Question

Water at 80°F and 30 psia is heated in a chamber by mixing it with saturated water vapor at 30 psia. If both streams enter the mixing chamber at the same mass flow rate, determine the temperature and the quality of the exiting stream. The enthalpy of water at 80°F is 48.065 Btu/lbm, and enthalpy of water at 30 psia is 1164.1 Btu/lbm. Use data from the steam tables.

Answer 1

Answer:

Quality=41%

Temperature=250.30 °F

Explanation:

Hello,

In this case, from the energy balance over the chamber at equal inlet flow rates, we have:

$mh_1+mh_2=2mh_{out}\\h_1+h_2=2h_{out}\\h_{out}=\frac{48.065+1165.1}{2}=606.08Btu/lbm$

Then, looking for such value in table A-5E at the same pressure (30 psia) since it does not change during the mixing process, we obtain:

$x=\frac{606.08Btu/lbm-218.93Btu/lbm}{945.21Btu/lbm}=0.409=41\%$

And the temperature at that pressure is just 250.30 °F

Best regards.