Question 15.7: Twenty lbm per minute of liquid water–ammonia solution at 15......
Twenty lbm per minute of liquid water–ammonia solution at 150 psia, 220 F, and concentration of 0.25 lbm ammonia per lbm of solution is mixed in a steady-flow adiabatic process with 10 lbm per minute of saturated water–ammonia solution at 150 psia and 100 F. Determine the enthalpy, concentration, and temperature of the mixture.
Chart 2 in Appendix E will be used to carry out the solution, and Fig. 15-24 shows the procedure. State 1 is a subcooled condition located on the diagram at t = 220 F and x = 0.25. State 2 is a saturated liquid and is located at the intersection of the 150 psia boiling line and the 100 F temperature line. States 1 and 2 are joined by a straight line. State 3 is located on the connecting line and is determined by the concentration x_{3} or enthalpy i_{3}. Using Eq. 15-18, we get
x_{3} = x_{1} + \frac{\dot{m}_{2}}{\dot{m}_{3}} (x_{2} – x_{1}) (15-18)
x_{3} = x_{1} + \frac{\dot{m}_{2}}{\dot{m}_{3}} (x_{2} – x_{1}) = 0.25 + \frac{10}{30} (0.72 – 0.25)
x_{3} = 0.41 lbm ammonia/lbm mixture
Because the resulting mixture state lies within the saturation region, it is a mixture of vapor and liquid. To determine the temperature t_{3} a graphical trial-and-error procedure shown in Fig. 15-24 is used. The fractional proportions of liquid and vapor in the mixture may also be determined from Chart 2 in a graphical manner:
\frac{\dot{m}_{l}}{\dot{m}_{3}} = \frac{\bar{l} _{3}}{\bar{l} _{v}} = 0.038 (15-20)
Therefore, the mixture is 3.8 percent vapor and 96.2 percent liquid.
