written 8.2 years ago by | • modified 3.4 years ago |
the cycle efficiency and mean effective pressure.
written 8.2 years ago by | • modified 3.4 years ago |
the cycle efficiency and mean effective pressure.
written 8.2 years ago by | • modified 8.2 years ago |
p1=0.1MPa,T1=35+273=308K,r=v1v2=8:1,heat supplied=2100KJ/Kg
Adiabatic compression process 1-2:
T2T1=v1v2(γ−1)=r(γ−1)=8(1.4−1)
T2=2.297×308=707.476K
p1vγ1=p2vγ2
p2p1=v1v2γ=81.4=18.379
p2=18.379×1=18.379 bar
Constant volume process 2-3:
Heat added during the process,
cv(T3−T2)=2100
0.72(T3−707.476)=2100
T3=3624.143K
Also,
p2T2=p3T3
p3=p2T3T2
=18.379×3624.143707.476
p3=94.14 bar
Thermal efficiency,
ηth=1−1r(γ−1)
=1−18(1.4−1)
ηth=.5674=56.74%
∴pressure ratio,rp=p3p2 =94.1418.379=5.122
The mean effective pressure is given by,
pm=p1r[(r(γ−1)−1)(rp−1)](γ−1)(r−1)
=(1×8[8(1.4−1)−15.122−1](1.4−1)(8−1)
pm=15.28bar
written 3.4 years ago by | • modified 3.4 years ago |
1.243457723×10^-3 this is my answer