Fluid Dynamics: Cylindrical Bucket, Water Flow & Height Increase

[momogiri]

A cylindrical bucket, open at the top has height 30.0cm and diameter 10.0cm. A circular hole with a cross-sectional area 1.35cm^2 is cut in the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of 2.20×10−4 m^3/s

How high will the water in the bucket rise?
Take the free fall acceleration to be = 9.80

What I've done is
A_1*v_1 (bottom) = A_2*v_2 (top)

So Q_2 = 2.2*10^-4 = A_2*v_2

(0.0135)v_1 = 2.2*10^-4

v_1 = 0.016296m/s
A_1 = 0.0135m^2
v_2 = 0.028011m/s
A_2 = 0.007853m^2

Then I did the bernoulli's principle, cancelling out some stuff
P_1 + 0.5ρv_1^2 + ρgh = P2 + 0.5ρv_2^2 +ρgh

It was here where I got confused, but here's what I did..
(0.5)(1000)(0.016296^2) = (0.5)(1000)(0.028011^2) + (1000)(9.8)h

Is this the correct way to do it? Because I'm totally unsure.. so thanks for your time

 

[Astronuc]

The idea is that the pressure head is related to the flow rate out of the bottom, and the height increases until the flowrate out matches the flowrate in. The use of Bernoulli's equation is not correct as applied.

Certainly the continuity equation applies (flow in = flow out in steady state, i.e. dh/dt = 0), but one must find the mass flow rate out in terms of the pressure head.

 

[m2003]

.

Yes, you are on the right track with your calculations. The Bernoulli's principle equation you used is correct, and you have correctly substituted in the values for the velocities and areas. However, there are a few things to note in your calculations:

1. The first thing to note is that the units for velocity should be in m/s, not cm/s. So when you calculated the velocity v1, it should be 0.016296 m/s, not 0.00016296 m/s.

2. The next thing to note is that the cross-sectional area A1 should actually be the area of the circular hole at the bottom of the bucket, not the top. So A1 should be 0.0135 m^2, not 0.007853 m^2.

3. When you substitute in the values for the velocities and areas into the equation, you should also include the density of water (ρ=1000 kg/m^3) in your calculations. So the equation should look like this:

P1 + 0.5ρv1^2 + ρgh = P2 + 0.5ρv2^2 + ρgh

4. Finally, when you solve for the height h, make sure to rearrange the equation correctly. You should end up with:

h = (P2 - P1 + 0.5ρv1^2 - 0.5ρv2^2) / (ρg)

Substituting in the values you calculated, the height h should be approximately 0.0022 m, or 2.2 cm.

So in conclusion, your calculations were mostly correct, but there were a few minor errors in the units and substitution of values. Keep up the good work!