Emptying a water tank (Inverted circular cone)

**Grimey** · September 20th 2009, 12:13 PM

ok, i drew out a diagram and I have no idea how or where to start this problem. Please help guide me so I can learn how to do it...

A water tank has the shape of an inverted circular cone with a diameter of 4 feet and a height of 6 feet. If the tank is full, find the work required to empty the tank through a pipe that extends 3 feet above the top of the tank. Use 62.5 as the weight-density of water. Give your answer in foot-pounds.

**Grimey** · September 20th 2009, 12:27 PM

just found a really helpful sample problem online. thanks anyway.

might be posting another problem here in a few

**skeeter** · September 20th 2009, 12:38 PM

Originally Posted by Grimey

ok, i drew out a diagram and I have no idea how or where to start this problem. Please help guide me so I can learn how to do it...

A water tank has the shape of an inverted circular cone with a diameter of 4 feet and a height of 6 feet. If the tank is full, find the work required to empty the tank through a pipe that extends 3 feet above the top of the tank. Use 62.5 as the weight-density of water. Give your answer in foot-pounds.

work = integral of WALT

W = weight-density in lbs per cubic foot

A = cross-sectional area of a representative horizontal "slice" of liquid in terms of y

L = lift distance of the representative "slice" in terms of y

T = "slice" thickness , dy.

let the vertex of the cone be at the origin ... one side of the cone is modeled by the linear equation ... $y = \frac{6}{2} x = 3x$

"slice" radius = $x = \frac{y}{3}$ , so the cross-sectional area of a representative "slice" is , $A = \pi \left(\frac{y}{3}\right)^2$

lift distance, $L = 9-y$

the liquid resides between $y = 0$ and $y = 6$ ... your limits of integration.

set up the integral and find the work required.

**Grimey** · September 20th 2009, 12:49 PM

thanks, ill be sure to remember WALT.

This is how I solved it, not 100% if its right but it seems it might be...

Weight of slab = density * volume
weight of slab = 500 pi dx

Distance = x + 3

Work = (x + 3)500pi dx

W = integral from 0 to 6 [(x+3)(500pi)]dx

W = 56,548.7 ft lbs

**skeeter** · September 20th 2009, 03:30 PM

Originally Posted by Grimey

thanks, ill be sure to remember WALT.

This is how I solved it, not 100% if its right but it seems it might be...

Weight of slab = density * volume
weight of slab = 500 pi dx

Distance = x + 3

Work = (x + 3)500pi dx

W = integral from 0 to 6 [(x+3)(500pi)]dx

W = 56,548.7 ft lbs

this would not be correct for the described cone.

$62.5 \pi \int_0^6 \left(\frac{y}{3}\right)^2 (9-y) \, dt = 2250\pi$ ft-lbs

**Grimey** · September 20th 2009, 04:05 PM

Originally Posted by skeeter

this would not be correct for the described cone.

$62.5 \pi \int_0^6 \left(\frac{y}{3}\right)^2 (9-y) \, dt = 2250\pi$ ft-lbs

Thanks.

So with finding A, the cross section area...
If its a disc, it will always be A = pi r^2?

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