Pipe diameter impedance

[karimwassef]

I'm looking for some easy math for pipe diameter to flow restriction.

Here's the application:

I have a 2" diameter 2ft pipe from my surge to my DT that has an 2" actuator that open and closes to release or stop the flow. The actuator is 5" long out of the 2ft drop.

There's enough water in the surge that 20gal flows out in about 10sec with gravity.

If I replace the 2" actuator (that's 5" long) with a 1" actuator that's 3" long and with the reducer connectors (2" each?), how much of a flow hit would I experience?

So

24" length of 2" diameter (19" straight + 5" actuator)
vs.
24" of mixed diameter - 17" length of 2" diameter + 2 x 2" length reducer (2" to 1" diameter) + 3" length of 1" diameter actuator.

The first is 2gal/sec.
What's the constriction impact? 1.5g/s?

 

[sleepydoc]

Hard to picture exactly what you're doing, but in general, pipe resistance is proportional to 1/r^4, so halving the diameter has a major impact on resistance. The other part of the equation is that resistance/impedance is dependent on flow, but flow is dependent on impedance (i.e. It's a differential equation)

Since A2 = 4A1, for a given flow, Q, the velocity in the 1" pipe will be 4x the velocity in the 2" pipe. Resistance is proportional to Q/r^4, and since Q1 = 4Q2, and r1^4 = 1/16 r2^4, the resistance of a given length of 1" pipe will be 64x the resistance of the same length of 2" pipe, assuming the same volumetric flow rate.

The actual flow rate depends on the pressure drop which I'm too tired to calculate right now, but it's safe to say the drop to a 1 inch pipe will have a significant effect on the flow.

 

[karimwassef]

Ok but the pipe is only constricted for ~3" of travel out of the total of 24".

It's a pinch point with higher resistance, but how much will that effect the total flow rate?

It's at 2gps or 7200gph now and only travels for 24" through a 2" pipe.

Are you saying the necking down for a small portion of the travel will reduce the flow to 1/64th ~ 113gph?

 

[Wazzel]

No he is not saying that. Only that the losses will go up by that factor. if your current system is oversized it should not be a big deal. The flow in the modified system will be a lower gpm but with more velocity than the original. A sketch or picture would help.

 

[karimwassef]

Not sure why it isn't clear. It's a straight pipe vs. a reduced section pipe

<a href="http://s1062.photobucket.com/user/karimwassef/media/8FFE3E62-AD93-46AE-8BA7-6567863CA2A7.png_zpsjrw1nwyh.jpeg.html" target="_blank"><img src="http://i1062.photobucket.com/albums/t496/karimwassef/8FFE3E62-AD93-46AE-8BA7-6567863CA2A7.png_zpsjrw1nwyh.jpeg" border="0" alt=" photo 8FFE3E62-AD93-46AE-8BA7-6567863CA2A7.png_zpsjrw1nwyh.jpeg"/></a>

 

[Wazzel]

Do you have any technical information on the two actuators?

Flow through the valve will be

GPM=Cv(delta P/G)^1/2

Cv = valve constant - should be in tech data
delta P = pressure difference - going to have to make some assumptions
G = specific gravity of fluid

 

[karimwassef]

Assume ball valves. They're DIYs with actuators.

So 2" ball valve vs. 1" ball valve with two 2" to 1" reducers (one before, one after)

 

[karimwassef]

The flow rate through the 2" ball valve now is 7200gph. That's the baseline case.

 

[Wazzel]

Cv, 1" - 94, 2" - 480 http://www.engineeringtoolbox.com/ball-valves-flow-coefficients-d_223.html
GPM 2" - 120 - your data
Delta P for intial case - 1.05 psi calculated

Assuming same pressure drop for mods
GPM 1" - 95 GPM (5700 gph) approximate

 

[karimwassef]

I like that answer, but not sure I follow.

GPM=Cv(delta P/G)^1/2

Since the (delta P/G)^1/2 is identical in both cases, isn't the ratio of GPM for the two cases equal to the ratio of Cv = 94/480 ~ 20% of the flow?

 

[Fishmommy]

You will experience additional losses at the expansion and contraction points, as well as loss due to reduced diameter and loss through the valve.

 

[Gorgok]

Is this accurate? http://www.beananimal.com/articles/hydraulics-for-the-aquarist.aspx

If so, assuming you have a 28" head height from the surge to the tank and a submerged pipe you would flow the ~7200 gph flow with a 2" minimum pipe orifice, and drop all the way to 1800 gph with a 1" minimum. The longer the small diameter is there the more additional loss there would be, from friction, but after that increasing the size won't 'add' flow just reduce the loss already present from the major flow restriction.

 

[sleepydoc]

That's what I get using the Hagen Poiseuille eqn:

This makes a bunch of assumptions and ignores the resistance/turbulence caused by the transition, so the actual flow would probably be less.

 

[karimwassef]

Thanks guys.

So it's ~1/3 to 1/4 of the original flow.

So even if I used 2 valves at 1", I would still be at half the current flow.

Thank you.

 

[laverda]

A 2" circle has an area of 3.14 Sq inches, Almost 4 times the area of a 1" circle at .79 Sq inches, so your flow will be reduced by about 75% unless your 2" pipe could handle more flow then it currently is.
A 2" drain pipe without any other restrictions will flow approximately 2350GPH according to the RC drain calculator. While a 1" will only flow 590GPH. It figures out to just about 1/4 the flow for a 1" pipe. Your getting more flow since it is a full syphon and you have the head pressure of 20 gallons of water, but still you should expect about a 75% decrease in flow.