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Is the outlet pressure of the gear pump related to the tank level?

2011-01-14View Original

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Assuming the tank liquid level is 5 meters, the operating pressure is at atmospheric level, the specific gravity of the liquid is around 1.0, and the TDH (total differential head, or head pressure) required for the gear pump is 80 meters (or 8.0 bar). Is the outlet pressure 8.5 barG or 8.0 barG? (The centrifugal pump is 8.5 barG). Assume the pressure drop in the inlet pipeline is ignored.
Reply #22011-01-14
Suddenly confused. I think, just like with centrifugal pumps, the outlet pressure should be around 8.5 barG.
Reply #32011-01-14
Relevantly, it can be calculated based on Bernoulli’s equation
Reply #42011-04-26
Reply to 1# chemwanghq: It’s roughly 8.0 bar, but the outlet pressure of the gear pump is determined by the characteristics of the piping; it’s better to have a slight margin, as well as more headroom in terms of motor power.
Reply #52011-04-27
The outlet pressure of a positive displacement pump is related to the characteristics of the pipeline at the outlet. As for what the original poster said about the effect of inlet pressure on outlet pressure being similar to two pumps connected in series. The relationship between the values requires specific analysis
Reply #62011-04-27
It can be calculated using Bernoulli’s equation
Reply #72011-04-27
Thank you all! Let the numbers speak; please have those with expertise in manufacturing provide an answer.
Reply #82011-04-27
The outlet pressure of a positive displacement pump is determined by the load; theoretically, the outlet pressure can be infinitely high as long as the motor and the equipment allow it
Reply #92011-04-27
It can be considered as 0.8Mpa
Reply #102011-04-27
8.5 Bar, calculated using Bernoulli’s equation; there is no difference between centrifugal pumps and gear pumps – they are the same.
Reply #112011-04-28
Gear pumps have a certain leakage rate, and they exert force on the fluid in a different manner; unlike centrifugal pumps, whose blades are used to exert force on the fluid. Personally, I don’t think it’s appropriate to simply treat it as a flow channel for calculation. Simple addition is incorrect

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