Head of centrifugal pump
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Is the head of a centrifugal pump approximately equal to the pressure at the pump outlet?
Pump outlet pressure = Pump inlet pressure + Head
For example, consider a football rolling towards you; it could originally roll 10 meters, but after you give it an extra push, it rolls 100 meters. Those 90 meters represent the additional kinetic energy provided by your push – which is the sum of the original kinetic energy and the kinetic energy you added.
If pipe resistance losses are not taken into account, it should be approximately equal to the pressure difference.
Not approximate! Head is the outlet pressure of the pump minus the inlet pressure, divided by ρg. It is also related to the density of the medium.
Not approximate! Head is the outlet pressure of the pump minus the inlet pressure, divided by ρg. It is also related to the density of the medium.
Yes, I think upstairs it was said that... the head of a centrifugal pump is independent of the liquid density, but the outlet pressure is related to the liquid density
Ignoring pipe resistance and other factors, the theoretical head is simply the height of transfer, and it has nothing to do with density; however, different densities result in different motor powers
The head of centrifugal pumps is usually tested by manufacturers using water as the medium, with atmospheric pressure at the inlet, under rated flow conditions； Therefore, when in use on-site, the outlet pressure must be calculated based on the specific gravity of the medium, the inlet pressure, and the relevant process conditions. Thank you!