Calculating the flow rate of water from a tap and on the surface of a water tank

In summary, the conversation discusses how to solve a problem involving an open rectangular water tank that is being emptied through a tap. It suggests using a related rates approach and Bernoulli's equation to calculate the velocity of the water surface in the tank and the velocity of the water coming out of the tap. The conversation provides a working solution using Bernoulli's equation to solve for the velocity of the water coming out of the tap.
  • #1
savva
39
0
I am really not sure how to go about this problem, can anybody please give me an indication on how I can start solving the problem?

Homework Statement


An open rectangular water tank 1.2 m x 1.2 m x 2.6 m (height) is completely filled with water. As the tap at the bottom is opened, the flow rate from the tap at that moment is
2.0 x 10-4 m3 /s.

At this moment, calculate the

(i) the velocity of the water surface in the tank.

(ii) the velocity of the water coming out of the tap.

Homework Equations


The Attempt at a Solution

 
Physics news on Phys.org
  • #2
The speed of the water surface in the tank is related to how much water that leaves the tank per second. What's the relation between the amount leaving the tank and the change in the height of water in the tank? The direction ("speed + direction" = velocity) is -y, or downwards, since the surface moves down.
 
  • #3
Are you familiar with Bernoulli's equation?
 
  • #4
For (a) I would recommend a "related rates" calculus approach, if you are familiar with calculus (it is not really required here because the container is simply rectangular).
The idea is to write the rate you are looking for as the product of two related rates. You are given dV/dt and you want dh/dt so you would write
dh/dt = dV/dt*d?/d? [chain rule]
It is easy to find the ?/? expression with the chain rule and to work out its value from the formula for the volume of the tank.

For (b), one approach is to realize that the pressure at the tap is sufficient to push the water up to the height of the surface in the tank. That would be a conversion to a gravitational potential energy. In the case of the open tap, the same energy appears as kinetic energy. Begin with PE = KE for a mass m of water, cancel the m's and solve for v.
 
Last edited:
  • #5
Ok, with regard to the related rates approach, I know how to apply it but I haven't it used in over a year so I need a bit of help with it. I've had a go at it, but I know I've done wrong or missed something along the way.

I have dh/dt=(dV/dt) x (dh/dV)

dV/dt = 2x10^-4 m^3/s

V=(h^2)*y

dV/dh = 2h

dh/dV = 1/2h

Sub-in dV/dt and dh/dV

dh/dt = 2x10^-4 x 1/2(1.2) = 8.33 x 10^-5 m/s

if dh/dV happened to equal 1/h^2 it would give the right answer but I can't get the mathematics to back it up at the moment.

--------------------------------------------------------------------------------------

I have had a try at the method delphi stated for the second part of the question, it got the correct answer, i'll post the working below:

PE=KE

mgh = 1/2m(v)^2

2gh = v^2

v = (2gh)^0.5

sub=in g=9.8 and h = 2.6m

v = (2(9.8)(2.6))^0.5 = 7.14 m/s
 
Last edited:
  • #6
dV/dh = 2h, dh/dV = 1/2h
can't be right; has wrong dimensions for one thing.
V = lwh so dV/dh = lw, dh/dV = 1/(lw)
 
  • #7
PE=KE

mgh = 1/2m(v)^2

2gh = v^2

v = (2gh)^0.5

sub=in g=9.8 and h = 2.6m

v = (2(9.8)(2.6))^0.5 = 7.14 m/s

The above is the same as the Bernoulli's equation.

h1 + .5V1^2/g = h2 + .5V2^2/g

V1 term is negligible. h2 = 0.

V2 = (2 * g * h1)^.5

The Bernoulli equation can be applied to this reservoir situation even though the points of application are not on the same streamline.
 
  • #8
Thanks for that, Lawrence. I was under the impression that Bernoulli's equation was something very complex, but it now looks like it may come from a statement of conservation of energy.
 

1. How is the flow rate of water from a tap calculated?

The flow rate of water from a tap can be calculated by dividing the volume of water that flows out of the tap in a certain amount of time by the time it takes to flow. This can be expressed as: Flow Rate = Volume/Time.

2. What factors affect the flow rate of water from a tap?

The flow rate of water from a tap can be affected by the diameter of the tap, the pressure of the water in the pipes, and any obstacles or restrictions in the pipes. The height of the water source also plays a role as it determines the potential energy of the water.

3. How is the flow rate of water on the surface of a water tank calculated?

The flow rate of water on the surface of a water tank can be calculated by measuring the volume of water that flows out of the tank in a certain amount of time and dividing it by the time it takes to flow. This can be expressed as: Flow Rate = Volume/Time.

4. What is the difference between flow rate and water pressure?

Flow rate is the amount of water that flows past a certain point in a given amount of time, while water pressure is the force exerted by the water on a surface. They are related, as a higher water pressure can result in a higher flow rate, but they are not interchangeable terms.

5. How can the flow rate of water be increased?

The flow rate of water can be increased by increasing the pressure of the water in the pipes, using a larger diameter tap or pipe, and minimizing any obstacles or restrictions in the pipes. Increasing the height of the water source can also increase the potential energy of the water, resulting in a higher flow rate.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
577
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Engineering and Comp Sci Homework Help
2
Replies
56
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
748
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
3K
Back
Top