# Fluid mechanics

1. Oct 24, 2012

### Saitama

1. The problem statement, all variables and given/known data
http://i50.tinypic.com/2m2tbaq.png

2. Relevant equations

3. The attempt at a solution
I suppose i have to apply the Bernoulli's equation here but i don't even have the slightest idea on how to apply it here. I am a dumb at fluid mechanics, any help would be appreciated.

2. Oct 24, 2012

### Basic_Physics

To solve for the velocity of flow you set Bernouli's equation up at the surface of the liquid and at the bottom of the opening. The pressure at both ends are atmospheric. The only point difficult to grasp is that the flow velocity at the surface is assumed to be zero, that is we assume that the container is very large and that the level is not dropping so that the water just beyond the entry point do not need to rush into the pipe due to the large source of water (this portion just outside of the top end of the pipe is still part of the streamlines).

3. Oct 24, 2012

### Saitama

If i take the surface of water as the reference, i set the Bernoulli's equation as:
$$P_o=P_o+ρgh+\frac{1}{2}ρv^2$$
where v=flow velocity, ρ=density of fluid, h=3.6m and Po is the atmospheric pressure.
I don't think i will get an answer using the above equation. Please tell me where i am wrong.

4. Oct 25, 2012

### Basic_Physics

The height should be negative because it is below the reference level. What answer do you get then? I would like to also add that the flow takes into account the whole of the system - that is the flow at the top is the water from the whole of the surface of the container. The speed of the water molecules would then be very low due to the large surface area contributing to the flow - which we take to be zero.

Last edited: Oct 25, 2012
5. Oct 25, 2012

### Saitama

Oops, absolutely forgot that. I get my answer as 6√2 m/s as mentioned in the answer key.
I am stuck at b) part. I guess to find the discharge rate of flow, i need to use the equation Q=Av, where Q is rate of flow, A is cross-section area and v is the velocity of liquid. If i plug in the values A=64*10^(-3) m^2 and v=6√2 m/s, i don't get the answer mentioned in the answer key. Please tell me where i am wrong.

Thanks for the help.

6. Oct 25, 2012

### Basic_Physics

All error I can see is that 1 cm is 10-2 m so it so 10-4 m2 for A. Also it is r2 so it should be divided by 2 since it is the diameter.

To get the pressure at the crest you set BE up at the crest and at the surface of the water. The speed of the water at the crest will be the same as the exit speed calculated in (a) - the water flows with the same speed throughout the pipe. In this case the height is positive (above the zero level). I think the density of water is 103 kg/m3?

Last edited: Oct 25, 2012
7. Oct 25, 2012

### Saitama

I came to edit my post but you beat me. :tongue2:

Using the BE, i get
$$P_o=P_A+ρgh+\frac{1}{2}ρv^2$$
P_A is the pressure at A. Is this equation right?

8. Oct 25, 2012

### Basic_Physics

Yes, and it should be /4 not 2.

9. Oct 25, 2012

### Saitama

If i plugin the values in my equation, i get the correct answer, thanks for the help Basic_Physics!