Answer to Question #231111 in Molecular Physics | Thermodynamics for Unknown346307

Question

A rigid cylinder of volume 0.028 m3 contains steam at 80 bar and 350°C.

The cylinder is cooled until the pressure is 50 bar. Calculate :

(i) The state of steam after cooling ;

(ii) The amount of heat rejected by the steam.

Answer 1

Gives

Volume (v)=0.028m³

$P_1=80$ bar

T=315°C

$P_2=50 bar$

Mass=$\frac{0.028}{0.02995}=0.935kg$

Internal energy

$$u_1=h_1-p_1V=2987.3-80\times10^5\times\frac{0.02995}{1000}$$

$u_1=2747.7Kj/kg$

State of steam after cooling

$$x_2=\frac{v_2}{v_{g2}}=\frac{0.02995}{0.0394}=0.76$$

amount of heat rejected by the steam

$$u_2=(1-x_2)u_{f2}+x_2u_{g2}\\u_2=(1-0.76)\times1149+0.76\times2597=2249.48Kj/kg$$