CAPITOL UNIVERSITY

College of Engineering
Mechanical Engineering Department
Thermal Power Examination
POWER PLANT ENGINEERING

NAME: Vincent Rey Olario

BSME – 5
Instruction:
1. Show complete solution as applicable.
2. Follow the standard presentation of solution.
Encircle the letter of the correct answer.
1) A Rankine cycle operates with a thermal efficiency of 40 % and the factor of evaporation of
the boiler is 1.15.
Determine the mass flow rate of steam if the cycle power output is 5.5 MW.
a) 5.3 kg/s b) 4.3 kg/s c) 3.5 kg/s d) 6.3 kg/s
Solution:
A. List of Equation used:
W net QA QA
Q= ; ms = =
e th h A−hB 2257(FE )
B. Figure that shows the cycle

C. Calculations
W net 5500 QA QA 13,750 ks
Q= = =13,750 kW ; ms= = = =5.3
e th 0.40 h A −h B 2257 ( FE) 2257(1.15) s
2) In a Rankine cycle, steam enters the turbine at 2.5MPa and a condenser pressure of 50KPa.
What is the quality
of steam at the turbine exhaust? Steam Properties:
@ 2.5Mpaa, h = 2803.1 kJ/kg & s = 6.2575 kJ/kg- O K;
@ 50kPaa, h f = 340.49 kJ/kg, h fg = 2305.4 kJ/kg,
v f = 0.00103 m 3 /kg, s f = 1.0910 kJ/kg- O K,
s fg = 6.5029 kJ/kg- O K.
a) 79.45 % b) 97.45 % c) 59.75 % d) 95.55 %
Solution:
A. List of Equation used:

x 2=
( s fg)
s2 −s f 2
100 %

B. Figure that shows the cycle

C. Calculations

x 2= ( )
s2 −s f 2
s fg
×100 %= (
6.2575−1.0910
6.5029 )
×100 %=79.45 %
3) It is a closed vessel intended for use in heating water of for application of heat to generate
steam or other vapor to be used externally from it. What do you call this pressure vessel?
a) Boiler or steam generator c) Unfired pressure vessel
b) Fired pressure vessel d) Pressurized tank
4) Wet saturated steam at 16 bar (h f = 859 kJ/kg , h g = 1935 kJ/kg) reducing valve and is
throttled to a pressure of 8 bar (h r = 721 kJ/kg, h fg = 2048 kJ/kg). Find the dryness fraction of
the reduce pressure steam.
a) 0.8833 b) 0.9933 c) 0.7733 d) 0.6633
Solution:
A. List of Equation used:
h1 =¿ h2=h w1 =hf 2+ x hfg
B. Calculations
For throttling process enthalpy remains constant
h1 =¿ h2=h w1 =hf 2+ x hfg
859 = 721 + X (2048)
X 2 =0.067

5) The water added to boiler feed to compensate for the water lost through blowdown, leakage,
etc
a) condensate b) Make-up c) sulfur d) economizer
6) A large paper plant uses its own cogeneration power plant to generate its electricity because
a) it permits the use of a more efficient generator.
b) the company can better control the voltage.
c) it does away with high power lines leading to the plant.
d) it provides a more efficient conversion of fuel.
7) A 6 MW steam turbine generator power plant has a full-load steam rate of 8 kg/kW-hr.
assuming that no-load steam consumption as 15% of full-load steam consumption, compute for
the hourly steam consumption at 75% load,in kg/hr.
a) 37,800 kg/hr b) 30,780 kg/hr c) 38,700 kg/hr d) 30,870 kg/hr
Solution:
A. List of Equation used:
y− y 1 y 2− y 1
m2=P ( FR ) ; =
x−x 1 x 2−x 1

C. Calculations
Steam Consumption at full load:
kg
m2=( 6000 ) ( 8 ) =48,000
hr
Steam Consumption at full load (m1 ¿ :

kg
m1=( 0.15 ) ( 48,000 )=7,200
hr
Use two-point form:
Where:
P 1 ( x 2− y 2 ) =P1 ( 0,7200 ) ; P 2 ( x 2− y2 ) =P 2 ( 6000,48,000 )

Then:
m−7200 48,000−7200
= ; m−7200=6.8 L
L−0 6000−0
m=6.8 L+7,200
At 75% load:
L=( 0.75 )( 6000 ) =4500 kW
m=6.8 ( 4500 ) +7,200=37,800 kg /hr
8) In a Rankine cycle, steam turbine with exhaust enthalpy of 2500 kJ/kg delivers 0.50 kg/sec of
steam. If the exit enthalpy of the condenser is 500 kJ/kg, what is the heat rejected?
a) 1000W b) 1200W c) 1700W d) 1500
Solution:
A. List of Equation used:
Qr = m x (h2-h3)

b. Calculations
Qr = m x (h2-h3)
Qr = 0.50 kg/s (2500 kj/kg – 500 kj/kg)
Qr = 1000 watts
9) The pressure and temperature entering the turbine are 1800 kPa and 380°C. The pressure
leaving the turbine is 20 kPa. The quality of the steam entering the condenser is 90%. What is the
turbine work?
Steam properties: At 1800 kPa and 380°C (h = 3207.2 kJ/kg)
At 20 kPa, h f = 251.4 kJ/kg, h g = 2606.7 kJ/kg
a) 1833.33 kJ/kg
b) 597.5 kJ/kg
c) 1333.8 kJ/kg
d) 833.33 kJ/kg

Solution:
A. List of Equation used:
h2 =hf + xhfg ; W T =h1−h2
C. Calculations
kJ
h2 =hf + xhfg=251.4 + ( .90 ) (2358.3 )=2373.87
kg
kJ
W T =h1 −h2=3207.2−2373.87=833.33
kg
10) Steam with an enthalpy of 800 kCal/kg enters a nozzle at a velocity of 80 m/sec. Find the
velocity of the steam at the exit of the nozzle if its enthalpy is reduced to 750 kCal/kg, assuming
the nozzle is horizontal and disregarding heat losses. Take g = 9.81 m/sec 2 and J constant = 427
kg m/kCal.
a) 452.37 m/sec
b) 651.92 m/sec
c) 245.45 m/sec
d) 427.54 m/sec
Solution:
A. List of Equation used:
h1 + KE 1 = h2

KE 2 - KE 1 = h1 - h2

B. Calculations
v 2−¿ v
(
kCal 4816 J
)
2 2
1
=( 800−750 ) ¿
2 kg 1 kCal
v 2−¿ 80
2 2

=416,600 ¿
2
m
v 2=651.92
s
11) Determine the brake power of the engine having a brake thermal efficiency of 35 % and uses
250 API fuel with fuel consumption of 40 kg/hr.
a) 160.67 kW b) 173.52 kW c) 174.52 kW d) 165.84 kW
Solution:

A. List of Equation used:

Brake Power x 3,600
Ƞ th =
mf x C .V

B. Calculations
Brake Power x 3,600
Ƞ th =
mf x C .V
 Brake Power x 3,600
0.35=
40 x 45000
B . P=173.52kW
12) It is based on the generation of 34.5 lbm/hr of steam from water at 212 deg F to steam at 212
deg F and equivalent to 33,500 Btu/hr.
a) One horsepower c) One kilowatt
b) One boiler horsepower d) None of the above
13) In a geothermal power plant, the enthalpies of the ground water and the turbine inlet are 1500
kJ/kg and 3500 kJ/kg respectively. If the enthalpy of the hot water in the flash tank is 700 kJ/kg
and the mass flow rate of steam is 15 kg/s, what is the mass flow rate of ground water?
a) 52.5 kg/s b) 55.2 kg/s c) 25.5 kg/s d) 35.2 kg/s
Solution:
A. List of Equation used:
H 2=H f +x (H g −H f )
ms =x mg

b. Calculations
H 2=H f + x ( H g −H f )
1500=700+ x ( 3500−700 )
x=0.286
ms =x mg

15=(0.286)(m g)
mg =52.5 kg/ s
14) The head required to produce a flow of the water.
a) static head b) velocity head c) pressure head D) Dynamic Head
15) A cylindrical pipe with water flowing downward at 0.03 m 3 /sec having top diameter of
0.08, bottom diameter of
0.04m and height of 1.5m. Find the pressure between the pipe.
a) 154.63 kPa
b) 197.93 kPa
c) 252.44 kPa
d) 243.92 kPa
Solution:
A. List of Equation used:
2 2
P1 V 1 P V
+ +Z 1= 2 + 2 + Z 2
y 2g y 2g
2 2
P 1−P2 V 2 −V 1
= +(Z ¿ ¿ 2−Z 1)¿
y 2g

C. Calculations
Z2 −Z 1=1.5 m

Z1 −Z 2=−1.5 m

0.03 m
V 1= =5.968
π 2 s
(0.08 )
4
0.03 m
V 2= =23.87
π 2 s
(0.04 )
4
P 1−P2 23.87 2−5.9682
= −15
9.81 2 ( 9.81 )
P1−P2=252.44 kPa
16) The percent rating of water tube boiler is 200 %, factor of evaporation is 1.10, and heating
surface is 400 ft 2 .Determine the rate of evaporation, in kg/hr.
a) 1831 b) 1831 c) 1138 d) 1813
Solution:
A. Calculations
Dev.Bo.Hp = ms ¿ ¿
37.18
Rated Bo.Hp. = = 40.85
0.91
Percent rating = Dev.Bo.Hp/Rated.Bo.Hp x 100
200 = Dev.Bo.Hp/40.85 x 100 = 81.7 Hp
81.7 = ms(1.10(2257))/35322
ms = 1162 kg/hr
A water tube boiler generates 7,300 kg of steam per hour at a pressure of1.4 MPa and a quality
of 98% when the
feed-water is 24°C. From Steam Table: h s = 2750.81 KJ/kg, h f = 100.71 KJ/kg

17) The Factor of Evaporation is equal to

a) 1.471 b) 1.174 c) 7.114 d) 1.971
Solution:
A. List of Equation used:
U 1−U 2
¿
H

B. Calculations
U 1−U 2
Factor of evaporation = ¿
H
2749.69−100.05
=
2260
= 1.172

18) The Equivalent Evaporation is equal to

a) 8571.44 kg/hr b) 5718.44 kg/hr c) 7851.44 kg/hr d) 1874.44 kg/hr
Solution:
A. List of Equation used:
Equivalent evaporation = ms (F.E)

C. Calculations

Equivalent evaporation = ms (F.E)

= 2.03 kg/sec x 1.172
= 2.379 kg/sec x 3600 sec/hr
 = 8571.44 kg/hr

19) The Developed Boiler Horsepower is equal to

a) 747.5 hp b) 477.5 hp c) 547.7 hp d) 775.4 hp
Solution:
A. List of Equation used:
ms ¿ ¿
C. Calculations
Developed Boiler Horsepower = ms ¿ ¿
= 7300 ¿ ¿

= 546.75 hp

20) The Rated Boiler Horsepower is equal to

a) 211hp b)311hp c) 411hp d) 511 hp
Solution:
A. List of Equation used:
Rated Boiler Power= ms/34.75

B. Calculations:
Rated Boiler Power= 7300/34.75
= 211hp