Question

The radiator of a steam heating system has a volume of 20 L and is filled with superheated vapor at 200 kPa and 150°C. At this moment both the inlet and exit valves to the radiator are closed. After a while the temperature of the steam drops 40C as a result of heat transfer to the room air. Determine the entropy change of the steam during this process.

Answer 1

Solution;

Given;

$V=20L =0.02m^3$

At $P_1=200kPa,T_1=150°c;$

$v_1=0.93m^3/kg$

$s_1=7.2kJ/kgK$

At $T_2=40°c$

$v_1=v_2$

$v_f=0.001m^3/kg$

$v_g=19.51m^3/kg$

Therefore;

$v_2=v_f+x(v_g-v_f)$

$0.93=19.51+x(19.51-0.001)$

$x=0.047$

$s_2=s_f+xs_{fg}$

$s_2=0.57+0.047(7.638)$

$s_2=0.93kJ/kgK$

Hence;

$\Delta s=s_1-s_2=7.2-0.93$

$\Delta s=6.27kgK$

$m=\frac{V}{v_1}=\frac{0.02}{0.93}=0.022kg$

$\Delta s=6.27×0.022$

$\Delta s=0.138kJ/K$