calculus Work problem

Question

A cylindrical aquarium with radius of 20 feet and height of 100 feet is filled with salt water (65 pounds/ cubic feet). Find the work done using integration and use calculator to calculate the integral.

a. How much work is done pumping all the water over the top edge of the tank?

b. How many hours will a 1 horsepower pump take to empty the tank? (1 horsepower pump works at a rate of 33,000 foot-pounds per minute)

Answer 1

The statement of the problem uses the word "pound" inconsistently.

In the specification of density "pound" refers to mass, while in the definition of power

"pound" refers to weight.

To disambiguate the situation I will use "lb" to be a pound of mass, and "glb" to be the weight of 1 lb.

That is, 1 gbl = g × 1 lb -- gravitational acceleration multiplied by the mass of 1 lb.

Let

r = 20 ft be the radius of the cylinder,

h = 100 ft be the height of the cylinder,

d = 65 lb/ft³ be the density of the salt water,

P = 33000 ft-glb/min be the power output of the pump,

g be gravitational acceleration.

The distance, s, from the surface to the rim changes from 0 to h.

At each point the pump must lift over the distance s an infinitesimal disk of water with thickness ds.

The disk has

volume πr²ds,

mass dπr²ds,

weight gdπr²ds.

The work, which is the integral of the force (equal to the weight) over the distance s:

W = ∫gdπr²sds, where the integral is taken from 0 to h.

W = gdπr²∫sds = gdπr²s²/2 from 0 to h = gdπr²h²/2

Substituting actual numbers

W = g × 65 × π × 20² × 100²/2 = g × 130000000π ft-lb = 130000000π ft-glb

The time required is

W/P = 130000000π ft-glb / 33000 ft-glb/min = 1237.6 min = 20.6 hours