Mechanical Engineering: Thermodynamics and Engine Analysis Homework

This document presents a detailed solution to a mechanical engineering homework assignment focusing on thermodynamics and engine analysis. The solution encompasses a variety of problems, including calculations related to adiabatic processes, internal energy changes, and work done in thermodynamic cycles. It analyzes engine performance, determining parameters such as mechanical efficiency, brake power, specific fuel consumption, and thermal efficiencies (indicated and brake). The document provides step-by-step solutions, including the application of relevant formulas and principles, offering a comprehensive understanding of the concepts involved. Furthermore, the solution covers the analysis of multi-cylinder engines, calculating key parameters such as the mean effective pressure and engine dimensions. The document provides a complete breakdown of the calculations and analysis, making it a valuable resource for students studying mechanical engineering.

4) i)
γ= C p
Cv
γ= 1
0.72 =1.389
Pbottom
1
γ V bottom=Ptop
1
γ V top
V bottom
V top
= 10
1
V bottom
V top
= Ptop
Pbottom
1
γ =¿ Ptop
Pbottom
1
γ =10
Ptop
Pbottom
=101.389=24.49
Pbottom=0.286 MPa
Similarly,
T bottom
1
γ−1 V bottom=T top
1
γ −1 V top
Ttop
Tbottom
1
γ −1 =10
Ttop
Tbottom
=100.389=2.45
T bottom=723.673 K =450.67℃
iii)
∆ U =Cv ∆T
∆ T = ( 450.67−1500 ) ℃=−1049.33 K
Cv=0.72 KJ /kgK
∆ U =0.72 × (−1049.33 ) =−755.515 KJ /kg
ii)
For adiabatic process,
∆ W =−∆U
∆ W =755.515 KJ /kg
iv)
Since it is an adiabatic process heat flow is 0.

5)
PV =mRT
V 1=0.0298
Similarly,
V 2=0.19=V 3
P3 ×V 3=P4 × V 4
220000 ×0.19=1400000× V 4
V 4 =0.0298
a)
First step is adiabatic process,

P1
1−γ T 1
γ =P2
1−γ T 2
γ
P1=1400 KN
m2 ,T 1=633 K , P2=100 KN
m2
T 2=14
1−γ
γ ×633
Next step is isochoric process,
T2
P2
= T 3
P3
T 3=633 K , P3 =220 KN
m2
(14 ¿¿ 1−γ
γ ×633)
100 = 633
220 ¿
14
1− γ
γ = 5
11
γ=1.426
b) Putting the value of γ=1.426
In T 2we get the value of T 2=287.727
γ= C p
Cv
=1.004
Cv
=1.426
Cv=0.704 KJ
kgK
Change in internal energy in adiabatic process
∆ U =m× Cv ×(T 2−T1 )
∆ U =−55.912 KJ
7)
a)
Number of cylinder, ncyl = 4;
No. of strokes per cycle for the engine, S = 2;

Power developed, I.P. = 36 kW;
Engine speed, N = 3000 r.p.m.;
Actual mean effective pressure, Pam = 900 KN/m2 = 9bar
Length of stroke, L = 1.4 D (bore);
No. of missed cycle, nmc = 0
d=¿
d=¿
d=56.66 mm
Length = 1.4 X d = 79.32 mm
b)
Mechanical efficiency = Brake Power/Indicative Power
Brake Power = 0.83 * 36 = 29.88 KW
Brake thermal efficiency= Brake Power
mf × CV
0.29= 29.88
mf × 45800
mf =0.00225
BSFC (brake specific Fuel Consumption) = mf
Brake Power
BSFC= 0.00225
29.88 × 3600=0.000075 ×3600=0.271 g /kWh
8)
Mean effective pressure of cylinder 1 = 10.15 bar
Mean effective pressure of cylinder 2 = 9.98 bar
Mean effective pressure of cylinder 3 = 9.96 bar
Mean effective pressure of cylinder 4 = 9.92 bar
10.0025 bar = Pam
d = 90mm = 0.09m

L = 95mm = 0.095m, N = 300 rpm. S = 4, Net Brake Load = 340 N
a)
I . P=
100 ×10.0025 ×0.095 × π
4 ×(0.09 ×0.09)×( 2 ×300
4 )× 4
60
I . P=6.045 KW
Effective brake circumference
b)
Reff =0.46
BP= 2 πNT
1000× 60 = 2 πN Reff (W −S )
1000 ×60 =2 π × 300× 0.46 ×340
60000 =4.913 KW
c)
Mechanical Efficiency = B.P/I.P = 81.28%.
d)
Fuel used per hour (mf )= 20 * 0.74 / 3600 = 4.11 ×10−3kg/sec
Specific fuel consumption = 4.11 × 10−3
4.913 ×3600=¿ 3.002
e)
CV =46500 KJ /kg
Indicated thermal efficiency = I . P
mf ×CV
I . P
mf ×CV = 6.045
4.11× 10−3 × 46500 =0.0316=3.16 %
f)
Brake thermal efficiency= Brake Power
mf × CV
Brake thermal efficiency= 4.913
4.11 ×10−3 × 46500 =0.0257=2.57 %