# Pump used to halve the time taken by fluid to flow out

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1. Jun 28, 2016

### Soren4

1. The problem statement, all variables and given/known data
The tank in picture has section $S_1$ and it's linked with a tube of section $S_2=\frac{1}{\sqrt{17}} S_1$. The tank is filled at an height $h$ and this level is kept constant by a sink $R$, in such way that the exiting volumetric flow rate is $Q=0.3 m^3/s$ and constant. Determine the power $P$ of a pump located in the tube that can halve the time taken by the fluid to come out of the tank if the density of the fluid is $\rho =1000kg/m^3$.

2. Relevant equations
Bernoulli equation

3. The attempt at a solution
What is unclear to me is how to interpret the condition of halving the time taken by the fluid to come out. I interpeted it as equivalent to double the coming out speed of the fluid, that is to have a volumetric flow rate of $2Q$ instead of $Q$. Is that possibly correct?

In that case I can write $$\rho g h (2Q) +P= \frac{1}{2} \rho (\frac{2Q}{S_2})^2 (2Q)$$ And find $P$ from here. Is the reasoning correct for this problem?

2. Jun 28, 2016

### haruspex

I agree with your interpretation (of a poorly worded question). But note the information you have regarding S1. Think about how that might be relevant.

Last edited: Jun 29, 2016
3. Jun 29, 2016

### rude man

I can think of no reasonable alternative interpretation. Sure!
You can write a second equation with P = 0 and eliminate h so that P = P(Q only).