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

• Combinatorics
In summary, the conversation discusses a problem where the flow rate in the lower tap is not equal to the flow rate in the upper tap due to a T-junction in the pipes. The solution involves using Bernoulli's equation and taking the reference level as the lower tap. The correct answer is A, with the flow rate being conserved for steady state flow passing through two places.
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!

Delta2
Combinatorics said:
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

Combinatorics
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?

BvU, Combinatorics and Steve4Physics
Combinatorics said:
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 and Delta2
BvU said:
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!

Delta2 said:
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!

Steve4Physics said:
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.

Delta2
Delta2 said:
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.

Delta2

## 1. How do you calculate the flow rate in a tap?

To calculate the flow rate in a tap, you will need to measure the amount of water that flows out of the tap in a certain amount of time. This can be done by placing a bucket under the tap and timing how long it takes to fill up a certain volume. Then, divide the volume by the time to get the flow rate in units of volume per time (e.g. liters per second).

## 2. What factors affect the flow rate in a tap?

The flow rate in a tap can be affected by several factors, such as the diameter of the tap, the pressure of the water, and any obstructions or blockages in the tap or pipes. The type of tap and the type of water being used can also have an impact on the flow rate.

## 3. How do you calculate the height of a reservoir?

To calculate the height of a reservoir, you will need to measure the distance from the top of the reservoir to the bottom. This can be done using a measuring tape or ruler. Alternatively, if you know the volume of the reservoir and the area of the base, you can use the formula height = volume/area to calculate the height.

## 4. Can the flow rate in a tap be increased by increasing the height of the reservoir?

Yes, the flow rate in a tap can be increased by increasing the height of the reservoir. This is because the higher the water level in the reservoir, the greater the potential energy and pressure of the water, resulting in a faster flow rate when the tap is opened.

## 5. How does the flow rate in a tap affect water usage?

The flow rate in a tap directly affects water usage. A higher flow rate means more water is being used in a given amount of time, resulting in higher water usage. By reducing the flow rate, you can conserve water and potentially save on water bills. Additionally, a lower flow rate can also help prevent water wastage and reduce the strain on water resources.

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