Question 8.5.5: Equilibrium Flash Vaporization A 30 wt% NH3 solution at 100 ......
Equilibrium Flash Vaporization
A 30 wt% NH_{3} solution at 100 psia is fed at a rate of 100 lb_{m}/h to a tank in which the pressure is 1 atm. The enthalpy of the feed solution relative to the reference conditions used to construct Figure 8.5-2 is 100 Btu/lb_{m}.
The vapor composition is to be 89 wt% NH_{3}. Determine the temperature of the stream leaving the tank, the mass fraction of NH_{3} in the liquid product, the flow rates of the liquid and vapor product streams, and the rate at which heat must be transferred to the vaporizer.
Basis: 100 lb_{m}/h Feed
From Figure 8.5-2,
x_{V} = 0.89 lb_{m} NH_{3}/lb_{m} \\ \boxed{T = 120°F} \\ \boxed{x_{L} = 0.185 lb_{m} NH_{3}/lb_{m}} \\ \hat{H}_{V} = 728 Btu/lb_{m} \\ \hat{H}_{L} = 45 Btu/lb_{m}
From Equation 8.5-8
\frac{L}{F} = \frac{x_{V} – x_{F}}{x_{V} – x_{L}} =\frac{\overline{AC} }{\overline{BC} } (8.5-8)
\frac{\dot{m}_{L}}{100 lb_{m}/h} = \frac{x_{V} – x_{F}}{x_{V} – x_{L}} \\ \left. \Large{\Downarrow} \right.\\ \dot{m}_{L} =(100 lb_{m}/h) \frac{0.89 – 0.30}{0.89 – 0.185} = \boxed{84 lb_{m}/h \text{ liquid product}} \\ \dot{m}_{V} =(100 – 84) lb_{m}/h = \boxed{16 lb_{m}/h \text{ vapor product}}
Energy Balance: \dot{Q} = \Delta \dot{H} =\dot{m}_{V} \hat{H}_{V} + \dot{m}_{L} \hat{H}_{L} – 100 \hat{H}_{F}
=[(16)(728) + (84)(45) – (100)(100)] Btu/h = \boxed{5400 Btu/h}
