Explanation:
Equation for mass balance is as follows.
[tex]\Delta m_{system} = m_{in} - m_{out} = 0[/tex]
[tex]m_{in} = m_{out}[/tex]
[tex]m_{1} + m_{2} = m_{3} = 2m[/tex] .......... (1)
[tex]m_{1} = m_{2} = m[/tex]
Equation for energy balance is as follows.
[tex]E_{in} - E_{out} = \Delta E = 0[/tex]
[tex]E_{in} = E_{out}[/tex]
Hence, [tex]m_{1}h_{1} + m_{2}h_{2} = m_{3}h_{3}[/tex] ......... (2)
When we combine both equations (1) an d(2) will be as follows.
[tex]m_{1}h_{1} + m_{2}h_{2} = 2mh_{3}[/tex]
or, [tex]h_{3} = \frac{(h_{1} + h_{2})}{2}[/tex]
Hence, putting the given values into the above equation as follows.
[tex]h_{3} = \frac{(h_{1} + h_{2})}{2}[/tex]
[tex]h_{3} = \frac{(33.08 + 1156.2)}{2}[/tex]
= 594.6 Btu/lbm
Therefore, [tex]T_{3} = T_{sat 20psi} = 228^{o}F[/tex]
And, [tex]x_{3} = \frac{h_{3} - h_{f}}{h_{fg}}[/tex]
= [tex]\frac{594.6 - 196.27}{1156.2 - 196.27}[/tex]
= 0.415