‫‪Model Answer for final exam‬‬
‫‪Mechanical Engineering Technology‬‬
‫‪Prepared by‬‬
‫‪Dr. Mohamed Saied Shehata‬‬
‫‪2010/2011‬‬
‫حل أمتحان تكنولوجيا الهندسة الميكانيكية‬
‫الفرقة ‪ :‬الثانية (مدني)‬
‫المعهد العالي للتكنولوجيا‬
‫الممتحن أ‪.‬د‪/‬محمد سعيد عطية شحاتة‬
‫يوم االمتحان‪ :‬األحد ‪6122-2-62‬‬
‫‪1‬‬
Benha University
Mechanical Eng. Tech. Dep.
‫بسم هللا الرمحن الرحيم‬
B.H.I.T.
Mechanical Eng. Technology
Second Year Civel Eng.
Final Exam June 2011 , Time 3 hr
Notes:- Allowable for using steam tables and charts. Assuming any missing data
First Question
PV
 const
T
V = 0.02271 m3/mol at 0 oC and 1 bar
2. For ideal gas (air):
Cv 
5
R =20.75 J/kmol K and C p  7 R =29.099 J/mol K for air
2
2
(a)
Initial (1 bar, 25 oC)

final ( 5 bar, 25 oC)
(b )
(a) Air is Cooled at constant pressure followed by heating at constant volume
(b) Air is heated at constant volume followed by cooling at constant pressure
For two cases (a) and (b) Find:
Q, W , U , H  ?
Solution:298.15
)  0.02479 m 3
273.15
1
 (0.02479)( )  0.004958 m 3
5
V1  (0.02271)(
1-
P
V2  V1 1
P2
 1bar (T=?)   
 (25C,5bar)
Air(25C,1bar)   
cooling
heating
P const
V const
1st step: const P
V
0.004958
T2  T1 2 , T2  (298.15)(
)  59.63 K
V1
0.02479
ideal gas
,
Q1)  H1)  C pT1)  ( 29.099)(59.63  298.15)  6,941 J
U1)  H   ( PV )  H  PV
2
 6,941  (1  105 )(0.004958  0.02479)  4,958 J
W1)  U1)  Q1)  4,958  ( 6,941)  1,983 J
2nd step : const V
Q2)  U 2)  Cv T2)  ( 20.785)( 298.15  59.63)  4,958 J
W2)  U 2)  Q2)  0  V  0
H 2)  U 2)  VP  4,958  0.004958(5  105  1  105 )  6,941 J
Entire process:
Q  Q1)  Q2)  6,941  4,958  1,983 J
U  U1)  U 2)  4,958  4,958  0
W  W1)  W2)  1,983  0  1,983 J
H  H1)  H 2)  6,941  6,941  0
:
U  Q  W  1,983  1,983  0
H  U   ( PV )  0  0  0
(entire process)
(entire process)
(b) Air(25C, 1 bar)    5 bar (T=?)
heating
V coonst
  
 (25C, 5 bar)
cooling
P const
T=(298.15) (
1
)  1,490.75 K
5
1ST step: const V
Q1)  U1)  Cv T1)  ( 20.785)(1,490.75  298.15)  24,788 J
3
W1)  U1)  Q1)  0
H1)  U1)  VP  24,788  0.02479(5  105  1  105 )  34,703 J
2nd step: const P
Q2)  H 2)  C pT2)  ( 29.099)( 298.15  1490.75)  34,703 J
U 2)  H 2)   ( PV )
 34,703  (5  105 )(0.004958  0.02479)  24,788 J
W2)  U 2)  Q2)  24,788  ( 34,703)  9,915 J
Entire process:
Q  24,788  34,703  9,915 J
W  0  9,915  9,915 J
ΔU  24,788  24,783  0 J
ΔH  34,703  34,703  0 J
Q & W : (a)  (b)
ΔU & ΔH : (a)  (b)
for I.G. U(T) , H(T)

Actual process: b.p. N 2  195.8 C  77.35 K

b.p. O 2  183 C  90.15 K
Second equation
2- An ideal dual engine takes in air at a pressure of 101 kPa and temperature of 50 Co. It is
compressed to 1:14 of its original volume. Heat is added in two stages, during the first stage
of heat addition, the pressure increases at constant volume to twice the pressure at the
clearance volume after that heat add at constant pressure to the end of heat addition process.
The air is then allowed to expand adiabatically to the end of the expansion stroke where it is
exhausted. If the cut of ratio is equal two, Calculate:
a. The temperature and pressure at the ends of each process of the cycle
b. The ideal thermal efficiency.
c. The power of the cycle if air mass flow is 2kg/s
4
Solution
:
T2 = T1 (14)y-1=323 (14) 0.4 = 930 K
3
P
γ
V 
1.4
P  P  1   (14)
x 101  4064 kPa
2 1 V 
 2
 P  2 x 4064  8128 Kpa
3
4
2
5
P
8128
T  T 3  930 x
 1860 K
3 2P
4064
2
V 
T  T  4  1860 x 2  3720 K
4 3V 
 3
Now, Expansion ratio =
T 
5
T
4
(V /V )
5 4
γ -1

1
V
V
5  14  7
V
2
4
3720
 1710 K
70.4
Heat added (Qadd) = CV(T3-T2) +Cp (T4-T3)
= 0.71 (1860-930) - 1.0035 (3720 – 1860)
= 2527 kJ/kg of air
Heat rejected (Qrej)= CV (T5-T1) = 0.71 (1710-323)
= 985 kJ/kg
Air standard efficiency (ηη 
Q
add.
Q
Q
Rej.
Rej.

2527  985
 0.61 or 61 %
2527
.
.
Power  m W  m* (Q
 Q )  2(2527  985)  3084 kW
net
add. Rej.
5
Third Question
3- The ideal air Brayton cycle operates with air entering the compressor at 95 kPa, 22 oC.
The pressure ratio rp is 6:1 and the air leaves the heat addition process at 1100 K.
Determine the compressor work and the turbine work per unit mass flow, the cycle
efficiency, the back work ratio, and compare the compressor exit temperature to the
turbine exit temperature.
6
7
Fourth Question
2-Consider the closed, rigid container of water shown below. The
pressure is 700 kPa, the mass of the saturated liquid is 1.78 kg, and the
mass of the saturated vapor is 0.22 kg. Heat is added to the water until
the pressure increases to 8 MPa. Find the final temperature, enthalpy,
and internal energy and heat added to the water. Does the liquid level
rise or fall?
Solution
For the closed system the total mass is
constant and since the process is one
in which the volume is constant, the
average specific volume of the
saturated mixture during the process
is given by
Saturated
Vapor,
mg, Vg
Saturated
Liquid,
v=V/m=constant
v2=v1
or
Now to find v1 recall that in the two-phase region at state 1
x1 
mg1
m f 1  mg1

0.22 kg
 0.11
(1.78  0.22) kg
then, at P = 700 kPa
v1  v f 1  x1 (vg1  v f 1 )
 0.001108  (0.11)(0.2728  0.001108)
m3
 0.031
kg
8
mf, Vf
State 2 is specified by: P2 = 8 MPa, v2 = 0.031 m3/kg
At 8 MPa = 8000 kPa,
vf = 0.001384 m3/kg
At 8 MPa,
vg = 0.02352 m3/kg
v2 = 0.031 m3/kg
Is v2  v f ? No


Is v f  v2  v g ? No
Is v g  v2 ? Yes
Therefore, State 2 is superheated.
Interpolating in the superheated tables at 8 MPa, v = 0.031 m3/kg gives
T2 = 361 C
h2 = 3024
kJ/kg
u2 = 2776
kJ/kg
Since state 2 is superheated, the liquid level falls.
9
√Fifth Question -For the following sentences chose the correct answer.
1
NO.
Sentence
Work and heat are state functions but internal energy is path function
2
Temperature is extensive property but mass is intensive property
3
Energy can be created and cannot be transformed from one form to other
4
Electrical energy, chemical energy, potential energy, kinetic energy wind energy all have
the same SI units
5
Cycle is defined as a sum of group of stats
6
7
Work is equal to area of enclosed cycle on T-S diagram and heat is equal to area on P-V
diagram
For closed cycle the net heat is equal the net work
8
The work for open system is equal the work for closed system
9
Compression ratio is equal (V2/V1) for diesel cycle and equal (V1/V2) for Otto cycle
10
System can be changed from open to closed be changing system boundary
11
For dual cycle cut of ratio is equal (V4/V3) and pressure ratio is equal (P3/P2)
12
For simple Brayton cycle compression ratio is equal to (P 2/P1)=(P3/P4)
13
Triple point is intersection of saturated liquid line with saturated vapor line.
14
Critical point is the point of maximum temperature and pressure of wet region
15
Dryness fraction of wet vapor is changed between (0 and 1 )
16
For power cycle heat is added to get work and for reversed cycle work done to transfer heat.
17
Oxygen and Nitrogen are pure substance
18
Saturated liquid line is the line of boiling points for different pressure
19
Saturated vapor line is the line of complete evaporation
20
Critical thickness is conditions of optimum isolation
21
Conduction heat transfer occurs in solid substances and convection in different fluids.
22
The arrangement of isolations affecting on thermal residence of system
23
For air reversed cycle the fluid change its phase from one compound to other through cycle.
24
For Brayton cycle steam is used as working fluid for energy transfer.
25
Chose of isolation is not effected by operating temperature
26
Compression ratio for actual diesel engine is higher than Petrol engine
27
28
For the same compression ratio and heat rejection, thermal efficiency for Petrol engine is higher
than diesel engine.
Specific heat at constant volume is always greater than specific heat at constant pressure
29
The boiler and pump are operate at constant pressure
10
True
False
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30
Compressor and turbine are used for power generation
31
Pumps are used for pressure head increases but condenser is used for heat transfer
32
Flow work is equal to work for open system and work for closed system
33
Open Brayton cycle used very clean fuel but closed Brayton cycle used very cheap fuel
34
36
For diesel cycle heat added at constant volume but for Otto cycle heat added at constant
pressure.
First law of thermodynamic is applied for obtaining general equation for conduction heat
transfer.
Conduction heat transfer and convection heat transfer can be found in one system
37
Enthalpy and heat are different form of potential energy
38
39
For compression ratio equal one, thermal efficiency for air stander cycles have maximum
values.
For Brayton cycle the optimum pressure ratio gives maximum cycle power
40
Brayton cycle is applied for air conditioning but reversed cycle is used for power generation
35
11
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