The pump in Fig. 8–14 is used to supply water through the 100-mm-diameter pipe at 3 m/s. If the measured power loss through the pipe is 3 kW, determine the power loss under similar conditions when a 75-mm-diameter pipe is used.

Question

Accepted Answer

By definition, power is the product of force and velocity, [latex] \dot{W}=F V [/latex].
Here the pump provides a pressure on the water to push it through the pipe, and since F = ΔpA, then

[latex] \dot{W}=\Delta p\left(\pi D^2 / 4 ight) V [/latex]

or

[latex] \frac{\dot{W}_1}{\dot{W}_2}=\left(\frac{\Delta p_1}{\Delta p_2} ight)\left(\frac{D_1}{D_2} ight)^2\left(\frac{V_1}{V_2} ight) [/latex] (1)

The relationship between the pressures and velocities is obtained from Euler number similitude, and between the velocities and the dimensions from Reynolds number similitude. For the same fluid (water), and using the pipe diameter D for the “characteristic length” L, we have

[latex] \begin{aligned} & \left(\frac{ ho V D}{\mu} ight)_1=\left(\frac{ ho V D}{\mu} ight)_2 ; \quad V_2=V_1\left(\frac{D_1}{D_2} ight)=(3 m / s )\left(\frac{100 mm }{75 mm } ight)=4 m / s \ & \left(\frac{\Delta p}{ ho V^2} ight)_1=\left(\frac{\Delta p}{ ho V^2} ight)_2 ; \quad \Delta p_2=\left(\frac{V_2}{V_1} ight)^2 \Delta p_1=\left(\frac{4 m / s }{3 m / s } ight)^2 \Delta p_1=1.7778 \Delta p_1 \end{aligned} [/latex]

Substituting these results into Eq. 1, we get

[latex] \begin{aligned} \frac{3 kW }{\dot{W}_2} & =\left(\frac{\Delta p_1}{1.7778 \Delta p_1} ight)\left(\frac{100 mm }{75 mm } ight)^2\left(\frac{3 m / s }{4 m / s } ight) \ \ \dot{W}_2 & =4.00 kW \end{aligned} [/latex]

Question 8.8: The pump in Fig. 8–14 is used to supply water through the 10...

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