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Differential Pressures

Differential Pressures

Differential Pressures

(OP)
This has to do with a private Fire Hydrant onsite loop feed by a Fire Pump delivering 175 psi. When a Fire Department pump (1,500) gpm at 150psi is drawing from this system how long would it take with a 10 inch underground to draw down to 135 psi to get the back flow to open up on the site and equalize pressure. In other words I have a system where the pressure is 175 on the inside loop and 135 psi on the supply side of the back flow and Fire Pump feeding the inside loop at a consistant presssure of 175 psi. How would one calculate this to determine what the draw down time would be with this type of pumper working against the on site fire pump pressure. The reason for the question is the pump does not make Fire Flow tied into the hydrant system and therfore would require another pumper to draw down enough to get the pressure to qualize with a 4 inch connection to the hydrant? This would allow the back flow to open to allow more water to flow onsite from the second connection. Otherwise the second connection to the city supply is being held shut by the fire pump pressure and no additonal forward flow can be obtained.      

RE: Differential Pressures

I'm not sure I understand your question, but I'll answer anyway:  Either almost instantly, or never.  That is, if the demand on the fire pump is enough to run the fire pump out on its curve to less than 135 psi then the backflow will open just as soon as the demand materializes.  If the demand on the fire pump never gets that big, then the backflow stays closed (and then what's the problem?).   

RE: Differential Pressures

The flowrate of your private firewater pump when delivering at 135 psig must be at least 1500 gpm, otherwise (if that flowrate is less) than, if you can meet fireflow initially, as you surmize, you will not be able to sustain it.  The time you can sustain the 1500 gpm flowrate at the fire company's pumper unit can initially be approximated by taking the volume contained in the 10" diameter underground pipe and dividing that by the difference between (1500 gpm - your pump's flowrate at 135 psig).

If your pipe's volume is 10,000 gallong and your pump's flow capacity is 1250 gpm when delivering at 135 psig + the pressure drop in the 10" pipe at that flowrate of 1250 gpm, then time =
10,000 gallons / ( 1500 gpm - 1250 gpm) = 40 minutes

It will not take long (probably just a second or two) for the system pressures to get to the point where you can use that approximation with pretty good accuracy.

 

From "BigInch's Extremely simple theory of everything."

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