This question is incomplete, the complete question is;
An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.
Answer:
a) the maximum pressure in the cycle is 30.01 Mpa
b) the heat transfer to air is 0.7058 KJ
c) mass of Air is 0.002957 kg
Explanation:
Given the data in the question;
We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)
so from the "ideal gas properties of air table"
Pr1 = 238
Pr4 = 2.379
we know that Pressure P1 is only maximum at the beginning of the expansion process,
so
now we express the relative pressure and pressure relation for the process 4-1
P1 = (Pr2/Pr4)P4
so we substitute
P1 = (238/2.379)300 kPa
P1 = 30012.6 kPa = 30.01 Mpa
Therefore the maximum pressure in the cycle is 30.01 Mpa
b)
the Thermal heat efficiency of the Carnot cycle is expressed as;
ηth = 1 - (TL/TH)
we substitute
ηth = 1 - (350K/1200K)
ηth = 1 - 0.2916
ηth = 0.7084
now we find the heat transferred
Qin = W_net.out / ηth
given that the net work output per cycle is 0.5 kJ
we substitute
Qin = 0.5 / 0.7084
Qin = 0.7058 KJ
Therefore, the heat transfer to air is 0.7058 KJ
c)
first lets express the change in entropy for process 3 - 4
S4 - S3 = (S°4 - S°3) - R.In(P4/P3)
S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa
= -0.1989 Kj/Kg.K = S1 - S2
so that; S2 - S1 = 0.1989 Kj/Kg.K
Next we find the net work output per unit mass for the Carnot cycle
W"_netout = (S2 - S1)(TH - TL)
we substitute
W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K
= 169.065 kJ/kg
Finally we find the mass
mass m = W_ net.out / W"_netout
we substitute
m = 0.5 / 169.065
m = 0.002957 kg
Therefore, mass of Air is 0.002957 kg
