Rotary (pos. displacement) pump and system

[Howard]

Right now I've got a rotary pump in a system where the discharge pressure is about 350psi, and the nominal max is 200psi. One of my supervisors tells me that it's bending shafts and whatnot, in addition to delivering poor flow at the end of the system (which is weird, because it's a rotary); now, the pipe is 2" stainless steel and the fluid is something about as viscous as Vaseline. He's considering moving to 3" to relieve the pressure at the pump discharge.

As a cheaper measure to reduce discharge pressure, one of his colleagues suggested doing something like this:

http://www.rootminus1.com/howard/pics/pump.png

where the parallel pipe is the suggested installation. The system resistance would probably drop a little, lowering the discharge pressure, but as far as I know it would be just like upping the cross-sectional area for a part of the pipe.

Can somebody help me get started with the calculations to find the drop in discharge pressure for moving from 2" to 3" pipe? I can get the length of pipe, minor losses, viscosity of fluid, etc. The nominal flow rate of the pump is 50GPM, but it doesn't sound like it's delivering that much.

Pointing me to the right formulas would be great.

 

[futuristicmonkey]

Might wanna register here and check out the mechanical forum. These guys know damned near everything between the lot of them.

-ben

 

[Howard]

Ah, totally forgot about that. Thanks!

 

[CycloWizard]

It's hard to say what the best way to calculate it might be based on what you've stated. Vaseline isn't a purely viscous solution, it's viscoelastic and probably shear-thinning. According to this,
vaseline's viscosity is about 64,000 cP (640 Pa-s). I'll assume that that is the viscosity in the Newtonian plateau (i.e. low shear rates) and that the flow is probably laminar based on the low flowrate and the large pipe diameter. Then, the Hagen-Poiseuille law may be used to relate the pressure drop to what you know.

dP=(8QvL)/(pi R^4),
where dP is the pressure drop per unit length
Q is the volumetric flowrate
v is the viscosity
L is the length of pipe you're using
R is the pipe radius

 

[Howard]

Hmm... with pressure drop, I can find the approximate power required for a given flow rate. What I'd like to know is if there's a relationship between power and discharge pressure for a positive displacement pump. If not, I can't see how I can use dP to estimate discharge pressure with the 3" line.

 

[CycloWizard]

What is the length of piping following the pump? What is the outlet pressure from this piping? Since you know the flowrate, it's just a matter of back-calculating the initial pressure.

edit: Oops, dP is really the net pressure drop across length L, not the differential pressure gradient. Sorry about that.

 

[Howard]

I think it's 25', but I'm not sure. I'll be able to get an exact figure tomorrow.

Assuming that flow rate, length and viscosity stay constant, I can plug in radius to find dP. Then, if I measure the outlet pressure of the current pipe, I can find the discharge pressure. But how will the outlet pressure change if the radius of the pipe changes? Or will it stay the same?

 

[CycloWizard]

Originally posted by: Howard
I think it's 25', but I'm not sure. I'll be able to get an exact figure tomorrow.

Assuming that flow rate, length and viscosity stay constant, I can plug in radius to find dP. Then, if I measure the outlet pressure of the current pipe, I can find the discharge pressure. But how will the outlet pressure change if the radius of the pipe changes? Or will it stay the same?

When the pipe diameter changes, R (the pipe radius) changes as shown in the Hagen-Poiseuille law. Everything else should be constant if you're just increasing the pipe diameter.

edit: Apparently I can't read tonight. The pressure at the pump should be the same as in case #1, so just calculate the outlet pressure using the same formula with a new R value.

 

[Howard]

So I can simply add the new pressure drop to the current outlet pressure and get the new discharge pressure?

 

[CycloWizard]

Originally posted by: Howard
So I can simply add the new pressure drop to the current outlet pressure and get the new discharge pressure?

I'm getting confused by what you're calling outlet and discharge.

Let's say that Pp is the pressure coming out of the pump and Po is the pressure leaving the pipe system. Then, dP=Po-Pp. You know Po for case #1 (where R=1"), so you can calculate Pp. Pp should be the same for case #2 (where R=1.5"), so you can calculate Po, the new expected effluent pressure.

 

[Howard]

Sorry about that. Yes, that just about explains all I need to know. Why should Pp be the same for case 2, though?

Also, were I to include the minor losses in the calculations, would I be correct in doing this:

1 - use Colebrook's equation to find friction factor
2 - find the overall head loss from friction factor and minor losses using delta h = hf + (sigma)hm
3 - plug into steady flow energy equation to find delta p

 

[CycloWizard]

Originally posted by: Howard
Sorry about that. Yes, that just about explains all I need to know. Why should Pp be the same for case 2, though?

All things being equal (i.e. constant upstream pressure, constant power), the pump should output the same pressure. If you want to decrease the pressure that the pump is putting out, decrease its power supply (if you have a fancy, variable-speed pump) or downgrade pumps to something less powerful. Since it sounds like you're on a tight budget though, the pump is probably a fixture so you'll just have to make do with what it gives you.

Also, were I to include the minor losses in the calculations, would I be correct in doing this:

1 - use Colebrook's equation to find friction factor
2 - find the overall head loss from friction factor and minor losses using delta h = hf + (sigma)hm
3 - plug into steady flow energy equation to find delta p

1. The Colebrook equation is, IIRC, only for turbulent flow (see here). There are any number of ways to find friction factors, but the easiest way (since you're just doing this once) is to use a plot of friction factor as a function of Reynolds number, which should be in any decent fluild mechanics book (it's in at least three of mine ). Or, since you're easily within the laminar region (assuming vaseline's viscosity and density of water), the friction factor is simply 16/Re, where the Reynolds number (Re) is about 0.062.
2. If I'm reading your notation right, that should be fine.
3. Make sure you're not using the inviscid form of Bernoulli's equation, which is only valid for inviscid flow (that is, flow where viscosity is negligibly small, which is certainly not the case here where viscous forces are dominating). There is a less-known form of Bernoulli's that deals with viscous flows that I'm less familiar with. If you have a copy of Bird, Stewart, and Lightfoot 2002 edition, they solve a problem very similar to yours on p. 205-207. I know there are similar sections in other books, but mine aren't at home at the moment.