# Flow Rate thru Nozzle

1. Sep 2, 2010

### jman123456789

It's been a while since I've had to do these calculations..

I'm looking for the flow rate at the wall of a 3.175 mm nozzle tip. The diameter of the delivery leading up to it is 19 mm. This is a vertical assembly and the source of the air at 70 psi is a 1/4" outer diameter tube that is approximately 2.5" above the exit. There is about 120 cfh available maximum for this setup.

Anybody looking to take a shot at this, I'd be interested in seeing your solution..

2. Sep 3, 2010

### minger

You can loosely solve this quickly by Bernoulli's equation. The equation can be written as:
$$\frac{p}{\rho} + \frac{V^2}{2} + gz = \mbox{constant}$$

So, having what you know, we can solve for the final velocity, and then with the area, get the flow rate.
$$\frac{70 \frac{lbf}{in^2}}{0.43 \frac{lbm}{in^3}} + 0.0 + (386.4 \frac{in}{s^2})(30 in) = \frac{V^2}{2}$$
Note here that I have assumed that the velocity in the tank is zero, and that the jet is discharging to ambient pressure (0 gauge). We multiply everything out for:
$$162\frac{lbf\,in}{lbm} + 11592\frac{in^2}{s^2} = \frac{V^2}{2}$$
Note the messed up term all the way on the left. That is the cause of much frustration using imperial units. In order to correlate a pound-mass with a pound-force, we need to use the gravitational constant. Now, note that everything here I'm using the inches-pounds-second (IPS) system. We add the gravitational constant:
$$\left(162\frac{lbf\,in}{lbm}\right)\left(386.4 \frac{lbm\,in}{lbf\,s^2}\right) + 11592\frac{in^2}{s^2} = \frac{V^2}{2}$$
We now have in2/s2 on both terms. We multiply and add through for:
$$8981481\frac{in^2}{s^2} = \frac{V^2}{2}$$
Which results in a velocity of 2997 in/s. Calculate the area, which should be pretty easy. I found the flow rate to be 36.6 in3/s.

Now, for a more accurate calculation, you'll want to calculate the head loss due to the nozzle and pipe bends and add it as an extra term in the equation above.