/*! 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 52 The volume flow rate of the wate... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 52

The volume flow rate of the water supplied by a well is $2.0 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s} .\( The well is \)40.0 \mathrm{m}$ deep. (a) What is the power output of the pump - in other words, at what rate does the well do work on the water? (b) Find the pressure difference the pump must maintain. (c) Can the pump be at the top of the well or must it be at the bottom? Explain.

Short Answer

Expert verified
Should the pump be at the top or bottom of the well? #tag_content# Answer: The power output of the pump is 78.48 watts. The pressure difference that the pump must maintain is 392400 Pa. The pump can be at the top or at the bottom of the well; the pressure difference that the pump must maintain is the same, but the actual pressure at the top or bottom will vary depending on the position of the pump.

Step by step solution

01

Find the mass of the water moved in one time interval.

To find the mass of water, we use the equation m = 蟻V, where 蟻 is the density of water and V is the volume of water moved . The density of water is 1000 kg/m^3, and the volume flow rate is given as 2.0 x 10^{-4} m^3/s. Using m = 蟻V, we have: m = (1000 kg/m^3) * (2.0 x 10^{-4} m^3/s) m = 0.2 kg/s
02

Calculate the work done by the pump.

The work done by the pump is given by W = mgh, where m is mass, g is the acceleration due to gravity (9.81 m/s^2), and h is the depth of the well. So we have: W = (0.2 kg/s) * (9.81 m/s^2) * (40 m) W = 78.48 J
03

Find the power output of the pump.

The power output of the pump can be calculated using the formula P = W/t, where W is the work done, and t is the time taken. Since we found the mass moved during one second (0.2 kg/s), the time interval is 1 second. Therefore, P = (78.48 J) / (1 s) P = 78.48 W The power output of the pump is 78.48 watts.
04

Determine the pressure difference using Bernoulli's equation.

Bernoulli's equation: P_1 + 1/2蟻v_1^2 + 蟻gh_1 = P_2 + 1/2蟻v_2^2 + 蟻gh_2 The well does not give any kinetic energy since the water is not flowing through it, so the 1/2蟻v^2 terms can be neglected. Taking the surface as a reference point (h=0), h_1=0 and h_2=-40m. P_1 = P_2 - 蟻g螖h P_1 - P_2 = 蟻g螖h 螖P = 蟻g螖h 螖P = (1000 kg/m^3) * (9.81 m/s^2) * (-40 m) 螖P = -392400 Pa The pressure difference that the pump must maintain is -392400 Pa or 392400 Pa (magnitude), which means the pump must maintain an increase in pressure of 392400 Pa.
05

Determine the pump position.

If the pump is at the top of the well, the pressure at the bottom will be maintained at atmospheric pressure. The pump must maintain a pressure difference of 392400 Pa. If the pump is at the bottom, it must maintain a greater pressure at the bottom. The pressure difference is still 392400 Pa, but since the pump would be at the bottom, the pressure would be greater than atmospheric pressure. In conclusion, the pump can be at the top or at the bottom of the well; the pressure difference that the pump must maintain is the same at 392400 Pa. However, the actual pressure at the top or bottom will vary depending on the position of the pump.

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