20 lb-mol/min of the compound enters a steady-state boiler as saturated liquid at P=1 bar. Find the rate at which heat is added if the exiting stream is:
A) Saturated vapor at P = 1 bar.
B) Vapor at P = 1 bar and T = 200°F.
C) Vapor at P = 0.7 bar and T = 200°F
A) Set an energy balance around the boiler
\begin{aligned}\left\{\mathrm{M}\left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\}=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms
\begin{aligned}& 0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{Q}}\\& \dot{\mathrm{Q}}=\dot{\mathrm{m}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)\end{aligned}
The compound is being vaporized in the boiler, but no temperature or pressure changes occur, therefore \widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=\Delta \widehat{\mathrm{H}}^{\mathrm{vap}}=125 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}
\begin{aligned}& \dot{\mathrm{Q}}=\dot{\mathrm{m}}\left(125 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right)\\& \dot{\mathrm{Q}}=\left(20 \frac{\mathrm{lbmol}}{\min }\right)\left(\frac{50 \mathrm{lb}_{\mathrm{m}}}{1 \mathrm{lbmol}}\right)\left(125 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right)= \mathbf{125,000} \frac{\mathrm{BTU}}{\mathbf{m i n}}\end{aligned}
B) Set an energy balance around the boiler
\begin{aligned}\left\{\mathrm{M}\left(\widehat{\mathrm{U}}+\frac{\mathrm{v}^{2}}{2}+\mathrm{gh}\right)\right\}=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+ \dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms
\begin{aligned}& 0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)+\dot{\mathrm{Q}}\\& \dot{\mathrm{Q}}=\dot{\mathrm{m}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}\right)\end{aligned}
The compound is now also being heated after being vaporized \widehat{\mathrm{H}}_{\text {out }}-\widehat{\mathrm{H}}_{\mathrm{in}}=\Delta \widehat{\mathrm{H}}^{\text {vap }}+\int_{609.65^{\circ} \mathrm{R}}^{659.65^{\circ} \mathrm{C}} \mathrm{C}_{\mathrm{p}}^{*} \mathrm{dT}
\begin{aligned}& \dot{\mathrm{Q}}=\dot{\mathrm{m}}\left(\Delta \widehat{\mathrm{H}}^{\mathrm{vap}}+\int_{609.65^{\circ} \mathrm{R}}^{659.65^{\circ} \mathrm{R}} \mathrm{C}_{\mathrm{p}}^{*} \mathrm{dT}\right)\\& \dot{\mathrm{Q}}=\left(20 \frac{\mathrm{lbmol}}{\min }\right)\left(\frac{50 \mathrm{lb}_{\mathrm{m}}}{1 \mathrm{lbmol}_{\mathrm{m}}}\right)\left(\left(125 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right)+\left(43.7 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right)\right)=\mathbf{1 6 8}, \mathbf{7 0 0} \frac{\mathbf{B T U}}{\mathbf{m i n}}\end{aligned}
C) The pressures in parts B and C are both low enough to assume ideal gas behavior for the vapor phase. Parts B and C have the same inlet stream, and the outlet streams are at the same temperature. The outlet streams are at different pressures, but pressure doesn’t affect the enthalpy of an ideal gas. Consequently, assuming ideal gas behavior, the answer will be the same as in part B.