how much work is required to pump all the water out over the side?

Question

a circular swimming pool has a diameter of 24ft, the sides are 5ft high, and the depth of the water is 4ft. how much work is required to pump all the water out over the side?Use the fact that water weighs 62.5lb/ft^3.

in order to find the work done i am finding each step from volume to mass to force and finally to work where i will set up the final integral.

i found the volume to be 144*pi*deltax
i then found the mass 62.5*144*pi*deltax

ok this is where i am stuck finding the Force
do i multiply 62.5*144*pi*deltax by 9.8m/s^2?

and the work will just be whatever i find in the previous step multiplied by x which is the distance (w=F*d)

the answer is 108,000pi ft-lb

Answer 1

water in the pool exerts hydrostatic pressure against which the pump has to work for lifting water across the wall of 5 feet; This pressure acts across the pi r^2 area.
pressure is given by

P = h *rho* g (h is free surface water, h=4 here, rho is density)

if you use rho in FPS system of units g need not be multiplied

Force = P A = pi r^2 *rho* h

now consider: an element of water between x and x+dx depth from bottom of (where x=0) tank. the work done in lifting this element will be

dw = pi r^2 *rho* x * dx

total work done W = ∫ pi r^2 *rho* x * dx

limits will be x = 1 to x =5 here it is quite clear that first most element from surface will be lifted across (5-4) 1 feet ie lower limit and farthest element will be (5-0) = 5 feet
had pool been full we would have used x=5 to x=0

W = pi r^2 *rho*(1/2) [x]^2
W = (1/2) pi r^2 *rho [25 - 1]
W = 0.5*pi*12*12*62.5*24 = 108,000 pi
Answer

Answer 2

The answer is zero. Fill a garden hose with water, put one end in the pool and the other on the ground, and go have a beer while the pool siphons itself empty. The mathematically interesting question is: how much work is required to lift the water over the side? The average depth of water while this is happening is 2.5 feet, so the work required is 2.5 x pi x 12 x 12 x 62.5 foot-pounds.

Answer 3

It relies upon on the way you do the pumping. the two way, you will could do some calculus. The good way is to syphon the water with a hose with one bring about the water and the different end on the floor outdoors. That way, you do no longer could do any artwork. in actuality, in case you needed to, you could set up a hydroelectric generator and get artwork decrease back. How lots potential is obtainable? think of a quantity of water in a skinny around disk with diameter 16m and intensity dx. dV = pi (8m)^2 dx. The water has ability potential dU=(dm)gx, the place m = (rho)dV. rho of water is 1000 kg/m^3. dU = (rho)(pi)r^2g xdx = consistent circumstances xdx Now combine dU from x=0 to the intensity (2.5m) and you could have your potential you could recuperate with your hydroelectric generator (in case you will get one hundred% performance). the incorrect thank you to try this's to pump all the water as much as a top of four meters. consequently, dU = (dm)g(4-x). combine it out the comparable way and you gets the quantity of artwork you're able to do to get all that water as much as a top of four meters. Your instructor probable meant so you might resolve it the stupid way and discover out how lots artwork you're able to do. teach him you're smarter than that by using doing it the effectual way and producing potential rather! Edit: the respond above can provide a brilliant shortcut to stay away from the calculus--placed all the mass at cm (a million.25 m off floor). word, nevertheless, that he solves for the artwork you do with the inefficient technique (lifting from a million.25m as much as 4m). rather, calculate the artwork you get decrease back using the sensible syphon (draining from a million.25m right down to 0m). Edit2: and yeah, if the pool have been underground (no longer the way I examine the subject with partitions 4m severe), the coolest way would not artwork.