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Net positive suction head, or NPSH is the amount of fluid energy at the pump suction in reference to absolute zero energy.
With centrifugal pumps, the Net Positive Suction Head required (NPSHr) is the amount of fluid energy at the pump suction at which the pressure at the eye of the impeller drops below the vapor pressure of the liquid and vapor bubbles begin to form.
Pump manufacturers test their pumps to determine the NPSHr over the range of operating flow rates for the pump.
The amount of fluid energy the piping system provides to the pump suction is called the Net Positive Suction Head available (NPSHa). NPSHa is a function of the following characteristics of the piping system on the suction side of the pump:
To prevent cavitation from occurring, the piping system must provide the fluid with enough energy to keep the pressure at the eye from dropping below the fluid’s vapor pressure.
Here is the Net Positive Suction Head Available (NPSHa) equation:
For a tank open to the atmosphere, Ptank in the above equation would be 0 psig. Consider the following system in which water must be lifted from the tank to the pump suction, called a suction lift configuration:
The NPSHa calculation would be as follows:
As long as the pump’s NPSHr is less than 26.6 feet, the pump should not cavitate.
However, there are some suction piping configurations that may cause the pump to cavitate even though NPSHa is greater than NPSHr.
The pump suction pressure in the above system would be below atmospheric pressure, or at a vacuum, because the pump must pull the liquid up from the tank, so the change in elevation reduces the fluid pressure.
In addition, the head loss between the tank and pump suction also reduces the fluid’s static pressure, causing the pressure at the pump suction to be at a vacuum. This is an application of the Bernoulli principle in which pressure head is converted into elevation head, resulting in a reduction in the fluid pressure.
The actual pressure at the pump suction can be calculated by re-arranging the Bernoulli equation below to solve for P2:
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