Answer to Question #230099 in Molecular Physics | Thermodynamics for Unknown346307

Question

Dry saturated steam at 100 bar expands isothermally and reversibly to a

pressure of 10 bar. Calculate per kg of steam :

(i) The heat supplied ;

(ii) The work done

Answer 1

Solution;

Given ,

Initial pressure,$p_1=100$ bar

Final pressure,$p_2=10$ bar

At 100 bar,for saturated steam,from steam tables ,we have;

$s_1=s_g=5.619kJ/kg$

$t_{s_1}=311°c$

At 10 bar and 311°c ,the steam is superheated,hence by interpolation;

$s_2=7.124+(\frac{311-300}{350-300})(7.301-7.124)=7.163kJ/kgK$

(i)The heat supplied;

$Q=T(s_2-s_1)$

$Q=584(7.163-5.619)$

$Q=901.7kJ/kg$

(ii)The work done;

Applying non-flow energy equation;

$Q=(u_2-u_1)+W$

$W=Q-(u_2-u_1)$

From steam tables;

At 100 bar,dry saturated steam;

$u_1=u_g=2545kJ/kg$

At 10 bar,311°c ,by interpolation;

$u_2=2794+(\frac{311-300}{350-300})(2875-2794)$

$u_2=2811.8kJ/kg$

$W=901.7-(2811.8-2545)$

$W=634.9kJ/kg$