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October 28th, 2014, 08:45 AM   #1
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Work required to pump water out of a triangular tank

Hi all,

Wish I had a picture, but here's the problem anyway. You have an upside-down triangular tank. The length of the tank is 15 feet, the width (on the top) is 2 feet, and the length of the side is 3 feet. I need to calculate work required to pump water out of the top (there is no spout of extra length).

What I did so far is use similar triangles: 2/3 = wi/3-y - > wi = 2-2y/3. Displacement = 3 - y.

15 * 9.8 * 62.4 of the Integral from 0 to 3 of (2-2y/3)(3-y) dy

= 55036

I think I did this problem wrong. Please advise on what I did incorrectly. Thanks.
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October 28th, 2014, 03:47 PM   #2
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I hope I am in the right direction (the time has passed here).

$\text{Work} = F\cdot \text{distance}$

$$\mathbb{d}W = \text{weight of slice}\times\text{distance the distance travels}= p\cdot\mathbb{d}V\times\text{distance the slice travels}$$

$$\mathbb{d}V= xp \cdot \mathbb{d}h\cdot 15=\frac{2\sqrt{2}-h}{\sqrt{2}}\cdot \mathbb{d}h\cdot 15$$

$$\mathbb{d}W = p\frac{2\sqrt{2}-h}{\sqrt{2}}\cdot \mathbb{d}h\cdot 15 \times \text{distance the slice travels}=\frac{2\sqrt{2}-h}{\sqrt{2}}\cdot \mathbb{d}h\cdot 15\cdot h$$

$$W = p\frac{15}{\sqrt{2}}\int\limits_{0}^{2\sqrt{2}}(2 \sqrt{2}-h)h\;\mathbb{d}h=\frac{15}{p\sqrt{2}}\cdot \frac{8\sqrt{2}}{3}=40p\approx2496$$
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Last edited by CRGreathouse; October 29th, 2014 at 08:38 AM. Reason: remove assumption that p = 1
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October 28th, 2014, 09:49 PM   #3
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When I worked this problem I used a density of $\large 62.4 \frac{\text{lbf}}{\text{ft}^3}$ and arrived at an answer of $\large 40 \text{ ft}^4 \cdot 62.4 \frac{\text{lbf}}{\text{ft}^3} =2496 \text{ ft}\cdot \text{lbf}$.

Quote:
Please advise on what I did incorrectly.
1) You multiplied by 9.8, presumably $\large \frac{\text{m}}{\text{s}^2}$, which is not desirable with these units. The factor of $\large 62.4 \frac{\text{lbf}}{\text{ft}^3}$ includes the mass to weight (force) conversion.

2) Your expression for the width is taken along the length of the side, $3$, but the integration will be taken along the direction of the height $\sqrt{3^2-1^1}=2 \sqrt{2}$. So if $y$ is the height, the width is $\large 2-\frac{y}{\sqrt{2}}$, so at y=0 the width is 2 and at $y=2\sqrt{2}$ the width is 0.

3) The displacement is y since this is the height that the sheets will need to be pumped. I mention this because in your equation, at y=0, the width is the greatest and the height to be pumped is greatest. In reality, the width is lowest when the height to be pumped is greatest.

Putting it all together, instead of

Quote:
15 * 9.8 * 62.4 of the Integral from 0 to 3 of (2-2y/3)(3-y) dy
It becomes:

$\displaystyle \large {15 \cdot 62.4 \cdot \int_0^{2\sqrt{2}}\left(2-\frac{y}{\sqrt{2}}\right)\cdot y \cdot dy}=2496$
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October 29th, 2014, 02:54 AM   #4
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Yes you are right jks, I used $p=1Kg/m^3$ (metric system), I forgot that we are using ft.
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