT3 53.3 3047 

 1.697 ft 3 lb
p3
664.7 144 
v3 
k 1
7. Determine the volumetric analysis of a mixture which consists of 56
percent nitrogen, 12 percent carbon dioxide and 32 percent oxygen as
calculated on a mass basis.


M1
 0.56;
Mm
M2
 0.12;
Mm
M3
 0.32
Mm
m1  28;
m2  44;
m3  32
Mass Analysis
N2 ,
0.56  28  0.02000
0.4
v 
 1.697 
T4  T3  3 
 3047 
  3047 0.44   1341  R
v
 13.2 
 4
T 
 1341 
p4  p1  4   15 
  37.6 psia
 535 
 T1 
CO2 , 0.12  44  0.00273
V2
0.00273

 8.34% CO2
Vm 0.03273
O2 ,
V3
0.01000

 30.55% O2
Vm 0.03273
0.32  32  0.01000
(C ) qR  cV T4  T1   0.171 1341  535   137.8 Btu lb
wk  net 
(D)
 qS  qR  352  137.8  214.2 Btu lb
J
q  qR 214.2
(E ) t  S

 60.9%
qS
352


(F ) W k 
wk  M
J  42.42 
(G) T3 '  T2 
V3 ' 

214.2 0.5  60
42.42
 151.5 hp
qS '
 1580  2083  3663  R
cp
RT3 ' 53.3 3663 R 

 2.04 ft 3 lb
p3 '
664.7 144 
k 1
  0.03273
8. Calculate the mass of moisture, in pounds, contained in 4000 ft3 of
atmospheric air having a temperature of 90° F when the
barometric pressure is 30.12 in. Hg
(A) if the air is saturated and
(B) if the relative humidity is 50 percent.
(C) What is the dew point of the atmosphere of part (B)?
(D) What is the specific humidity of the atmosphere of part (B),
grains/lb dry air?

PB = 30.12 in. Hg;
0.4
V '
 2.04 
T4 '  T3 '  3 
 3663 
  1736 R
 13.2 
 V4 ' 
qR '  cV T4 ' T1   0.171 1736  535   205.4 Btu lb
t 
Volumetric Analysis
V1
0.02000

 61.11% N2
Vm 0.03273
qS ' qR ' 500  205.4 294.6


 58.9%
qS '
500
500
5. The mixture of gases shown in Figure has a pressure of one
atmosphere. Using the volumetric analysis shown in the figure,
calculate the partial pressures of the individual constituents.
t = 90° F
(A) VS  Vg  467.7 ft 3 lb  Steam Tables at 90 F 
4000
 8.55 lb
467.7
(B) pS  0.6988 psia  Steam Tables at 90 F 
MV 
pV   pS   0.5 0.6988   0.3494 psia
144  0.3494  4000   4.26 lb
pVVm

RV Tm
85.8 550 
MV 
(C ) dew po int  68.8 F  Steam Tables for 0.3494 psia
(D) PB  0.49130.12   14.79 psia

 4354 0.3494   105.3 grains lb dry air
4354 pV

pB  pV
14.79  0.3494
V 


1.
9. Atmospheric air has a temperature (dry bulb) of 80° F and a wet bulb
temperature of 60° F when the barometric pressure is 14.696 psia.
Determine:

1 atm = 14.696 psia
V3
V1
V2
=0.7;
=0.1;
=0.2
Vm
Vm
Vm
p1 = pm
V1
= 14.696   0.7  =10.287 psia
Vm
p 2 = pm
V2
= 14.696   0.1 =1.470 psia
Vm
p 3 = pm
V3
= 14.696   0.2  =2.939 psia
Vm
(A) the dew point, °F;
(B) the relative humidity, %; and
(C) the specific humidity, grains/lb dry air.


PB = 14.696 psia;
tD = 80° F;
tW = 60° F
( A) P 'S  0.2563 psia (Steam Tables for 60 F )
6. For the mixture shown in Figure, constituent 1 is nitrogen constituent
2 is carbon dioxide and constituent 3 is oxygen. Calculate the analysis
(percentage) by mass of the mixture.
PB  tD  tW 
(B)PS  0.5073 psia  Steam Tables for 80 F 

V1
 0.7;
Vm
m1  28;
V2
 0.1;
Vm
m2  44;
Volumetric Analysis
N2 ,
14.696 80  60 
 0.2563 
 0.1474 psia
2700
2700
dew po int  45 F (SteamTables)
PV  P 'S 
0.70  28  19.6
V3
 0.2
Vm
m3  32
Mass Analysis
M1
19.6

 64.47% N2
Mm 30.4
CO2 , 0.10  44  4.4
M2
4.4

 14.47% CO2
Mm 30.4
O2 ,
M3
6.4

 21.06% O2
Mm 30.4
0.20  32  6.4
  30.4
 
PV
0.1474

 100  29.1%
PS
0.5073
(C ) wV 
 43540.1474   44.1 grains lb dry air
4354PV

PB  PV
14.696  0.1474
10. Calculate the enthalpy, Btu/lb dry air, for an atmosphere having a
temperature (dry bulb) of 80° F and a specific humidity of 42.1
grains/lb dry air.


t = 80° F;
wV = 42.1
hg = 1096.4 Btu/lb (Steam Tables at 80° F)
hA = 0.24t + Vhg
= 0.24 80  +
 42.11096.4 
7000
= 25.8 Btu lb dry air
11. One hundred pounds of air per minute are to be heated from 60° F
and 55° F wet bulb temperatures to a final temperature of 110°F.
There is no change of total moisture during the process. Determine
the heat required for the process, Btu/min:
(A) by the analytical methods developed
(B) by use of the psychrometric chart. Barometric pressure is 29.92 in.
Hg.

13. Ten pounds of air at a dry bulb temperature of 50° F with a specific
humidity of 40 grains/lb dry air are mixed with 25 lb of air having a
temperature of 85° F and a specific humidity of 90 grains/lb dry air.
Calculate:
(A) the specific humidity of the mixture, grains/lb dry air;
(B) the dry bulb temperature, °F; and
(C) the enthalpy of the mixture, Btu/lb dry air.







PB = 29.92 in. Hg;
tD2 = 110° F;
( A) pV  p 'S 
wV 1
tD1 = 60° F;
V1 = V2;
pB  tD  tw 
tw1 = 55° F
M = 100 lb air/min
 0.4359 
29.92 60  55 
 0.3805 in. Hg abs
2700
2700
4354  0.3805 

4354 pV


 56.1 grains
dryair
lb
pB  pV
29.92  0.3805
hA  0.24tD  V hg  0.24tD  V 1061  0.45tD 
hA1  0.24tD1  V 1 1061  0.45tD1 
V4 = 40 grains/lb dry air
V2 = 90 grains/lb dry air
tD4 = 50° F;
tD2 = 85° F;
M4 V 4   M2 V 2 
M1

M4 V 4   M2 V 2 
10  40   25 90 
M4  M2
 75.7 grains lb dryair
10  25
(B) Since the process is adiabatic, and no moisture is gained or lost, the
hA2  hA1   0.24  0.45v1   tD2  tD1 

0.45 56.1  110  60  12.2 Btu lb dry air
  0.24 

 
7000



Q12  M  hA2  hA1   100 12.2   1220 Btu min
energy balance for the process is as follows:
H A1  H A2  H A4
M1  hA1   M2  hA2   M4  hA4 
By substitution and the elimination of the quantities M1 and w V 1, the energy
equation can be solved for t D1. Thus,
(B)From psychrometric chart , hA1  23.3; hA2  35.6

(A) V 1 
V 1 
hA2  0.24tD2  V 2 1061  0.45tD2 

M4 = 10 lb;
M2 = 25 lb;

Q12  M  hA2  hA1   100 35.6  23.3   1230 Btu min
12. One hundred pounds of air per minute at a temperature of 100° F
with a relative humidity of 60 percent are cooled and dehumidified to a
final temperature of 50° F. Using the psychrometric chart, determine:
M4  0.24  0.45V 4   tD1  tD 4   M2 0.24  0.45V 2  tD2  tD1 
0.45  40 
0.45  90

10  0.24 
tD1  50   25  0.24  7000  85  tD1 
7000 



tD1  75.1 F
(c) hA1  0.24tD1  V 1 1061  0.45tD1 
 0.24 75.1 
75.7 1061  0.45  75.1
(A) the heat abstracted by the process, Btu/min, and
(B) the moisture removed, lb/min
7000
 29.9 Btu lb dryair
14. Using the psychrometric chart, determine (a) the enthalpy, Btu/lb of
dry air, and (b) the specifc humidity, grains/lb dry air, for the mixture
described in the above example.






From psychrometric chart:
V1 = 175.5
Va = 175.5
V2 = 53.5
hA1 = 51.7;
hAa = 47.4;
hA2 = 20.2;
M’ = 100 lb air per min






( A) Q1a  M  hA1  hAa   100 51.7  47.4   430 Btu
min
Qa2  M  hAa  hA2   100  47.4  20.2   2720 Btu
min

Q12  Q1a  Q a2  430  2720  3150 Btu
min

M V 1  V 2 
7000
100 175.5  53.5
7000
 1.743 lb
tD1 
tD4 = 50° F
tD4 = 85° F
M4tD 4  M2tD2 10 50   25 85

 75 F
M4  M2
10  25
Connecting points 4 and 2 with a straight line crossing the

(B) dehumidification 
M4 = 10 lb;
M2 = 25 lb;
min
75 F ordinate at point 1, the state point of the mixture,
then from the chart:
(A) hA1  29.9 Btu lb dry air
(B) V 1  75.7 grains lb dry air
15. The heat losses from a group of compartments have been
determined to be 420,000 Btu/hr. Air is furnished to the compartments
at a temperature of 100° F and leaves the spaces with a temperature
of 70° F and a relative humidity of 50 percent. Assuming the system to
use 100 percent outdoor air at a temperature of 20° F with 100
percent relative humidity, determine:
(A) the mass of air which must be circulated, lb/hr;
(B) the capacity of the preheating coil, Btu/hr;
(C) the capacity of the reheating coil, Btu/hr; and
(D) the water vapor absorbed from the washer, lb/hr.

QD2 = 240,000 Btu/hr;
tDC = tWC = 45° F;
tDA = 83° F;
tD2 = 80° F;
tWA = 71° F
 = 40%;
VD = 51
From psychrometric chart:
V2 = 61 grains/lb dry air
VA = 95;
VC = 44 grains/lb dry air
hA2 = 28.9;
hAD = 20; hAA = 34.9
hAC = 17.6 Btu/lb dry air
•
(A) M =

•
QD2
240,000
=
=27,000 lb hr
hA2 -hAD 28.9-20
•
(B) moisture absorbed=
(C) F =
M  w V2 -w VD 
7000
=
27,000  61-51
7000
=38.6 lb hr
VD - VC
51 - 44
=
= 0.1372 lb lb
VA - VC
95 - 44
•
•
• 
(D) air through coil = MC = MD - F  MD 


•
MC = 27,000 -  27,000   0.1372  = 23,300 lb hr

(E) refrigeration 
QDE =420,000 Btu/hr;
tDE = 70° F;
From psychrometric chart:
hAD = 33.0;
hAC = hAB = 20.5;
VC = VE = VD = 55;

( A) M 
tDA = 20° F;
E = 50%;
A = 100%
tDD = 100° F
hAE = 25.4
hAA = 7.2 Btu/lb dry air
VA = VB = 16 grains/lb dry air

QDE
420,000

 55,260 lb hr
hAD  hAE
33.0  25.4




(B) Q AB  M  hAB  hAA   55,260 20.5  7.2   735,000 Btu hr
(C ) QCD  M  hAD  hAC   55,260 33.0  20.5  690,750 Btu hr

(D) humidification  M wVC  wVB  
55,260 55  16 
7000
 308 lb hr
16. For an air-conditioning system such as that described in this article
and illustrated in FIGURE, the freshened air (state A) has dry and wet
bulb temperatures of 83° F and 71° F, respectively, the temperature
leaving the cooling coils (state C) is 45° F, the specific humidity after
remixing is 51 grains/lb dry air and the temperature and relative
humidity leaving the conditioned spaces (state 2) are 80° F and 40
percent, respectively. If the total sensible and latent heat gain from the
spaces is 240,000 Btu/hr, calculate:
(A) the mass of air that must be circulated, lb/hr;
(B) the amount of moisture that can be absorbed from the conditioned
spaces, lb/hr;
(C) the fraction of air that bypasses the cooling coil, lb/lb; (d) the air that
flows through the coil, lb/hr; and
(E) the capacity of the refrigerating plant required for this system, tons.

M C  hAA  hAC 
60  200

23,300  34.9  17.6 
12,000
 33.6 tons
1. The specific volume of steam at atmospheric pressure and 212° F is
26.80 ft3/lb. Find (A) its density and (B) its specific weight.

1


g
 gC
1
3
26.80 ft

 0.03731 lbf
lb
ft 3
g
gC

= 0.03731 lb
1 lbf sec 

32.2 ft sec   32.2
lbft 
2
ft 3
2
2. If the density of mercury is 13,600 kg/m 3, find (A) its density in lb/ft3
and (B) its specific weight in N/m3.

A . Using the appropriate conversion factors
g
L
gC
1000 kg
m3  62.4278 lb
ft 3
kg
3
m
16.0185
lb 3
ft

32.1740 ft 2
s
p  62.4278 lb 3 
ft  32.1740 lbft
lbf s2

 2116.21 lbf 2
ft
 
A.  
B.  
(d) p  


(e) p 
2116.21 lbf
2
144 in
ft 2  14.6959 lbf
ft 2

 33, 8985 ft




in2
ρ = (13,600 kg/m3)(0.4535924 kg/lb)-1(0.3048 m/ft)3
or (13,600 kg/m3)[16.01846 (kg/m3)/(lb/ft3)]-1
= 849.0 lb/ft3
 

9.81 m 2
g
s
 13,600 kg 3
m
gC
B.
1.0 kg m
N  s2
  133.4 kN 3
m
 


.
3. A pressure gauge connected to a turbine inlet reads 400 psi. A
vacuum gauge connected in the exhaust trunk of the same turbine
reads 28.0 in. Hg. The barometer reads 30.1 in. Hg. Find: (A) the
absolute pressure at the turbine inlet, psia, and (B) the absolute
pressure in the turbine exhaust trunk, in. Hg abs.
5. The difference in height of the water legs in the open manometer
shown in the figure A is 20 in. (dimension in y), find the absolute
pressure at point A in psia. The specific weight of water is 62.4 lbf/cu.
ft.

Starting from point A, and writing an equation of equilibrium through
the manometer:
pA  p  gas leg  p  water leg  p(atm)  0
p  gas leg is a negligible numerical quantity
p  water leg    L;   62.4 lbf
p  62.4
ft 3
; L  20 in. 
20
ft
12
lbf 20
x
ft  104 lbf 2
ft
ft 3 12
104
 0.72 lbf 2 ( psi )
in
144
p(atm)  14.7 psi (assumed since atmospheric pressure

(a) 1 in. Hg = 14.7/29.92 = 0.491 psi
30.1 in. Hg = 30.1 x 0.491 = 14.8 psi
p = 14.8 + 400 = 414.8 psia
is not specifically given)
(b)
p A  p  water leg  p(atm)  0.72  14.7  15.42 psia
p(gauge) = 28.0 in. Hg
p(atm) = 30.1 in. Hg
p(abs) =p(atm) – p(vac)
p(abs) = 30.1 – 28.0 = 2.1 in. Hg abs
4. In the barometer shown in figure C, the mercury level in the vertical
tube is exactly 760 mm above the mercury level in the reservoir (y =
760 mm). The following data also apply: “water” and “mercury” can be
considered to be incompressible fluids having densities of 1000 kg/m3
and 13,595.1 kg/m3, respectively. The standard local acceleration of
gravity is equal to 9.80665 m/s2 or 32.1740 ft/s2. Find the barometric
pressure in:
(a) in. Hg
(b) ft H2O
(c) pascals, Pa(N/m2)
(d) lbf/ft2
(e) psi (lbf/in2)
p 
6. In the differential manometer shown in the figure B, points A and B are
at the same height. The mercury used as a manometric fluid stands 9
inches higher in the B leg than in the A leg (y = 9 in.). Find the
difference in pressure between points A and B in psf. Specific weight
of mercury is 849 lbf per cu. ft.

Writing the equilibrium equation starting from point A,
pA   (y in. water )    y in. Hg   pB  0
pA  pB   (y in. Hg)    y in. water 
p water    L  62.4 
9
 636.8 lbf 2
ft
12
lbf
pA  pB  636.8  46.8  590
ft 2
1.
7. In the figure, what is the difference in potential energy per pound mass
of fluid at B and at A?
p  Hg    L  849 


A.
L = y = 7 60 mm = 0.76 m = 76 cm
L = 76 cm/(2.54 cm/in) = 29.9213 in. Hg
P.E. 
B.
ρ1 = 1000 kg/m3 (water)
ρ2 = 13,595.1 kg/m3 (mercury)
g
g
p  1L1
 2L2
gC
gC

 13,595.1 
 1 in.   1 ft 
L1  2 L2  
 76 cm   2.54 cm   12 in 
1
 1000 



 33.8985 ft H2O
C. p  
g
L
gC
g
Z ft lbf
lb
gC
P.E.B 
g
ZB  30 ft lbf
lb
gC
P.E.A 
g
Z A  10 ft lbf
lb
gC
P.E.B  P.E.A  30  10
 20 ft lbf
lb
8. Steam is flowing in a pipe at a velocity of 100 ft per sec. What is the
associated kinetic energy per pound of steam flowing?

 9.80665 m 

s2  0.76 m 
 13,595.1 kg/m3 

kg

m
 1.0

N  s2 

 101,325 N 2  101.325 kPa
m

9
 46.8 lbf 2
ft
12

K.E 
V2
2gC
ft lbf
gC  32.2 lbft
lb
lbf sec2
100
K.E. 
2
  32.2
; V  100 ft
2
 155.3 ft lbf
lb
sec
14. Find the temperature, specific volume, internal energy, enthalpy and
entropy of saturated saturated steam at a pressure of 1 in. Hg
absolute.
9. The work delivered by a turbine is 400 Btu per lb of steam supplied
when the steam flow is 20,000 lb per hour. Find:
(a) the specific work in ft·lbf/lb
(b) the power delivered in hp
(c) the power delivered in kW

(a)

Saturated steam p = 1 in. Hg abs = 0.491 psia
Entering Table 2, an interpolation between 0.48 and 0.50 psia
wk12
 400 Btu
lb
J
wk12
is indicated.
 400  778  311,200 ft lbf
t = 78.32 + (11/20)(79.56 – 78.32) = 79.00° F
lb
20,000 lb
min
60
1 hp  2545 Btu
 33,000 ft lbf
hr
min

(b) M  20,000 lb
hr

Interpolation could be continued in Table 2 in the same manner for the
remaining quantities. But there is a convenient shortcut thanks to the
saturation temperature turning out to be a whole number. Simply enter
Table 1 with 79° F. Then:
υg = 652.7 ft3/lb
ug = 1036.7 Btu/lb
hg = 1096.0 Btu/lb
sg = 2.0384 Btu/lb·°R


wk12
Wk
M 
 20,000  400  8  106 Btu
hr
J
J
6
8  10

 3143 hp
2545
or
15. Find the temperature, specific volume, internal energy, enthalpy and
entropy of saturated steam at 850 psia.


Wk
20,000
 M wk12 
 311,200  103.73  106 ft lbf
min
J
60
6
103.73  10

 3143 hp
33,000
(c) 1 kW  3413 Btu

W k  8  10 Btu
6
hr

Saturated steam
850 psia
Using Table 2 and interpolating between 840 and 860 psia:
hr
 1.341hp
t = 524.01 + ½ (526.76 – 524.01) = 525.39° F
υg = 0.540 – ½ (0.5400 – 0.5264) = 0.5332 ft3/lb
 3143 hp
or

8  106
3143
Wk 
 2344 kW or W k 
 2344 kW
3413
1.341

υg = ½(0.540 + 0.5264) = 0.5332 ft3/lb
ug = 1114.1 – (0.5/2) =1113.9 Btu/lb
hg = 1198.0 – 0.3 = 1197.7 Btu/lb
or
10. In a certain non-flow process, the internal energy decreases 20
Btu/lb, and 77,800 ft·lbf/lb of work is done on the substance. Find the
heat added or abstracted.
sg = 1.4106 – ½(1.4106 – 1.4080) = 1.4093 Btu/lb·°R

u2  u1     20 Btu
lb
;wk12     77,800 ft lbf
wk12
u1  q12  u2 
J
wk12
q12  u2  u1 
J
77,800
 20 
    120 Btu
lb
778
16. The pressure and temperature of steam in a line are determined to
be 65 psia and 298° F. Since these data are inconclusive, a sample of
this steam is passed through a separating calorimeter and 0.5 lb of
water is collected in 5 minutes. From orifice data the rate of flow of the
dry vapor is found to be 0.4 lb/min.
Find (A) the quality, (B) the specific volume, (C) the entropy, (D) the
enthalpy and (E) the internal energy of the steam in the line.
lb

 abstracted 

U2  U1  5000 J;
M  2 kg
5000
u2  u1 
 2500 J
kg
2
wk12  q12  u2  u1      2500 J
kg
(on)
12. At a pressure of 100 psia and 400° F, the specific volume of steam is
4.934 ft3/lb, and the specific enthalpy is 1227.5 Btu/lb. Find the internal
energy at the given state, Btu/lb.

p  100psia;
  4.934 ft
3
lb
;
MDV
mass of dry vapor MDV


mass of mixture
MM
MDV  MH2O
MH2O  0.5 lb
MDV  0.4 lb
 5 min  2.0 lb
min
2.0
x 
 0.8  80%
2.0  0.5
(B) m  1  x   1  0.8  0.2
  mf  xhg
Extracting values for 298F from Table 1 or for 65 psia
from Table 2 :
3
  0.2  0.017  0.8  6.657  5.329 ft
lb
(c) s  sg  ms g  1.6380  0.2  1.2035  1.3973 Btu
lbR
(d) h  hg  mh g  1179.6  0.2  911.9  997.2 Btu
lb
(e) u  ug  mu g  1099.5  0.2  832.1  933.1 Btu
lb
( A) x 
11. In another non-flow process involving 2 kg of working substance
there is no heat transferred, but the internal energy increases 5000
joules. Find the work done on or by the substance, J/kg.
q12  0;
hg = 1197.4 + 0.3 = 1197.7 Btu/lb
h  1227.5 Btu
p
hu
;
J
p
u h
J
100  144  4.934
u  1227.5 
 1136.2 Btu
lb
778
lb
17. Steam at 210 psia and 386° F has an enthalpy of 1173 Btu/lb as
determined by a throttling calorimeter.
Find (A) the quality and (B) the specific volume of this steam.

p  210 psia;
t  386F ;
From Table 1 for t  386 F :
p  210.06 psia;
hg  1200.0 Btu
13. Find the pressure, specific volume, internal energy, enthalpy and
entropy of saturated water at 300° F.
From Table 2 for p  210 psia :

(a) h  hg  mh g ;
From Table 1 for 300° F:
p = 66.98 psia
υf =0.017448 ft3/lb
uf =269.52 Btu/lb
hf =269.73 Btu/lb
sf =0.43720 Btu/lb·°R
h  1173 Btu
hg  1200.0 Btu
t  385.97 F ;
m
hg  h
lb
lb
h g
Using either table, h g  839.9 and h g  1200.0
1200.0  1173
Then,
m
 0.032
839.9
x  1  m  1.000  0.032  0.968  96.8%
(b)   g  m g    
lb
For Table 1 for 386 F :
g  2.183;   0.018443
3
  2.183  0.032(2.183  0.018)  2.114 ft
From Table 2 for 210 psia :
g  2.184;
f  0.018443
lb
3
  2.184  0.032(2.184  0.018)  2.115 ft
lb
18. Find (a) the superheat and (b) the enthalpy of steam at 900 psia and
535° F.

(a)
(b)
p = 900 psia; t = 535° F
S.H. = t – tsat = 535 – 532.12 =2.88° F
h = 1193.6 for 900 psia and 530° F
h = 1204.8 for 900 psia and 540° F

p1 = 600 psia;
t1 =740° F;
Then, for t = 535° F
h = 1193.6 + ½(1204.8 – 1193.6) = 1199.2 Btu/lb
From Table 3:
h1 = 1373.7
s1 = s2 = 1.6067
For points that fall midway between tabulated values, as
frequently occurs, calculation is simplified by adding the two
appropriate entries and dividing by 2, as follows:
h = (1193.6 + 1204.8)/2 = 1199.2 Btu/lb
From the Mollier Chart entering with s2 and p2:
h2 = 897
h1 – h2 = 1374 – 897 = 477 Btu/lb
2.
19. For steam at 600 psia and 850° F, find (a) the superheat, (b) the
specific volume, (c) the internal energy, (d) the enthalpy and (e) the
entropy.

(a)
(b)
(c)
(d)
(e)
p = 600 psia ;
t = 850° F
tsat = 486.33
S.H. = t – tsat =850 – 486.33 = 363.67° F
At 840° F, υ = 1.2353 and at 860° F, υ = 1.2577
At 850° F, υ = (1,2353 + 1.2577)/2 = 1.2465 ft3/lb
u = (1292.7 + 1301.2)/2 = 1297.0 Btu/lb
h = (1429.8 + 1440.9)/2 = 1435.4 Btu/lb
s = (1.6517 + 1.6601)/2 = 1.6559 Btu/lb·°R
20. In a steam propulsion plant, feed water leaves the feed heater and
enters the main feed pump at 270° F and a gauge pressure of 35 psi.
Find for the water: (a) the specific volume, (b) the internal energy, (c)
the enthalpy and (d) the entropy.
23. Steam initially at 10 psia and a quality of 90% drops to a pressure of
2.5 psia in a non-flow reversible constant volume process. Show the
process on p-v and T-s coordinates and find:
(a) the final quality
(b) the heat transferred, Btu/lb


(a) p1  10 psia;
(a)
(b)
(c)
(d)
p = 35 + 14.7 = 49.7 psia;
υƒ = 0.017170 ft3/lb
uƒ = 238.82 Btu/lb
hƒ = 238.95 Btu/lb
sƒ = 0.39597 Btu/lb·°R
x1  0.9;
From Table 2 :
1   0.9  38.42   0.1  0.02   34.58
2  1  34.58
p2  2.5 psia;
146.40  135.79
g 
 141.10 (interpolating)
2
   0.016267 (by eye)
x2 
p = 1200 psia;
t = 275° F
Compressed liquid ----- Use Table 4
2   
34.58  0.02

 0.245  24.5%
g    141.10  0.02
(b) u1  q12  u2 
wk12
Equation Non-flow
J
2
wk12
  pd  0 since 1  2
1
J
Then q12  u2  u1
h = 220.61
h = 271.46
h = (220.61 + 271.46)/2 = 246.04
p1  10 psia;
At 1500 psia:
t = 250° F;
h = 221.65
t = 300° F;
h = 272.39
t = 275° F;
h = (221.65 + 272.39)/2 = 247.02
At 1200 psia:
t = 275° F;
h =264.04 + (2/5)(247.02 –
246.04)
h = 246.04 + 0.39 = 246.43 Btu/lb
22. Steam is admitted to a turbine at 600 psia and 740° F, and exhausts
to a condenser at a pressure of 1 psia. Assuming the process to be
isentropic, sketch the process on T-s and h-s coordinates and find the
drop in enthalpy, Btu/lb.
m1  0.1
1  2
1  x1g  m1 
t = 270° F
21. Water from the feed pump in the earlier example enters the boiler at
1200 psia and 275° F. Find the enthalpy of the feed water.
3.

At 1000 psia:
t = 250° F;
t = 300° F;
t = 275° F;
p2 = 1 psia
s1 = s2
m1  0.1
u1  ug  m1u g  1072.2  0.1  911.0  981.1
p2  2.5psia; x2  0.245
100.81  103.85
 102.33
2
953.1  951.0
u 
 952.05
2
u2  u  x2 u g
u 
4.
 102.33  (0.245  952.05)  335.6
q12  u2  u1  335.6  981.1     645.5 Btu
lb
(abstracted)
24. Four pounds of steam initially dry and saturated expand
isentropically in a non-flow process from an initial pressure of 275 psia
to a final pressure of 125 psia. Show the process on p-v and T-s
coordinates and find:
A. the final quality
B. the work done, Btu

(a) p1  275 psia  saturated vapor ;
From Table 2
s1  1.5192  s2 ;
s1  s2
u1  1117.5
p2  125psia
m2 
sg  s2
s g

1.5853  1.5192
 0.061
1.0893
x2  1  m2  1  0.061  0.939  93.9%
(b) u1  q12  u2 
q12 
Then

2
1
wk12
J
Tds  0 sin ce s1  s2
wk12
 u1  u2
J
u2  ug  m2 u g
 1108.8  0.061  793.3   1060.4
wk12
 1117.5  1060.4  57.1 Btu
(by )
lb
J
M  4lb
Wk12
wk12
M
 4  57.1  228.4 Btu (by)
J
J
1. One pound of air (consider here a perfect gas ) with an initial
temperature of 200F is allowed to expand without flow between
pressures of 90 and 15 psia. Which of the three processes, pv =
c, pvk = c or pv1.5 = c will produce the maximum work with
minimum heat supplied.















P1 = 359 kPa abs
T1 = 326 K
R = 287 J/kg.K
(A) V 
P1 = 90 psia
P2 = 15 psia
T1 = 660R
Process 1-2 isentropic(reversible adiabatic, n = k )
1 – 2’ isothermal ( T = C, n = 1.0)
1 – 2” polytropic , n = 1.5
Work Done:
(A) Work Isothermal Process:
w k12'
RT1  P1  53.3(660)  90 
Btu

ln   
ln 
     81 lb
J
J
P
778
15


 2
(B) Work Isentropic Process:
k 1
1.4 1
T2  P2  k
 15  1.4
 
 T2   660  
 396R

T1  P1 
 90 
Wk12 P2V2  P1V1
R(T2  T1 ) 53.3(396  660)
Btu
=


    45.2
J
J(1  k )
J(1  k )
778(1  1.4)
lb
(C) Work Polytropic Process: n = 1.5
n 1
n
1.5 1
1.0
P 
 15 
T2 '  T1 '  2 
 660 
 363R

P
 90 
 1
Wk12 P2 ' V2 ' P1 ' V1 ' R(T2 ' T1 ') 53.3(363  660)
Btu
=


    40.7
J
J(1  n)
J(1  n)
778(1  1.5)
lb
Heat Transfer:
(A) Isothermal Process:
wk12 '
 0  81 Btu/lb     81 Btu/lb
J
(B) Isentropic Process:
q12  u2 ' u1 '
q  0(by definition)
(C) PolytropicProcess:
wk12 '
w '
 C v T2 ' T1   k12
J
J
= 0.171 363  360   40.7
q12 '  u2 ' u1 '
=  50.8  40.7 = (-)10.1 Btu/lb
2. One kilogram of a perfect gas (air) is used as a working
substance in a Carnot power cycle. At the beginning of isentropic
compression, the temperature is 326K and the absolute pressure
is 359 kPa. The absolute pressure at the end of the isentropic
compression is 1373 kPa. For this cycle, the isothermal
expansion ratio (v3/v2) is 2.0. For the cycle, calculate :
(A) The pressures, temperatures and specific volumes at each
process termination point.
(B) The heat supplied, kJ/kg
(C) The heat rejected, kJ/kg
(D) The net work done by the T-s and p-v area method
(E) The thermal efficiency, %

P2 = 1373 Kpa
V3/V2 = 2.0
RT1 (287)(326)

 0.2606 m3 /kg
P1
359 x 103
T2  P2 
 
T1  P1 
k 1
k
1.4 1
 1373  1.4
T2  326  
 478.3K

 359 
R(T2 ) 287  (478.3)
V2 

 0.100 m3 /kg
P2
1373 (103 )
V 
1
P3  P2  2   1373     686.5kPa(abs)
2
 V3 
T3  T2  478.3K
V3  2V2  2  0.100   0.200 m3 / kg
T4 = T1 = 326 K
k
1.4
 T  k 1
 326 1.4 1
P4 = P4  4 
 686.5 
 179.5 kPa

T
 478.3 
 3
287  (326)  0.5212 m3 /kg
R(T4 )
V4 

P4
179.5 (103 )
V 
(B) qs = P2 V2 ln  3  = 1373   0.100  ln  2  = 95.17 kJ/kg
 V2 
V 
 0.5212 
(C) qR = P4 V4 ln  4  = 179.5   0.5212  ln 
 = 64.85 kJ/kg
 0.2606 
 V1 
(E) w k(net)  qs  qR  95.17  64.85 = 30.32 kJ/kg
3. Calculate the available energy in Btu/lb for a Carnot cycle with a
source temperature of 3460R, a sink temperature of 520R and an
energy supply as heat of 100 Btu to one pound of a working
substance. Then calculate the reduction of available energy,
Btu/lb, for a similar Carnot cycle in which all conditions remain the
same as before except that the working substance is limited to a
maximum temperature of 1960R.


T1 = 3460 R
T1’ = 1960 R
To = 520R
Qs = 100 Btu/lb (aBCb or aB’C’c)

For the heat addition process B-C:
q
q
100
Btu
ds 
and sC  sB  s 
 0.0289
T
T1
3460
lb.R
Unavailable Energy = To  S BC  520  0.0289  = 15
A.E.  qs  To  S   100  15 = 85 Btu/lb
Btu
lb
In this non-flow system the non-flow energy equation will apply to
each phase of the liquid involved. By writing the equation separately
for each fluid and equating through the heat transferred terms, a heat
balance is formed.
From :Equation:
qs  qR T1  To
T
q

and 1 - R  1  o
qs
T1
qs
T1
T 
 520 
qR  qs  o   100 
  15 Btu/lb
 3460 
 T1 
wk
 qs  qR  100  15  85 Btu/lb
J
for 100 Btu supplied to one pound of substance at 1960R rather than
3460R:
q
100
Sc ' SB '  s 
 0.0510 Btu/lb.R
T1 ' 1960
Unavailable energy = To  S B ' C ' = 520(0.0510) = 26.5 Btu/lb
A. )
Ms
 h1 ' h2   Mw  h2  h1 
(1)(1150.5 - h2 )  9(h2  28.08)
h2 = 140.32 Btu/lb
Reduction in A.E. = 85 - 73.5 = 11.5 Btu/lb
t2 = 172.4F
4. Using the same source and receiver temperatures as in the
above example (3460 R and 520R ) , assume first that 100 Btu of
energy as heat supplied reversibly to a cycle such as a34b and
3460R. The entropy change for heat addition is:
B.) 1 (s2  s1 ')  0.25113  1.7567  1.5056 Btu/R

6. In an ideal steam cycle, heat is added at a constant pressure of
200 psia. Water entering the boiler has an entropy of 0.0555
Btu/lb.R, and steam leaving the superheater has an entropy of
1.7395 Btu/lb.R. Superheated steam temperature is 1500F, and
heat added is 1744.5 Btu/lb ( equals h2 – h1 ). Heat rejection is
carried out in a condenser at a constant temperature of 60F.
Average temperature of the combustion gases in the boiler
furnace is 300F. Calculate:
A. The available energy of combustion gases with respect to
the sink temperature of 520R, Btu/lb
B. The available energy of an ideal cycle receiving its energy at
the constant source temperature equal to the superheated
steam temperature, Btu/lb
C. The available energy of the ideal steam cycle, Btu/lb
D. The mean effective temperature of the working substance
during heat receipt in the original steam cycle, degree F
S4  S3 
Q34
100

 0.0289 Btu/lb.R
T3
3460
C.) 9 (s2  s1 )  9(0.25113  0.05555)  1.7602 Btu/lb
D.) Net entropy change for system = (+) 1.7602 - 1.5056 = (+) 0.2546 Btu/R

s1 = s4 = 0.0555 Btu/lb.R
s2 = s3 = 1.7395 Btu/lb.R
P2 = P1 = 2000 psia
t2 = 1500F t3 = t4 = 60F
t2’ = combustion temperature
= 3000 F
qs = 1744.5 Btu/lb
From the diagram it is clear that (SB – S1 ) = (S4 – S3). Therefore,
the unavailable portion of Qs is:
U.A.E. = T1 (SB – S1) = 520 ( 0.0289) = 15 Btu
Now assume the same quantity of heat is added reversibly, but at a
constant temperature of 1960R. The resulting cycle are is
represented by a26c on the figure, and the change in entropy is:
S6  S2 
Qs
100

 0.0501 Btu/lb.R
T2
1960
The unavailable portion, as determined by To(S) is:
U.A.E. = T1 ( S7 – S1 ) = T1 (S6 – S2) = 520 ( 0.0510) = 26.5
5. In a perfectly insulated cylinder and piston arrangement, 1 pound
of saturated steam at 212F is to be mixed with 9 pounds of water
at 60F. The entire heat exchange process is to take place at
standard atmospheric pressure. Calculate the following:
A.
B.
C.
D.
The final temperature of the mixture:
The change in entropy of steam, Btu/R.
The change of entropy of the water, Btu/R
The net change of entropy of the system, Btu/R
1. The first stage of a 50% reaction groups receives steam with a
negligible approach velocity at a pressure of 300 psia with a
temperature of 520F. The available energy to the stage is 10
Btu/lb, the fixed blade efficiency is 96%, the velocity coefficient for
the moving row is 0.88, the reactive effectiveness is 90%, and the
relative inlet and absolute exit velocities are 146 ft/s and 160 ft/s,
respectively. Assume the stage efficiency is the same as the
combined blade efficiency and calculate the following:
(A)
(B)
(C)
(D)
(E)
The steam velocity leaving the fixed blades, ft/s
The relative exit velocity from the moving blades, ft/s
The stage work, Btu/lb
The combined blade efficiency, %
The enthalpy of the steam entering the succeeding stage,
Btu/lb for complete velocity carryover

P0 = 300 psia
t o = 520°F
(AE)st  10 Btu/lb
en = 0.96 ;
Cm = 0.88 ;
eR = 0.90
Vr1  146 ft/s
V2 = 160 ft/s
(A)  AE n   AE R 
 Cm Vr2 
2
2gc J
=
2gc J
 Cm Vr1 
Vr2 

2
2
= eR  AE R

r1=1.625 in.;
  2g c J  eR  AE R
2
 492 ft/s
(C)
 wk 
J
=
V
2
1
 
- V2 2 + Vr2 2 - Vr12

 490  - 160   +  492 - 146   = 8.69 Btu/lb
2
2
2  32.2  778 
Btu/lb
h2 = 26 Btu hrg ft 2 g°F
k tube = 26;
k scale = 0.6 Btu hrg ftg°F
(A) For clean tube:
r r
1
1
 3 2 
 r2 
 rt  h2
h1   kt  
 r3 
 r3 
1
0.375
1



 2.0 
 2.1875  26
1700   2.375  12  26   2.375 




 0.0007  0.0013  0.0385  0.0405
1
U
 24.69 Btu hr  ft 2  F  external area 
0.0405
(C ) For tube and scale (R' = resistance/ft 2 area):
L
1
R'scale  R'1 

 rS 
 r1 
k   h1  
 r3 
 r3 
0.375
1

 0.0691
 1.8125 
 1.625 
12  0.6   2.375  1700  2.375 




 0.0691  0.0013  0.0385  0.1089

R'total
U 

1
1

 9.18 Btu hr ft 2 F
R 'total
0.1089
(D) For clean tube:
h0  1269.4 Btu lb ;
wk st
J
wk st
(B) st 
h1= 1700;
1

U
2. To illustrate the effect of tip leakage, it is assumed that 5 percent
of the steam supplied to the stage of the above example, flows
past the stage without performing useful work. Assume there are
no other stage losses and calculate:
A. the stage work, Btu/lb
B. the stage efficiency, %
C. the reheat due to leakage, Btu/lb
D. the actual enthalpy entering the succeeding stage for complete
velocity carryover, Btu/lb
J
r3 = 2.375 in
t water = 500° F
2
 w k st
8.69
(D) ecb =
=
= 0.869 = 86.9%
J  A.E.st
10
 w k st  v 2 2
h 2 = ho 

J  A.E.st 2g c J
2
160 
h2 = 1269.4 - 8.7  1260.2
2  32.2  778 
( A)
r2 = 2.0 in.;
t gas = 2100° F;
L
rt = r2 + = 2.0 + 0.1875 = 2.1875 in.
2
L
rS = r1 +
= 1.625 + 0.1875 = 1.8175 in.
2
2gc J
2
=
1.
2.
3.
4.
5.
6.
7.
8.
= 490 ft/s
 0.88 146     2 32.2 778   0.90 5 

the overall coefficient of heat transfer for the clean tube,
Btu/hr·ft2·°F;
(B) the overall heat transfer coefficient including the effect of the
scale, Btu/hr·ft2·°F, and
(C) the external tube surface temperature before and after the
scale has formed.
10
 5 Btu/lb
2
 2  32.2  778    0.96  5 

 Vr2 
2
(A)
 2gc J en  AE n
V1 
(B)
 AE st
3. The internal diameter of a boiler generating tube is 4 in. and the
wall thickness is 0.375 in. In operation the external surface film
coefficient is 26 Btu/hr·ft2·°F, the internal film coefficient is 1700
Btu/hr·ft2·°F and a 0.375 in. thickness scale having a conductivity
of 0.6 Btu/hr·ft2·°F is deposited on the inner surface of the tube.
The metal conductivity is 26 Btu/hr·ft2·°F, the furnace gas
temperature is 2100° F and the temperature of the water is 500° F.
Calculate:
 A.E.st
 8.69 Btu lb ;
 10 Btu lb
h2  1260 Btu lb
 0.95 8.69   8.26 Btu lb
wk st
J  A.E.st

8.26
 82.6%
10
(C ) leakage reheat  8.69  8.26  0.43 Btu lb
(D) hx  1260.2  0.43  1260.6 Btu lb
 R '2 
 0.0385 
t  
  t1  t2   
 2100  500   1521 F
 0.0405 
 R 'total 
tube surface temperature  2100  1521  579 F
For tube with scale:
 R '2 
 0.0385 
t  
  t1  t2   
 2100  500   566 F
R
'
 0.1089 
 total 
t surface of tube = 2100  566  1534 F
4. A steam condenser with a net heat transfer area of 23,500 ft 2 has
a design value of U of 486 Btu/hr·ft2·°F for a log mean temperature
difference of 32.4 and a design operating pressure of 2.5 in. Hg
abs. The water consumption at rated capacity is not to exceed
40,500 gpm with a temperature rise of 18.3°F. Measured
temperatures of fresh water at this flow rate during an acceptance
test were 71.5°F and 90°F at entrance and exit, respectively, and
the average condenser pressure was 2.45 in. Hg abs. Assuming
no subcooling of the condensate, calculate the overall heat
transfer coefficient established by the acceptance test, and
compare this with the manufacturer’s design value given above.
6. The following is an ultimate analysis of a typical naval fuel oil, ash
and moisture free:
Carbon = 0.8663 lb
Hygrogen = 0.1127 lb
Oxygen = 0.0019 lb
Nitrogen = 0.0028 lb
Sulfur
= 0.0163 lb
1.0000 lb
Calculate the following for “complete” or “theoretical” combustion
with air, in pounds per pound of fuel:
A.
B.
C.
D.
E.
F.
oxygen required from air
nitrogen required from air
air required
water formed from combustion
carbon dioxide formed
total mass of flue gases

Fuel constituents
pounds per pound

Twater (entering) = 71.5° F;
tleaving = 90° F
Psteam = 2.45 in. Hg abs
water quantity = 40,500 gpm
Anet = 23,500 ft2
m’ = (40,500)(8.34)(60) = 20.27 x 106 lb/hr


Q  20.27  106
C = 0.8663
H = 0.1127
O = 0.0019
N = 0.0028
S = 0.0163
 1 90  71.5  375  10
6
Btu hr
36.5  18

 26.2 F
36.5
ln
18
m
A.
B.

U 
Q
375  106

 609 Btu hr  ft 2  F
A m
23, 500 26.2
C.

t1 in = 266° C;
t3 in = 593° C;

Ma’ = 1.8 kg/sec
Ma’ = 1.88 kg/sec
t2 out = ?;
t4 out = 310° C;

(A) Q  M g c pg  t  g  1.88 kg s  1090 J kg  C  283 C 
 579.92 kJ s




M a c pa  t a  M g c pg  t g
(B) Q a  Q g ;


M a c pa  t2  t1 a  M g c pg  t3  t4 g

t2  t1 
Q

M a c pa
 266 
579920
 586.6 C
1.8 1005 
  2  t4  t1    t3  t2 
(C)  m  1


t  t1 
ln 1
ln 4
2
t3  t2 

310  266   593  586.6 
44
ln
6.4
 19.5 C
0.8663(2.667) = 2.3104
0.1127(8.000) = 0.9016
reduces external O2 = -0.0019
inert = 0.0000
0.0163(1.00) = 0.0163
lb O2 required from air = 3.2264
oxygen required = 3.2264 lb/lb fuel
nitrogen from air = 3.2264 (0.768/0.232)
= 10.6805 lb N2/lb fuel
air required
= 3.2264 + 10.6805
= 13.9069 lb air/lb fuel
Element + Oxygen = Combustion product
C 0.8663 + 2.3104 = 3.1767 lb CO2/lb fuel
H 0.1127 + 0.9016 = 1.0143 lb H2O/lb fuel
O 0.0019 + --------- = -----------------------------N 0.0028 + 0.0000 = 0.0028 lb N2/lb fuel
S 0.0163 + 0.0163 = 0.0326 lb SO2/lb fuel
Σ = 4.2264 lb combustion products per lb fuel
5. An exhaust gas regenerator (counter-flow heat exchanger) for a
marine gas turbine handles 1.8 kg kg/sec of air from its
compressor and heats it by means of 1.88 kg/sec of hot exhaust
gas. Exhaust gas enters the regenerator at 593° C and leaves at
310° C. Compressed air enters the regenerator at 266° C. For this
temperature range a constant pressure specific heat for the
exhaust gas may be estimated at 1090 J/kg·C. Assume no heat
transfer other than between the regenerator fluids. Calculate:
A. the energy exchanged as heat by the two fluids, kJ/sec
B. the air temperature leaving the generator, C
C. the log mean temperature difference for the exchanger, C
Pounds of O2 required from air
per pound of fuel constituent
D.
E.
F.
water from combustion = 1.0143 lb/lb fuel
carbon dioxide formed = 3.1767 lb/lb fuel
total mass of flue gases = 4.2264 + 10.6805
= 14.9069 lb/lb fuel
7. Calculate the following quantities in pounds per pound of fuel for
combustion with 50 percent excess air using the same fuel
analysis as in the example of the preceding article:
A. oxygen supplied from air
B. nitrogen accompanying oxygen
C. air supplied
D. water formed from combustion
E. carbon dioxide formed
F. total mass of flue gases

Fuel constituents
Pounds of O2 required from
per pound of fuel
C = 0.8663
H = 0.1127
O = 0.0019
0.8663(2.667) = 2.3104
0.1127(8.000) = 0.9016
reduces external O2 = -0.0019
N = 0.0028
S = 0.0163
1.0000
0.0163(1.00) = 0.0163
3.2264
A. oxygen supplied with 50% excess air = 1.5(3.2264)
= 4.8396 lb/lb fuel
B. nitrogen from air = 4.8396(0.768/0.232) = 16.0207 lb/lb fuel
C. air supplied
= 4.8396 + 16.0207 = 20.8603 lb/lb fuel
Element + Oxygen
= Combustion product
C 0.8663 + 2.3104
= 3.1767 lb CO2
H 0.1127 + 0.9016
= 1.0143 lb H2O
O 0.0019 + (Appears
= 1.6132 lb O2 (Excess O2 = 0.5 x 3.2264)
with other
constituents)
N 0.0028 + 0.0000
= 0.0028 lb N2/lb fuel
S 0.0163 + 0.0163
= 0.0326 lb SO2/lb fuel
Σ = 4.2264 lb combustion products per lb fuel
D. water formed
= 1.0143 lb/lb fuel
E. carbon dioxide formed = 3.1767 lb/lb fuel
F. total mass of flue gases = 5.8396 + 16.0207
= 21.8603 lb flue gases/lb fuel
8. An analysis of the flue gases of a combustion process, percent by
volume, is as follows:
CO2 = 10.0%;
CO =2.0%;
O2 = 8.0%;
N2 = 80.0%

Mols of Constituent
per mol of flue gas
Mol wt.
Mass per mol
flue gas
CO2
0.11

44
=
4.84
CO
0.02

28
=
0.56
O2
0.045

32
=
1.44
N2
0.825

28
=
22.40
=
29.94

Percent
mass
4.84
= 16.17%
29.94
0.56
= 1.87 %
29.94
1.44
= 4.81%
29.94
23.10
= 77.15%
29.94
 =100.00%
(A) mass of dry flue gas per lb of fuel:
CF
0.8700
lb gas

 16.69
12
12
12
12
lb fuel
CO2 
CO
0.1617  
0.0187 


44
28
44
28
(B) mass of air actually supplied:
MG =
 N2 
 0.7715 
 0.768   CF 
 0.768   0.8700 
lb air




MA 

 16.77
12
12
12
12
lb fuel
CO2 +
CO
0.1617  
0.0187 


44
28
44
28
(C) air required for ideal combustion of one pound of fuel:
carbon 0.8700(11.49) = 9.9963 lb of air for carbon
hydrogen 0.1200(34.48) = 4.1376 lb air for hydrogen
sulfur 0.0020(4.31) = 0.0086 lb air for sulfur
14.1425 total air required, lb
Since the fuel contains 0.006 pound of oxygen, assume the hydrogen combines
with it and reduces the external oxygen required:
 lb O2   lb air  
4.1376  
   4.1376  0.006   4.31

 lb fuel   lb O2  
 4.1117 lb air for hydrogen

Consti tuent
Volume
fraction
CO2
0.10
mols CO2
lb CO2
lb
 44
= 4.40
mol gas
mol
mol gas
4.40
= 14.71
29.92
CO
0.02
mols CO
lb
lb CO
 28
= 0.56
mol gas
mol
mol gas
0.56
= 1.87
29.92
O2
0.08
mols O2
lb O2
lb
 32
= 2.56
mol gas
mol
mol gas
2.56
= 8.56
29.92
N2
0.80
mols N2
lb N2
lb
 28
= 22.40
mol gas
mol
mol gas
22.40
= 74.86
29.92

Mol wt.
= 29.92
Percent mass
of constituent
lb gas
mol gas
Apparent molecular weight o f gas mixture = 29.92
R for the mixture =
 =100.00
lb gas
mol gas
1545
ft lbf
 51.64
29.92
lb°R
A fuel oil has the following analysis on an ash-and-moisture-free
basis and yields, on burning, the following Orsat analysis:
Fuel analysis,
lb/lb fuel
C = 0.8700
H = 0.1200
N = 0.0020
S = 0.0020
O = 0.0060
Orsat Analysis
CO2 = 11.0 %
CO = 2.0 %
O2 = 4.5 %
N2 = 4.5 %
Calculate:
A. the mass of dry flue gas, lb/lb fuel
B. the mass of air actually supplied, lb/lb fuel
C. the air required for ideal combustion, lb/lb fuel
D. the percent excess air supplied

Then, ideal combustion of one pound of fuel requires
14.1425  0.0259  14.1166 lb air/lb fuel
(D) % excess air:
16.77  14.12
 18.8%
14.12
The following data were observed during an oil-fired boiler test:
1. Duration of test
1 hr
2. Steam delivered by boiler
200,000 lb
3. Average steam temperature at
superheater outlet
760° F
4. Average steam pressure at
superheater outlet
600 psia
5. Feed water temperature
240° F
6. Feed water pressure
700 psia
7. Fuel fired (dry basis)
15,385 lb
8. Flue gas temperature leaving
last heat transfer passage
450° F
9. Dry bulb temperature of air
supplied for combustion
80° F
10. Wet bulb temperature of air
supplied for combustion
70° F
11. Barometric pressure at test
location
29.92 in. Hg
12. Temperature of fuel supplied
to burners
80° F
13. Ultimate analysis of fuel on an as-fired basis:
Carbon
0.8095 lb/lb fuel
Hydrogen
0.1143 lb/lb fuel
Nitrogen
0.0048 lb/lb fuel
Sulfur
0.0143 lb/lb fuel
Oxygen
0.0095 lb/lb fuel
Moisture
0.0476 lb/lb fuel
Ash
0.0000 lb/lb fuel
1.0000 lb fuel
14. Volume analysis of flue gases in percent (Orsat):
CO2 = 11.34%
CO = 00.71%
O2 = 5.06%
N2 = 82.89%
100.00%
15. Higher heating value of fuel is 19,500 Btu/lb “dry” fuel.
Calculate an energy balance for the tested boiler.

As-fired basis
Carbon
0.8095
Hydrogen
0.1143
Nitrogen
0.0048
Sulfur
0.0143
Oxygen
0.0095
Moisture
0.0476
÷ (1 – 0.0476) =
Dry basis
0.8500 lb/lb
0.1200 lb/lb
0.0050 lb/lb
0.0150 lb/lb
0.0100 lb/lb
1.000
Flue Gas Analysis:
Percent
volume dry
Mol wt.
Pounds per
100 mols gas
CO2
=
11.34

44
=
498.96
CO
=
0.71

28
=
19.88
O2
=
5.06

32
=
161.92
N2
=
82.89

28
=
2320.92
 =
3001.68
Percent
mass
498.96
= 16.62
3001.68
19.88
= 0.66
3001.68
161.92
= 5.40
3001.68
2320.92
= 77.32
3001.68
 =100.00
(17.1) A boiler consumes 16,800 pounds of feul/hr when producing
210,000 lb of steam per hour at 620 psia and 800 deg F from
feed water at 650 psia and 800F from feed water at 650 psia
and 300F. The heating value of fuel is 18,500 Btu/lb and the
furnace volume is 1250 ft3 . Calculate :
(A)
(B)
(C)
(D)
The boiler Capacity (in mB/hr) Ans: 238.6 mB/hr
The factor of evaporation
Ans: 245,700 lb/hr
The equivalent evaporation
Ans: 1.17
The furnace heat-release rate, Btu/hr.ft3
Ans: 248,600 Btu/hr,ft3
(E) The boiler efficiency, %
Ans: 76.8%
Energy balance:
(1) Energy absorbed by water and steam in boiler, Btu per pound of dry fuel:
Q1  MS  h2  h1  
200, 000
1385.1  209.9  *
15, 385
= 13 1175.2   15, 277.6 Btu lb fuel (dry)
* where
h2 at 600 psia and 760 F = 1385.1 Btu lb
h3 at 700 psia and 240 F = 209.9 Btu/lb
(2) Energy loss to dry flue gases, Btu per pound dry fuel:
Q2  MG c p t2  t1   17.65 0.24   450  80 
 1567.32 Btu/lb dry fuel
CF
0.8500

12
12
12
12
CO2 +
CO
0.1662  

0.0066 
44
28
44
28
= 17.65 lb dry gas/lb fuel
where MG 
c p  0.24 (average specific heat of flue gases)
t1  entering air temperature = 80 F
t2  leaving flue gas temperature = 450 F
(3) Energy loss due to moisture from burning hydrogen, Btu per pound
(17.2) A fuel oil has the following ultimate analysis ash and moisture
free:
Carbon
= 0.85 lb
Hydrogen = 0.14 lb
Oxygen
= 0.00 lb
Nitrogen
= 0.01 lb
Sulfur
= 0.01 lb
1.00 lb
For complete combustion with air, calculate the following in pounds
per pound of fuel:
(A) Oxygen required from air
Ans: 3.32
(B) Nitrogen accompanying oxygen from air
Ans: 10.98
(C) Air required
Ans: 14.30
(D) Water formed from combustion of hydrogen Ans: 1.176
(E) Carbon dioxide formed
Ans: 3.23
(F) Total mass of flue gases
Ans: 15.30
of dry fuel:
9 12 
9H
 h4  h3   100 1265.0  48.1
100
= 1314.25 Btu/lb dry fuel
Q3 
where H = percent hydrogen in 1 lb dry fuel
h4 = 1265.0 (Table 3, Steam Tables at 1 psia and 450 F)
h3 = hf at 80 F = 48.1 Btu/lb
(4) Energy loss to moisture accompanying one pound dry fuel:
Q4 
MM
0.0476
 h  h3   1  0.0476 1265.0  48.1
1  MM 4
= 0.05 1216.9   60.85 Btu/lb dry fuel
In spite of calculating on the dry fuel basis, account must be taken of the moisture
(16.2) Air at 45F (DB) and 41F (WB) is heated and humidified to
72F (DB) and 59F (WB). To what temperature should the air
be heated before humidification?
Ans: 64F
(16.3) Air is to be conditioned from td = 39F and  = 80% to td =
74F and  = 70%.
(A) To what temperature should the air be heated before
humidifying F? Ans:102.5F
(B) How much moisture in grains is added during
humidification ? Ans: 61 grains/lbda
accompanying the fuel as fired:
MM  mass of moisture in 1 lb of fuel as received = 0.0476 lb
MM
 0.05 lb of moisture/lb dry fuel
1  MM
(5) Energy loss to moisture in air supplied for combustion:
Q5  M A MW  h4  h5   17.77 0.0134  1265.0  1096.6 
= 40.1 Btu/lb dry fuel
 0.7732 
N2

 0.8500 
CF 
 0.768 
0.768
where MA 

12
12
12
12
CO2 
CO
0.1662   28 0.0066 
44
28
44
= 17.77 lb air supplied/lb dry fuel
N2 , CO2 , CO = percents by mass in flue gases
MW = 0.0134 lb moisture/lb dry air (from psychrometric chart at 80 F
dry bulb and 70 F wet bulb)
(15.1) It is desired to produce a mixture of helium and hydrogen
which will have a specific heat of 1.0 Btu/lb.F at constant
volume . What must be the volumetric percentage of helium?
Gas
Cv
Helium
0.750
Hydrogen
2.440
Ans: 74.2%
(15.2) A gaseous mixture has the following mass analysis:H2 ,
10%; CO2 , 5% ; N2 , 85%, Cp for common gases at room
temperature : O2, 0.217; H2 3.42; N2, 0.247; CO, 0.243; CO2,
0.205. Find the:
(A) specific heat at constant pressure, Ans: 0.562 Btu/lb.F
h5  1096.6 Btu/lb (enthalpy of superheated steam at 80 F, read most
easily from Mollier Chart; partial pressure has relatively small
effect on enthalpy at this temperature evidenced by practically
horizontal 80 F line on chart)
(6) Energy loss to incomplete combustion, Btu per pound of dry fuel:


CO
Q6  CF 
 10,100 
 CO2  CO 
0.71
= 0.85
10,100   505.84 Btu/lb dry fuel
11.34  0.71
CO2 , CO = volume percentages from Orsat analysis of flue gases
CF  0.8500 lb carbon/lb dry fuel
10,100 = Btu loss/lb of carbon burned to CO rather than to CO2
(7) Energy loss to radiation and unaccounted-for losses:
Q7  H.H.V .  (Q1  Q2  Q3  Q4  Q5  Q6 )
 19, 500  15, 277.6  1567.3  1314.3  60.8  40.1  505.8 
 19, 500  18, 765.9  734.1 Btu/lb dry fuel
(B) volumetric analysis, % H2 = 61.4%,CO2 =1.4%,N2 = 37.2%
(C) Partial pressure of N2 is psia if the barometer is standard
and the mixture is at barometric pressure: Ans: 5.47 psia
(15.3) A tank contains air at 50 psia. Air maybe assumed to consist
of 79.1 percent of N2 and 20.9% O2 by volume. Calculate:
(A) the partial pressure due to the oxygen, Ans:10.45 psia
(B) the partial pressure due to the nitrogen, psia.
Ans: 39.55 psia
(15.4) A mass analysis of gases in a compartment shows the
following: O2 , 20 lb; N2 , 140 lb; CO2 , 15 lb; H2O, 4 lb. Find
the volumetric analysis of the gases, %.
.
Ans. O2 = 10.10% ; N2 = 80.81 %; CO2 = 5.50 % ;
H2O = 3.59%
(15.6) Given, for atmospheric air: temperature, 82F; barometric
pressure, 29.92 in. Hg; partial pressure of water vapor ,
0.3632 psia. What is the dew point temperature?
.
Ans. 70 F
(15.7) A room 14 ft x 16 ft x 10 ft contains atmospheric air at 72F.
The partial pressure of the water vapor in air is 0.2140 psia.
Barometer is standard. Calculate :
(A) the mass of dry air in the room, lb;
Ans: 164.7 lb
(B) the mass of water vapor in the room
Ans: 1.51 lb
(C) the dew point temperature
Ans: 55F
.
(15.13) For atmospheric air, given: dry bulb temperature, 85F ; wet
bulb temperature , 70F; barometer , standard. The air cooled at
constant pressure to 50F. Using psychrometric chart, calculate:
(A) the water vapor condensed
Ans: 32 grains
(B) the heat rejected
Ans: 13.7 Btu/lbda
(14.1) A heat exchanger receives oil having specific heat of 0.45
Btu/lb and a temperature of 160F at the rate of 40,000 lb/hr. Fresh
water at an initial temperature of 160F at the rate of 40,000 lb/hr.
Fresh water at an initial temperature of 60F flows through the
apparatus at the rate of 120,000 lb/hr. Assume unlimited heat
transfer area and calculate :
(A) the common temperature which the fluids will reach under
parallel flow, F. Ans: 73F
(B) the heat transferred by the heat exchanger
Ans:1,560,000 Btu/hr
(14.3) Calculate the heat transfer area required by a parallel flow oil
cooler which removes 1,524,000 Btu/hr from the oil while cooling it
from 160F at inlet to 75F at discharge, when the cooling water
temperatures are 60F and 72.7F, respectively. U for the heater is
52 Btu/hr-ft2.F.
Ans: 1132 ft2
(13.1) Calculate the rate of heat flow, in Btu/hr, through a 10-in wall
of solid concrete which is 20 ft long by 8 ft high. The thermal
conductivity of concrete is 0.10 Btu/hr.ft.F, the external surface
temperature is 5F and the internal surface temperature is 45F.
Ans: 7680 Btu/hr
(13.2) For the wall of the previous problem , the outside film
coefficient is 6.0 and the inside is 1.5 Btu/hr.ft 2.F, the outside air
temperature is - 3F and the inside air temperature is 77F. Calculate
(A) the overall heat transfer coefficient, Btu/hr.ft 2.F
Ans: 0.60 Btu/hr.ft2
(B) the rate of heat flow through the wall
Ans: 7680 Btu/hr
(12.1) In a simple impulse stage, the blade speed is 150 m/s and the
nozzle angle is 18 degrees. The velocity of the steam leaving the
nozzle is 320 m/s. The bucket entrance and exit angles are both 33
degrees. The bucket velocity coefficient is 0.89. Find the:
(A) relative velocity of entering the buckets;
Ans: 184 m/s
(B) relative velocity of leaving the buckets
Ans: 163.8 m/s
(C) total change of velocity relative to and in the direction of motion
of the buckets;
Ans: 291.7 m/s
(D) the bucket work, kJ/kg
Ans: 43.75 kJ/kg
(E) the absolute exit velocity
Ans: 90.1 m/s
(12.2) Steam enters the buckets of a simple impulse wheel at an
absolute velocity of 1200 ft/s and leaves with an absolute velocity of
245 ft/s. The relative entering velocity is 640 ft/s, and the relative exit
velocity if 540 ft/s. Determine:
(A) available energy to the buckets
Ans: 22,360 ft-lbf/lb
(B) the bucket loss , ft-lb/lb
Ans: 1830 ft-lb/lb
(C) the unused kinetic energy in the leaving jet Ans: 930 ft-lb/ft
(D) the bucket work
Ans: 19,600 ft-lb/lb and 25.2 Btu/lb
(E) the diagram efficiency
Ans: 87.7%
(F) Power developed by the bucket s if steam is supplied at the
rate of 5000 lb/hr
Ans: 49.5 hp
(12.3) Steam enters the nozzles of a simple stage with negligible
velocity of 190 psia and 500 F. The velocity of the steam leaving the
nozzles is 1175 ft/s, and the stage pressure is 140 psia. The steam
leaves the bucket s with an absolute velocity of 300 ft/s and an
enthalpy of 1245.5 Btu/lb. Assume the stage efficiency is equal to the
nozzle-bucket efficiency and that there is negligible velocity carryover
to the next stage. Find :
(A) the available energy to the stage, Btu/lb
Ans: 29.8 Btu/lb
(B) the nozzle efficiency
Ans: 92.5%
(C) the nozzle reheat
Ans: 2.2 Btu/lb
(C) the blade reheat
Ans: 3.2 Btu/lb
(D) the exit reheat
Ans: 1.8 Btu/lb
(E) the enthalpy of steam entering the next stage Ans: 1247.3 Btu/lb
(F) the stage work
Ans: 1247.3 Btu/lb
(G) the diagram efficiency of the buckets
Ans: 82%
(H) the nozzle-bucket efficiency.
Ans: 75.8%
(12.5) In a 50% reaction stage, the carryover velocity from the
preceding stage is 55 m/s, the stage available energy is 50 kJ/kg, the
fixed blade (nozzle) efficiency is 0.94 and the fixed blade velocity
coefficient is 0.90. Find the velocity of the steam leaving the fixed
blades, m/s.
Answer: 222.4 m/s
(12.6) In the turbine stage with 30% reaction, the steam enters the
moving blades with relative velocity of 350 ft/s , the stage available
energy is 25 Btu/lb, the reactive effectiveness is 0.90 and the moving
blade velocity coefficient is 0.92. Find the relative velocity of the
steam leaving the moving blades, ft/s.
Answer: 665 m/s
(12.7) Steam enters a 50% reaction stage in a low pressure turbine
with negligible carryover velocity at 12 psia and a moisture content of
0.06. The steam leaves the fixed blades with a velocity of 670 ft/s.
The relative velocities entering and leaving the moving blades are
230 ft/s and 670 ft/s respectively. The steam at exit from the moving
blades has an absolute velocity of 230 ft/s and a pressure of 9.0 psia.
Assume the combined blade efficiency and the stage efficiency are
equal and find the
(A) stage work
Ans: 15.8 Btu/lb
(B) the available energy to the stage Ans: 19.2 Btu/lb
(C) the combined blade efficiency
Ans: 82.3%
(12.8) The rotative speed of a high pressure turbine at full power is
6000 rpm. The first stage is of the simple impulse configuration, and
the nozzles receive steam at 900 pisa and 900F with negligible
velocity. The pitch diameter of the first stage buckets is 30 inches
and the nozzle angle is 17 degrees. Find these conditions
(A) Blade speed
Ans: 785 ft/s
(B) The ideal blade speed-steam speed ratio
Ans: 0.478
(C) The corresponding absolute (“sprouting”)
Velocity of the steam leaving the nozzles
Ans: 1642 ft/s
(D) The enthalpy drop across the nozzles assuming an
isentropic process.
Ans: 53.8 Btu/lb
(E) The entropy and enthalpy of the steam leaving the nozzles
and its estimated pressure and temperature from the
Mollier Chart.
Ans: S = 1.6257 Btu/lb
Ans: h = 1398.1 Btu/lb
Ans: P = 620 psia
Ans: T = 790F
(8.15) A split-shaft gas turbine has its power unit receiving 140 lb/s
combustion products at 45 psia and 1600 R. Exhaust temperature
and pressure for the power turbine are 1245R and 0.20 in Hg
gauge. Barometric pressure is 29 in. Hg absolute, while the ambient
temperature is 59F. Using air tables determine:
(A) isentropic turbine outlet temperature, R Ans: 1194R
(B) the isentropic turbine work, Btu/lb
Ans: 106 Btu/lb
(C) the actual turbine work, Btu/lb
Ans: 92.9 Btu/lb
(D) the isentropic turbine efficiency, %
Ans: 87.7%
(E) the turbine internal power, hp
Ans: 18,400 hp
(8.16) A split-shaft gas turbine has its power turbine supplied with
150 lb/sec of gas at 50 psia and 1100 F and exhausts the gas from
the power turbine at 16 psia and 800F, respectively. The power
turbine exhaust is led through a counter-flow regenerator where the
gas temperature is lowered another 40F. The compressor pressure
ratio is 12, and the inlet conditions are 14.6 psia and 60F.
Compressor discharge temperature is 740F. Using air tables
calculate:
(A) The power turbine efficiency %
Ans: 74.5%
(B) The compressed air temperature leaving the
regenerator, F
Ans: 780F
(C) the regenerator effectiveness, Btu/lb Ans: 66.7%
(8.17) A Brayton cycle aircraft gas turbine engine has an axial flow
compressor which provides a pressure ratio of 10 to 1. Material
design conditions limit the temperature of the working substance
entering the turbine to 2200F. The engine is designed to handle 90
lb of air /sec at static conditions and 120 lb/sec at a flight speed of
400 knots when sea level ambient pressure and temperature are 15
psia and 530R, respectively. Estimate:
(A) the sea level static thrust, lbf
Ans: 7300 lbf
(B) the thrust , lbf
Ans: 7559 lbf
(C) the propulsive efficiency, % for a flight speed of 400 knots
at sea level . (1 knot = 1.69 ft/s)
Ans: 40%
(E)
(8.17) A Brayton cycle aircraft gas turbine engine has an axial flow
compressor which provides a pressure ratio of 12 to 1. Material
design conditions limit the temperature of the working substance
entering the turbine to 2000F. The engine is designed to handle 90
lb of air /sec at static conditions and 120 lb/sec at a flight speed of
500 knots when sea level ambient pressure and temperature are 15
psia and 530R, respectively. Estimate:
(A) the sea level static thrust, lbf
Ans: 6635 lbf
(B) the thrust , lbf
Ans: 6045 lbf
(C) the propulsive efficiency, %
Ans: 51%
(D) cycle thermal efficiency , % for a flight speed of 500 knots
at sea level . (1 knot = 1.69 ft/s)
Ans: 55.8%
Ans: 9180 lbf
Ans: 44.5%
hg - h2 '
m2 ' =
hfg

1174.4  1085.9
 9.58%
924.2
2 '= g  m2 'fg  8.518  0.0958(8.501)  7.704 ft 3 /lb
'
dexit 
4 M2

V
4(60.1)(7.704)
144 
 60    2727 
dexit = 0.720 in.
/2
When equipped for after burning, nozzle inlet temperature is limited
to 2500R. Assuming all stated performance parameters remain
constant, estimate:
(E)the thrust, lbf and
(F) the thermal efficiency, %
h2 '= h1 - KE = 1234.4 - 148.5 = 1085.9 Btu/lb
rexit
rthroat
L
  r - r
tan   = exit t
L
2
r r
0.362 - 0.25
L  exit t =
= 1.05 in.
0.1051
 
tan  
2
(f)
A converging-diverging nozzle receives steam at a pressure of 380
psia with temperature of 480F and expands it to a pressure of 50
psia. Assuming the velocity coefficient of 0.98 for the
supersaturated throat condition and an overall nozzle efficiency of
92 percent, calculate the following:
A.
B.
The actual throat velocity , ft/s
Ans: 1582 ft/s
The mass rate of flow for a throat diameter of 0.50 in
Ans: 60.1 lb/min
C. The actual kinetic energy available at the nozzle exit
Ans: 148.5 Btu/lb
D. The actual nozzle exit velocity, ft/sec
Ans: 2727 ft/sec
E. The required nozzle exit diameter, inches and
Ans: 0.720 in.
F. The length of the divergent section of the nozzle, in inches, for
an included angle of 12 degrees between nozzle sides.
Ans: 1.05 in.

From: Steam table: At P = 380 psia and temperature of 480F.
h1 = 1234.4 Btu/lb
S1 = 1.5220 Btu/lb.R
Pt = Pc = 0.55P1 = 0.55(380) = 209 psia
st(ideal) = s1 ; m =
sg - s t
sfg
1.5427 - 1.5220
=
= 2.08%
0.9941
ht(ideal)  hg  mhfg  1199.9   0.0208  840.3  = 1182.4 Btu/lb
(A) Vt = Cn Videal  223.38Cn h1  htideal
= 223.8  0.98  1234.4  1182.4 = 1582 ft/s
(B) enozzle '
=
cn 
2
 (0.98)2  0.96
h1 - ht actual = en'  h1 - hideal   0.9604(1234.4  1182.4)
= 49.9Btu/lb
hactual = h1 - 49.9 = 1234.4 - 49.9 = 1184.5 Btu/lb
mtactual =
hg - htactual 1199.9-1184.5
=
=1.83%
hfg
840.3
actual = g  mtfg  2.194  0.0183  2.176 
= 2.154 ft 3 / lb
0.1964 2
At =
ft
144
A V 0.1964(1582)  60 
m' = t t =
= 60.1 lb/min
t
(144)(2.154)
(C) s2ideal  s1; m2 
sg  s2ideal
sfg

1.6589  1.5220
 10.97%
1.2476
h2ideal = hg - m2hfg = 1174.4 - 0.1097(92.4) = 1073.0 Btu/lb
KEactual = h1- h2' = en (h1  h2 ' ) = 0.92(1234.4-1073) = 148.5 Btu/lb
(D) Vactual = V2 '= 223.8 h1 - h2'
= 223.8 148.5 = 2727
ft
2
11-1 Air enters an ideal converging-diverging at a pressure of 73.5
psia with temperature of 1400F and negligible approach velocity/.
For isentropic expansion to an exit pressure of 14.7 psia, calculate:
(A) The temperature of the air leaving the nozzle F
Ans: 714F
(B) The kinetic energy of the air leaving the nozzle, Btu/lb
Ans: 164.6 Btu/lb
(C) The velocity of the air leaving the nozzle.
Ans: 2870 ft/s
11-2 The pressure of the air entering the ideal convergent nozzle is
73.5 psia, the temperature is 1400F and the velocity of approach is
negligible. The nozzle discharges against a pressure of 14.7 psia.
What is the nozzle exit velocity when expansion in the nozzle exit is
isentropic, ft./sec?
Answer: 1930 ft/sec
11-3 Air enters a convergent-divergent nozzle having stagnation
conditions of 65C and 285 kPa. The Mach number of the throat is
0.77. For isentropic expansion through the nozzle, calculate:
(A) The throat pressure, kPa
(B) The throat temperature, K
(C) The throat velocity m/s
Ans: 192.5 kPa
Ans: 302.2 K
Ans: 268.3 m/s
11-6 Air enters a diffuser of a jet engine with a velocity of 1800 ft/s
relative to the aircraft. The intake pressure is 1.05 psia and intake
temperature is - 70F. Assuming isentropic compression in the
diffuser, calculate:
(A) The sonic velocity at the inlet condition Ans: 968 ft/s
(B) The inlet mach number
Ans: 1.86
(C) Stagnation temperature
Ans: 659.8R
(D) The stagnation pressure
Ans: 6.61 psia
5-4 Steam leaves the boiler at 600 psia and 750F at the rate of
75,000 lb/hr through the main steam line, which has a cross sectional
area of 0.322 ft2. Find the velocity of the steam in the lin, ft/s
Answer: 73.3 ft/s
5-5 Steams leaves the boiler at 6550 kPa absolute and 510 C at the
rate of 45,400 kg/hr through the main steam line, which has cross
sectional area of 0.030 m2. Determine the velocity of the steam in the
line ., m/sec. The specific volume of the steam is 0.0525 m3/kg.
Answer: 22.1 m/sec
5-6 An air compressor takes in 50 ft3/ min of air at 14.7 psia and
60F. The air is discharged at 100 psia and 260F. Find:
(A) Mass flow rate
Ans: 3.82 lb/min
(B) Volume flow rate at the discharge
Ans: 10.18 ft3/min
5-7 Steam enters the first stage nozzles of a large turbine with
negligible velocity at a pressure of 540 psia and a temperature of
800F. The pressure at the nozzle exit is 220 psia. If the process is
isentropic ; find:
(A) The final enthalpy
Ans: 1302 Btu/lbs
(B) The kinetic energy at exit
Ans: 1081 Btu/lb
(C) The velocity at the exit
Ans: 2327 ft/s
A mass of 1.0 kg is moving at a velocity of 5 m/s. Determine the
kinetic enegy on a unit mass basis:
A.
B.
m'5 =
1000 1164.3  
(104,000)  218.9   (100,000)  48.1   5000  68.1
1178.7 - 218.9 
m'5 = 19,566 lb/hr
In SI Unit, J/kg
In Engineer’s unit, ft-lb
Inserting this result into the mass balance equation gives:
Solution:
(A) KE =
m’4 = m’1 + m’3 + m’5 – m’2
V2
2gc
m’4 = 100,000 + 5000 + 19,566 – 1000
1  1 N.s2  2 m2
= 
  5  2 = 12.5 J/kg
2  kg.m 
s
m’4 = 123,566 lb/hr

J   1.0 kg   778 ft-lbf 
(B) KE =  12.5  
kg   2.205 lb   1055 J 

ft-lb f
= 4.182
lbm
A Rankine regenerative steam cycle employs two stages of steam
extraction for feed water heating. Boiler pressure and temperature
are 1200 psia and 1050 F respectively. Saturated steam at
condenser pressure has a temperature of 79 F.
Steam has a value of internal energy u = 1171.9 Btu/lb at a
temperature of 500F and pressure of 150 psia. Determine the
values of the above in their SI equivalences.
Solution:
Internal Energy (SI)
 2.326 kJ/kg 
3
u = 1171.9 Btu/lb 
 = 2.726 x 10 kJ/kg
 1.0 Btu/lb 
Pressure (SI)
 101.325 kPa 
3
P = 150 psi 
 = 1.034x10 kPa
 14.7 psia 
The Universal Gas Constant:
ft-lbf
J
R = 1545
 8314
lbmol R
kgmol .K
Dearating feed water heater similar to the fig shown operates at 30
psia shell pressure with the following flow rates and properties:
No
1
2
3
4
5
ITEM
Condensate (liquid)
Vent (Vapor)
Drains (liquid)
Feed ( liquid)
Exhaust Steam
Flow Rate
(lb/hr)
100,000
1,000
5,000
?
?
Temp.
(F)
80.00
250.34
100.00
250.34
280.00
Enthalpy
49.10
1164.3
68.1
218.9
1178.7
Determine the exhaust steam flow rate required to operate the heater
under these conditions. What quantity of boiler feed will be available
from the heater (lb/hr) ?
  m'h in   m'h out  Energy Balance
m'1h1 + m'3h3 + m'5h5 = m'2h2 + m'4h4
  m' in   m' out  Mass Balance
m'1 + m'3 + m'5 = m'2 + m'4
Since m’4 is a function of m’5, a direct solution is achieved by solving
the mass balance equation explicitly for m’4 and then substituting that
the mass balance equation explicitly for m’5 as the only unknown, its
magnitude may be determined:
m’4 = m’1 + m’3 + m’5 – m’2
m'1h1 + m'3h3 + m'5h5 = m'2h2 + (m'1+m'3 +m'5 - m'2 )h4
m'5  h5 - h4  = m'2h2 + (m'1+m'3 - m'2 )h4  m'1h1  m'3h3
m5 
m'2h2  (m'1+m'3 - m'2 )h4  m'1h1  m'3h3
h5 - h4 
Calculate the following :
(A) The optimum extraction pressures to the nearest pound per
square inch.
(B) The mass of steam removed from the turbine at each
extraction point, pounds per pound of throttled steam
(C) Pump work (Btu/lb)
(D) The heat supplied to the cycle, Btu/lb
(E) The heat rejected by the cycle, Btu/lb
(F) The turbine work, Btu/lb throttled steam
(G) The net cycle work, Btu/lb throttle steam
(H) The thermal efficiency of the cycle, %
In a PWR nuclear plant, the primary loop coolant water, pressurized
to 2000 psia, enters the reactor at 500F and leaves at 600F. It
then enters the steam generator (boiler) section at the same
temperature where energy in the form of heat is transferred to the
secondary loop. The secondary loop operates on simple Rankine
cycle without regenerative feed heating or any external heat supply.
Steam leaves the boiler at 420 psia with a moisture content of 1 %
and enters the turbine at the same conditions. The condenser
pressure is 2 psia. The pressure of the water leaving the feed pump
and entering the steam generatot is 500 psia. The flow rate in the
steam plant (secondary ) side is 500,000 lb/hr. Assuming no losses
other than the pressure drop in the steam generator given above,
find:
(A) The required heat exchange rate in the steam generator,
Btu/hr,
(B) The primary coolant flow rate necessary to satisfy the heat
transfer rate of part (A)
(C) The heat supplied , Btu/lb
(D) The heat rejected, Btu/lb
(E) The thermal efficiency, %


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
A large turbine receives steam at the throttle at 560 psia and 800F
at the rate of 100 lb/sec. The pressure after the throttle is 500 psia.
At the exhaust flange the absolute pressure is 1.5 in of mercury, the
steam velocity is 1000 ft/sec and the moisture content is 10% . Find:

A.
B.
C.
D.
E.
F.
G.
H.
The availability energy to the turbine, Btu/lb
The throttling loss, Btu/lb
The leaving loss, Btu/lb
The enthalpy at the exhaust point, Btu/lb
The internal turbine work, Btu/lb
The internal engine efficiency
Other internal losses not
accounted for by (b) and (c) above, Btu/lb
The internal power developed, hp








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
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
A Rankine regenerative steam cycle employs two stages of steam
extraction for feed water heating. Boiler pressure and temperature
are 1200 psia and 1050F, respectively. Saturated steam at
condenser pressure has a temperature of 79F. Sketch the cycle on
T-S coordinates and calculate:
A 60-kW auxiliary generator operates with dry saturated steam at
200 psi. It has three nozzles; one is always open, and the other two
are fitted with manually operated nozzle control valves. With both
nozzle control valves closed, the capacity of the machine is 30 kW.
Speed control is obtained by use of a throttle valve actuated by the
mechanical governor. At quarter load, the pressure in the steam
chest is 115 psia when both hand valves are closed. At the same
load with both hand valves open, the chest pressure becomes 62
psia referred to an exhaust pressure of 2.4 psia, find for this load
condition:

A. The throttling loss which must be accepted for control purposes
B. The additional and unnecessary throttling loss if both hand nozzle
valves are inadvertently opened.











A.
B.
C.
D.
E.
F.
G.
H.
The optimum extraction pressure to the neared pound per
square inch
The mass of steam removed from the turbine at each
extraction point, pounds of throttle steam
Pump work, Btu/lb
The heat supplied to the cycle, Btu/lb
The heat rejected by the cycle , Btu/lb throttle steam
The turbine work, Btu/lb throttle steam
The net cycle work, Btu/lb throttle steam
The thermal efficiency on the cycle, %

Turbine A receives steam at the throttle at 850 psia and 940F and
exhausts to the condenser at 0.70 psia; the throttle steam flow is
164,000 lb/hr. Turbine B receives steam at 620 psia and 900F,
exhausting at 1.0 psia, and the throttle steam flow is 187,000 lb/hr.
Under the stated conditions, both turbines deliver 30,000 shaft
horsepower with mechanical efficiency of 96%. Find:
A. The water rate, the heat rate , the shaft engine efficiency and the
internal engine efficiency for the turbine A
B. The same quantities for turbine B










1. A refrigerating plant for an air-conditioning system removes
10,000 Btu/min from the air. The plant circulates 170 lb of
refrigerant/min and the internal power delivered by its
compressor is 60 horsepower. The refrigerant evaporation
temperature is 40F, and its condensation temperature is
100·F. Calculate :
From: Freon table Attached :Appendix
P1= P2 = 26.51 psia
P3 = P4 = 107.9 psia
t 2 = 14F

A. the capacity of the plant, tons;
B. the refrigerating effect, Btuflb;
C. the coefficient of performance of the actual plant; and
D . the coefficient of performance of the equivalent Carnot cycle.
B. m' =

C. Wnet
heat absorbed(Btu/min)
10,000
=
= 50 tons
200 Btu/min-ton
200
heat absorbed(Btu/min)
10, 000
B. RE =
=
= 58.8 Btu/lb
mass of Refrigerant (lb/min)
170
D. COP =
J x RE
58.8
COP =
=
= 3.93
Wnet
14.97
T
(40  460)
D. COPcarnot = L 
 8.33
TH -TL (100  460)  (40  460)
2. A r e f r i g e r a t i n g plant circulates 23 lb Freon-12 per
minute and is assumed to operate on a cycle similar to that
of the figure . The pressure in the evaporator coil is 50 psia,
the temperature of the Freon-12 entering the compressor is
50" F, the pressure in the condenser is 120 psia and the
temperature of the liquid refrigerant entering the
expansion valve is 86F. Calculate or determine:
A. the evaporation temperature, • F;
B. the condensation temperature, F;
C. the refrigerating effect, Btu/lb;
D. the capacity of the plant, tons;
E. the power required to compress the Freon-12, hp; and
F. the coefficient of performance.
.
heat absorbed 200xCapacity 50(200)


 183.6
RE
RE
54.48
= h3 - h2 = 91.13 - 80.04 = 11.09 Btu/lb
h -h
RE
54.48
= 2 1 =
= 4.91
Wnet
h3 - h 2
11.09
m'freon-12   Wnet 
E. Wnet/ton =
x 60
2545 x capacity
F. QR   m'freon-12  ( h3 - h4 ) = 183.6 (91.13 - 25.56) = 12,040 Btu/min
m'freon-12  2   where : 
G. PD =
Capacity
183.6  1.516 
PD =
2
 1.516 ft 3 /lb
= 5.57 ft 3 /min.ton
50
4. An air compression refrigeration system is to have an air
pressure of 100 psia in the brine tank and an allowable air
temperature increase of 60F. For standard vapor
compression cycle temperatures of 77F entering the
expansion cylinder and 14 F entering the compression
cylinder, calculate:
A. the coefficient of performance;
B. the mass of air circulated per ton of refrigeration;
C. the required piston displacement of the compressor
cylinder, neglecting volumetric efficiency.
P

P=C

S=C
S=
S=C


T
P=C 
P

183.6 x 11.09 x 60
 0.96
2545 x 50
=
E
Wnet
(Hp)(2545 Btu/Hp-min)
60 x2545

=
= 14.97Btu/min 
J
mass of Refrigerant x 60
170(60)
C.
A. RE = h2 - h1 = 80.04 - 25.56 = 54.48 Btu/lb
PRESSUR
A. Capacity =
t 4 = 77F
S2  S3  0.17317 Btu/lb ; h2 = 80.04 and h3 = 91.13 Btu/lb

P=C
C
P=

C


P=C
V



h

h4 = h1 = 27.72 Btu/lb
h2 = 84.24 Btu/lb
S2 =S3 =0.17187
h3 =91.31 Btu/lb
See Attached Appendix:
A. Evaporation temperature = 38.3F
B. Condensation temperature = 93.4 F
C. RE = h2 – h1 = 84.24 – 27.72 = 56.52 Btu/lb
m'(RE)
23(56.52)
=
= 6.5 tons
200
200
m'  h3 - h2 
23  56.52 
E. Wnet =
=
= 3.83 Hp
42.42
42.42
h -h
RE
84.24 - 27.72
F. COP =
= 2 1 =
= 7.99
Wnet
h3 - h2
91.31 - 84.24
D. Capacity ' =
3. A cooling plant using Freon-12 as the refrigerant is to have
a capacity of 50 tons when operating on the refrigerant
rating cycle. For this ideal plant cycle determine:
D.
E.
F.
G.
.
T4 = 77 + 460 = 537R
T2 = 14 + 460 = 474R
T2 - T1 = 60F
T1 = 474 - 60 = 414R
Since: Processes 3-4 and 1-2 are constant pressure:
Use:Attached Figure
A.
B.
C.
S
the refrigerating effect, Btu/lb;
the rate of Freon-12 circulation, lb/min;
the net work required per:pound of Freon-12 circulated,
Btu/lb;
the coefficient of performance;
the power required per ton of refrigeration, hp/ton;
the heat rejected by the condenser, Btu/min; and
the compressor
piston displacement, ft3/min·ton of
refrigeration.
P3 P4

and
P2 P1
 P4 
 
 P1 
k 1
k
=
T
T4
= 3
T1
T2
T 
 537 
T3 = T2  4  = 474 
 = 615°R
 414 
 T1 
Cp  T2 -T1 
60
A. COP =
=
= 3.33
Cp  T3 -T4  - Cp  T2 - T1   615-537  - 60
B. RE - Cp  T2 -T1  = 0.24(60) = 14.4 Btu/lb
200 Btu/min.ton
=13.9 lb/min.ton
14.4 Btu/lb
m'RT2
C. Piston Displacement = V2 =
P2
m'air =

13.9(53.3)(474)
= 24.4 ft 3 /min.ton
144 x 100
5. A simplified line diagram and TS plot for one section of a
cooling system for a large aircraft are shown below. When it
is used for cooling on the ground, the following Fahrenheit
temperatures are experienced at the numbered points on the
diagrams:
T
1
3
8. A refrigerating plant for an air-conditioning system is to have a
capacity of 10 tons and a coefficient of performance of 2.50
when operating With a refrigerating effect of 61.4 Btu/lb of
refrigerant. Calculate :
A. the refrigerant flow rate, lb/min;
B. the work done on the refrigerant by the compressor, Btu/lb
C. the compressor internal horsepower, hp; and
D. the rate of heat rejection from the system, Btu/min.
2
5
S
t (• F)
1
2
3
4
5
342
142
252
145
35
For a situation where the air flow rate through the system
is 65 lb/min, specific heat of the air is assumed constant
and the compressor and expander processes are assumed
isentropic, calculate:
(A) the heat transferred to the atmospheric air supply,
Btu/min;
(B) the power developed by the expander, hp; and
(C) the heat transferred from auxiliary power unit compressor
bleed, expressed in tons of refrigeration.

A. Q1-4 =m'Cp  t1 - t 2    t 3 - t 4  
= 65(0.24)  342 - 142    252 - 145  
= 4789.2 Btu/min
m'  h4 - h5 
65(0.24)(145 - 35)
B. Wnet =
=
= 40.45 hp
42.42
42.42
Q
4789.2
C. REFRIG = 1-4 =
= 23.95 tons
200
200
6 . An ideal Freon-12 refrigerating system has a capacity of 50
tons.
The condenser pressure is 180 psia, and the Freon12 temperature leaving the condenser is 120° F.
The
pressure leaving the expansion valve is 44 psia, and the
temperature of the Freon-12 leaving the succeeding coil is 40°
F. Circulating water enters the condenser at a temperature
of 100° F and leaves it at 1 1 5  F. Determine:
the mass of Freon-12 circulated, lb/hr;
the compressor power for isentropic compression, Btu/hr
the heat capacity of the system, Btu/hr
the mass of water circulated through the condenser and
heating system, lb/hr
E. the useful heat furnished per Btu of compressor work (heating
performance ratio)
A.
B.
C.
D.

See Freon-12 table s attached Appendix/Figure:
h1 = h4 = 36.16 Btu/lb
h2 = 83.03 Btu/lb
S2 = S3 0.17142 Btu/lb
h3 = 94.31 Btu/lb
12,000 x capacity
A. m =
h2 -h1
'
12,000(50)
= 12,800lb/hr
83.03-36.16
B. Compressor Power = m'Wnet '
=
= m'  h3 - h2  = 12,800(94.31-83.03)
= 144,400 Btu/hr
C. Heating Effect : QR = h3 - h4
= 94.31 - 36.16 = 58.15 Btu/lb
Heat Capacity = HC = m'(RE)
= 12,800(58.15) = 744,300 Btu/hr
HC
744,300
D. mcirculating water =
=
= 49,620 lb/hr
C  t out - tin 
1(115  100)
E. Heating Performance =
A. 50 percent;
B. 25 percent; and
C. 12.5 percent.
Answers: (A) 4.71 hp; (B) 1.57 hp; (C) 0.673 hp
4
point
7. Calculate the horsepower required per ton of refrigeration
produced by the reversal of a Carnot cycle having a thermal
efficiency of
HC
744, 300

 5.15
Compressor Power 144, 400
Answers: (A) 32.6lb/min; (B) 24.6 Btu/lb; (C) 18.9 hp; (D) 2800
Btu/min
9. A refrigeration system has a capacity of 25 tons and rejects heat
at the rate of 6560 Btu/min. Calculate:
(A) the rate of heat absorption by the refrigerant, Btu/min;
(B) the power required as input to the system, Btu/min; and
(C) the coefficient of performance for the system.
Answers: (A) 5,000 Btu/min; (B) 1560 Btu/min; (C) 3.2
10. A modified Rankine refrigerating cycle operates with an
evaporator pressure of 21.4 psia and a condenser pressure of
141 psia. Refrigerant is Freon-12 circulating through the
system at 30 lb/min. Liquid refrigerant at 141 psia and 100"
F enters the expansion valve, and surerheated vapor at 21.4
psia and 5" F enters the compressor.
Calculate:
(A) the refrigerating effect, Btu/lb;
(B) the plant capacity in tons of refrigeration;
(C) the power required to compress the refrigerant, hp; and
(D)
the plant coefficient of performance.
Answers: (A) 47.86 Btu/lb; (B) 7.18 tons; (C) 10.66 hp; (D) 3.17
11. In an ideal (reversed Joule cycle) air-refrigerating system
the temperature of the air entering the compression
cylinder is 50F, the temperature entering the after-cooler
is 160° F. and the temperature entering the brine tank is
0F. Calculate:
(A) the temperature of the air leaving the after-cooler
(B) the coefficient of performance;
(C) the mass of air which must be circulated per ton of
refrigeration, lb/min.
Answers: (A) 99F (B) 4.54 (C) 16.7 lb/min
CNS 04
MDSP/MESL
(A) p1  14.7 psia; T1  50  460  510 R; M = 900 lb/min
1. In an ideal Brayton cycle, air enters the compressor at 15 psia and
75˚ F. The temperature of the air at turbine inlet is 1600˚ F. For
maximum theoretical net work, find
(A) the temperature of the air leaving the compressor, F;
(B) the pressure ratio;
(C) the net work, Btu/lb; and
(D) the thermal efficiency.
p2  162 psia; T2a  648  460  1108R

p 
(C ) T2S  T1  2 
 p1 
If the end conditions remain unchanged but the pressure ratio is
increased to 15, find:
(E) the net work, Btu/lb, and (f) the thermal efficiency.
C 
S


 k 1 / k 


 162 
 510 

 14.7 
1.4 1 /1.4 
 1012.4


c p T2S  T1
wkS
1012.4  510


 0.84 or 84%
wka
1108  510
c p T2a  T1

(D) HPS  M c p (T1  T2S )
P1 = 15 psia
T1 = 75 +460 = 535 R
 k / k 1



(D) t max work   1  T1 / T3  1  535 / 2060  0.49  49%
(E ) rp  15
 k 1 / k 
 k 1 / k 
T2
  rp 
; T2  T1  rp 
T1
 1160 R
TT
535  2060
T4  1 3 
 950 R
T2
1160
wk
 c p T3  T4   c p T1  T2 
J
= 0.24(2060 - 950) + 0.24(535 -1160)
= 266.4 - 150.0 = 116.4 Btu/lb
wk 116.4
(f) t 

 0.539  53.9%
or
qs
216.0
1
 rp 
 k 1 / k 
 1

1
15
0.4 /1.4 

p1 = 105 kPa abs ; p2 = 1260 kPa abs
t1 = tA = tB = tE = 25˚ C
(A) best intercooler pressure:
p1  p2  105  1260  363.7 kPa abs
(B) work of ideal cycle:
 k 1 / k  

p 

Wk t  2c pT1 1   i 


 p1 
1.4 1 /1.4 

 363.7 
=  2 1.0048 25  273  1  


 105 


3. An axial flow compressor discharges 900 lb/min of air. Inlet
conditions are 14.7 psia and 50 F, while the actual discharge
conditions are 162 psia and 648˚ F. For the process:
(A) sketch the T-s diagram and then calculate
(B) the capacity, ft3/min
(C) the isentropic compression efficiency
(D) the isentropic compression power, hp
(E) the actual compression power, hp
    3045 hp (on)
S
(A) the discharge pressure,
(B) the discharge temperature (actual), and
(C) the work inpt per kg f air.

p 
(A) rp   2  ; p2  rp p1
 p1 
p2  3  100 kPa = 300 kPa abs
(B) discharge temperature:
P 
 2
T1
 P1 
T2S
 k 1 / k 

; T2S  298  3 


0.4 /1.4 
 407.9 K

wk a  c p T1  T2a  c p T1  T2S / CS
T2a 


T1 CS  1  T2S
C

C 
S
wk a 


c p T1  T2S
wkS

wka
c p T1  T2a

c p T1  T2S
C
298 0.7  1  407.9
0.7
S
(C) compressor work:
 455 K


  1.0048 298  407.9      157.8 kJ/kg
0.70
S
2.
5. A split shaft gas turbine has a power turbine rated at 15,000
internal horsepower. Typical operating conditions for the unit are:
compressor inlet 14.5 psia and 60˚ F; compressor discharge 174
psia and 716˚ F; compressor turbine inlet 171 psia and 1630˚ F;
and power turbine exhaust 14.8 psia and 760˚ F. The compressor
turbine at the above rating has an 85 percent isentropic turbine
efficiency. For these rated conditions,
(A) sketch the T-s diagram for the cycle, and then calculate:
(B) the compressor isentropic efficiency, %
(C) the compressor turbine discharge pressure and temperature,
psia, ˚R
(D) the power turbine isentropic efficiency, %
(E) the compressor turbine power output, hp
(F) the cycle thermal efficiency
    255.2 kJ/kg

42.42
4. A 2 kW centrifugal compressor operates with suction conditions of
100 kPa abs and 25˚ C. The pressure ratio for the unit is 3 and
C  0.70 Determine:
 53.9%
2. Determine :
(A) the best intercooler pressure and
(B) the work required per kilogram of air for an ideal two-stage
compressor operating between a suction pressure of 105 kPa abs
and a discharge pressure of 1260 kPa abs. The suction
temperature is 25˚ C.
1.
    2558 hp (on)

900 0.24 510  1108 
 10.6
= 0.24 2060 + 535 -2 535  2060  118.9 Btu/lb
t  1 

(E) HPa  M c p T1  T2a
1.4 / 0.4 

T2  535(15)
42.42


T 
p
 1050 
(B) rp  2   2 


p1  T1 
 535 
Wk  max 
(C )
=c p T3  T1  2 T1T3
J
0.4 /1.4
900  0.24 510  1012.4 

T3 = 1600 + 460 = 2060 R
(A) T2  T1T3  535  2060  1050 R = 590 F
p1 
MRT1  900  53.3 510 

 11,557 ft 3 / min
p1
14.7 144 

(B) V 
fuel
2
Combustion
Chamber
3’
3
Power
Turbine
compressor
1
From
atmosphere
Compressor
Turbine
Split Shaft Open Brayton Cycle
(A) TS Diagram
4
to
atmosphere
Output
6. A split-shaft gas turbinehas a power turbine rated at 15,000
internal horsepower. Typical operating conditions for the unit are:
compressor inlet, 14.5 psia and 60˚ F; compressor discharge, 174
psia and 716˚ F; compressor turbine inlet, 171 psia and 1630˚ F;
and power turbine exhaust, 14.8 psia and 760˚ F. The compressor
turbine has an 85% isentropic turbine efficiency at the above
power rating.
(A) Sketch the T-s diagram for the cycle; then, using the Air
Tables, calculate:
(B) the compressor isentropic efficiency, %
(C) the compressor turbine discharge pressure and temperature,
psia, ˚R
(D) the power turbine isentropic efficiency, %
(E) the compressor turbine power output, hp
(F) the cycle thermal efficiency, %


p1 = 14.5 psia
p2 = 174 psia
p3 = 171 psia
p4 = 14.8 psia
T1 = 60 + 460 = 520˚ R
T2a = 716 + 460 = 1176˚ R
T3 = 1630 + 460 = 2090˚ R
T4a = 760 + 460 = 1220˚ R
p1 = 14.5 psia
p2 = 174 psia
p3 = 171 psia
p4 = 14.8 psia
T1 = 60 + 460 = 520˚ R
T2a = 716 + 460 = 1176˚ R
T3 = 1630 + 460 = 2090˚ R
T4a = 760 + 460 = 1220˚ R
(A) TS - Diagram
p 
(B) T2S  T1  2 
 p1 
k 1
k
 174 
 520 

 14.5 
0.4
1.4
 1056.6 R
isentropic compressor work T2S  T1

actual copressor work
T2a  T1
C 
S
1057.6 - 520
 0.8195  82.0%
1176 - 520
=


(C ) W k12   W k33a '





M c p T1  T2a   M c p T3  T3a '

T3a '  T1  T2a  T3  520  1176  2090  1434 R
t
actual turbine work

isentropic turbine work
T3  T3 '
 0.85
T3  T3S '
S
(B) From the Gas Tables:
 T3  T3a ' 
T3'S  T3  

 0.85 
 2090  1434 
 2090  
  1318 R
0.85


State
1
1.4
 1318  0.4
p3' = 171 
  34.06 psia
 2090 
(D) For the power turbine:
State
2S
k 1
0.4
P  k
 14.8 1.4
T4S  T3a '  4 
 1434 
  1130 R
P
'
 34.06 
 3 
actual expansion work
 tS 
isentropic expansion work
T3 ' T4a
1434  1220
 tS  a

 70.4%
T3a '  T4S
1434  1130
C 
S
State
2a
T
1048



 12,389 lb/min

Wk 



P = W k  M c p T2a  T1
P=
42.42
wk33a ' wk3 4 ' wk12a
t =
a a
1434 - 1220 
 2090 - 1176 
vr
26.62
pr
22.28
u
204.59
vr
19.556

T
2090
h
529.75
pr
208.06
u
386.48
vr
3.721

M h  T1  h  T2a     M h  T3  h  T3'a  
 45,980 hp (by) or
h  T3'a 
 h  T1  h  T2a  h  T3'a 
= 124.27 - 285.20 + 529.75 = 368.82
12,389 0.240 1176  520 
q23
u
181.11
252.95  124.27
 0.7996  80.0%
285.20  124.27

 45,980 hp (by)
(F) Thermal efficiency:
t 
h
285.20

42.42

pr
14.584
a
State
3
12,389 0.240 2090  1434 

vr
158.58
(C) W k12   W k33'
For the compressor turbine:

u
88.62
or
15,000  42.42 
0.240  1434  1220 
W k  M c p T3  T3a '
h
252.95
T
1176
S
W k  M c p (T3a ' T4a )
M 
pr
1.2147
h < T2S > - h <T1 
isentropic compression work

actual compression work
h  T2a  h  T1 
C 
(E ) For the power turbine:

h
124.27
p 
 174 
pr2  pr1  2   1.2147 
  14.576
 14.5 
 p1 S
k
 T3 '  k 1
p3 '  p3  S 
 T3 

T
520

wk3 4 '
a a
q23

 0.234  23.4%

c T


c p T3a ' T4a
p
3
 T2a
State
3’a
T
1499
h
368.91
pr
55.71
u
266.14
vr
9.967

T3a '  1499 R
t 
S
h < T3 > - h <T3a ' 

Tamb= Ta = 530˚ R
T3 = 2000˚ R
pa = 15 psia
actual expansion work
isentropic expansion work
 0.85
h  T3  h  T3S ' 
h  T3S ' 
M = 116 lb/sec
Va = 850 ft/sec
h  T3  h  T3a ' 
 h  T3  
0.85
529.75  368.91
 529.75 
0.85
 340.53
T
1391
State
3’S
h
340.55
pr
41.84
u
245.20
vr
12.315
 pr 3 ' 
 41.84 
p3 '  p3  S   171 
  34.39 psia
 208.06 
 pr 3 
p 
 14.8 
(D) pr4  pr 3 '  4   55.71 
  23.98
 34.39 
 p3' S
a
State
4a
State
4S
t 
S

T
h
pr
u
vr
1220
296.1
25.53
212.78
17.700
1200
291.30
24.01
209.05
18.514
h < T3a ' > - h <T4a 
actual expansion work

isentropic expansion work h  T3a '  h  T4S 
368.91  296.41
 0.934  93.4%
368.91  291.30
Ram compression:
V2
V2
ha  a  h1  1  h1  0
2gC J
2gC J
(E) For the power turbine:



W k  M h  T3a '  h  T4a 
15,000 42.42

M
368.91  296.41

or
 8,780 lb/min


W k  M h  T3  h  T3a ' 

M

or

k
or

W k  M h  T2a  h  T1 


p2  12p1  261.6 psia
42.42
p 
T2  T1  2 
 p1 
W k  33,300 hp (by)
a a
q23



wk3 4 '
0.4
 590 12 1.4  1200 R
Combustor:
a a
q23
h < T3a ' > - h <T4a 
h  T3  h  T2a 
368.91  296.41
 29.6%
529.75  285.2
7. A turbojet engine is equipped with a 12-stage, axial flow
compressor having a mechanical compression pressure ratio of 12
to 1. The engine is designed to handle 90 lb of air per second
under sea level static conditions and 116 lb of air per second at a
flight velocity of 850 ft/sec at sea level. Material design
considerations limit the temperature of the working substance
entering the turbine to 2000˚ R at 7950 rpm. The ambient
temperature and pressure are 15 psia and 530˚ R.
(A) Estimate the thrust, propulsive efficiency and thermal
efficiency of the engine, assuming all processes to be ideal and
the engine frictionless, for a flight velocity of 850 ft/sec at sea level.
(B) When the above engine is equipped for afterburning, the
nozzle inlet temperature is 2500˚ R. Assuming all of the stated
performance parameters remain constant, estimate the thrust with
afterburning for a flight velocity of 850 ft/sec at sea level.





k 1
k
h2 + q23 = h3
(F) Thermal Efficiency:
wk 33a ' wk3 4 ' wk12a
1.4
Compressor:
8780  285.20  124.27 

t 
 0.24 T1  530 
 T  k 1
 590  0.4
p1  pa  1 
 15 
 21.8 psia

T
 530 
 a


850 
2 32.2 778 
T1  590 R
8,780  529.75  368.91
 42.42 
M  33,300 hp (by)
Va2
 c p T1  Ta 
2gC J
2
For the compressor turbine:

Va2
 h1  ha
2gC J
q23 = h3 – h2 = cp(T3 – T2)
= 0.24 (2000 – 1200) = 192.0 Btu/lb
p3 = p2 = 261.6 psia
Turbine:
h3 = h4 +
wk 34
J
wk 34
= h3 - h4
J
wk(turbine)
wk(compressor)
=
J
J
thus; h2 - h1 = h3 -h4
Cp (T2 - T1 ) = Cp ( T3 -T4 )
T4  T3  (T2  T1 )
T4 = 2000 - (1200 - 590)  1390R
k
1.4
 T  k 1
 1390  0.4
P4 = P3  4 
 261.6 
 = 73.2 psia
 2000 
 T3 
Nozzle:
h4 
Vj2
V42
 hj 
2gC J
2gC J
Vj2
2gC J
P 
Tj  T4  j 
 P4 
 h4  hj = Cp (T4 -Tj )
k 1
k
Tj  883.7 R
Vj  2gc JC p (T4  T j )
 2(32.2)(778)(0.24)(1390  883.7)
Vj = 2467 ft/s
Thrust;
F t=
M
116
2467  850  = 5825 lbf
Vj  Va   32.2
gc
Propulsive Efficiency:
p 
2
2

 51.3%
Vj
 2467 
1
1

Va
 850 
Thermal Efficiency:


qR = Cp Tj - Ta = 0.24(883.7 - 530) = 84.9 Btu/lb
q -q
q
84.9
th = s R = 1 - R = 1 = 55.8%
qs
qs
192
Part (b)
k-1
0.4
 P'  k
 15 1.4
Tj '  T4'  j   2500  
  1589R
P
 73.2 
 4
Vj ' = 2gc JC p (T4 ' T j ')
 2(32.2)(778)(0.24)(2500  1589)
Vj ' = 3309 ft/s
Ft =
M
116
3309  850 = 8860 lbf
V ' Va   32.2
gc j
0.4
 15 1.4
 1390  

 73.2 
1. At the beginning of compression an ideal dual combustion cycle
T1  75F
1
2
State
2S
(A) The pressures and temperatures at the end of isentropic
compression
(B) The pressure and temperature at the end of each heat
addition process
(C) The temperature at the beginning of heat rejectionR
(D) the heat rejected, Btu/lb
(E) the net work, Btu/lb
(F) The thermal efficiency, %
(G) The horsepower developed by an ideal engine operating on
the cycle using 0.50 lb of air per second.

P1 = 15 psia
T
1391
pr
41.84
u
245.20
(B) u T3  u T2  q23  245.20  176  421.2
State 3
T
2253
h
575.53
pr
283.0
u
421.09
T 
 2253 
P3 = P2  3   467.80 
  757.7 psia
T
 1391 
 2
P4 = P3 = 757.7psia
h T4  h T3  q34  575.53  176  751.53
State
4
T
2866
u
555.08
 
 13.2 
(C ) r 5  r 4  5   1.3700 
  12.92

 1.4 
 4
k 1
State 5
T 
 2475 
P3 = P2  3  =486 
 =832 psia
 1446 
 T2 
P4 = P3  832 psia
T
1367
h
pr
334.31
39.16
T5  1367R
u
240.60
(D) qR  u T5  u T1
 240.60  91.19  149.41Btu / lb
w k (net )
(E)
 qs  qR  352  149.41  202.59 Btu/lb
J
q  qR 202.59
(F) et  s
=
 57.6%
qs
352
 T  13.2  3208 
(C) 4  3  4  
 1.426 ft 3 / lb
12  2475 
 T3 
 
 1.426 
T5  T4  4   3208 
 1317R


 13.2 
 5
(D ) qR  Cv T5  T1   0.1711317  535  = 134 Btu/lb
0.40
(G) w k ' 
202.59(0.50)(60)
 143 hp
42.42
3. A boiler produces 250,000 pounds of steam per hour at 1200 psia
and 1050 F from feed water entering the boiler at 1500 psia and
300F. Fuel oil having a higher heating value of 18,000 Btu/lb is
supplied to the burners at a rate of 20,500 lb/hr. Furnace volume is
1500 cubic feet. Calculate:
w(net)
(E)
 qs  qR  352  134  218 Btu/lb
J
q  qR 218
(F) e t  s

 61.9%
qs
352
218(0.5)(60)
 154.2 hp
42.42
2. At the beginning of compression an ideal dual combustion cycle
using air has a pressure of 15 psia, a temperature of 75F and a
specific volume of 13.2 ft3 per pound. For a compression ratio of
12 and a heat addition of 176 Btu/lb at constant volume and 176
Btu/lb at constant pressure. Calculate the following (using air
table):
(A)
(B)
(C)
(D)
(E)
Boiler capacity, MB/hr
Factor of Evaporation
Equivalent Evaporation, lb/hr
Furnace heat-release rate, Btu/hr.ft3
Boiler efficiency, %

from: Steam Tables
hsteam = 1528.9 Btu/lb
h for water = hf 300 Fand 1500 psi = 272.39 Btu/lb
(A) The pressure and temperature at the end of isentropic
compression
(B) The pressure and temperature at the end of each heat
addition process
(C) The temperature at the beginning of heat rejection process
(D) the heat rejected, Btu/lb
(E) the net work, Btu/lb
(F) The thermal efficiency, %
(G) The horsepower developed by an ideal engine operating on
the cycle using 0.50 lb of air per second.

Note: This solution is based on table attached with
interpolation to the nearest degree.
1  13.2 ft 3 / lb
1 / 2  12.0
Heat Added: 176 Btu/lb at constant volume
176 Btu/lb at constant pressure
T
535
h
126.78
pr
1.3416
vr
12.931
P 
 39.16 
P5  P4  r 5   757.7 
  38.3 psia
P
 774.9 
 r4 
q
176
T4  T2  34  2475 
 3208R
Cp
0.24
State
1
vr
1.3700
4  3 
 
0.40
T2  T1  1   535 12 
 1446R

 2
q
176
(B ) T3  T2  23  1446 
 2475R
Cv
0.171
k 1
h
pr
751.54
774.9
T4  2866R
 T4  13.2  2866 
 1.40 ft 3 /lb

12  2253 
 T3 
k
 
(A) P2  P1  1   15(12)1.4 = 15(32.4) = 486 psia
 2 
T1  75F
vr
29.49
T3  2253R
5  1  13.2 ft 3 / lb
1/2 = 12.0
P1 = 15 psia
vr
12.315
P 
 41.84 
P2s  p1  r 2   15 
  467.8 psia
 1.3416 
 Pr 1 S
See Figure:
'
h
340.55
T2s  1391R
Heat Added:
176 Btu/lb at constant volume
176 Btu/lb at constant pressure
(G ) w k 
 2 
 1 
  147.72 
  12.310

 12 
 1 S
r  r 
using air has a pressure of 15 psia, a temperature of 75F and a
specific volume of 13.2 ft3 per pound. For a compression ratio of
12 and a heat addition of 176 Btu/lb at constant volume and 176
Btu/lb at constant pressure. Calculate the following:
u
91.19
vr
147.72
(A) Capacity = 250,000(1528.9 - 272.4) = 314.1mB/hr
h - h 1528.9 - 272.4
(B) FE = out in =
= 1.295
hfg
970.3
(D) Equivalent Evaporation =1.295(250,000)
 323,750 lb/hr
250,000(18,000)
(D) Furnace heat release rate =
1500
= 246,000 Btu/hr.ft 2
20,500(1528.9-272.4)
(E) eboiler 
 85%
 20,500 18,000 
4. A Rankine steam power cycle operates with steam at 600 psia and
(A)
850 F from the boiler and a condenser pressure of one inch of
mercury absolute. Sketch the cycle on h-s and T-s coordinates and
determine for the cycle:
(A) Enthalpies for the steam leaving the boiler, leaving the
turbine, leaving the condenser and leaving the pump, Btu/lb
(B) The pump work, Btu/lb
(C) Heat supplied , Btu/lb
(D) Heat Rejected, Btu/lb
(E) Net work, Btu/lb
(F) The turbine work, Btu/lb
(G) The thermal efficiency, %
(H) The average temperature of heat receipt as determined by the
heat added divided by the change of entropy during addition
of heat, F
NA = 240 rpm
(SHP)A = 35,000hp
em = 0.95
IHP A =
SHP 35,000
=
=36,840hp
em
0.95
(MLHP)A =IHP-SHP=36,840-35,000=1840hp
2
N 
 122 
(MLHP)B =(MLHP)A  B  =1840 
 =475hp
 240 
 NA 
(B) SHP=4750, MLHP=475,
IHP=4750+475=5225
SHP
4750
em =
=
= 0.909 = 90.9%
IHP
5225
2
(C) Refer to figure 9-10: P1 = 1250 psia ; t1  940F ; Pa=0.70psi
h1 = 1462.6 ; s1 = sa = 1.5994
ha = 876.0
(A.E.)t = h1 - ha = 1462.6 - 876 = 586.6 Btu/min
wk  shaft 
2545 xSHP 2545 x 4750
=
= 376 Btu/lb
m'
32150
 Wshaft   376  0.641  64.1%
ees 
J(A.E.)t 586.6
J
(D )
(wk )t 2545 xIHP 2545 x 5225


= 413.6 Btu/lb
J
m'
32150
wk t 413.6
eei 

 0.705  70.5%
J ( AE )t 586.6
eei 

from: Steam Tables and Mollier Chart;
h1 = 49.4 (from table 4)
h2 = 1435.4 ; s2 =1.6559
h3 = 890
h4 = 47.1 ;
s 4 = 0.09146 = s1
w k (pump)
= h1 - h4 = 49.4 - 47.1 = 2.3 Btu/lb
J
(C) qs = h2 - h1 = 1435.4 - 49.4 = 1386.0 Btu/lb
(B)
(D) qR = h3 - h4 = 890 - 47.1 = 842.9 Btu/lb
(E) Net Work = qs - qR  1386.0  842.9 = 543.1 Btu/lb
(F) Turbine Work = h2  h1  1435.4  890.00 = 545.4 Btu/lb
q  qR 1386.0  842.9
(G ) e t  s

 39.2%
qs
1386.0
et 
(H) t AV
output Wturbine  Wpump
545.4 - 2.3
=
=
= 39.2%
input
qs
1386.0
q
1386.0
 s  460 
 460  426F
s
1.5644
5. A geared turbine propulsion unit delivers 35,000 shaft horsepower
at full power with a shaft speed of 240 rpm. The mechanical
efficiency of the unit under this conditions is 95%. At 122 rpm the
unit delivers 4750 shp, receiving steam at the throttle at 1250 psia
and 940F at the rate of 32,150 lb/hr and exhausting to a condenser
at 0.7 psia. Assuming the mechanical losses vary as the square of
the rotative speed, find the 122-rpm condition:
A. The estimated mechanical loss, hp
B. The mechanical efficiency,%
C. The shaft engine efficiency,%
D. The internal engine efficiency, %



ees 64.1

 0.705  70.5%
em 90.9
6. In a simple impulse stage, steam leaves the nozzles with a velocity
of 1200 ft/s. The nozzle angle is 15 deg. Assume the bucket
entrance and exit angles are to be the same and that the bucket
velocity coefficient is 0.88. The wheel speed is 580 ft/s, and steam
is supplied to the turbine at the rate of 6000 lb/hr. Find
(A)
(B)
(C)
(D)
(E)
The required bucket entrance angle for the given conditions
The bucket work, ft-lb/lb and Btu/lb
The power developed in the buckets, hp
The available energy to the buckets, ft-lb/lb and Btu/lb
The diagram efficiency.

Refer to Fig. 12-2:
V1 = 1200 ft/s
 = 15
Vb = 580 ft/s
Cb  0.88
1   2
V1 sin 
1200 sin15
(A) tan 1 

V1 cos   Vb 1200 cos15  580
1 =  2  28.2
V cos   Vb 1200 cos15  580
Vr1  1

= 657.1 ft/s
cos 1
cos 28.2
Vr2  Cb Vr1 = 0.88(657.1) = 578.2 ft/s
(B)  wk b 
580  657.1cos 28.2  578.2cos 28.2 
32.2
 Btu 
 wk b  19,610 ft-lb 
 = 25.2 Btu/lb
 778 
(C) Wk ' =m'(w k )  where : m '= 6000 lb/hr = 100 lb/min
Wk ' = 100(19,610) =
100(19,610)
= 59.4 hp
33,000
or:Alternate Solution:
Wk ' = 6000 x 25.2 Btu/hr =
(D) (A.E.)b =
6000 x 25.3
 59.4 hp
2545
V12 (1200)2

= 22,360 ft-lb/lb
2g c 2(32.2)
22,360
= 28.74 Btu/lb
778
(w k )b
19,610
(E) eb 

 0.877  87.7%
 A.E.b 22,360
or: ( A.E )b =
7. Steam enters a simple impulse bucket wheel with an absolute
velocity of 450 m/s and a relative velocity of 270 m/s. It leaves the
blades with relative velocity of 230 m/s and an absolute velocity of
105 m/s. Find:
(A)
(B)
(C)
(D)
(E)
The bucket velocity coefficient
The available energy, kJ/kg
The bucket loss, kJ/kg
The unused kinetic energy at exit, kJ/kg
The diagram efficiency.
V22
2g c J
 350 
2  32.2  778 
2
= 1184.7 
= 1187.1 Btu/lb
(E) P2 ' = 140 psia
h g - h 2'
m2 ' =


hfg
1193.8  1187.1
 0.77%
868.7
9. The first stage of a 50% reaction groups receives steam with a
V1 = 450 m/s
Vr2 = 230 m/s
Vr1 = 270 m/s
V2 = 105 m/s
negligible approach velocity at a pressure of 300 psia with a
temperature of 520F. The available energy to the stage is 10
Btu/lb, the fixed blade efficiency is 96%, the velocity coefficient for
the moving row is 0.88, the reactive effectiveness is 90%, and the
relative inlet and absolute exit velocities are 146 ft/s and 160 ft/s,
respectively. Assume the stage efficiency is the same as the
combined blade efficiency and calculate the following:
1   2
(A) Cb 
Vr2 230

 0.85
Vr1 270
V12 (450)2 x10 3

= 101.25 kJ/kg
2g c
2
(B) (A.E.)b =
Vr12  Vr22  270    230 

x10 3  10 kJ/kg
2g c
2
2
(C) bucket loss 
(E) eb 
2
V
2
1
 
 V2 2  Vr 12  Vr 2 2
 450

or : eb 
2
3
V2 2 105  x10

= 5.51 kJ/kg
2g c
2
(D) KE 2 
eb
D) h2' = h2 
2
1
V
2
 105 
2
 450 
  230 
2
The steam velocity leaving the fixed blades, ft/s
The relative exit velocity from the moving blades, ft/s
The stage work, Btu/lb
The combined blade efficiency, %
The enthalpy of the steam entering the succeeding stage,
Btu/lb for complete velocity carryover


   270 
(A)
(B)
(C)
(D)
(E)
2
2
  0.847  84.7%
w k b  AE  b losses 101.25  10  5.51


 84.7%
( AE )b
101.25
 AE b
P0 = 300 psia
t o = 520°F
(AE)st  10 Btu/lb
en = 0.96 ;
Cm = 0.88 ;
eR = 0.90
Vr1  146 ft/s
V2 = 160 ft/s
(A)  AE n   AE R 
temperature of 420F when the stage pressure is 140 psia. Under
these conditions the available energy to the stage is 38.5 Btu/lb,
the nozzle-bucket efficiency is 85% and the absolute blade
entrance and exit velocities are 1350 f/s and 350 ft/s respectively.
Assume the stage efficiency is equal to the nozzle-bucket
efficiency and calculate:
(A) The enthalpy of the steam after isentropic expansion to the
stage pressure, Btu/lb
(B) The enthalpy of the steam entering the blades, Btu/lb
(C) The enthalpy of the steam leaving the blades
(D) The enthalpy of the steam leaving the stage if there is
negligible carryover velocity, Btu/lb
(E) The state of the steam leaving the stage, pressure (psia) and
moisture (%) for the conditions postulated (D) above


Refer to Fig. 12-2:
 AE st = 38.5 Btu/lb
nozzle-bucket efficiency = 85%
V1 = 1350 ft/s ; V2 = 350 ft/s
1350 
V1
 1219.8 
= 1183.4 Btu/lb
2g c J
(2)(32.2)(778)
2
J
= emeb  AE st = 0.85(38.5) = 32.7 Btu/lb
h2  ho 
2
= eR  AE R
  2g c J  eR  AE R
 0.88 146     2 32.2 778   0.90 5 
2
 wk 
(C)
J
=
V
2
1
 
- V2 2 + Vr2 2 - Vr12
 490  - 160   +  492 - 146   = 8.69 Btu/lb
2
=
(D) ecb =

2gc J
2
2
2
2  32.2  778 
 w k st
J  A.E.st
=
8.69
= 0.869 = 86.9%
10
 w k st  v 2 2

J  A.E.st 2g c J
2
160 
= 1269.4 - 8.7  1260.2
2  32.2  778 
h2
on one side of the wall is 50C, the surface temperature on the
opposite side is 15C, k for the brick is 1.32 W/m.C and the wall is
20 cm thick. Calculate the heat transfer for a wall surface are of 10
m2:

A = 10 m2
2gc J
 350 
2  32.2  778 
2
= 1219.8 - 32.7 -
Btu/lb
t a  50C ; t b  15C
 w k st  V2 2

J
2gc J
= 490 ft/s
10. A solid brick wall separates two rooms. The surface temperature
(A) ha = ho - (AE)st = 1219.8 - 38.5 = 1181.3 Btu/lb
 w k st
2
 492 ft/s
t o = 420°F
P1 = P2 = P2' = 140 psia
(C)

 Cm Vr2 
=
 Cm Vr1 
Vr2 
10
 5 Btu/lb
2
 2  32.2  778    0.96  5 
2
2gc J

h2 = ho 
P0 = 220 psia
(B) h1  ho 
 Vr2 
(B)
2
 2gc J en  AE n
V1 

8. The impulse stage of a turbine receives steam at 220 psia with a
 AE st
= 1187.1 Btu/lb
Q=
; k = 1.32 W/m.C ; L =0.20 m
kA  ta  t b 
L

(1.32)(10)(50  15)
= 2310 W
0.20
11. A typical furnace side wall is constructed of a 1-in layer of
diatomaceous earth insulating block, and 2 ½ in high temperature
insulating brick faced with 4 ½ in. Firebrick with a 1/8-in steel
casing. Average values of thermal conductivities are:
diatomaceous earth insulating block, 0.063; insulating brick, 0.62 ;
firebrick, 4.0; and steel casing 26 Btu/hr.ft.F. Average film
coefficients are 3.0 and 2.2 Btu/hr.ft.2F for inner and outer surface
films, respectively. The wall area is 50 ft2 , the average gas
temperature is 2100F and the ambient air temperature is 100F.
Calculate :
(A) heat transfer coefficient, U, for this furnace wall, and
(B) the heat transferred by the wall because of the conduction.
(C) Calculate the temperature at the interface between the
insulating brick and diatomaceous insulating block.

(A) k fb = 4.0 ; k IB = 0.62; k DE = 0.063
k s = 26 Btu/hr.ft.F
h1 = 3.0 ; h2 = 2.2 Btu/ft 2.hr.F
1 1 L fb LIb LDE LS 1
= 


 +
U h1 k fb kIb k DE k s h2
1
=
U
1
0.30

 1.45 
 1.60 
 40  
 12  26   4.25 
4.25




1.50
1.0
1


+
 2.42 
 3.73  1.2
12  0.08  
 12  0.04   4.25 
 4.25 


U = 0.1659 Btu/lb.ft 2 F(external area)
'
 2 r4 
(B) Q  UA  t1  t 2   U 
x  (t 1 t 2 )
 12 
 2  4.25 

Q =  0.1659  
180   (800  100)  46,516 Btu/hr
12


13. Calculate the overall heat transfer coefficient for tubular heat
exchanger wherein the liquid carried in the tubes is heated by
steam which surrounds them. The tubes are 5/8 in. Admiralty
metal with a wall thickness of 0.049 in. The steam surface film
coefficient is 1250 Btu/hr.ft2.F and the liquid surface film
coefficient is 20 Btu/hr.ft2.F.
1
1
4.5
2.5
1
0.125
1
=





U
3.0 12  4  12  0.62  12  0.063  12  26  2.2
U = 0.394 Btu/hr.ft 2 .F

'
(B) Q  UA  t1  t 2   (0.394)(50)(2100  100)
Q = 39,400 Btu/hr
(C) for parallel Surfaces; A = A 1= A FB = A IB
R1=
0.333
0.094
0.336
2.54
; RFB =
; RFB =
; Rt =
A1
A FB
A IB
A
t1 = 2100F
La
LP
Lm
1
1



+
r
r 
 
r 
r  h
h1  1  k p  p  k a  a  k m  m  2
r
r
r
 4
 4
 4
 r4 
1
=
U
and t 2 = 100F
 R + RFB + RIB 
t to interface =  1
  t1  t 2 
Rt


A  0.333 + 0.094+ 0.336 
= 
  2100  100   600F
A
2.540

Temperature at interface = 2100 - 600 = 1500F
12. A high temperature steam line is covered with two successive
layers of insulation. The layer in contact with the pipe is 1 ½ in
thickness of asbestos for which k is 0.08 Btu/hr.ft.F. The asbestos
is covered with a 1-in thickness of magnesia insulation, which has
a value of 0.04. The internal pipe diameter is 2.90 in., the pipe wall
thickness is 0.30 in and k for the pipe is 26 Btu/hr.ft.F. The steam
temperature is 800F, and the internal surface film coefficient is 40
Btu/lb.ft2 .F, while the ambient outer temperature is 100F and the
outer surface film coefficient is 1.2.Calculate :
(A) the value of U based upon the external area of magnesia
covering,
(B) the heat loss from the steam for a length of 180 feet of pipe,
Btu/hr


h1=20; h2 =1250 Btu/hr.ft 2.F
r2 = 0.3125 in. ; r1 = 0.2635 in; k = 70 Btu/hr.ft.F
r1 + r2
0.2635 + 0.3125
=
=0.288
2
2
1
1
L
1
=

+
U
 r1 
 rt  h2
h1   k t  
 r2 
 r2 
rt =
1
0.0490
1

+
= 0.0602
 0.2635 
 0.288  1250
12
70
 20  




 0.3125 
 0.3125 


1
U =
= 16.61Btu/ hr.ft 2 .F
0.0602
=
14. In the first stage of gas turbine, air enters a group of nozzles at
1200 F and leaves at 950F. The entering velocity, ft/s is
negligible.
A. Find the kinetic energy; Btu/lb.
B. The velocity of air leaving the nozzle.

(A) KE2 
V22
 h1  h2 ; where: h1- h2 = Cp (T1-T2 )
2g c J
KE2 = Cp (T1 - T2 ) = 0.24(1660 - 1440) = 60 Btu/lb
(B)
V22
 h1  h2  Cp (T1-T2 )
2g c J
V2  2gc JCp T1  T2   2  32.2  778  0.24  250  = 1734 ft/s
h1 = 40
ka = 0.08
t1 = 800F
15. A propulsion turbine receives steam at the throttle at 875 psia
and 940F at the rate of 100,000 lb/hr. After an irreversible
expansion process, the steam exhausts from the turbine at a
pressure of 0.60 psia with a moisture content of 10%.Assume
difference between the entrance and exit kinetic energies is
negligible and find:
A. Work done, Btu/lb.
B. The power developed, hp
2
h2 = 1.2 Btu/hr.ft .F
km = 0.04
t2 = 200 F
kp = 26 Btu/hr.ft.F
x = 180 ft length
(A) rp = pipe mean radius =
r2 - r1
1.75-1.45
=
= 1.60 in.
 r2 
 1.75 
ln
ln  
 1.45 


 r1 
ra = pipe asbestos radius =
rm = mean magnesia radius =
r3 - r2
3.25 -1.75
=
= 2.42 in.
 r3 
 3.25 
ln
ln  
 1.75 


 r2 
r4 - r3
4.25 -3.25
=
= 3.73 in.
 r4 
 4.25 
ln 
ln  

 3.25 
 r3 

P1 = 875 psia ; t1 = 940F
P2 = 0.60 psia ; m2  0.10
from:steam tables: h1 = 1475.6 and
h2 = 1098.6 - 0.1(1045.4)=994.1
w
(A) k12 =  h2 - h1  = 1475.6 - 994.1 = 481.5 Btu/lb
J
m'wk12 10,000(481.5)
(B) W'k12 

 18,920 hp
2545
2545
1. A propulsion turbine receives steam at the throttle at 875 psia and
 A TH 
940F at the rate of 100,000 lb/hr. After an irreversible expansion
process, the steam exhaust s from the turbine at a pressure of
0.60 psia with a moisture content of 10%. Assume the difference
between the entrance and exit kinetic energies is negligible and
find:
P2  P1

1 g
32.2
where:  =  
=
= 58.82 lb f /ft 3
0.017(32.2)
 1  gc
144 1200-29.8 
TH =
= 2864.8 ft
58.82
(A) The work done, Btu/lb
(B) The power develop, hp
'
B 

P1 = 875 psia
T1  940F
m2  0.10
P2 = 0.60 psia
From Steam Table:
h1  1475.6 Btu/lb
WHP 
Wk12 ' 
M'  wk12 
2545J

10,000(481.5)
= 18,920 hp
2545(1)
2. A boiler receives feed water at 1200 psia and 250F and delivers
steam from the superheater at 900 psia and 950F. Find the heat
added, Btu/lb. Refer to Fig. 5-3

P1 = 1200 psia
P2 = 900 psia
t1 = 250F
t 2 = 950F
M TH  g
550g c

125,000(2864.8)(32.2)
= 180.9 hp
3600(550)(32.2)
WHP
180.9
=
= 301.5 hp
ep
0.60
(C) BHP =
(D) Interpolating in Table 4 of the steam tables with:
s1 = s2  0.36772 and P2 = 1200 psia
h2  1098.6  (0.10)(1045.4)  994.1
wk
(A) 12 = h1 - h2 = 1475.6 - 994.1
J
= 481.5Btu/lb
(B) M' = 100,000 lb/hr ; 1 hp = 2545 Btu/hr
;
h2 = 222.30
wk12
= h1 - h2  218.59  222.30 = (-) 3.17Btu/lb
J
wk12(actual)
wk12(isentropic) (-) 3.17
(E)
=
=
= (-) 6.18 Btu/lb
J
0.60
J  epump 
wk12
= 218.59 + 6.18 = 224.77 Btu/lb
J
Entering Table 4 and interpolating between 1000 and 500 psi gives:
t
h
250C
221.03
300C
271.83
224.77 - 221.03
t =
x 50  3.68
271.83 - 221.03
t 2 = 250 + 3.7 = 253.7F
(F) h 2 = h1 -
q12 = h2 - h1
2
(1.04) = 221.0
5
From Table 3: h2 = 1480.5 Btu/lb
From Table 4: h1 = 220.61 +
q12 = 1480.5 - 221 = 1259.5 Btu/lb
3. If, in the preceding example, the feed water entering has a velocity
of 3 m/s and the steam leaving the super heater has a velocity of
50 m/s. Find:
(A) The additional heat required to accommodate the change in
kinetic energy across the boiler, J/kg
5. A water cooled reciprocating air compressor takes in air at 15 psia
and 60F and discharges it at 60 psia and 200F. Heat is removed
in the amount of 21.4 Btu/lb. Assume steady flow conditions and
the work done, Btu/lb.

P1 = 15 psia
T1 = 40+ 460 = 520R
P2 = 60 psia
T2 = 200 + 460 = 660R
q12 =
 -  21.4 Btu/lb
(B) The percentage error introduced by neglecting the kinetic
energy change.
wk12
= q12 + h1 - h2
J
where :  h1 - h2   Cp T  0.24(520  660)  33.6

wk12
= -21.4 - 33.6 =  -  55 Btu/lb (on)
J
(A) q12 = h2 - h1 -
V22 -V12 (50)2  (3)3

 1245 J / kg
2gc
2(1)
Btu  
J / kg 

(B)q12 =  1259.2
2326
= 2930 x 10 3 J/kg
lb  
Btu / lb 

1245.5 x 100
Error =
 0.043%
2930 x 103
4. Saturated water at 250F enters a centrifugal main feed pump and
is discharges at 1200 psia. The pump efficiency is 60% and the
delivery rate is 125,000 lb/hr. Find:
(A)
(B)
(C)
(D)
(E)
(F)

The total head developed by the pump, ft
The water horsepower, WHP
The brakepower, BHP
The ideal (isentropic) pump work, Btu/lb
The actual pump work, Btu/lb
The estimated temperature of the water at discharge.
; P1 = 29.8 psia
1  0.017001 ft 3 / lb
; P2  1200 psia 
M'=125,000 lb/hr
V1  V2 ; Z1  Z2
pressure is 215 psia and passed through an Ellison throttling
calorimeter. The calorimeter thermometer reads 250 F and the
barometer is standard. Find for the line steam:
A.
B.
C.
D.

The enthalpy
The quality
The entropy
The temperature to the nearest whole degree

P1 = 215 psia
P2 = 14.696 psia
T2 = 250F
(A) from table 3:
h2 = 1168.8
h1 = h2 = 1168.8 Btu/min  throttling(h=constant)
(B) from table 2 for 215 psia: h g  1200.3; hfg  838.1

t1 = 250F (sat.water)
h1  218.59 Btu/lb
6. A sample of steam is removed from a steam line where the
h1  hg - m1hfg
; epump  60%
m1 
hg - h1
hfg

1200.3  1168.8 31.5

 0.0376
838.1
838.1
x1 = 1 - m1 = 1 - 0.0376 = 0.9624 = 96.24%
(C) s1  sg  m1sfg
 1.5403  0.0376(0.9887)  1.5031 Btu/lb.R
(D) t1  387.97  388F
7. In a lube oil cooler, oil enters at 140F and leaves at 100F, at the
rate of 400 lb/min. The cooling medium is sea water , which enters
at 60F. The average specific heat of the oil is 0.50 Btu/lb.F and the
salt water is 0.94 Btu/lb.F. If the flow of the sea water is at the rate
of 500 lb/min, find the overload discharge temperature.






t 2  100F
t1 = 40°F

t 3  60F

M'A = 400 lb/min; C A = 0.50 Btu/lb.F

M'B = 500 lb/min; CB  0.94 Btu/lb.F

M'A  h1 - h2   M'B  h4 - h3  Btu / min

M'ACA  t1 - t 2   M'BCB  t 4 - t 3  Btu / min
400  0.50  140-100 
 t4
- t3  =
 t4
- 60  = 17  t 4 = 77°F
500  0.94 
M'o = 60,000 lb/hr
Co  0.50 Btu/lb.F
= 17
A = 258 ft 2
A. Q' = M'oCo  t1  t 2   (60,000)(0.50)(145  120)
= 750,000 Btu/hr
 
55  45
m  1 2 
 49.8F
 1 
 55 
ln
ln  
 45 


 2 
8. Steam enters the condenser of propulsion plant at 0.50 psia and a
quality of 89 percent at the rate of 100,000 lb/hr and with a velocity
of 1000 ft/s. It leaves the condenser hotwell as saturated liquid
without any change in pressure but a velocity of 10 ft/s. The salt
water inlet ( injection ) temperature is 70F and the discharge
(overboard ) temperature is 85F. Sea water has a specific heat of
0.94 Btu/lb.F and a density of 64 lb/ft3 . The injection and
overboard velocities are substantially equal. Calculate the
following:
(A) The rate at which energy is extracted from the condensing
steam as heat,
(B) The flow of sea water required, gallons per minute (gpm)





B. For parallel flow:
Q'  750,000 Btu/hr

U = 58.4Btu/hr-ft 2  F
 
70  30
m  1 2 
 47.2F
 1 
 70 
ln   ln 

 30 
 2 



P1 = P2  0.50 psia
t 3  70F
m1  1  0.89  0.11
t 4  85F
V1  1000 ft/s
A=
CB  0.94 Btu/lb.F
10. A steam superheater has a net heat transfer area of 1620 ft 2
B = 64 lb/ft 3
V2  10 ft/s
M' A = 100,000 lb/hr ; 1 ft 3 = 7.481 gal
(A) q12 = h2 - h1 
V22  V12
2g c J
h1  1096.2  0.11 (1048.6)  980.9 Btu/lb
h2  47.7
10   1000 
2  32.2  778 
2
q12 = 47.7  980.9 +
M'A q12 =
2
=    953.2
Btu
lb
M'B 

   1,588,700 Btu/min
60
 M 'B xCB  t 4  t 3 
Q '34
1,588,700

= 112,670 lb/min
CB  t 4  t3  0.94(85  70)
flow, (gpm) = 7.481

gal  M ' 


ft 3   B
7.481112,670 
64
= 13,170 gpm
9. A counter-flow lubricating oil cooler with a net heat transfer area of
258 ft2 cools 60,000 lb of oil per hour from a temperature of 145F
at inlet to 120F at discharge. The temperatures of the cooling
water are 75F and 90F respectively, and the specific heat of the oil
is 0.50 Btu/lb.F. Calculate:
B.
and a design capacity of 221,000 lb of steam/hr when receiving
saturated steam at 650 psia and discharging at 850F with a
pressure drop if not of not more than 25 psi and through heater.
The design heat transfer coefficient is 30 Btu/hr.ft2.F. In operation
of superheater receives 220,000 lb of saturated steam per hour at
a pressure of 650 psia and discharges against a pressure of 630
psia with a temperature s at entrance and exit are 2100F and
1430F, respectively. Calculate the operating heat transfer
coefficient using counter flow log mean temperature difference.
100,000  953.2 
(b) (-)Q'12  Q '34
A.
Q'
750,000

= 272 ft 2
Um  58.4  47.2 
The value of the overall heat transfer coefficient under these
operating conditions, Btu/hr.ft2F,
and the required area for a parallel flow device having the
same capacity under identical operating conditions.




Q '   Ms  hs    220,000 1434.2  1203.1  51x10 6 Btu / hr
m 
U
1250  935
 1085F
 1250 
ln 

 935 
Q'
51x106

 29 Btu / hr ft 2F
A 1620 1085 