Repost - PSI at Periphery of Centrifugal Pump Impeller

That is not an easy task.

For that high of a pressure you will be looking at a multi-stage centrifugal.

What is your required flowrate?

Does it really need to be a centrifugal? Or can it be something else such as a gear pump?

This isn't homework is it?

You wouldn't have access to a really sophisticated CFD program would you?

Lastly, why are you trying to design a pump from the ground up - why not have the experts (suppliers/manufacturers) help you?

This sounds like homework, of one kind or another.

"fully flooded but sealed from discharge" is not terminology normally associated with pump design.

"I'm looking for a formula that I could could plug in the variables and get the pressure".

Go Fish!

Yeah - I had a quick look at my 2nd year Uni fluids book and it sounds like a Euler Turbomachinery Equations problem.

Right

No single answer to this question. 0.7 sounds like pumping ethanol or similar. Bad bad bad for cavitation in centrifugal pumping.

Multi-stage for sure as the energetic gain between in and out on an single impeller would boil the liquid in between which is why you get cavitation in the first place.

submersible variable displacement multi-piston pump would be a better bet unless your just fishing for homework help. If that is the case, pick a different career engineering is not a good feild for slackers. Try politics....

Beg, borrow, buy, or steal a copy of a "Cameron Hydraulic Data" and read chapter 1 three or four times. There is not a single formula for the variables you list, but with enough patience I think you can set up a spreadsheet to give what you want.

Cameron's used to be published be Ingersoll Rand, I'm sure Ebay has used copies. One of my more valuable reference books.

greetings from shankar bangalore india

my reference book for pumps is one by carter and karassik, 'pump questions and answers'(ancient but very informative) from which i quote the following:

knowing the diameter of an impeller and its operating speed, how can one determine approximately the head that a pump can develop?

the head in feet developed by an impeller at the capacity at which maximum efficiency is obtained can be determined from the formula:

H =(Db/1840Ku)^2

where H= head, D = impeller diameter,inches, n = speed rpm, Ku=pump-design co-efficient

the value of Ku may be between .95 and 1.15 for normal impeller designs. Generally, using 1.03 is accurate enough in which case

H=(Dn/1900)^2

only power demand varies with liquids of different specific gravity - not head and discharge

shankar