Answer:
[tex]\dot m_{2} = 0.199\,\frac{kg}{s}[/tex]
Explanation:
The mixing chamber can be modelled by applying the First Law of Thermodynamics:
[tex]\dot W_{in}+\dot m_{1}\cdot h_{1} +\dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0[/tex]
Since that mass flow rate of water at inlet 1 is the only known variable, the expression has to be simplified like this:
[tex]\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - z\cdot h_{3} = 0[/tex]
Besides, the following expression derived from the Principle of Mass Conservation is presented below:
[tex]1 + y = z[/tex]
Then, the expression is simplified afterwards:
[tex]\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - (1+y)\cdot h_{3} = 0[/tex]
[tex]\frac{\dot W_{in}}{\dot m_{1}} +h_{1} - h_{3} + y\cdot (h_{2}-h_{3}) = 0[/tex]
Specific enthalpies are obtained from steam tables and described as follows:
State 1 (Superheated vapor)
[tex]h = 2994.3\,\frac{kJ}{kg}[/tex]
State 2 (Saturated liquid)
[tex]h = 1008.3\,\frac{kJ}{kg}[/tex]
State 3 (Liquid-Vapor mixture)
[tex]h = 2444.22\,\frac{kJ}{kg}[/tex]
The ratio of the stream at state 2 to the stream at state 1 is:
[tex]y = \frac{\frac{\dot W_{in}}{\dot m_{1}}+h_{1}-h_{3}}{h_{3}-h_{2}}[/tex]
[tex]y = \frac{\frac{10\,kW}{0.5\,\frac{kg}{s} }+2994.3\,\frac{kJ}{kg}-2444.22\,\frac{kJ}{kg} }{2444.22\,\frac{kJ}{kg}-1008.3\,\frac{kJ}{kg} }[/tex]
[tex]y = 0.397[/tex]
The mass flow rate of the saturated liquid is:
[tex]\dot m_{2} = y\cdot \dot m_{1}[/tex]
[tex]\dot m_{2} = 0.397\cdot (0.5\,\frac{kg}{s} )[/tex]
[tex]\dot m_{2} = 0.199\,\frac{kg}{s}[/tex]
