A) Work = Force × Distance = Mass × Acceleration × Distance
Distance upward that the water is pumped is changing as time passes. Initially it is 20 meters: 5+15.
When there is 1 kg of water in the tower, the water level may be found by considering the density and dimensions of the water tower.
(1 \mathrm{~kg})\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}}\right)\left(\frac{1 \mathrm{~cm}^{3}}{1 \mathrm{~g}}\right)=1000 \mathrm{~cm}^{3}
Volume = Area of Base of Water Tower × Height =\frac{\pi}{4} Diameter² × Height
1000 \mathrm{~cm}^{3}=\frac{\pi}{4}(300 \mathrm{~cm})^{2} \times Height
This is the change in water level that accompanies adding 1 kg of water to the tower. It is so small that we will consider it negligible; the “distance” is 20 meters both at the beginning and at the end of the process of pumping 1 kg of water into the tower.
Work =1 \mathrm{~kg} \times 9.8 \frac{\mathrm{m}}{\mathrm{sec}^{2}} \times 20 \mathrm{~m}=\bf196 \mathbf{Nm}=\bf 196 \mathbf{~J}
B) The cylinder is 4 meters tall so the height is now 24 meters.
Work = Force × Distance = Mass × Acceleration × Distance
Work =1 \mathrm{~kg} \times 9.8 \frac{\mathrm{m}}{\mathrm{sec}^{2}} \times(20 \mathrm{~m}+4 \mathrm{~m})=\bf 235.2 \mathbf{Nm}=\bf 235.2 \mathbf{~J}
C)
Set up an energy balance. The control volume is as shown in the figure- it includes the faucet, the pipes leading to the faucet, and a portion of the water in the tower- but a portion of the tower, the top, is outside of the system. Thus water leaving the faucet is leaving the system, and the top layer of water is entering the system as the tower drains. This definition allows us to model the system as a steady state system. Of course, once all the water above the system has drained into the system, the system can no longer remain at steady state if water continues to flow out.
The maximum velocity occurs when the water tower is full. To find the absolute maximum velocity, we will assume the height of the water entering the system is 19 m off the ground- in other words, the “layer” of water that is above the system is very thin.
Assume steady state conditions yield maximum velocity, and frictionless piping.
\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}_{\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, use the height of the faucet at a baseline (h=0), this means h=16 at the top of the tower.
\begin{aligned}\mathrm{gh}_{\mathrm{in}} & =\left(\frac{\mathrm{v}_{\text {out }}^{2}}{2}\right) \\\mathrm{v}_{\text {out }} & =\sqrt{2\left(9.8 \frac{\mathrm{m}}{\mathrm{sec}^{2}}\right)(16 \mathrm{~m})} \\\mathrm{v}_{\text {out }} & =\bf17.7 \frac{\mathbf{m}}{\mathbf{sec}}\end{aligned}
D) The energy balance is identical to that in the previous problem, except here, the difference in height between the faucet and the top of the tower is now 24 meters (5+15+4).
\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}+g h_{\mathrm{in}}\right)-\dot{\mathrm{m}}_{\text {out }}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\text {out }}^{2}}{2}+g h_{\mathrm{out}}\right)+\dot{\mathrm{W}}_{\mathrm{S}}+ \\& \dot{\mathrm{W}}_{\mathrm{EC}}+\dot{\mathrm{Q}} \\& \mathrm{gh}_{\mathrm{in}}=\left(\frac{\mathrm{v}_{\text {out }}^{2}}{2}\right) \\& \mathrm{v}_{\text {out }}=\sqrt{2\left(9.8 \frac{\mathrm{m}}{\sec ^{2}}\right)(24 \mathrm{~m})} \\& \mathrm{v}_{\text {out }}=\bf21.7 \frac{\mathbf{m}}{\bf sec }\end{aligned}
