Integration - work required to empty the tank

**anonymousperson** · April 30th 2011, 12:02 PM

A tank in the shape of an inverted right circular cone has height 11 meters and radius 20 meters. It is filled with 1 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is:

$\delta = 1480kg/m^3$

To find the area of the circle: r/h = 20/11. Solved for r: 20h/11. The thickness is taken care of by dy in the integral. Gravity taken into account and density is in there.

$\int_{1}^{11}14504{N/m^3}\pi({\frac{20h}{11}})^2dy$

The integral is from, I think, 1 to 11. I don't know if I am on the right track, or what to do from here. Taking the antiderivative of this leaves me with an h and other stuff, but I don't think the integral is correct to begin with.

Thanks.

**skeeter** · April 30th 2011, 02:35 PM

Originally Posted by anonymousperson

A tank in the shape of an inverted right circular cone has height 11 meters and radius 20 meters. It is filled with 1 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is:

$\delta = 1480kg/m^3$

To find the area of the circle: r/h = 20/11. Solved for r: 20h/11. The thickness is taken care of by dy in the integral. Gravity taken into account and density is in there.

$\int_{1}^{11}14504{N/m^3}\pi({\frac{20h}{11}})^2dy$

The integral is from, I think, 1 to 11. I don't know if I am on the right track, or what to do from here. Taking the antiderivative of this leaves me with an h and other stuff, but I don't think the integral is correct to begin with.

Thanks.

Work = integral of WALT

W = weight density = $\delta g$

A = horizontal cross-sectional area = $\pi \cdot \left(\dfrac{20}{11} y\right)^2$

L = lift distance for a representative cross-section = $(11-y)$

T = cross-sectional thickness = $dy$

cross-sections lie within the bounds y = 0 to y = 1 ... i.e. your limits of integration

**anonymousperson** · April 30th 2011, 02:58 PM

Thanks. I got it right.

Actually, does this WALT idea apply to other situations? It worked with circular things and cones, but for this trough problem I am not sure:

A trough is 2 meters long, 3 meters wide, and 1 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isosceles triangle (with height 1 meters, and base, on top, of length 3 meters). The trough is full of water (density 1000 kg/m^3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g=9.8 m/s^2 as the acceleration due to gravity.)

Setting this up:

$\int_{0}^{1} 1000kg/m^3*9.8m/s^2*1.5m^2(1-y)dy$

It says that the trough is 1 meters deep, and that the trough is filled with water. The isosceles triangles are parallel to the ends of the trough. So wouldn't that mean if I took an infinitesimally small thickness that that would be with respect to x? But that can't be, since we are trying to get over the top of the trough.

Also, if possible, don't actually set up the integral, I want to do that. Can you just say if I am close, or forgetting to do something? Thanks!

**skeeter** · April 30th 2011, 06:12 PM

WALT works for any horizontal slice lifted a vertical distance ...

for the trough problem ...

width = $2x = \dfrac{4}{3}y$

length = $2$

thickness = $dy$

finish it.

**anonymousperson** · April 30th 2011, 07:52 PM

For the width I got $3-3x$ . I'm not sure how you got $2x$ , or even if that would work. Going down, from the top, a depth of $x$ creates similar triangles. The width of the small triangle is $w$ , and the height is $1-x$ . Setting that up as: $\frac{(1-x)}{w}=\frac{1}{3}$ . Solving for the width gives me $3-3x$ . The height that slab would have to travel is $x$ .

So the final integral:

$\int_{0}^{1}9800(6-6x)(x)dx$

I think this is right. My book only gives me the odd numbers, but this seems like it has to be right.

Integrating it:

$9800\left [3x^2-2x^3\right]_{0}^{1}= 9800J$

**skeeter** · May 1st 2011, 05:05 AM

Originally Posted by anonymousperson

For the width I got $3-3x$ . I'm not sure how you got $2x$ , or even if that would work. Going down, from the top, a depth of $x$ creates similar triangles. The width of the small triangle is $w$ , and the height is $1-x$ . Setting that up as: $\frac{(1-x)}{w}=\frac{1}{3}$ . Solving for the width gives me $3-3x$ . The height that slab would have to travel is $x$ .

So the final integral:

$\int_{0}^{1}9800(6-6x)(x)dx$

I think this is right. My book only gives me the odd numbers, but this seems like it has to be right.

Integrating it:

$9800\left [3x^2-2x^3\right]_{0}^{1}= 9800J$

I misinterpreted the dimensions of the trough, so let's try this again. look at the sketch ...

volume of a single horizontal slice of water,

$dV = 2x \cdot 2 \cdot dy$

since $x = \dfrac{3}{2}y$ ...

$dV = 6y \, dy$

$\displaystyle W = \int_0^1 \delta g \cdot 6y (1-y) \, dy$

this gets your same result.

**anonymousperson** · May 1st 2011, 10:10 AM

Okay, that makes sense. Thanks for the help!

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