Answer:
i) Heat rejection = 1.0288 KJ
Network production = 1.212 kJ
ii) Thermal efficiency = 54.08%
mean effective pressure = 353.5 kPa
Explanation:
Given data :
Compression ratio ( r ) = 7
P1 ( initial pressure ) = 90 kPa,
T1 ( initial temperature ) = 27°C + 273 = 300 K
V1 = 0.004 m^3
Max cycle Temperature = 1127°C for each repetition of the cycle
i) Determine the heat rejection and net work production
considering that process 1-2 is an Isentropic compression
Find ; T2 = T1 ( r )^1.4-1
T2 = 300 ( 7 )^0.4 = 653.371 K
P2 = P1( r )^k
= 90 ( 7 )^1.4 = 1372.081 kPa
considering that process 3-4 is an Isentropic expansion
T4 = T3 / r^k-1
= 1400 / 7^(1.4 -1 ) = 642.82 K
Next ; Calculate the value of m
m = P1V1 / RT1 = 90(0.004) / 0.287 ( 300 )
m = 4.18 * 10^-3 kg
Finally :
amount of heat rejected = mCv ( T4 - T1 )
= 4.18 * 10^-3 ( 0.718 ) ( 642.82 - 300 )
Qout = 1.0288 KJ
amount of heat added ( Qin ) = mCv ( T3 - T2 )
= 4.18 * 10^-3 ( 0.718 ) ( 1400 - 653.371 )
hence Qin = 2.2408 kJ
Network production( Wnet) = Qin - Qout
= 2.2408 - 1.0288 ) KJ
= 1.212 kJ
ii) Determine the thermal efficiency and mean effective pressure of the cycle
Thermal efficiency = 1 - Qout / Qin
= 1 - ( 1.0288 / 2.2408 ) = 54.08%
mean effective pressure = Wnet / V1 ( 1 - 1/r )
= 1.212 / 0.004 ( 1 - 1/7 )
= 353.5 kPa