Question

Steam at 7 bar and dryness fraction 0.95 expands in a cylinder behind a

piston isothermally and reversibly to a pressure of 1.5 bar. The heat supplied during the process

is found to be 420 kJ/kg. Calculate per kg :

(i) The change of internal energy ;

(ii) The change of enthalpy ;

(iii) The work done.

Answer 1

$$P_1=7bar=7\times10^{5}N/m^2\\P_2=1.5bar=1.5\times10^{5}N/m^2\\Q=420kj/kg$$

Internal energy

$u=(1-x)u_f+xu_g$

$$u=(1-0.95)\times696+(0.95\times2573)\\u_1=2479.15KJ/kg\\u_2=2580+\frac{25}{50}\times(2856-2580)\\u_2=2602.8kj/kg$$

Gain

$u_2-u_1=123.65kj/kg$

Enthalpy

$h_1=h_f+x_1h_{fg1}$

$$h_1=697.1+095\times2064.9=2658.75kj/kg\\h_2=2802.69kj/kg$$

Enthalpy

$$h=h_2-h_1=2802.69-2658.75=143.94kj/kg$$

Work done

$Q=(u_2-u_1)+w$

$$w=420-(123.65)=296.35kj/kg$$