Question 15.8: Five lbm per minute of saturated liquid aqua–ammonia at 100 ......
Five lbm per minute of saturated liquid aqua–ammonia at 100 psia and 220 F is mixed with 10 lbm/min of saturated vapor aqua–ammonia at 100 psia and 220 F. Heat transfer from the mixture is 4200 Btu/min. Find the enthalpy, temperature, and concentration of the final mixture.
The concentration x_{3} of the mixture is given by Eq. 15-18. Using Chart 2 in Appendix E, we get
x_{3} = x_{1} + \frac{\dot{m}_{2}}{\dot{m}_{3}} (x_{2} – x_{1}) (15-18)
x_{3} = 0.214 + \frac{10}{15} (0.845 – 0.214) = 0.635 \frac{lbm ammonia }{lbm of mixture}
The total mass flow rate \dot{m}_{3} is the sum of the two inlet flow rates and
\frac{\dot{q}}{\dot{m}_{3}} = \frac{4200}{15} = 280 Btu/lbm
Then using Chart 2, point 3 is located at x_{3} = x_{3^{′}} = 0.635 and
i_{3} = i_{3^{′}} – \frac{\dot{q}}{\dot{m}_{3}} = 578 – 280 = 298 Btu/lbm
Using a straightedge and the equilibrium construction line, we find that the final temperature t_{3} is about 148 F.
