Answer:
Work done = 96.15 KJ
Heat transferred = 24 KJ
Explanation:
We are given;
Mass of air;m = 1.5 kg
Initial pressure;P1 = 100 KPa
Initial temperature;T1 = 17°C = 273 + 17 K = 290 K
Final volume;V2 = 0.5V1
Since the polytropic exponent is 1.3,thus it means;
P2 = P1[V1/V2]^(1.3)
So,P2 = 100(V1/0.5V1)^(1.3)
P2 = 100(2)^(1.3)
P2 = 246.2 KPa
To find the final temperature, we will make use of combined gas law.
So,
(P1×V1)/T1 = (P2×V2)/T2
T2 = (P2×V2×T1)/(P1×V1)
Plugging in the known values;
T2 = (246.2×0.5V1×290)/(100×V1)
V1 cancels out to give;
T2 = (246.2×0.5×290)/100
T2 = 356.99 K
The boundary work for this polytropic process is given by;
W_b = - ∫P. dv between the initial boundary and final boundary.
Thus,
W_b = - (P2.V2 - P1.V1)/(1 - n) = -mR(T2 - T1)/(1 - n)
R is gas constant for air = 0.28705 KPa.m³/Kg.K
n is the polytropic exponent which is 1.3
Thus;
W_b = -1.5 × 0.28705(356.99 - 290)/(1 - 1.3)
W_b = 96.15 KJ
The formula for the heat transfer is given as;
Q_out = W_b - m.c_v(T2 - T1)
c_v for air = 0.718 KJ/Kg.k
Q_out = 96.15 - (1.5×0.718(356.99 - 290))
Q_out = 24 KJ
