Explanation:
As it is given that stream A is super heated. From stream tables, we get that specific enthalpy, ([tex]h_{A}[/tex]) is 3054.29 kJ/kg.
For stream B, it is saturated water at 25 degree celsius. It's [tex]h_{B}[/tex] is 2546.54 kJ/kg.
Stream C, data will be as follows.
P = 200 kPa, [tex]\chi[/tex] = 0.9
So, [tex]h_{c}[/tex] = [tex]h_{f} + \chi \times h_{fg}[/tex]
= 504.47 + 0.9 \times 2201.7
= 2486 kJ/kg
Now, energy balance formula will be as follows.
[tex]m_{A}h_{A} + m_{B}h_{B}[/tex] = [tex](m_{A} + m_{B})h_{c}[/tex]
[tex]3 \times 3054.29 + m_{B} \times 2546.54[/tex] = [tex](3 + m_{B}) \times 2486[/tex]
[tex]m_{B}[/tex] = 28.16 kg/s
Hence, inflow of saturated liquid is 28.16 kg/s. According to steam table, temperature of wet steam is [tex]120.23^{o}C[/tex]
Mass flow rate of out flow is 31.16 kg/s.
