Calculate flow rate in a tap and the height of a reservoir

[Combinatorics]

Homework Statement:

A water reservoir is located on a hill of height $H$ above sea level, and open to the atmosphere. The water flow from the reservoir to the buildings in the nearby city. It is known that in one of the buildings, in the first floor, which is in height $2 \, \text{m}$ above sea level, a tap is open with a flow rate of $S=0.0145 \frac{m^3}{s}$. In the same building, in the last floor, which is of height $24 m$ above sea level, a tap that is connected to the tap in the first floor is open. The area of the tap opening of both taps is $A=0.0003m^2$. What is the flow rate in the last floor and what is $H$?

Relevant Equations:

Bernouli equation: $p+\rho g h + \frac{1}{2} \rho v^2 = \text{constant}$

Flow rate is $\text{area}\times \text{velocity}$

The possible answers are: (I do not know what is the right one)
A. $H=120m$ , $S=0.0131\frac{m^3}{s}$
B. $H=60m$ , $S=0.0231\frac{m^3}{s}$
C. $H=120m$ , $S=0.0231\frac{m^3}{s}$
D. $H=240m$ , $S=0.0231\frac{m^3}{s}$
E. $H=60m$ , $S=0.0131\frac{m^3}{s}$
F. $H=240m$ , $S=0.0131\frac{m^3}{s}$

I first cannot understand how is it possible that the flow rate in the lower tap is not equal to the flow rate in the upper tap, given that the two are connected. Isn't the flow rate conserved?

As for the solution:
*) Using the information on $A$ and $S$, we can deduce that the velocity in the lower tap is $\frac{S}{A}=48.3\frac{m}{s}$.

*) Given that the reservoir is open, we have that the LHS of Bernoulli equation is
$P_{atm} + \rho g H$ where $P_{atm}$ is the atmospheric pressure.

*) Equating with the information on the lower tap, we obtain
$P_{atm} + \rho g H = P_{lower} + \rho g 2 + \frac{1}{2}\rho 48.3^2$.

*) Equating with the information on the uppwer tap, we obtain
$P_{atm} + \rho g H = P_{lower} + \rho g 2 + \frac{1}{2}\rho 48.3^2 = P_{atm} + \rho g 24 + \frac{1}{2} \rho v_{upper}^2$
where $v_{upper}$ is the velocity in the upper tap.
The problem is that these are two equations, with three unknowns - $P_{lower}, v_{upper}, H$ and I have no idea how to find another equation that relates some of them. In addition, I am not sure why isn't the flow rate the same in the lower and upper taps.

Will be happy if you will be able to help me out here.

Thanks!

 

[BvU]

I first cannot understand how is it possible that the flow rate in the lower tap is not equal to the flow rate in the upper tap, given that the two are connected. Isn't the flow rate conserved?

Flow rate is conserved for a steady state flow that passes through two places. That isn't happening here. The higher tap 'sees' a lower pressure on the inflow side and therefore delivers a smaller flow.

Perhaps you are supposed to use a relationship between $S, p$ and $A$ ? As in orifice flow ? See flow factor here

 

[Delta2]

Not sure at all but I think the problem wants us to take $P_{atm}=P_{lower}$, since the taps are open the pressure of the fluid equals the atmospheric pressure. Why not?

 

[Steve4Physics]

The problem is that these are two equations, with three unknowns - and I have no idea how to find another equation that relates some of them. In addition, I am not sure why isn't the flow rate the same in the lower and upper taps.

I agree with @Delta. $P_{lower}$ is simply atmospheric pressure. This follows by thinking about the physics - the flows are entirely driven by pressure-differences arising from the height-differences (the $\rho gh$ terms). If there were no height-difference, the flow rate would have to be zero because the entry and exit pressures are both atmospheric.

The taps are not in series - the output of one tap is not the input to the other tap. There will be a ‘T-junction’, so the incoming water-pipe splits somewhere into two pipes - one pipe going to the lower tap and the other pipe going to the upper tap. So the flow rates will be different.

Handy hint. You can simplify the maths a little bit if you take the reference level as the lower tap (take h=0), so the upper tap has h=22m and the reservoir surface is at height h=X (then at the end, H = X+2).

 

[Combinatorics]

Flow rate is conserved for a steady state flow that passes through two places. That isn't happening here. The higher tap 'sees' a lower pressure on the inflow side and therefore delivers a smaller flow.

Perhaps you are supposed to use a relationship between $S, p$ and $A$ ? As in orifice flow ? See flow factor here

Thank you BVU for helping me understand my misunderstanding regarding the flow rate!

Not sure at all but I think the problem wants us to take $P_{atm}=P_{lower}$, since the taps are open the pressure of the fluid equals the atmospheric pressure. Why not?

Great. This yields answer A as the closest to what I obtain in my calculations. Thank you!

I agree with @Delta. $P_{lower}$ is simply atmospheric pressure. This follows by thinking about the physics - the flows are entirely driven by pressure-differences arising from the height-differences (the $\rho gh$ terms). If there were no height-difference, the flow rate would have to be zero because the entry and exit pressures are both atmospheric.

The taps are not in series - the output of one tap is not the input to the other tap. There will be a ‘T-junction’, so the incoming water-pipe splits somewhere into two pipes - one pipe going to the lower tap and the other pipe going to the upper tap. So the flow rates will be different.

Handy hint. You can simplify the maths a little bit if you take the reference level as the lower tap (take h=0), so the upper tap has h=22m and the reservoir surface is at height h=X (then at the end, H = X+2).

Thank you Steve! My original thought was indeed that the taps are connected in series. It is much clearer now.

 

[BvU]

Not sure at all but I think the problem wants us to take $P_{atm}=P_{lower}$, since the taps are open the pressure of the fluid equals the atmospheric pressure. Why not?

Looking at the 0.0003 m I can only agree: 1 cm diam is wide open and virtually no constriction in the flow conduit.