- #1

Combinatorics

- 36

- 5

- 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!

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!