Question
A pump provides pressure to the lower end of a long pipeline that supplies water from a reservoir to a house located on a hill 150 m vertically upward from the lower end of the pipe (where the water is initially at rest before being pumped). The pipeline has a constant diameter of 3.5 cm, and the upper end of the pipeline is open to the atmosphere. What pressure must the pump provide for water to flow from the upper end of the pipeline at a rate of 5.0 m/s?
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Solution video
OpenStax College Physics for AP® Courses, Chapter 12, Problem 4 (Test Prep for AP® Courses)
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This is College Physics Answers with Shaun Dychko. There is a pump here at the end of a very long pipeline, it's going to pump water up this hill to this house such that the nozzle at the end where the house is has water gushing out at a speed of 5.0 meters per second and this nozzle or this opening is open to the atmosphere and so the pressure at this position here where the house is is atmospheric pressure and this house is a height—150 meters— compared to where this pump is. So this number 2 denotes anything to do with the destination where the house is and the subscript 1 is labeling things at this initial position here where the pump is and so h 1 is 0 meters—we'll take that to be the reference level— and the speed of the water here we are told is 0 meters per second— the water is initially at rest. The fact that the pipeline is very long tells us what to assume for the pressure in this pipe. Now of course the reservoir— wherever it is... some pool of water here— is gonna be atmospheric pressure here but as this water goes through the pipe, I guess there's so much pipe and so much friction as the fluid flows along that by the time it gets to here, we are to presume that there is zero absolute pressure at this position. So that's what we'll take... so we'll take P 1 to be the pressure that the pump needs to provide, it has to provide all the pressure that's going to exist at this position. So Bernoulli's principle says that the absolute pressure at this first position plus one-half times the density of the fluid times its speed there squared plus density of the fluid times g times its height equals P 2 plus the same terms for the second position and v 1 is 0 and h 1 is also 0 so these two terms are respectively zero and P 1 is a pressure due to the pump; pressure at position two is atmospheric pressure we are told and it's gonna have some speed v 2 at some height h 2 and well, this is just kind of saying the same thing on this line here but without the 0's and we'll factor out ρ—density of water—from these two terms and we have the pressure of the pump then is atmospheric pressure plus density of water times its speed at the house squared divided by 2 plus gravitational field strength times the height of the house. So that's 1.013 times 10 to the 5 pascals— atmospheric pressure— plus 1000 kilograms per cubic meter— density of water— times 5.0 meters per second squared divided by 2 plus 9.81 newtons per kilogram times 150 meters and this gives 1.58 times 10 to the 6 pascals.