Solved Assignments and Study Material for Desklib

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°C
iii)
∆ U =Cv ∆T
∆ T = ( 450.67−1500 ) °C=−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,