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:

$W = \frac{P_2V_2 - P_1V_1}{1 - γ}$

$W = \frac{2 × 0.11545 - 5 × 0.06}{1 - 1.4} × 10^5$

$W = 0.17275 × 10^5 J$

For constant Pressure process, For Q = 0.

Q - W = Δh

Therefore, -W = Δh

-W = 80 kJ

Net Work done = -80 + 17.275

= -62.725 kJ