A tank has a shape of a cone with a radius at the top of 3 m and a height of 5 m. The tank also has a 1 m spout at the top of the tank.

Question

The tank is filled with water up to a height of 2 m. Find the work needed to pump all the water out the top of the spout. (Use 9.8 m/s² for g and the fact that the density of water is1000 kg/m³.)

Answer 1

Work = force•distance where, in this case, "force" is the weight of the water to be lifted and "distance" is "y"

The weight of the water is the (volume of the water)•(density of water)•(acceleration due to gravity or "g")

The Volume of the water (V) is the Area•thickness = πr²•(dy) but "r" changes as the height of the water changes. You can use similar triangles to get the ratio: 3/5 = r/h or 5r = 3h or r = 0.6h

So V = πr²•(dy) = π(0.6h)²•(dy) = 0.6πh²•(dy) = 1.885h²•(dy)

Since the height of the spout is 4m from the bottom, then h = 4 - y so now we can write volume as:

V = 1.885(4 - y)²•(dy)

Weight of the water = V•density•g = 1.885(4 - y)²•(dy)•1000•9.8 = 18473(4 - y)²•(dy)

Multiplying this out we get:

Weight of water = 18473(16 - 8y + y²)•(dy) = 295561 - 147781y + 18473y²•(dy)

Work = force • distance = weight • distance = (295561 - 147781y + 18473y²•(dy))•(y)

Work = 295561y - 147781y² + 18473y³•(dy)

To find all the work we integrate from y = 2 to y = 4 (y = 2 is where the water is at the start and y = 4 is when the tank is empty):

Work = ₂∫⁴ 295561y - 147781y² + 18473y³ dy

You can do this manually fairly easily or use a calculator. The units of work are Nm (Newton meters) which is the same as joules.