Answer to Question #217158 in Molecular Physics | Thermodynamics for Mhyzta

Question

0.05kg of steam at 10 bar, dryness fraction of 0.84 is heated reversibly in a rigid vessel until the pressure is 20 bar. Calculate the change in the total entropy, the heat supplied and the average temperature at which heat is added. Show the area which represents the heat supplied on a T-s diagram 

Answer 1

$m_s = 0.05kg\\$

At 10 bar,

$\begin{aligned} s_1 &= s_{f1} + x_1s_{fg1}\\ &= 2.138 + 0.84(4.448)\\ &= 5.874 kJ/J.kg \end{aligned} \\ ​$

At 10 bar,

$v_{g1} = 0.1944m³/kg \\ v_1 = x_1(v_{g1}) v_1 = 0.84×0.1944 = 0.163m³/kg\\ \dfrac{V_1}{V_2} = (\dfrac{P_1}{P_2})^{\frac{1}{2}}\\ V_2 = 0.163(10/20)^{1.1} = 0.163 × 0.533 = 0.087m³/kg x_2 = v_2/v_{g2} = 0.082/3.992 = 0.021 \\ s_2 = s_{f2} + x_2s_{g_2} = 1.026 + 0.021(6.646) = 1.166 kJ/kg.K$

Change of entropy = $s_2 - s_1 = 1.166 - 5.874 = -4.708 kJ/kg.K$