/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 210 \( \mathrm{A}\) pump draws water... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 210

\( \mathrm{A}\) pump draws water from a reservoir through a 150-mm-diameter suction pipe and delivers it to a \(75-\mathrm{mm}\) diameter discharge pipe. The end of the suction pipe is \(2 \mathrm{m}\) below the free surface of the reservoir. The pressure gage on the discharge pipe \((2 \mathrm{m}\) above the reservoir surface) reads 170 kPa. The average speed in the discharge pipe is \(3 \mathrm{m} / \mathrm{s}\). If the pump efficiency is 75 percent, determine the power required to drive it.

Short Answer

Expert verified
The power required to drive the pump is approximately 39.7 kW.

Step by step solution

01

Identify the pressure at the suction pipe end

This will be atmospheric pressure as it is open to atmosphere. By convention, we commonly consider atmospheric pressure as a reference, i.e., consider it as zero.
02

Identify the pressures at different points in the system

The gage pressure at the discharge pipe is given as 170 kPa. And it's above the reservoir surface, so we have to subtract the pressure equivalent of this height (which is the liquid column height). The pressure at this point will be \[170000 - (\rho * g * 2)\]. We need to remember that \(g\) is the acceleration due to gravity and \(\rho\) is the density of the water. So, considering the density for water as 1000 kg/m鲁 and acceleration due to gravity as 9.81 m/s虏, the actual pressure at the discharge pipe is \[170000 - (1000 * 9.81*2) = 170000 - 19620 = 150380 \, Pa\].
03

Use Bernoulli's equation

With this information, we can set up Bernoulli's equation from the surface of the reservoir (where the velocity is approximately 0) to the exit point in the pipe: \[P_1 + \frac{1}{2} * \rho * V_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} * \rho * V_2^2 + \rho * g * h_2\]. Here \(P_1\) and \(P_2\) are the pressures at the start and the end, \(V_1\) and \(V_2\) are the velocities and \(h_1\) and \(h_2\) are the heights. Substituting the values, \(0 + 0 + 1000*9.81*0 = 150380 + \frac{1}{2} * 1000 * 3^2 + 1000*9.81*2\]
04

Calculate the power required

By solving the equation of step 3, the left-hand side equals to right-hand side we are getting the total energy as \(3820.1 \, J/kg\). Now we know energy, we can compute power as \(Power = Energy * Efficiency * FlowRate\). Given, Efficiency = 0.75 and the Flow rate can be calculated with \(Rho * A * V\) (where \(A\) is the cross-sectional area of pipe.). Now using area \(A = \pi*(D/2)^2\) Diameter being given as \(0.075m\). The power is then calculated as: \(Power = 3820.1 * 0.75 * 1000 * \pi*(0.075/2)^2 * 3 = 39718.5 \, W\) or \(39.7 \, kW\)

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