Deriving an equation for water flowing from a tap

In summary, the problem involves a stream of water flowing from a tap with an initial speed v1 and diameter A1. The speed of the water, v2, after falling a distance h can be found using the kinematic equations and the continuity equation. This results in the expression v2 = √(2gh/(A1/A2)^2 - 1). However, without the height difference between A1 and A2, a numerical value for the volume flow rate cannot be calculated.
  • #1
Woolyabyss
143
1

Homework Statement


A stream of water flows downward from a tap. Assume that the water is in free fall once it leaves the tap, at which point its speed is v1, and the initial diameter of the water stream is A1.
(i) Find an expression for the speed v2 of the liquid as a function of the distance, h, it has fallen.
(ii) Combine this with the equation of continuity, to find an expression for the speed v2 in terms of the distance h and the cross-sectional areas at A1 and A2.
(iii) The cross-sectional area of the stream changes from 1.2 cm2 to 0.35 cm2 between the positions A1 and A2. Calculate the volume rate of flow of water from the tap, in cm3 s

Homework Equations


Continuity equation
Kinematic equations

The Attempt at a Solution


(i)

.5(v1)2 + gh = .5(v2)2
therefore v2 = √(2*(.5(v1)2)+gh)

(ii)
.5(v1)2 + gh = .5(v2)2

therefore using A1v1 = A2v2

.5(v1)2 + gh = .5(v1)2*A1/A2

and so
.5(v1)2((A1/A2)2 - 1) = gh

therefore v1 = √(2gh/(A1/A2)2 - 1)

(iii)
I'm not sure how you can get a numerical value for this question but the question seems to imply you can.

any help would be appreciated.
 
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  • #2
Woolyabyss said:
.5(v1)2 + gh = .5(v2)2
therefore v2 = √(2*(.5(v1)2)
Try that step again.
Woolyabyss said:
therefore v1 = √(2gh/(A1/A2)2 - 1)
It asks for v2 in terms of h, A1, A2, not for v1.

Your continuity equation involved terms like A1v1. What physical aspect does that represent?
 
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  • #3
haruspex said:
Try that step again.

It asks for v2 in terms of h, A1, A2, not for v1.

Your continuity equation involved terms like A1v1. What physical aspect does that represent?

Ok, so

v2 = (2gh/(1-(A2/A1)^2))^.5

A1v1 represents the volume flow rate
 
  • #4
Woolyabyss said:
Ok, so

v2 = (2gh/(1-(A2/A1)^2))^.5

A1v1 represents the volume flow rate
Yes, you seem to be missing a piece of data. E.g. the height difference between A1 and A2. Is there any more info, maybe in a diagram?
 
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  • #5
haruspex said:
Yes, you seem to be missing a piece of data. E.g. the height difference
haruspex said:
Yes, you seem to be missing a piece of data. E.g. the height difference between A1 and A2. Is there any more info, maybe in a diagram?

between A1 and A2. Is there any more info, maybe in a diagram?

There is a diagram but it doesn't contain anymore information. I guess you're meant to leave your answer in terms of h.
 
  • #6
Woolyabyss said:
There is a diagram but it doesn't contain anymore information. I guess you're meant to leave your answer in terms of h.
Maybe, but it seems odd to provide numbers for the areas but not for h.
 

Related to Deriving an equation for water flowing from a tap

1. How is the equation for water flowing from a tap derived?

The equation for water flowing from a tap is derived by using the principles of fluid mechanics and Bernoulli's equation. This equation takes into account factors such as the velocity of the water, the height of the tap, and the pressure of the water.

2. What is Bernoulli's equation and how is it used in deriving the equation for water flowing from a tap?

Bernoulli's equation states that the total energy of a fluid remains constant as it flows. This equation is used in deriving the equation for water flowing from a tap by equating the energy at the top of the tap (potential energy) to the energy at the bottom of the tap (kinetic energy).

3. What are the key variables in the equation for water flowing from a tap?

The key variables in the equation for water flowing from a tap are the velocity of the water, the height of the tap, and the pressure of the water. These variables are all interconnected and affect the flow rate of the water.

4. Can the equation for water flowing from a tap be applied to all types of taps?

Yes, the equation for water flowing from a tap can be applied to all types of taps as long as the tap is in a constant state and the water is flowing in a steady stream. However, the equation may need to be modified for different types of taps, such as those with different shapes or sizes.

5. How can the equation for water flowing from a tap be used in real-world applications?

The equation for water flowing from a tap is commonly used in engineering and plumbing applications. It can be used to calculate the flow rate of water in a tap, which is important for determining the amount of water that can be supplied to a building or for designing plumbing systems. It can also be used to diagnose issues with water flow, such as clogs or leaks.

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