Deriving an equation for water flowing from a tap

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SUMMARY

The discussion focuses on deriving equations related to the flow of water from a tap, specifically addressing the speed of the water stream as it falls. The continuity equation and kinematic equations are utilized to express the speed v2 in terms of the height h and the cross-sectional areas A1 and A2. The final expression derived for v2 is v2 = √(2gh/(1 - (A2/A1)²)). Additionally, the volume flow rate is represented by A1v1, but the height difference between A1 and A2 remains unspecified, indicating that the answer should be left in terms of h.

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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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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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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
 
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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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.
 
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.
 

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