The relationship between the outlet pressure and head of a centrifugal pump
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Dear sea friends: Our facility has a domestic water pump with an outlet pressure of 8 kilograms, an inlet pressure of 2 kilograms, and a head of 60 meters. Since the head represents the energy difference between the pump’s inlet and outlet, why isn’t this head equal to the pressures at the pump’s inlet and outlet? Thank you. Please explain why, the relationship between head and pressure.
The data you provided are correct; the head multiplied by density gives a pressure value. The pressure corresponding to 60 units is approximately 6 kg/cm2, and adding the 2 kg/cm2 at the pump inlet results in a pressure of 8 kg/cm2 at the pump outlet.
The inlet pressure also needs to be reduced; that is the work done by the pump
The theoretical head is 80m, but due to pipeline losses, it is not possible to reach 80. The outlet pressure represents the theoretical head; the actual head is obtained by subtracting the pipe losses.
It refers to the pump outlet pressure. In fact, due to conservative design, it is more common for the actual pressure at the pump outlet to be higher than the pressure specified in the design.
Pumping: Head = Outlet pressure - Inlet pressure
Head = (Outlet pressure – Inlet pressure) * 102 / Specific gravity of the medium. Head in meters, pressure in MPaG; specific gravity is the density divided by 1000. 102 represents 1000/9.8