Work integral for pumping water from a tank

[DrunkEngineer]

Homework Statement

Homework Equations

$W = \int F dx$

For density of water i used
D = 62.4lb/ft^3
g = 32.2 ft/s^2

$g(D = \frac{m}{V})$

$gDV = mg$

$gDV = F$

The Attempt at a Solution

i use limit from 0 to 6
$W = \int_{0}^{6} F dx$

$W = \int_{0}^{6} gDV dx$

$W = \int_{0}^{6} (32.2)(62.4)(\pi{6}^{2}{15}) dx$

$W = 20451979.29$

help check this pls

 

[Mark44]

Homework Statement

Homework Equations

$W = \int F dx$

For density of water i used
D = 62.4lb/ft^3
g = 32.2 ft/s^2

$g(D = \frac{m}{V})$

$gDV = mg$

$gDV = F$

The Attempt at a Solution

i use limit from 0 to 6
$W = \int_{0}^{6} F dx$

$W = \int_{0}^{6} gDV dx$

$W = \int_{0}^{6} (32.2)(62.4)(\pi{6}^{2}{15}) dx$

$W = 20451979.29$

help check this pls

This is incorrect. You are assuming that all of the water has to be lifted 15 feet, and this is not the case. Mentally divide the water into cylindrical slices, each of which is $\pi$*36*$\Delta y$ in volume. From the volume, you can get to the weight of each slice, but the distance a slice of water has to be pumped depends on where the slice is in the tank. The layers (slices) at the top require almost no effort to pump out, since they don't need to be lifted very far. The ones at the bottom require a lot more work, since the have to be lifted much farther.

 

[DrunkEngineer]

found the answer thanks for the advice

slice of cylindrical element anywhere inside cylinder
$dV = 36{\pi}dy$

$dF = 32.2(62.4(\pi{36})dy)$

work done from this differential element to the top: then (distance = 15 - y)
$dW = 32.2(62.4(\pi{36})dy)(15-y)$

$W = \int_{0}^{15} dW$
etc.

$W = 25564974$lb-ft

$W = 793943$slugs-ft

 

[Mark44]

I didn't verify your numbers, but what you have looks to be set up correctly.