Answer:
Total work: -5.25 kJ
Total Heat: 52 kJ
Explanation:
V0 = 0.15
P0 = 350 kPa
t0 = 150 C = 423 K
P1 = 105 kPa (isentropical transformation)
Δh1-2 = 52 kJ (at constant pressure)
Ideal gas equation:
P * V = m * R * T
m = (R * T) / (P * V)
R is 0.287 kJ/kg for air
m = (0.287 * 423) / (350 * 0.15) = 2.25 kg
The specifiv volume is
v0 = V0/m = 0.15 / 2.25 = 0.067 m^3/kg
Now we calculate the parameters at point 1
T1/T0 = (P1/P0)^((k-1)/k)
k for air is 1.4
T1 = T0 * (P1/P0)^((k-1)/k)
T1 = 423 * (105/350)^((1.4-1)/1.4) = 300 K
The ideal gas equation:
P0 * v0 / T0 = P1 * v1 / T1
v1 = P0 * v0 * T1 / (T0 * P1)
v1 = 350 * 0.067 * 300 / (423 * 105) = 0.16 m^3/kg
V1 = m * v1 = 2.25 * 0.16 = 0.36 m^3
The work of this transformation is:
L1 = P1*V1 - P0*V0
L1 = 105*0.36 - 350*0.15 = -14.7 kJ/kg
Q1 = 0 because it is an isentropic process.
Then the second transformation. It is at constant pressure.
P2 = P1 = 105 kPa
The enthalpy is raised in 52 kJ
Cv * T1 + P1*v1 = Cv * T2 + P2*v2 + Δh
And the idal gas equation is:
P1 * v1 / T1 = P2 * v2 / T2
T2 = T1 * P2 * v2 / (P1 * v1)
Replacing:
Cv * T1 + P1*v1 + Δh = Cv * T1 * P2 * v2 / (P1 * v1) + P2*v2
Cv * T1 + P1*v1 + Δh = v2 * (Cv * T1 * P2 / (P1 * v1) + P2)
v2 = (Cv * T1 + P1*v1 + Δh) / (Cv * T1 * P2 / (P1 * v1) + P2)
The Cv of air is 0.7 kJ/kg
v2 = (0.7 * 300 + 105*0.16 + 52) / (0.7 * 300 * 105 / (105 * 0.16) + 105) = 0.2 m^3/kg
V2 = 2.25 * 0.2 = 0.45 m^3
T2 = 300 * 105 * 0.2 / (105 * 0.16) = 375 K
The heat exchanged is Q = Δh = 52 kJ
The work is:
L2 = P2*V2 - P1*V1
L2 = 105 * 0.45 - 105 * 0.36 = 9.45 kJ
The total work is
L = L1 + L2
L = -14.7 + 9.45 = -5.25 kJ
