Pumping water upwards in a column

[cc94]

Homework Statement

Suppose you have a tank 10 m high. The tank is currently filled 5 m high with 1000 kg of water. How much energy does it take to pump in another 1000 kg of water from the bottom to fill the column to 10 m? (Ignore friction, etc.)

Homework Equations

E = m*g*h

The Attempt at a Solution

I'm always confused by what we're supposed to consider in fluid mechanics problems. Is the answer simply

1000*9.8*5

since the waterline is currently at 5 m and needs to be raised another 5 m? I would think that the more water is on top of the pump, the harder it would get to pump even more water inside.

 

[scottdave]

So 9.8*5*1000 would be if you are pumping 1000kg up by a height of 5m. But that isn't exactly what happens is it?

 

[kuruman]

As the tank fills, it takes increasing amounts of work to add height $dy$ to what is already there because there is increasing pressure to work against. Your answer would be correct if it took the same amount of work to do that, i.e. if the force pushing down on the water you are adding did not increase as the height increases. You need to set up an integral.

 

[cc94]

So 9.8*5*1000 would be if you are pumping 1000kg up by a height of 5m. But that isn't exactly what happens is it?

Yeah I feel like I'm pumping a little water to a height of 5 m, then a little to 5.1 m, etc. My other idea was that I need to perform an integral over the height:

$$E = \int_5^{10} \rho gr\,dr$$

where $\rho$ is the weight of water per $dr$.

 

[Dr Dr news]

I think you have the right idea almost. The work is defined as the integral of F dot dx which in this case is the integral of p dV = p A dy. p at the bottom, if the tank is open to the atmosphere at the top is p(atm) + rho g h. The cross-sectional area of the tank you can get from the initial condition that the tank contains 1000 kg of water, 5 m deep which means the tank is 0.20 m^2 cross-sectional area. So the work is the integral from 5 m to 10 m of the quantity (p atm + rho g y) A dy

 

[cc94]

I think you have the right idea almost. The work is defined as the integral of F dot dx which in this case is the integral of p dV = p A dy. p at the bottom, if the tank is open to the atmosphere at the top is p(atm) + rho g h. The cross-sectional area of the tank you can get from the initial condition that the tank contains 1000 kg of water, 5 m deep which means the tank is 0.20 m^2 cross-sectional area. So the work is the integral from 5 m to 10 m of the quantity (p atm + rho g y) A dy

Ah, so I'm also forgetting to add on atmospheric pressure. Thanks! So one final question, now if I have a shape where the area changes as a function of height, I can just substitute in A(y) and the rest stays the same, correct? I seem to recall that pressure at depth doesn't depend on the shape of water above it, only the height.

 

[haruspex]

Ah, so I'm also forgetting to add on atmospheric pressure.

No no! @Dr Dr news seems to think the water at the bottom is not exposed to the atmosphere. I would assume it is, so atmospheric pressure cancels out.
Think about mass centres. What is the height of the mass centre of the pumped water before it is pumped? What about after?

 

[Dr Dr news]

cc94: You are right. p does not rely on the shape but A(y) gives you V(y)

 

[cc94]

@haruspex I might have either situation. Thanks!

@Dr Dr news Cool, that's just what I needed. Thanks!