Benha University
Mech. Eng.Tech. Department
First Semester
Higher Institute of Tech.
Third year of Mechanical Eng.
Finial exam 2010/2011
Course No.( Thermodynamics (2) MEP(212)
Third year of Mechanical Engineering
M3ohamed4@yahoo.com
Model Answer for final exam
Prepared by
Dr. Mohamed Saied Shehata
‫حممد سعيد عطية شحاتة‬/‫د‬
2010/2011
1
Exam solution
First Question
3. An ideal diesel cycle. Air enters at 101kPa and 20 oC. The compression ratio is 18
and constant pressure energy addition ceases at 10% of the swept volume (stroke
volume). The engine uses 100 m3 of air/h. if  = 1.4 .Determine:
a) The maximum temperature and pressure in the cycle
b) The thermal efficiency of the engine
c) The indicate power of engine.
Solution
Let clearance volume = V2
Swept volume = V1-V2=V2(r-1)=V2(18-1)=17V2
10% of the swept volume = 17V2 x 0.1 = 1.7V2
Constant pressure energy addition ceases at V2 + 1.7V2 = 2.7V2
rc=V3/V2 = 2.7V2/V2 = 2.7
Thermal efficiency
 γ

1  rc  1 
1
η 1 
1 
γ  1   (r  1) 
1.4
18  1
r
 c

 2.71.4 1 

 1  0.40  0.6 or 60%
1.4 (2.7  1) 
T  2  2 73  293 K,
P 101 kpa.
1
1
γ
V 


1
P P x
101x (18)1.4  5777 kpa.
2 1 V 
 2
P  P  5777 kpa.
3 2
γ 1
V 
T T x 1 
 293 x (18)1.4  1
2 1 V 
 2
 293 x 3.175  930 K
P V P V
2 2 3 3
T
T
2
3
V
T  T 3  930 x 2.7  2510 K or 2237 oC
3 2V
2
The maximum temperature and pressure of the cycle are 2237Co and 5777 kPa
respectively.
v1=RT1/P1=0.287*293/101=0.8326 m3/kg
v2=v1/r=0.8326/18=0.04625 m3/kg
2
v3=2.7v2=2.7*0.04625=0.1249 m3/kg
v4=v1=0.8326 m3/kg
P4=P3 (v3/v4)1.4 = 5777 x (0.1249/0.8326)1.4=405 kPa
T4=P4v4/R=405*0.8326/0.287=1175 K
q add=CP(T3-T2) =1.0035(2510-930)=1585.53 kJ/kg
qrej.=CV(T4-T1) = 0.71(1175-293)=626 kJ/kg
Wnet=qadd - qrej.=1585.53-626=959 kJ/kg
ρ1=P1/RT1=101/(0.287*293)=1.2 kg/m3
mair= ρ*Vair=1.2*100/60*60=0.0334 kg/s
Indicate Power = mair*Wnet=0.0334*959=32 kW
Second Question
3. Isentropic efficiency of each compressor
80 %
Total pressure ratio of each compressor
2:1
Total pressure loss in the intercooler
6.896 kPa
Total pressure loss for each side of the heat exchanger
10.3 kpa
Thermal ratio of heat exchanger
0.75
Total pressure loss in the combustion chamber
13.7
Combustion efficiency
98%
Isentropic efficiency of compressor turbine
87%
Mechanical transmission efficiency
99%
Total pressure loss in the re-heater
10.3 kPa
Combustion efficiency of re-heater
98 %
Isentropic efficiency of power turbine
80%
Ambient air temperature and pressure
15 oC and 101 kpa.
Maximum cycle temperature
1000 K
Air mass flow
22.7 kg/s
Heating value of fuel
43054 kJ/Kg.
During compression Cp = 1.0035 kJ/kg K and  = 1.4
During heating and expansion Cp = 1.15 kJ/kg K and  = 1.33
Determine:
a. Net power output., B- Specific fuel consumption., C- Overall thermal efficiency.
Solution
1) The output from the compressor turbine must be just sufficient to
derive the two coupled compressors.
2) The pressure and temperature at intake to the first compressor is
equal to the ambient pressure and temperature.
3
8
Inter Cooler
2-3
2
1
Heat exchanger
4-9
3
4
Combustion
Chamber 6-7
Combustion
Chamber
5-9
6
9
10
1 –2
First
Compressor
3-4
Second
Compressor
5-6
First
Turbine
T
5
8
7
5
7-8
Second
Turbine
7
6
8
4
2
3
10
1
S
Fig. (2-20)
Thus

T  288 K
1
and
P 101
1
Kpa
288  0.286 
(T  T ) 
2
 1  79

2 1 0.80 
T  288  79  367 K and
P  2 X 101  202 Kpa
2
2
Because of pressure loss in inter-cooler equal 6.896 Kpa, so,
P2 = 202-6.896 = 195 kpa
4
Electrical
Generator
For perfect inter-cooling then
T  T  288 K
3
1
288  0.286 
T T 
2
 1  79

4
3 0.8 
T  288  79  367 K
4
P  2 x 195  390 kPa
4


and
The work required to drive compressor per kg of air/sec will be
W  C (T  T )  (T  T   1.0035{( 367  288)  (367  288)}
C
p  2 1
4
3) 
1.0035 X 79 X2 158.6 kJ/kg
Power output from the compressor turbine will be
158.6
W 
160 KJ/kg
T1
0.99
Now C (T  T ) 160
P 5
6
160
T T 
138.5  T   875.87 K
5
6 1.155
6






1


and also
T  T  0.87 X T 1 
5 6
5T
γ 1 


 (r ) γ 
p








1


138.5  0.87x1000 1 
 r  2.015
p
γ 1 


 (r ) γ 
p


From where
But
(rp)T1 = 2.015
P5 = P4– [ Pressure loss in H.E. (air side)
+ Pressure loss in combustion chamber]
= 390 – (10.3 + 13.7) = 366 Kpa

366
P 
181.6
6 2.015
kPa
Allowing the pressure loss during reheat
P7 = 181.6 – 10.3 = 171.3 Kpa
5
The exhaust pressure would have been 1.03 kg/cm2 if there is no pressure
loss in the gas side of heat exchanger.
 Pressure ratio of power turbine R
T2
will be
171.3
(r ) 
1.54
p T2 111.3
Therefore power turbine output








1
1
  0.80 x 1000 1 
T  T  η T 1 

7
8
T2 7
γ 1 
1 - 1.33



 (1.54) 1.33
 (r ) γ 

p









From where [Gas is reheated in the reheated up to the same max.
temperature]
T7T2 – T8T2 = 80
Out put/kg/ sec =CP(T7T2 – T8T2)= 80 X 1.15 = 92 kJ/Kg/S
WT2(kW/kg sec) = 92 kJ/kg/s
Net power output =mair*WT2= 92 x 22.7
= 2088 KW
Now to determine fuel consumption, temperature rise in the combustion
chamber and re-heater must be found.
Temperature rise during second combustion chamber
 T  T 1000  (1000  138.5) 138
7
6
Temperature rise in first combustion chamber = T5-T 9
T T
4
Heat Exchanger effectiven ess ( )  9
T T
8
4
So,
(T - T ) = 0.75 (T - T )
9 4
8 4
6

T  367  0.75 (1000  80  34)
8
 781 K

T  T  1000  781  219
5 9
A heat supplied/k g of air
 1.15 (219  138.5)
 411.1 kj/kg
If mfuel be the amount of fuel/kg of air then
mfuel x 0.98 X 43054 = 411 kJ/kg
or
mfuel = 0.00974 Kg fuel/kg air
0.00974 X 22.7 X 3600
 0.3812 kg fuel/kW her
2088
92
Overall efficiency 
411/0.98
 22%

The design performance of any gas turbine plant supplying shaft power
can be determined on the basis of the method illustrated above.
Third Question
2. Consider the cogeneration plant shown blew. Steam enters the turbine
at 7 MPa and 500 oC. Some steam is extracted from the turbine at 500
kpa for process heating. The remaining steam continues to expand to
5kpa. Steam is then condensed at constant pressure and pumped to the
boiler pressure of 7Mpa. At times of high demand for process heat,
some steam leaving the boiler is throttled to 500 kpa and is routed to
the process heater. The extraction fractions are adjusted so that steam
leaves the process heater as a saturated liquid at 500 kpa. It is
subsequently pumped to 7 Mpa. The mass flow rate of steam through
the boiler is 15kg/s. Disregarding any pressure drops and heat losses in
the piping and assuming the turbine and the pump to be isentropic,
determine: (a) The maximum rate at which process heat can be
supplied, (b) The power produced and the utilization factor when no
process heat is supplied, and (c) The rate of process heat supply when
10 percent of the steam is extracted before it enters the turbine and
70percent of the steam is extracted from the turbine at 500kpa for
process heating.
7
Schematic and T-S Diagram
Solution:
The schematic of the cogeneration plant and the T-s diagram of the cycle
are shown blew the work input to the pumps and the enthalpies various
states are as follows:
Wpumpl,in.= 8(P9-P8)=(0.001005m3/kg)[(7000-5)kpa] = 7.03kj/kg
Wpumpl,in.= 8(P9-P8)=(0.001005m3/kg)[(7000-5)kpa] = 7.03kj/kg
Wpumpll,in=7(P10 - P7)=(0.001095m3/kg)[(7000-500)kpa] = 7.12kj/kg
h1=h2=h3=h4=3410.3kj/kg
h5=2738.2 kj/kg (x5=0.995)
h6=2071.9kj/kg (x6=0.798)
h7=hf
500kPa=640.23
kj/kg
h8= hf 5kPa=137.82 kj/kg
h9=h8+wpumpl.in=(137.82+7.03)kj/kg=144.85kj/kg
h10=h7+Wpumplt.in=(640.23+7.12)kj/kg=647.35kj/kg
(a) The maximum rate of process heat is achieved when all the steam
leaving the boiler is throttled and sent to the process heater and none
is sent to the turbine (that is, m`4=m`1=15 kg/s and m`3=m`5=m`6=0).
Thus
8
Qp,max=m`1(h7-h4)=(15kg/s) [(3410.3-640.23)kj/kg]=41.551kw
The utilization factor is 100 percent in this case since no heat is
rejected in the condenser and heat losses from the piping and other
components are assumed to be negligible.
(b) when no process heat is supplied, all the steam leaving the boiler
will pass through the turbine and will expand to the condenser
pressure of 5 kpa(that is, m`3=m`6=m`=15kg/s and m`=m`=m`=0)
maximum power will be produced in the mode, which is determined
to be
W`turb.out=m`1(h3-h6)=(15kg/s)[(3410.3-2071.9)kj/kg]=20.076 kw
Wpump.in=W`pumpl.in+W`pumpll.in=m`1Wpumpt.in
=(15 kg/s)(7.03kj/kg)=105kw
W`net.out=W`turb.out-Wpumpl.in=(20.076-105)kw=19.971kw
Q`in=m`1(h1-h11)=(15kg/s)[(3410.3-144.85)kj/kg]=48.982kw

 ` Q
W
net
p (19.971  0)kj/kg
Thus, ε 

 0.408(or40.8%)

u
48.982kj/k g
Q
in
That is, 40.8 percent of the energy is utilized for a useful purpose. Notice
that the utilization factor is equivalent to the thermal efficiency in this
case.
(c) Neglecting any kinetic and potential energy changes, an energy
balance on the process heater yields.
 W
 h  m
 h
Q
  m
e e
i i

 h m
 h m
 h
Q m
7 7
4 4 5 5
Where m`4=(0.1)(15kg/s)=1.5kg/s
M`5=(0.7)(15kg/s)=10.5kg/s
Thus, Q`=(12kg/s)(640.23kj/kg)-(1.5kg/s)(3410.3 kj/kg)
-(10.5 kg/s)(2738.2kj/kg) = -26.184 kw
or
Q`p=26.184kw
9
That is, 26,184 kw of the heat transferred will be utilized in the process
heater. We could show that 10.299 kw of power is produced in this case,
and the rate of heat input in the boiler is 42.951kW. Thus the utilization
factor is 84.9 percent.
Fourth Question
3. Refrigerant 12 is used as the working fluid in vapor refrigeration
cycle. Saturated liquid leaves the condenser at pressure of 1.4 Mpa.
Saturated vapor enters the Compressor at 12 oC. If the mass flow rate
of the refrigerant is 0.008 kg/s. Determine:a) Compressor power in kW, refrigeration capacity in ton ,
(C.O.P)ref
b) If the compressor has an efficiency of 80% and the liquid
leave the condenser at 48 oC, find compressor power ,
refrigeration capacity and (C.O.P)ref.
Solution
1. at point 1 where inlet compressor at saturated vapor at temperature
=12 Co
h1=192.56 kJ/kg , s1=0.6913 kJ/kg.
From 1-2 isentropic compression, So, s2=s1 = 0.6913 kJ/kg
state 3 is saturated liquid at 1.4 MP h3 = h4 =91.46KW
The refrigeration capacity is equal
Qin=m( h1- h4 )= 0.008*60 (192.56 – 91.46) kJ/kg
kw/ton 
3.516
3.516

 0.699
(C.O.P)
5.08
ref.
h
2s
1
η

Com. h  h
2 1
h
h
h  h  ( 2s
2
1
z
h
1 )  192.56  (212.71  192.56)  217.75 Kj/kg
0.8
c
B) State 2 is determine by P2 =1.4 and P2 =1.4 Mpa
From table s2 = 0.706 J/kg .
To determine state 3
10
At T= 48 oC pressure = 1.639 bar, hf =82.83 kJ/kg .
From the equ. of determination of properties in liquid region
h3 =hf(48)+vf(P3–Psat)= 82.83+(0.000819) (1.4–1.1639)x 106 = 83.02 kJ/kg
s3=sf(48c) =0.2973 kJ/kg,
h3 = h4 =82.03 kJ /kg
h h
83.02  47.26
X  4 4f 
 0.2461
4 h h
145.3
g4
4f
S  S  x (S  S )  0.1817  (0.2461)(0 .6913  0.1817)  0.3071kj/k g
4
4f
4 4g f4
W  m(h  h )  0.008 * 60(192.56 - 83.02) = 2.579 kJ/min.
c
1 4
= 52.579/211 = 0.249 ton refrigerat ion
C.O.P.)
ref.

h h
1
4  192.56  83.02  4.3
h h
217.75  192.56
2
1
Condenser
3
2
Throttle Valve
Compressor
4
1
Evaporator
T
T
2s
2
2
3
3
h3=h4
4
h3=h4
1
1
4
S
S
Case (A)
Case (B)
11
12