Question

Water is pumped from a depth of 10 m and delivered through a pipe of cross section 10–2 m2 upto a height of 10 m. If it is needed to deliver a volume 0.2 m3 per second, find the power required.

Answer 1

Answer on Question#56928 - Physics - Mechanics - Relativity

Water is pumped from a depth of $h_d = 10 \, \text{m}$ and delivered through a pipe of cross section $A = 10^{-2} \, \text{m}^2$ up to a height of $h = 10 \, \text{m}$. If it is needed to deliver a volume $Q = 0.2 \, \text{m}^3$ per second, find the power required.

Solution:

The pressure difference between the ends of the pipe is given by

$$\Delta p = \rho g (h + h _ {d}),$$

Where $\rho = 1000\frac{\mathrm{kg}}{\mathrm{m}^3}$ – is the density of the water, and $g = 9.8\frac{\mathrm{m}}{\mathrm{s}^2}$ – is the acceleration due to gravity. The speed of water stream is

$$v = \frac {Q}{A}$$

The total power (power to overcome pressure $\Delta p$ and power to accelerate water to the speed $v$) is given by

$$\begin{array}{l} P = \Delta p Q + \frac {\rho Q v ^ {2}}{2} = \rho g (h + h _ {d}) Q + \frac {\rho Q ^ {3}}{2 A ^ {2}} = \\ = 1000 \frac {\mathrm {kg}}{\mathrm {m} ^ {3}} \cdot 9.8 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} (10 \mathrm {m} + 10 \mathrm {m}) + \frac {1000 \frac {\mathrm {kg}}{\mathrm {m} ^ {3}} \left(0.2 \frac {\mathrm {m} ^ {3}}{\mathrm {s}}\right) ^ {3}}{2 \cdot (10 ^ {- 2} \mathrm {m} ^ {2}) ^ {2}} = 236 \mathrm {kW} \end{array}$$

Answer: 236 kW.

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