# Calcualting Residual pressure

1. Dec 26, 2009

### sprinkyrob

Is there a formula to calculate the residual pressure needed to flow a specific amount of Water through a fixed orifice size.

I am up in Canada and cannot do a residual flow test on a fire hydrant until next spring.

I do however have water flow information from the fire dept in US gallons per minute. When they fill thier pumper truck through a storz connection with an internal dimension of 3-1/2" they are able to flow 825 US gal per minute. Is there a formula that would be able to tell me what the residual pressure would be to achieve this flow rate?

Rob

2. Dec 27, 2009

### spacester

"Orifice" usually refers to small holes (less than 1/4 inch), so forget those formulas.

What you need to find is the friction of the hose and some help from Mr. Bernoulli.

I do not know what you mean by "residual" pressure. I presume you mean the static pressure at the fire hydrant that pushed that amount of water thru that particular hose at that particular rate, yes?

So find the type of hose and its length. Use standard tables or e/d (see the link) to get the friction factor for the hose and then solve for the delta-P in a standard flow equation. Account for any fittings in the line by adding equivalent feet of hose. Assume Pressure is zero (atmospheric pressure, gage pressure=zero) at the tank and then your delta-P is the static pressure at the hydrant.

http://www.efunda.com/formulae/fluids/calc_pipe_friction.cfm

3. Dec 27, 2009

### FredGarvin

This is going to be tough because your terminology is different than an engineer's. Is there any way you can provide a quick sketch of the flow circuit? I am not sure if you want the flow at the end of a fire hose or the like which was the assumption of the above post, or if you want the flow through a flow restriction like a nozzle. Either way, there isn't a quick and dirty way to do this properly. Whatever else you can provide would be a big help.

4. Dec 27, 2009

### sprinkyrob

I am trying to calculate the residual pressure at the discharge outlet of a 6" fire hydrant.

The fire dept connects to a 4" outlet on the hydrant to fill their pumper truck. The static pressure in the line piping to the hydrant is approx 72 PSI and is supplied by a 10" underground feed main.

The "Storz Fire Dept Connection” is attached directly to the 4" outlet on the hydrant and has an internal diameter of 3-1/2".

The pumper truck has a capacity of 3,000 US Gallons and takes exactly 3.5 minutes to fill when completely empty. That is where I came up with the average flow rate of 857 US Gallons per minute.

What I need to know is if there is a formula using Static pressure, water flow and orifice size that can calculate an average or approximate residual pressure that would be at the hydrant during these flow conditions.

Hope this makes it a little clearer,

Rob

5. Dec 27, 2009

### spacester

OK, so is my previous post on target, except for the issue of static pressure versus residual pressure?

Where did you get the term 'residual pressure'?

It is not familiar to the engineers here so far. If you made it up, then meet us in the middle and adopt our terminology, OK?

Why do you keep talking about orifices? A 3-1/2" hose is about as far away from an orifice size calculation as you can get in flow calcs. Orifices are tiny little holes.

I can get you to your answer if you work with me here.

6. Dec 27, 2009

### sprinkyrob

Hi Spacester

Sorry about the mix up in terminology.

We always just refer to any opening in a pipe, fitting or sprinkler head as its orifice size. In this case however, I was referring to the internal diameter (3.5") of the pipe used to fill the pumper truck.

The term residual pressure means the force exerted on the inside walls of a pipe when water (or any fluid) is flowing through it, as opposed to Static pressure which measures the pressure when the water is not flowing or "static".

Normally we find the residual pressure by attaching a pressure gauge down stream while we flow water through a hydrant and measure GPM with a pitot tube. Unfortunately the weather for the next 4 months will not allow for this type of flow test, which is why I am seeking an alternative way to find the pressure at this hydrant under flow conditions.

Thanks

7. Dec 27, 2009

### FredGarvin

Just to clarify, what you are talking about is still static pressure. Moving fluids exert a static pressure. When there is no flow, the static pressure will increase to equal the total pressure in the system. The situation is usually referred to as "dead-headed."

That makes it very tough to give you a once off equation to plug and chug. You need to have a very controlled set up which will allow you to have the proper knowledge of the frictional drop per unit length of hose your gauge is away from the pump source. It is definitely not impossible. The tough part is going to be the friction of the hose. Since you are looking for the pressure drop, you could take a first pass estimate using

$$\Delta P = f \frac{L}{d} \frac{\rho v^2}{2}$$

http://www.engineersedge.com/fluid_flow/pressure_drop/pressure_drop.htm

http://www.pressure-drop.com/Online-Calculator/index.html

Last edited: Dec 27, 2009
8. Dec 27, 2009

### spacester

Aha! That is as I thought. But I am not finding the perfect link to back me up here.

(FredGarvin or others: feel free to check me on this.)

Here's the key to this problem:

The Total Pressure at a point in a flow circuit (hose) is made up of two components: The Velocity pressure and the Static pressure. As the fluid flows, some of the total energy shows up as velocity pressure, that which doesn't remains as static pressure.

http://water.me.vccs.edu/courses/CIV240/lesson4.htm

{A decent link but it seems to be in error IMO when it says:
"Static pressure = pressure that is measured when water is at rest."
Static Pressure is commonly measured in flowing systems. See the pitot tube.

IMO this sentence should be corrected to "Total pressure = pressure that is measured when water is at rest. Under no-flow conditions, Static Pressure = Total Pressure"}

So it seems to me that what you call residual pressure is exactly the same thing I call static pressure. BTW, the term 'residual' actually makes sense to me now.

So you know the no-flow pressure at the hydrant (72 psi), this is good. I will assume that the total pressure remains the same under flow and no-flow conditions. This assumption seems reasonable because you have a 10-inch main serving the hydrant, so the system should be able to maintain the system (total) pressure entering the hydrant. We are basically ignoring the friction from going thru the hydrant itself.

This means all you have to do is find the velocity pressure under flow conditions and subtract that from 72 psi and you have your answer. You have the flow rate and the cross-sectional area so you have the velocity and Velocity pressure = 1/2 * density * velocity^2

Just remember that your calculated value is not as dependable as an actual measured value.

9. Dec 27, 2009

### FredGarvin

While they are technically correct in saying that static pressure is the measurement of the pressure of a fluid at rest, it leads some to think that it means there must be no flow in the entire system if one wants to measure static pressure, which is not true.

10. Dec 27, 2009

### tiny-tim

Hi spacester!

Isn't static pressure just the pressure measured across a surface which moves with the flow?

11. Dec 27, 2009

### spacester

Um, no. BTW I remember you from Laugh-In :-)

The static pressure is the pressure as measured perpendicular to the flow, as in a probe thru the wall of a pipe.

Total pressure is the pressure seen by a probe pointed directly upstream. Since the static pressure is the same in all directions, you subtract the static from the total to get the velocity pressure.

Certain probes are built so that you can measure this difference directly (manometer method). It is a very accurate way to find the velocity and thus flow rate, with a major IF: the flow must be "fully developed" and laminar, which just means that the velocity is uniform across all cross-sections of pipe at that location. If this is not the case, one must take readings at several points in the cross-section, and use math to find the average velocity. Done correctly, this is still quite accurate.

12. Dec 27, 2009

### FredGarvin

The flow does not need to be laminar. As a matter of fact, the turbulent velocity profile is much more even across the diameter. However, for the sake of accuracy, a traverse is always the best method.

13. Jan 12, 2010

### Artman

First off I am familiar with the term "residual pressure", it is not a made up term.

You also need to subtract the difference in height of the truck's tank fill level or pipe discharge height above the point where the pressure was measured because this is an open system. If the truck was ten feet tall to the pipe discharge, you would have to subtract approximately 4.3 psi from your static pressure of 72 as well as your hose losses.

14. Jan 12, 2010

### spacester

@ FredGarvin - re. turbulent flow: good correction, that makes sense. Good to know, thanks.

@ Artman: Thanks for that info. That is the same thing I was taught to call 'Head', which I always thought was a stupid term. Good to know, thanks.

15. Jan 13, 2010

### Artman

Yes "head". It's an easy item to forget, but can make a big difference because it is a fixed quantity and cannot be corrected by a change of pipe size, like friction losses.

I find it is good to make a sketch of the entire system from inlet to outlet, showing net changes in elevation, length of pipe runs, fittings, and required minimum residual pressure to operate whatever equipment you are designing (sprinkler system, kitchen equipment, whatever).

16. Jan 16, 2010

### CarlAK

I believe you can use orifice flow formula to approximate the static pressure necessary to achieve a given flow through an opening. A 3.5" diameter orifice still behaves similar to a smaller orifice, it may have a little different Cd (coefficient of discharge) than other sizes.

As I recall velocity at an orifice is V=Cd (2 gh)^0.5
For Q=825 gpm, D=3.5", Cd=0.9, V=27.5 ft/sec, then p is about 6 psi.
there are a few on-line calculators, this seems to be in the ballpark. This is a pressure difference across the orifice. To have adequate pressure to overcome hose losses you'd want more like 20 psi minimum at the outlet from the hydrant.

However this is all theoretical, it has nothing to do with what the system can actually deliver. It's possible the hydrant can flow anywhere from 50 to 5000 gpm at 20 psi residual, depending on the system capability.

17. Nov 19, 2010

### firavia

Can we say that 'residual pressure' or the pressure required for a certain fixture is the pressure loss that will occur in this fixture before all the fluid is discharged to the atmospher??