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


by savva
Tags: flow, rate, surface, tank, water
savva
savva is offline
#1
Nov14-11, 12:15 PM
P: 39
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?

1. The problem statement, all variables and given/known data
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.


2. Relevant equations



3. The attempt at a solution
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PerUlven
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#2
Nov14-11, 12:30 PM
P: 8
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.
LawrenceC
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#3
Nov14-11, 12:47 PM
P: 1,195
Are you familiar with Bernoulli's equation?

Delphi51
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#4
Nov14-11, 01:13 PM
HW Helper
P: 3,394

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


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.
savva
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#5
Nov14-11, 03:47 PM
P: 39
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
Delphi51
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#6
Nov14-11, 05:11 PM
HW Helper
P: 3,394
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)
LawrenceC
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#7
Nov15-11, 08:21 AM
P: 1,195
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.
Delphi51
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#8
Nov15-11, 12:58 PM
HW Helper
P: 3,394
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.


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