Course: heat engines & combustion (a)
Third-year Mechanical Power Engineering
Exam date 4-1-2012
Course Instractor : Prof. Dr. Ramadan Youssef Sakr
1a) environmental impact for steam power station is summarized in
i- The temperature rise in the cooling water required fir condenser which affect the
ecological life for the nearby river.
ii- The exhaust gases discharged from chimney due to fuel combustion.
iii- Noise due to exhaust steam.
1b) A steam turbine receives superheated steam at 20 bar. A part of the steam (6.5 tons/hr) is
extracted to be used in an industrial process. At the point of extraction the pressure is 3 bar
and the temperature is 180 °C. The remaining steam continues to expand down to the
condenser vacuum pressure which is 69cm Hg. The isentropic efficiency in each of the two
parts of the turbine is 75 %. The indicated power of the turbine is 2500 kW. Find the
following:
i) The initial temp, of the steam supplied to the turbine.
ii) The total amount of steam to be supplied to the turbine per hour.
1
T
Turbine
1
2
Industrial process
2s
2
3
3s
3
Condenser
s
Given : p1=20 bar mprocess =6.5 tons/h = 1.805 kg/s
P2= bar, t2=180 C, p3=69 cm Hg vacuum = (1-69/72) = 0.1 bar, isentropic efficiency of the
turbine in both stages 75% and IP=2500 kW
Required: T1 & ms
Solution:
To get the temperature of the state point 1, use trial and error process as follows:
No.
1
2
3
(t1)
350
355
354
(h1)
3138
3150
3148
T1=354 C h1= 3148
WT = ms(h1-h2)+(ms-mprocess) (h2-h3)
2000 = ms(3148-2825)+(ms-1.805)(2825-2423.8)
So, ms= (2000-729.581)/724.2=3.7617 kg/s = 13.5 tons/hr
(h2s)
2710
2718
2716
(hpt)
73.4%
75.5%
75%
2a) The use of multistage compression with intercooling and multistage expansion with
reheating in gas turbine cycle both reduced the power required to drive the compressor and
increase the power output from the turbine so increase the thermal efficiency of the cycle and
in the limit the cycle will approach the air standard Striling cycle.
2b) A gas-turbine plant operates on the regenerative Brayton cycle. The isentropic efficiency
of the compressor, the effectiveness of the regenerator, the air-fuel ratio in the combustion
chamber, the net power output, the back work ratio, the thermal efficiency, the second law
efficiency, the exergy efficiencies of the compressor, the turbine, and the regenerator, and the
rate of the exergy of the combustion gases at the regenerator exit are to be determined.
Solution:
Properties The properties of air at 500ºC = 773 K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K,
R = 0.287 kJ/kg·K, and k = 1.357.
(f) The second-law efficiency of the cycle is defined as the ratio of actual thermal efficiency
to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature
for the cycle can be taken to be the turbine inlet temperature. That is,
3a) Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 55°C at
a rate of 0.018 kg/s and leaves at 750 kPa subcooled by 3°C. The refrigerant enters the
compressor at 200 kPa superheated by 4°C. Determine (a) the isentropic efficiency of the
compressor, (b) the rate of heat supplied to the heated room, and (c) the COP of the heat
pump. Also, determine (d) the COP and the rate of heat supplied to the heated room if
this heat pump operated on the ideal vapor-compression cycle between the pressure limits
of 200 and 800 kPa.
3b) Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 55°C at a
rate of 0.018 kg/s and leaves at 750 kPa subcooled by 3°C. The refrigerant enters the
compressor at 200 kPa superheated by 4°C. Determine (a) the isentropic efficiency of the
compressor, (b) the rate of heat supplied to the heated room, and (c) the COP of the heat
pump. Also, determine (d) the COP and the rate of heat supplied to the heated room if
this heat pump operated on the ideal vapor-compression cycle between the pressure
limits of 200 and 800 kPa.
4-a) An air compressor receives air from the surroundings at 100 kPa and 21 °C. There is a
5.0 kPa drop through the intake valves, and the temperature at the end of intake is 38 °C.
The discharge pressure is 475 kPa and there is a 25 kPa pressure drop through the
discharge valves. Determine: (a) volumetric efficiency, ideal and actual; (b) the power if
the piston displacement is 14.0 liters and the ideal intake volume is 11.2 liters. The
compressor is operating at 200 rev/min and n = 1.35.
Solution:
p1 = 100-5 = 95 kPa po = 100 kPa
p2 = 475 + 25 = 500 kPa , To = 294 K, Ti= 311 K
to verify that p2> discharge pressure
 vol 
11.2
 0.8
14
 95  294 

  0.72
 100  311 
 vol actual  0.8
wcycle =n/(n-1) p1(v1 -v4 ) [(p2/p1)(n-1)/n -1]
wcycle 
n
p1 v-actual VPD [1 - (p 2 /p 1 ) (n -1)/n ]
(n - 1)
wcycle 
1.35
100 (0.72) (0.014) [1 - (500/95) (0.35)/1.35 ]
(1.35 - 1)
= -2.1 kJ/cycle
2.1x 200
W 
7
60
kW
4b) In the constant volume cycle the pressure and temperature at the beginning of the
compression are 1 bar and 43 °C and at the end of compression the temperature is 323 °C
respectively. Determine the air standard efficiency and the compression ratio. A four
stroke petrol engine of six cylinders, with the above compression ratio develops 30
indicated kW power at 2400 r.p.m. The air fuel ratio is 14 volumetric efficiency is 0.82,
and bore to stroke is 0.85. The specific gravity of the fuel is 0.78 and its calorific value is
40500 kJ/kg. Determine the bore and stroke, indicated thermal efficiency and the ratio
between the efficiency of the engine and the efficiency of the air standard cycle.
Solution:
Constant volume cycle (Otto)
P1=1 bar
T1= 43+273 = 316 K
T2=323+273 = 596 K
Find A.S
4 stroke z = 6 IP=30 kW N=2400 rpm A/F=14, vol =0.82 L/D=0.85 s.g of fuel =0.78
CV=40500 kJ/kg
Determine D, L, ith , rel
T2=T1(rv)k-1 so, rv =4.88
A.S = 1-1/(rv)k-1 = 46.96%

2400
Va  D 2 (0.85 D)
.x0.82 x0.85 x6  65.69 D 3
4
2 x60
a 
p
10 5

 1.1
RT 287 * 316
ma  72.26 D 3
Assume: V fuel  10l / h = 2.78x10-6 m3/s
mf = 2.78x10-6 x 780 = 2.168x10-3 kg/s
ma = 14x 2.168x10-3 = 30.352x10-3 kg/s
30.352x10-3 = 72.26 D3 so, D = 7.5 cm L=6.4 cm
IP
30
 ith 

 0.34
m f CV 2.168 x10 3 x40500
 rel 
 ith
0.34

 0.72
 A.S 0.4696