Fluid Mechanics: Flow rate required to achieve a constant height

[WhiteWolf98]

Homework Statement

A surface water drain causes your basement to flood at the steady rate of $2.5~cm/hour$. The basement floor area is $121~m^2$. At what flow rate (in $m^3/s$ should a pump operate to keep the water accumulated in your basement at a constant level? (give your answer in $m^3/hour$).

Relevant Equations

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Some thoughts that I've had on the question are saying the volume flow rate ($Q$) in, must equal the volume flow rate out. If that's the case, then:

$Q_{in} = Q_{out}$

$A_1V_1=A_2V_2$

But... no areas have been given. And height doesn't enter this equation at all.

Then I thought it could have something to do with Torricelli's Law.
$\Delta t = \frac {2A} {a \sqrt {2g}} (\sqrt {h_1} - \sqrt {h_2}$

But again, still, no areas are given. Also, if the height is constant, then:

$\sqrt {h_1} - \sqrt {h_2} = 0$

So the whole equation becomes zero. Besides of which, velocity isn't in that equation at all.

Finally, I thought Bernoulli; that's just out of the question though. There's no streamline.

I know I need to link the height with velocity and somehow area, but I can't find a relationship.

 

[haruspex]

the volume flow rate (Q) in, must equal the volume flow rate out

Right, so what is the flow rate in in this case?

 

[WhiteWolf98]

Do you mean the flow rate of the water into the basement? How can I calculate the flow rate in without knowing the area of the water drain?

I do know the area of the basement however... So, first I converted the speed of the water from $\frac {cm} {hour}$ to $\frac m s$. I then multiplied this speed by the area of the basement to obtain a flow rate. Which, I guess is still the flow rate in...?

 

[haruspex]

Which, I guess is still the flow rate in...?

Yes.

the speed of the water

To be clear, it is the rate of rise of the water if nothing is flowing out.

 

[WhiteWolf98]

Oh. So I've pretty much solved the question... There's not really anything more to do. The flow rate out has to be equal to what I've calculated.

Guess the answer was simpler than I thought. Thank you