The tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank. The tank shown is a hemisphere with r = 5 ft. The water is to be pumped out at the top.
First I solved for ri, ri / (5 - xi) = 5 / 5
ri = 5 - xi
Vi = pi*(ri)^2*delta x
Vi = pi*(5 - xi)^2*delta x
mi = density x volume
mi = 62.5*pi*(5 - xi)^2*delta x
Fi = mi x g, g = 9.8 m/s^2 or 32.152231 ft/s^2
Fi = (32.152231 ft/s^2)*(62.5*pi* (5 - xi)^2*delta x)
Fi = 2009.5*pi*(5 - xi)^2*delta x
Wi = (Fi)*(xi) = (2009.5*pi)(xi)(5 - xi)^2*delta x
W = INTEGRAL from 0 to 5 of: 2009.5*pi*(xi)(5 - xi)^2*dx
W = 2009.5*pi INTEGRAL xi*(5 - xi)^2*dx, from 0 to 5.
W = 2009.5*[12.5x^2 - (10x^3)/3 + (x^4)/4] evaluated at 5 and 0
W = 2009.5*[12.5*(5)^2 - (10*(5^3))/3 + (5^4)/4]
W = 100220 ft-lb
I am not sure which part I did wrong. Could you please point me in the right direction? Thanks!
For Further Reading
* Calculus - bobpursley, Friday, June 22, 2007 at 9:22pm
Your calculation of force included g. But you used lbs for mass. Lbs is a force unit (it really isnt mass). So delete the 32.15 ft/s^2 from the force.
Work=NTEGRAL from 0 to 5 of: 62.5*pi*(xi)(5 - xi)^2*dx
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I am really sorry to post this again but I still cannot come up with the right answer...
Work=NTEGRAL from 0 to 5 of: 62.5*pi*(xi)(5 - xi)^2*dx
=(62.5*pi)[(12.5)(5^2)-(10(5^3))+(5^4)/4]
=10226.53859 ft-lb
This is wrong though...am I in the wrong units? Please help. Thanks.
If you are performing a vertical integration with x as the variable of integration, it seems to me that relationship between x (the height above the base) and r (the radius of the circular volume element of thickness dx), is
r^2 + x^2 = 5^2
NOT ri = 5 - xi
That might be the casue of your error.
When you integrate for total work, the integral should be
(Integral of) pi r^2 * (rho) g * (5-x) dx
x = 0 to 5
Substitute 25 - x^2 for r^2. Then you will just have terms in x in the integrand.
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