Pipe Flow Rate 2

[Chairs]

I feel like this should be easy but I'm having a good amount of trouble figuring it out. I'm given the inlet pressure, the length and the inner diameter of the hose and I'm asked to find out the flow rate coming out. I'm also assuming the outlet pressure is atmospheric. Without being given a velocity how would I go about doing this?

P=110 psi
ID=2"
L=48"
Media= 70 deg Water

The hose is straight and rigid so it might as well be a pipe but I'd also be interested to know how it differs when bent.

The point of this is to help outside sales compare the flow rate of smooth bore to convoluted hose.

 

[Chairs]

To make it clearer we can use an outlet of 0 psig. The only specifics to the question are there is an inlet pressure and a pipe full of water. Solve for the flow rate. I choose the outlet pressure and length arbitrarily.

 

[BigInch]

Aside from the fact that you cannot solve the typical 3 variable (Q, D, [Δ]P) problem by arbitrarily assuming two of those values, in your case Q and [Δ]P. You can only play with one of those at a time. In a frictional flow equation, assuming an outlet pressure of 0 means that the friction force provided by the pipe at your assumed flow rate is roughly (inlet pressure + outlet pressure)/2 x cross sectional area. In this case that could be very far from the truth.

I hate Windowz 8!!!!

 

[katmar]

My vote goes with the explanation by 77JQX.

If the 95 psi is across the hose only and we can ignore all losses other than friction I get a flow rate of 1692 USGPM and a velocity of 173 ft/s.

If the 95 psig is measured inside the start of the hose and we can ignore the inlet loss but include for the exit loss (K=1) then Q=957 USGPM and V=98 ft/s.

If the 95 psig is measured inside the main that supplies the hose, and we include inlet and exit losses then Q=827 USGPM and V=84 ft/s.

All assuming ID is exactly 2" and roughness is 0.002" (typical smooth fire hose).

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[iainuts]

How about you're all correct?

To do this properly, you have to account for the pressure loss due to the change in velocity at the entrance* plus the pressure loss/gain due to any elevation changes, plus the losses due to frictional flow, plus the exit losses. This isn't a straight pipe flow equation (Darcy-Weisbach) and it's not a straight kinetic energy/velocity calculation. You have to apply Bernoulli's equation to the entire flow and account for other losses:

Chairs, I don't know why you can't get the convoluted hose papers off that web page I pointed to. I can reattach them here if you're still having trouble.

Dave.

*I'm not sure if the entrance loss (ie: K=1) accouunts for the kinetic energy loss at the entrance, but I don't think it does. Anyone know for sure?

 

[katmar]

iainuts, the entrance loss would have K=0.5 for a typical flush entrance, but can be reduced to being virtually nothing by rounding the inlet sufficiently - a bell-mouth entrance. This loss is only the loss of getting the fluid into the pipe, and does not include for any kinetic energy.

The kinetic loss, which is actually consumed at the entrance, is termed the exit loss because that is where it is finally lost (or at least, not recovered). This causes a lot of confusion. One of the best explanations of this was given by PingPong in another engineering forum. See

http://www.cheresources.com/invision/topic/19822-exit-loss-coefficient/

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[ione]

When I made reference to exit losses in my previous post I just wanted to focus Chairs’ attention on the fact that sometimes those, which are defined minor losses, actually are not. And this is certainly the case.
Exit losses refer specifically to that energy which is lost due to stream entering the pipe (acceleration with subsequent pressure decrease) and that it’s not recovered as the fluid decelerates just because expansion takes place in the atmosphere, outside the pipe. I must admit that terminology could appear to be misleading, but this is what has been conventionally established and we have to live with it.

By the way I’ve got practically the same results as those presented by Katmar.