# Simple Fluid Flow

1. Oct 17, 2009

Doing some review here:

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

A fire hose has an inner diameter of 5 inches and water is flowing at 600 gal/min. The flow exits through a nozzle contraction with
diameter dn. For steady flow, what should dn be in inches in order for the exit velocity to be 25 m/s.

My only question here is really do I even need the given inlet diameter = 5 in ?

As far as I can see. Since density is constant, the volume flow rates at inlet and exit should be equal. That is,

$$Q_1=Q_2$$

$$\Rightarrow Q_1 = \frac{500\,\text{gal}}{\text{min}} = A_2V_2$$

$$(\frac{\pi}{4}d_n^2)V_2=\frac{500\,\text{gal}}{\text{min}}$$

Now assuming I convert my units properly, this should be correct? It's just that given inlet diameter that's bugging me.

Thanks.

2. Oct 17, 2009

### rl.bhat

Can you convert gal/min to A*V?

3. Oct 17, 2009

I would have to say yes.

$$\frac{gal}{min}\equiv\frac{Volume}{time}$$

$$Area*V\equiv\frac{dx}{dt}*A\equiv\frac{Volume}{time}$$

4. Oct 17, 2009

### rl.bhat

Inner diameter is given. Find the velocity of inflow. Velocity of the outflow is given. Find dn

5. Oct 17, 2009

I am sorry rl, I still do not see what I am supposed to use the inner diameter for. I thought that we just said.

Q1=Q2=A2V2

I am given Q1, why should I bother finding V1?

6. Oct 17, 2009

### rl.bhat

Yes. You are right.

7. Oct 17, 2009