What restricts the water flow?

[eey01]

If I want to measure the flow at the bottom of a rain water pipe, then want to add a nozzle... how can I calculate the flow at the nozzle? will it be the same?

I have experimentally measured Q at the big pipe.

I want to design a Pelton turbine so I have to know how to calculate the flow and how the flow in affected as I decrease the area of the nozzle.

and how can I calculated the power of the system if the efficiency of the turbine is 80% and the height of the tube is 15ms.

will the jet velocity increase as we decrease the nozzle size? and if it increases to what limit...

thanks a lot!

 

[chrisbaird]

If I understand you right, you want Bernoulli's principle.

 

[Andy Resnick]

Adding a 'smooth' nozzle at the end of a pipe is a standard problem- I'm looking in Streeter's "Fluid Mechanics". From continuity, the flow velocities into (V_in) and out of (V_out) the nozzle are related by the two areas A_in and A_out:

V_out = (A_in/A_out)V_in

To get one of the velocities, you need to know the pressure at the nozzle inlet, as the pressure at the outlet is atmospheric. Most likely, you could set the pressure equal to the pressure required to generate the flow Q out of the open pipe. Then use Bernoulli's equation and you can solve for the velocities.

If the contraction is sudden, things are very different due to viscous effects. There will be a pressure drop within the nozzle ('minor losses') which depends on the ratio of the inlet and outlet diameters, the specific fluid,and the velocity at the inlet. If I did everything correctly, the pressure drop is

dP = (1/C -1)^2 V_out^2/(2*d), where 'C' is the 'contraction coefficient' and was first determined (for water) by Weisbach. 'd' is the density.

The advantage is that Bernoulli's equation can still be used- this head loss is simply another pressure term:

$P_{0} + \frac{V_{in}^{2}}{2ρ} = \frac{V_{out}^{2}}{2ρ} + dP =[1+(1/C-1)^{2}] \frac{V_{out}^{2}}{2ρ}$, and V_in and V_out are related by continuity as before.

Broadly speaking, decreasing the exit diameter will result in an increased velocity but decreased Q.

 

[eey01]

thank you :)