Calculating Work Required to Pump Water from a Hemispherical Tank

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SUMMARY

The discussion focuses on calculating the work required to pump water from a hemispherical tank with a radius of 4 meters to a point 2 meters above the tank. The work is calculated using the formula w = V*d*g*h, where V is the volume of water, d is the density (1000 kg/m³), g is the acceleration due to gravity (9.8 m/s²), and h is the height to which the water is lifted. The final expression for work yields approximately 1463466π Joules. The participant expresses concern about the accuracy of this energy calculation, given the varying heights of water layers.

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  • Knowledge of the properties of hemispherical shapes and their geometric equations.
  • Basic understanding of density and its application in calculating mass.
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Homework Statement



A tank in the form of a hemispherical bowl of radius 4 m is full of water. Find the work required to pump all of the water to a point 2 m above the tank.


Homework Equations



w = fd = mgd
density = 1000L/m2
1L = 1kg

The Attempt at a Solution



x = height

w = V*d*g*h
= (π)(r^2)(x)*(1000)*(9.8)*(x)*dx

I think my problem is trying to show how the height a strip of cylinder varies with the radius. I've tried using similar triangles but it was obviously incorrect. Is there a similar property I can use for hemispheres?
 
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Think about this in terms of horizontal layers of water. It's probably most convenient to put the origin at the top level of the hemispherical tank, so that \Delta y ranges from -4 to 0. Each layer has to be lifted (pumped) from its position in the tank to a point 2 m. above the top of the tank. You are using x in your integral expression; I would use y, and I would also find the equation of the semicircular cross section of the tank so that I could exploit a relationship between x and y at a point (x, y) on the semicircular boundary of the tank. A sketch will be very helpful if you haven't already done one. A drawing of a half-circle will do just fine, as long as you realize that the water layers are three-dimensional.
 
w = V*d*g*h <--new

V = pi*r^2*dh
r^2 = 8h-h^2
d = 1000
g = 9.8
h = 6-h

V = 9800pi ∫ [(8h-h^2)*(6-h)] dh
V = 9800pi*(24h^2 - (14/3)h^3 + (1/4)h^4)
V ~ 1463466pi Joules

This is the first time I've done a question like this and the energy seems a little high to me, is it correct?

I know I can't check it with the potential energy equation because you need different amounts of energy to lift the water at different heights, is there anyway I can check that this is correct?
 

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