A circular swimming pool has a diameter of 10 meters, the sides are 1.5 meters high and the depth of the water is 1.2 meters. How much work is required to pump all of the water over the side? (Density of water $= 1,000kg/m^3$)
So far all I have is what I think the force is $$ F = mg = 9.8 * 25,000\pi \Delta{x}$$
Desperately need help on this one.
[Edit]
The answer I got was: $$W = \int_{0.3}^{1.5} 245,000\pi(0.3 + x)dx$$ $$ W = 1.11 * 10^6 J$$
Is this right?
Best Answer
There are a number of acceptable choices of "vertical coordinate" for this problem (or related "hydraulic work" problems).
The one Ron Gordon is using is perhaps the one most often used: we take $ \ y \ $ as being measured upward from the bottom of the pool, and the integration must be taken from $ \ y \ = \ 0 \ $ (the lowest point in the water) to $ \ y \ = 1.2 \ \text{m.} $ (the surface of the water).
The choice made by Will Brothers is to start at the "lip" of the swimming pool ( $ x \ = \ 0 $ ) ; in this case, the integration over the depth of the water must be taken from $ \ x = 0.3 \ $ to $ \ x = 1.5 \ $ .
Yet another choice is to integrate over the depth of the water itself, from the surface ( $ z \ = \ 0 $ ) downward to the lowest point ( $ z \ = \ 1.2 $ ) .
The infinitesimal work to be done in lifting an infinitesimal layer of water is
$$ dW \ = \ dm \ \cdot \ g \ \cdot \ \text{(lifting distance)} $$
$$ = \ A_{cs} \ \cdot \ \rho g \ \cdot \ d \text{(vertical coordinate)} \ \cdot \ \text{(lifting distance)} \ \ , $$
with $ \ A_{cs} \ $ being the cross-sectional area of the layer and $ \ \rho \ $ being the volume mass density of the fluid.
[Note: using metric units, $ \rho \ $ and $ \ g \ $ may be entered separately; however, in English units, since weight is a force, $ \ \rho g \ $ must be entered as a single quantity, the "weight density" ($ \ \rho g \ = \ 62.5 \ $ lb./cu.ft. for water -- the volume mass density of water being 1.94 slugs/cu.ft.).]
In any event, the portion with physical quantities is calculated correctly above as
$$ A_{cs} \ \cdot \ \rho \ g \ = \ \pi \cdot 5^2 \cdot 1000 \cdot 9.81 \ \approx \ 245,000 \pi \ \frac{\text{N}}{\text{m}^2} \ . $$
The differences in the infinitesimal work come about from the choice of vertical coordinate. For $ \ y \ , $ the water is to be lifted by the distance of the layer from the "lip" of the pool, which is $ \ 1.5 - y \ , $ as Ron Gordon has said. For Will Brothers' choice, the distance to lift the water is simply $ \ x \ , $ as this is the vertical distance to the "lip". In the last choice, since the water's surface is 0.3 meters below the "lip" of the pool, each layer must be lifted a distance $ \ z + 0.3 \ $ meters.
The work integrals are then
$$ W \ = \ 245,000 \pi \ \int_0^{1.2} \ (1.5 - y) \ \ dy $$
$$ = \ 245,000 \pi \ \int_{0.3}^{1.5} \ x \ \ dx $$
$$ = \ 245,000 \pi \ \int_0^{1.2} \ (0.3 + z) \ \ dz \ \ , $$
all of which give $ \ 245,000 \pi \cdot \ 1.08 \ = \ 264,600 \pi \ \approx \ 831,300 \ $ J.
It is a useful fact (and one generally used by engineers) that, when the work function is linear, the total work is simply the mass of fluid to be lifted multiplied by the vertical distance by which the centroid of the volume must be lifted (assuming uniform density for the fluid). Thus,
$$ W \ = \ \pi \cdot 5^2 \cdot 1.2 \cdot 1000 \cdot 9.81 \ \text{N} \ \cdot \ \left(\frac{1.2}{2} \ + \ 0.3 \right) \text{m.} \ \approx \ 264,900 \pi \ \text{J} \ \ . $$
[The difference from the earlier result is due to using 9.81 for $ \ g \ $ , rather than just rounding to 9.8.] This approach works because the work integral actually involves a "first-moment integral" of the weight with respect to the level of the "lip" of the pool.