0.06 m3 air at 5 bar and $200^oC$ expands isentropically until its pressure becomes 2 bar. It is then heated at constant pressure until the enthalpy increase during this process is 80 KJ.

Given:

$v = 0.06 m^3$; P1 = 5 bar; T1 = 473K; P2 = 2 bar; Δh= 80 KJ

Solution:

For Process 1-2,

$P_1V_1^y = P_2V_2^y$

$2 × V_2^{1.4} = 5 × 0.06^{1.4}$

$V_2 = 0.11545m^3$

Now, work done is given as:

$W = \frac{P_2V_2 - P_1V_1}{1 - γ}$

$W …