- #1
chingel
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Homework Statement
Calculate the work done by air, when it pushes water out of a barrel (cylindrical), that is 1m in diameter. The starting water level was 80 cm from the bottom of the barrel and the water level at the end was 20 cm from the bottom. The water was pushed out through an inverted-U-shaped tube.
Homework Equations
[tex]A=PV[/tex]
The Attempt at a Solution
Since [itex]A=PV[/itex] and I know that [itex]V=π*0,5^{2}*0,6=0.15π[/itex], all I need to do is calculate the average pressure. What I find problematic is that neither the height of the barrel, nor the length of the tube is given. Am I correct in assuming that those variables affect the pressure that is needed to pump the water out of the barrel and therefore also affect the answer? Or is there some sort of a siphon effect that makes these variables unnecessary?
I also assumed that the pressure on the outer end of the inverted-U shaped tube is air pressure. First the gas needs to push the water up the tube, work equaling the volume of the tube to the top times the pressure created by the water column half-way to the top (half-way to get average pressure) plus air pressure. Then as the water goes down the tube, I would have to subtract the height of the water column in the tube from air pressure to which I add the height of the water column half-way from the current water level to the 20cm level, all of which I multiply by the volume of water needed to be pumped out to reach 20 cm water level.
Is my reasoning correct for the steps needed to solve the problem, and that without being given the height of the barrel or the tube, an answer could not be calculated?
A solution to the problem was also posted, which I don't think is correct:
Work done by the air pressure is equal to the increase of potential energy of the center of mass of the water (0,3 m below starting level).
[tex]A=E=mgh=ρVgh[/tex]
[tex]A=1000*0,15π*10*0,3=1410J[/tex]