Work problem -- lifting water out of tanks

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SUMMARY

The discussion focuses on calculating the work required to lift water from an inverted conical tank with a height of 6 meters and a base radius of 1.5 meters. The formula used is w = ∫ab pgA(y)D(y)dy, where A(y) represents the cross-sectional area and D(y) is the distance the water must be lifted. The participant incorrectly assumed D(y) to be a constant value of 6m instead of a function of y, which led to incorrect results. The conversation also raises a question about whether it requires more work to pump water from the bottom or the top of the tank.

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  • Understanding of calculus, specifically integration techniques
  • Familiarity with the concept of work in physics
  • Knowledge of fluid mechanics, particularly hydrostatic pressure
  • Ability to derive equations for geometric shapes, such as cones
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  • Study the derivation of the work formula for lifting fluids in tanks
  • Learn about hydrostatic pressure and its application in fluid mechanics
  • Explore the concept of variable distance in work calculations
  • Investigate the differences in work required for various tank shapes
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Students in physics or engineering, educators teaching fluid mechanics, and professionals involved in hydraulic systems or water resource management would benefit from this discussion.

iPhotonHQ
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< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >[/color]

I've encountered 2 problems in a row that involve lifting water out of tanks and finding the work needed. I am getting the incorrect answer.
w = ⌠ab pgA(y)D(y)dy

here is one of the problems:
A water tank is shaped like an inverted cone with a height 6m and a base radius of 1.5m...
a) if the tank is full, how much work is required to pump the water to the level of the top of the tank and out of the tank?

*Attempt*:
I drew a graph and a line from (0,0) to (1.5, 6) and found that the equation of that line is y = 4x or x = (1/4)y
cross-sectional volume of a slice: πr²h
r = y/4
V = π(y/4)²dy = A(y)
D(y) i think = 6m
g = 9.8 m/s²
p = 1000kg/m3
061000(9.8m/s²)(πy²/16)(6m)dy

I am getting the incorrect answer...
 
Last edited by a moderator:
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Your D(y) must be a function of y ...
 
Here's something interesting about this problem.

Does it take more work to pump water from bottom of tank or from top of tank as you slowly lower the hose?
 

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