Fluids: Water runs into a fountain

1. Nov 13, 2007

jago-k1

1. The problem statement, all variables and given/known data
Water runs into a fountain, filling all the pipes, at a steady rate of 7.52×10−2 m^3/s

How fast will it shoot out of a hole 4.53cm in diameter?

2. Relevant equations
A1V1 = A2V2

3. The attempt at a solution

A1V1 = 7.52x10-2 m^3/s = A2V2
Divide by A2, which would be 4.53/2 (to get Radius) = 2.265 --> .02265m or 2.265x10-2

Now, for A2, am I suppose to square it to get m/s as my answer? I'm assuming so since V1=m/s so that must mean A1 = m^2.

If I do all of that, I get (7.52x10-2 m^3/s) / (2.265x10-2 m)^2 = 146 m/s which I think is wrong. Thanks for the help.

2. Nov 13, 2007

mgb_phys

Just try and picture it instead of hitting equations.
Imagine a sausage coming out a hole 4.53cm diameter (work out the area in m)
Now what length would that sausage have to be to have a volume of 7.52E-2 m^3
If that length comes out every second - that is your speed.

3. Nov 13, 2007

jago-k1

Ok, I can imagine the .0453m diameter hole, also the sausge coming out lol Now i'm trying to figure out the length.

So I would have Pi x r2 x h = Volume
H=L
L=7.52E-2 m^3 / pi*r^2 ?

4. Nov 13, 2007

mgb_phys

Volume of a cylinder is just area * length.
Area = pi (0.0453/2)^2 = 0.00644m^2
The volume/sec = 7.52E-2 m^3
So length = volume/area = 7.5E-2/6.44E-3 = 11.6 m/s

(unless I got the sums wrong! )

5. Nov 13, 2007

jago-k1

Ok
Volume of Cylinder = Area * Length
Area = pi x r^2
Divide .0453 by 2 to get r
Area = pi x (.0453/2)^2=.0016117077

V/A = L

7.52E-2 m^3 / 1.61E-3 = 46.7 m/s

EDIT: Thanks for your help. You seemed to sum something up wrong lol I got the right answer. Thanks again.!

Last edited: Nov 13, 2007
6. Nov 13, 2007

mgb_phys

oops you're right I forgot to divide the diameter by 2!
Always check your calculations - or at least always check mine!!!!!

7. Nov 13, 2007

jago-k1

lol I also tried what I had put before[L=7.52E-2 m^3 / pi*r^2 ] and got the same # lol