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Question 14.9: The radial-flow pump in Fig. 14–19a is used to transfer wate......

The radial-flow pump in Fig. 14–19a is used to transfer water from the lake at A into a large storage tank, B. This is done through a 100-mm-diameter pipe that is 100 m long and has a friction factor of 0.015. The manufacturer’s data for the performance of the pump is given in Fig. 14–19b. Determine the flow if this pump is selected with a 175-mm impeller to do the job. Neglect minor losses.

Fig. 14-19
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Fluid Description. We assume steady, incompressible flow while the pump is operating.
System Equation. We can relate the pump head to the flow Q by applying the energy equation between the water levels at A and B. The control volume contains the water within the pipe and a portion of the water in the lake and tank. Since the friction factor has been given, we do not need to obtain its value from the Moody diagram.

\frac{p_{A}}{\gamma}+\frac{V_{A}^{2}}{2g}+z_{A}\,+\,h_{\mathrm{pump}}=\frac{p_{B}}{\gamma}+\frac{V_{B}{}^{2}}{2g}+\,z_{B}\,+\,h_{\mathrm{turb}}\,+\,h_{L}\\\left(h_{\mathrm{pump}}\right)_{\mathrm{act}}=25\,\mathrm{m}\ +\ (0.015)\left(\frac{100\,\mathrm{m}}{0.1\,\mathrm{m}}\right)\left[\frac{V^{2}}{2(9.81\,\mathrm{m}/\mathrm{s}^{2})}\right]

Also,

Q = V [π(0.05 m)²]

When the above two equations are combined, we get

\left(h_{\mathrm{pump}}\right)_{\mathrm{act}}=\left[25+12.394(10^{3})\,Q^{3}\right]\mathrm{m}                               (1)

A plot of this equation is shown in Fig. 14–19b as Eq. 1. It shows the actual pump head (h_{pump})_{act} that must be supplied by the pump to provide a flow Q_{req’d} for the system. Along with this curve are the manufacturer’s curves for the pump. (For convenience Q is reported in m³/s; however, in practice it is commonly in liters/min.) We see that for the pump with a 175-mm-diameter impeller, the operating point O is where the performance and system head curves intersect, graphically at about Q = 0.0295 m³/s and (h_{pump})_{act} = 36.0 m. Therefore, if this pump is selected, then the efficiency for this flow is determined from the efficiency curves to be about η = 74%.* By comparison, this is far from the best efficiency point for the pump (86%), and so the choice of this pump with a 175-mm impeller would not be appropriate. Instead, performance curves for other pumps would have to be considered to match a more efficient pump to the system.
It is interesting to note that if the elevation difference between the water levels in the lake and storage tank were 40 m, rather than 25 m, then Eq. 1 would plot as the dashed curve in Fig. 14–29b. In this case, the operating point O’ gives a flow of about 0.0225 m³/s; however, the operating efficiency is about η = 86%, making the pump the best choice for this condition.

^*As noted by the curves, use of a smaller diameter impeller would give a higher efficiency.

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