Answer:
16.46 kg/s
Explanation:
We solve this by writing an energy balance around the open feed water heater:
E,in=E,out
We know that heat transfer to surroundings and kinetic and potential energy are zero.
[tex]m_1h_1+m_2h_2=m_3h_3[/tex]
We can use the thermodynamic tables for saturaed liquid and steam to solve for the enthalpies.
For inlet 1:
50°C, 10 bar
[tex]h_1=209.31 kJ/kg[/tex]
For inlet 2:
steam, 10 bar, 200°C
[tex]h_2=2778.1 kJ/kg[/tex]
For outlet 3:
Saturated liquid at 10 bar
[tex]h_3=762.5 kJ/kg[/tex]
We can also perform a mass balance:
[tex]m_1+m_2=m_3[/tex]
We can solve the energy balance:
[tex]60\cdot{209.31}+m_2\cdot{2778.1}=(60+m_2)\cdot{762.5}[/tex]
[tex]12558.6+2778.1\cdot{m_2}=762.5\cdot{m_2}+45750[/tex]
Solve for m₂
[tex]m_2=33191.4/2015.6=16.47[/tex]