Answer:
a) [tex]T = 151.83^{\textdegree}C[/tex], b) [tex]\dot m_{2} = 58.61\times 10^{3}\,\frac{kg}{h}[/tex]
Explanation:
A feedwater heater is a mixing chamber that pre-heats water prior to compressor, improving the efficiency of the power cycle. This system is modelled after the First Law of Thermodynamics:
[tex]\dot m_{1}\cdot h_{1} + \dot m_{2}\cdot h_{2} - (\dot m_{1}+ \dot m_{2})\cdot h_{3} = 0[/tex]
The properties at inlets and outlets are, respectively:
Inlet 1 (Subcooled Liquid)
[tex]h_{1} = 188.44\,\frac{kJ}{kg}[/tex]
Inlet 2 (Superheated Vapor)
[tex]h_{2} = 3106\,\frac{kJ}{kg}[/tex]
Outlet 3 (Saturated Liquid)
[tex]h_{3} = 640.09\,\frac{kJ}{kg}[/tex]
a) The temperature at the exit is:
[tex]T = 151.83^{\textdegree}C[/tex]
b) The mass flow rate at inlet 2 is:
[tex]\left(3.2\times 10^{5}\,\frac{kg}{h} \right)\cdot \left(188.44\,\frac{kJ}{kg} \right) + \dot m_{2}\cdot \left(3106\,\frac{kJ}{kg} \right) - \left(3.2\times 10^{5}\,\frac{kg}{h}+\dot m_{2}\right)\cdot \left(640.09\,\frac{kJ}{kg} \right)= 0[/tex][tex]2465.91\cdot \dot m_{2} -144528000 = 0[/tex]
[tex]\dot m_{2} = 58.61\times 10^{3}\,\frac{kg}{h}[/tex]
