Question

A tank with dimensions of 12x4x8 ft contains 288 ft^3 of water. determine work work required to pump water out of the tank if the base is 12x4 ft.
I think I set up the integral a few too many times/different ways, so just curious what the actual format should be to determine work....

Answer 1

**** Also it's being pumped out the top of the tank, so it has to be lifted to the top.....

Answer 2

\[\int\limits_{0}^{?} g*\rho *48(8-y)dy\]

Answer 3

upper limit is the height of the water intially ie volume of water initially divided by 'area'

Answer 4

Well.... our format is integral of [a, b] density * volume * distance
.... we don't use gravity....
I'm just curious where the 288 cubit feet come into play for this equation...... unless it's something with density, because we're using density of water = 62.5 lb/ft^3

Answer 5

yeah 'weight' density of water is density * g, so that's done for you already, I guess.
288 is used to find the upper limit of integration

Answer 6

ok?

Answer 7

Umm...... but to find the density of the 288 cubic feet of water.... do we multiple it by the general density of water 62.5? Because 288 cubic feet is not a measure of density....

Answer 8

wut

Answer 9

the limits of integration here are y values, heights. y starts at 0 (bottom of the tank) and goes to the initial height of the water in the tank.

Answer 10

I know the basic integral form... but I now realize that the density part of it is what I'm stuck on .... because our problem is just giving me "There is 288 ft^3 in the tank". As in the tank is not full so just multiplying the density of water throughout the equation would not work.... so I'm wondering if I multiple the density of water by the volume of water... or if there's a different method. I know the water level goes up to 6 ft if that's any clue....

Answer 11

Eh, nevermind... I think I can try to figure it out through google....

Answer 12

\[\int\limits_{0}^{6} 62.5*48(8-y)dy\]

Answer 13

I'm thinking you're not quite seeing what's going on here conceptually.
you divide the water into slices, you find the volume of a slice, you find the force required to lift the slice and then multiply that by the distance the slice has to be lifted.
That's the work required to pump a single slice out of the top of the tank. You integrate to find the work required to lift every slice between the top of the water (y=6ft) and the bottom of the water (y=0ft)

Answer 14

So I'm only integrating from 0 to 6 then? Not adding another integration to go to 8?

Answer 15

nope because the water stops at y=6ft :)

Answer 16

mkay, thank you

Answer 17

sure!