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Chapter 12: Problem 17

Water at \(40^{\circ} \mathrm{C}\) is pumped from an open tank through \(200 \mathrm{m}\) of \(5 \mathrm{C}\) -mm-diameter smooth horizontal pipe as shown in Fig. \(P 12.17\) and discharges into the atmosphere with a velocity of \(3 \mathrm{m} / \mathrm{s}\). Minor losses are neglaible. (a) If the efficiency of the pump is \(70 \%\), how much power is being supplied to the pump? (b) What is the NPSH at the pump inlet? Neglect losses in the short section of pipe connecting the pump to the tank. Assume standard atmospheric pressure.

Short Answer

Expert verified
The solution should present the numerical values for power supplied to the pump and the NPSH at the pump inlet, given by the previous steps. For more precision, use up to two decimal places.

Step by step solution

01

Use the Bernoulli's Principle

Let's first denote the top of the tank as point 1, and the outlet into the atmosphere as point 2. Based on the Bernoulli’s principle, we can formulate the following: \[ P_{1}+ρgh_{1}+0.5ρv_{1}^{2}+P_{pump} = P_{2}+ρgh_{2}+0.5ρv_{2}^{2} \] where \( P_{1} \) and \( P_{2} \) are the pressures at points 1 and 2, respectively, \( h_{1} \) and \( h_{2} \) are the heights at points 1 and 2, \( v_{1} \) and \( v_{2} \) are the velocities at points 1 and 2, and \( P_{pump} \) is the pressure added by the pump.
02

Apply Given Conditions and Solve

As we assume standard atmospheric pressure, \( P_{1} \) and \( P_{2} \) will be \( 101325 \) Pa. Also, because the pipe is horizontal, \( h_{1} = h_{2} \), and \( v_{1} \) will be negligible compared to \( v_{2} \) because of the larger cross-sectional area of the tank. The density of water at \( 40^{\circ} \mathrm{C}\) is \( 992.2 \) kg/m^3. Thus, by simplifying we get, \( P_{pump} = 0.5ρv_{2}^{2}\), which substituting the values and solve, we will obtain the value for \( P_{pump} \) in watts.
03

Obtain the Efficiency of the Pump

The efficiency of the pump is given. So, the power supplied to the pump can be calculated using the relation: \[ Power_{supplied} = \frac{P_{pump}}{Efficiency} \] Substitute the calculated \( P_{pump} \) and given efficiency value (0.7) into the equation and solve for \( Power_{supplied} \).
04

Calculate NPSH

The Net Positive Suction Head (NPSH) can be given by the formula: \[ NPSH = \frac{P_{1} - P_{vapor}}{ρg} \] where \( P_{1} \) is the pressure at the pump inlet which is same as atmospheric pressure and \( P_{vapor} \) is the vapor pressure of water at \( 40^{\circ} \mathrm{C}\) which is \( 7370 \) Pa. Substituting these values and solving yields the NPSH in meters.

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