Question 7.12: Problem: An insulated mixing chamber has two inlets: the fir...
Problem: An insulated mixing chamber has two inlets: the first has saturated water entering at 100 kPa with a flow rate of 4 kg / s, while the second has 2 kg / s of superheated steam at 100 kPa and 150 °C. The mixture leaves the chamber at a pressure of 100 kPa. Find the rate of entropy generation in this process.
Find: Rate of entropy generation \dot{s}.
Known: Inlet pressure P_1 = 100 kPa, inlet temperature T_1 = 500 °C, mass flow rate \dot{m}_1 = 4 kg /s, inlet pressure P_2 = 100 kPa, inlet temperature T_2 = 150 °C, mass flow rate \dot{m}_2 = 2 kg /s, exit pressure P_3 = 100 kPa.
Diagram:
Governing Equations:
Energy rate balance \dot{Q}+\dot{W} =\sum\limits_{e}{\dot{m}_eh_e } -\sum\limits_{i}{\dot{m}_ih_i}
Entropy rate balance \dot{S}_{gen}+\underbrace{\dot{S}_{heat}}_{=0} =\sum\limits_{e}{\dot{m}_es_e } -\sum\limits_{i}{\dot{m}_is_i}
From a mass balance,
\dot{m}_1+\dot{m}_2=\dot{m}_3 \ , \\ \dot{m}_3=4 \ kg/s+2 \ kg/s=6 \ kg/s.
From an energy balance for a control volume, assuming negligible changes in kinetic and potential energy, and that the chamber is adiabatic and that there is no work done in it
\dot{m}_1h_1+\dot{m}_2h_2=\dot{m}_3h_3 \ .
For saturated water at 100 kPa (Appendix 8b), specific enthalpy of liquid h_1 = h_f = 417.46 kJ/kg. For superheated steam at 100 kPa and 150 °C, specific enthalpy h_2 = 2776.4 kJ / kg. So the exit enthalpy is
4 \ kg/s \times 417.46 \ kJ/kg+2 \ kg/s \times 2776.4 \ kJ=6 \ kg/s\times h_3 \ , \\ h_3=1203.8 \ kJ/kg.
Since h_f< h_3< h_g , we know that the exit state is a saturated vapour–liquid mixture, with quality
x_3=\frac{h_3-h_f}{h_g-h_f}=\frac{1203.8 \ kJ/kg-417.46 \ kJ/kg}{2675.5 \ kJ/kg-417.46 \ kJ/kg}=0.348 \ 24.
For saturated fluid at 100 kPa, specific entropy of liquid s_1 = s_f = 1.3026 kJ / kgK and vapour s_g = 7.3594 kJ / kgK. For superheated steam at 100 kPa and 150 °C, specific entropy s_2 = 7.6143 kJ / kg, so entropy at the exit is
s_3=s_f+x_3(s_g-s_f)=1.3026 \ kJ/kgK+0.34824 \times (7.3594 \ kJ/kgK-1.3026 \ kJ/kgK), \\ s_3=3.4118 \ kJ/kgK.
From an entropy balance,
\dot{S}_{gen}=\dot{m}_3s_3-(\dot{m}_1s_1+\dot{m}_2s_2) , \\ \dot{S}_{gen}= 6 \ kg/s \times 3.4118 \ kJ/kgK-(4 \ kg/s \times 1.3026 \ kJ/kgK+2 \ kg/s \times 7.6143 \ kJ/kg), \\ \dot{S}_{gen}= 0.031 \ 800 \ kW/kgK.
Answer: Entropy is generated at a rate of 0.0318 kW / kgK.