A) Do an energy balance on the waterfall, “System A” in the figure.
Energy balance on System A:
\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\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}_{\mathrm{out}}^{2}}{2}+g h_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms: this is a steady state process. Assume adiabatic and assume specific enthalpy is unchanged as the water falls:
\begin{aligned}& 0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\frac{\mathrm{v}_{\text {in }}^{2}}{2}+g h_{\text {in }}\right)-\dot{\mathrm{m}}_{\text {out }}\left(\frac{\mathrm{v}_{\text {out }}^{2}}{2}+g h_{\text {out }}\right) \\& \frac{0}{\dot{\mathrm{m}}}=\frac{\dot{\mathrm{m}}_{\text {in }}\left(\frac{\mathrm{v}_{\text {in }}^{2}}{2}+g h_{\text {in }}\right)-\dot{\mathrm{m}}_{\text {out }}\left(\frac{\mathrm{v}_{\text {out }}^{2}}{2}+g h_{\text {out }}\right)}{\dot{\mathrm{m}}}\\& 0=\left(\frac{\mathrm{v}_{\text {in }}^{2}}{2}+g h_{\mathrm{in}}\right)-\left(\frac{\mathrm{v}_{\text {out }}^{2}}{2}+g h_{\mathrm{out}}\right)\\& \frac{\mathrm{v}_{\text {in }}^{2}}{2}=\frac{1}{2}\left(2 \frac{\mathrm{m}}{\mathrm{s}}\right)^{2}=2 \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}}\\& \mathrm{gh}_{\mathrm{in}}=\left(9.8 \frac{\mathrm{m}}{\mathrm{sec}^{2}}\right)(50 \mathrm{~m})=490 \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}}\\& \mathrm{gh}_{\mathrm{out}}=0\end{aligned}
Rearranging to solve for our velocity at Point 2
\begin{aligned}& \frac{\mathrm{v}_{\text {out }}^{2}}{2}=\frac{\mathrm{v}_{\mathrm{in}}^{2}}{2}+g h_{\mathrm{in}}-\mathrm{gh}_{\mathrm{out}}\\& \frac{\rm v_{\text {out }}^{2}}{2}=2 \frac{\mathrm{m}^{2}}{\sec ^{2}}+490 \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}}-0\\& \mathbf{v}_{\mathbf{out }}=\bf 31.37 \frac{\mathbf{m}}{\mathbf{sec}}\end{aligned}
B) Now do an energy balance on the pool, system B in the figure:
\begin{aligned}\frac{\mathrm{d}}{\mathrm{dt}}\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}_{\mathrm{out}}^{2}}{2}+g h_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+\dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}}\end{aligned}
Cancelling terms-here again the process is steady state. Assume the water enters and leave the system at the same height, since the depth of the stream leaving the pool is unspecified:
0=\dot{\mathrm{m}}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\mathrm{in}}^{2}}{2}\right)-\dot{\mathrm{m}}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}\right)
Note:\mathrm{h}_{\text {out }} and \mathrm{h}_{\mathrm{in}} are identical, therefore, they cancel each other when \mathrm{m}_{\mathrm{in}}=\mathrm{m}_{\text {out }}.
\begin{aligned}& \frac{0}{\dot{\mathrm{m}}}=\frac{\dot{\mathrm{m}}_{\text {in }}\left(\widehat{\mathrm{H}}_{\text {in }}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}\right)-\dot{\mathrm{m}}_{\text {out }}\left(\widehat{\mathrm{H}}_{\text {out }}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}\right)}{\dot{\mathrm{m}}} \\& 0=\left(\widehat{\mathrm{H}}_{\text {in }}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}\right)-\left(\widehat{\mathrm{H}}_{\text {out }}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}\right)\\& \frac{\rm v_{\text {out }}^{2}}{2}=0 \\& \frac{\rm v_{\text {in }}^{2}}{2}=492.04 \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}}\\& \left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)=\left(-492.04 \frac{\mathrm{m}^{2}}{\mathrm{sec}^{2}}\right)\left(\frac{1 \mathrm{~J}}{1 \,\mathrm{Nm}}\right)\left(\frac{1 \mathrm{~N}}{\frac{1 \,\mathrm{kgm}}{\mathrm{s}^{2}}}\right)\left(\frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}\right)\\& \left(\widehat{\mathrm{H}}_{\mathrm{in}}\right)-\left(\widehat{\mathrm{H}}_{\mathrm{out}}\right)=-0.49204 \frac{\mathrm{kJ}}{\mathrm{kg}}\\& C_{P}\left(T_{\text {in }}-T_{\text {out }}\right)=-0.49204 \frac{\mathrm{kJ}}{\mathrm{kg}}\\& \left(4.19 \frac{\mathrm{kJ}}{\mathrm{kg}^{\circ} \mathrm{C}}\right)\left(T_{\text {in }}-T_{\text {out }}\right)=-0.49204 \frac{\mathrm{kJ}}{\mathrm{kg}}\\& \bf T_{\text {out }}=25.12^{\circ} {C}\end{aligned}
