A swimming pool is 20 m long and 10 m wide with a bottom that slopes uniformly f

Question

A swimming pool is 20 m long and 10 m wide with a bottom that slopes uniformly from a depth of 1 m at one end to a depth of 2 m at the other end. Assuming the pool is full of treated water whose density is delta, how much work is required to pump the water into a reserve tank of 0.2 m above the top of the pool? Break this problem up into two parts. The first part is highest-most meter of water where the length of a "slice" is a constant 20 m. The second part is the lower half where the length varies linearly from 20 m to 0 m. You'll need a function that computes the length of a "slice" as a function of the depth. After the pool is cleaned, the water is released from the reserve tank back into the pool. How much work is required to fill the pool? (This is a trick question.)

Explanation / Answer

In any case, the work required to move an object of mass m a distance h up against gravity is mgh. Split the pool into infinitesimal slices of height dy. The question isn't clear (maybe it came with a picture?) but I'll assume the distance between the lowest points and the highest points on the bottom of the pool is 20m (rather than 10m). The width of each slice is then constant at 10m, while the length varies. Let y denote the distance from the surface of the water to a given slice. If y <= 1, the length is just 20m. If y > 1, the length is 20*(2-y) since this function is linear and is correct at y=1 (where it gives 20) and at y=2 (where it gives 0).

The volume of water in such a slice is then
V(y) = dy * 10 * 20 if y <= 1
V(y) = dy * 10 * 20 * (2-y) if y > 1

A slice at position y needs to be transported a distance y+0.2 up. If the density of water is D, the energy to lift it is
mgh = DV(y) g (y+0.2)

Integrating this expression for y between 0 and 2 will give us the total work required. Split it into two pieces since V(y) is piecewise:

integral of DV(y) g (y+0.2) between 0 and 2
= integral of 200Dg (y+0.2)dy between 0 and 1
+ integral of 200Dg (2-y)(y+0.2)dy between 1 and 2
= ...basic calculus...
= 140Dg + 460Dg / 3
= 880Dg/3

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