# Homework Help: Work for Tank Drainage Review

1. Jan 24, 2009

### daytripper

Firstly, I'd like to quickly note that this isn't a homework problem, but rather a problem that I'm about to make up. I've done Calc IV already but recently my friend asked me for help on a problem involving the drainage of a tank and I got a little stuck in thought. I'd like to re-educate myself on this.

1. The problem statement, all variables and given/known data

let's say you have a cubic tank (sides = 4 feet). There's a pipe sticking up 1 foot off the top of the cube. The tank is filled up to the 3 foot mark. For these equations, I will be using the variable h to represent the distance from the top of the pipe to the water's surface.

2. Relevant equations

$$Work\; =\; \int_{init}^{final}{density\; \cdot \; volume\; \cdot \; dist\; \cdot \; dh}$$

3. The attempt at a solution

So an infintesimal cross section of this cubic tank would be 16*dh and it would weight 62.4*16*dh. We're moving it a total distance of h (where h varies from 2 to 5).
So then the equation would be:
$$Work\; =\; 62.4\cdot 16\int_{2}^{5}{h\; dh}$$
which evaluates to 62.4 * 16 * 10.5 = 10483.2... and I'm not sure which units this would be in.

To me, it seems that I did this problem correctly. I put it here under "coursework questions", as it's a typical problem you'd run into in a calculus class. The thing that's throwing me off is the fact that gravity isn't taken into consideration. I mean.. it requires more work to move something uphill than downhill, right? So why didn't I have to say something along the lines of " + 9.8*62.4*16*10.5" (that was sloppy deduction, but you get the idea).
Also, if someone could help me out with what units this is in, that'd be helpful as well.

Thanks for all the help. I can never say how happy I am this site exists. =]

-DT

Last edited: Jan 24, 2009
2. Jan 24, 2009

### Dick

That looks just fine. You questions about it are tied in with not knowing the units. The 62.4 is lbs/ft^3 or force/volume (the force per cubic foot for water on the earth). The 'g' is already built in. The other way do this with a visible 'g' in metric units is writing g*rho as the constant out front, where g=9.8m/s^2 and rho is the mass density of water 1000 kg/m^3. If you multiply those together you get units of (kg*m/s^2)/m^3. Since kg*m/s^2 is a newton, the metric force unit, then g*rho is again a force/volume. If you continue on, the rest of the calculation has units of volume*length. So, (force/volume)*(volume*length)=force*length. A newton*meter=joule. The unit of work. Or lb*ft in the daft unit system.