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!
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
It seems like you made a small mistake in your calculations. When calculating force, you included the acceleration due to gravity, but you used pounds (force) as the unit for mass instead of pounds (mass).
Since pounds is actually a force unit, you don't need to include the acceleration due to gravity. So, the correct expression for force (Fi) should be:
Fi = 2009.5 * pi * (5 - xi)^2 * delta x
And for the work (Wi) equation, you can use the correct expression for force (Fi):
Wi = Fi * xi = (2009.5 * pi * (5 - xi)^2 * delta x) * xi
To find the total work required to pump the water out of the tank, you can integrate this equation with respect to x:
W = INTEGRAL from 0 to 5 of: (2009.5 * pi * xi * (5 - xi)^2) dx
Evaluating this integral should give you the correct answer for the work required to pump the water out of the tank.