The amount of head developed in the impeller is approximately :

where V is the velocity at the tip of the impeller (ft/s) and g is the acceleration of gravity (32.2 ft/s2).

The velocity at the impeller tip can also be expressed as :

where rpm is the pump speed (in revolutions per minute) and D is the impeller diameter (in inches).

Substituting second Equation for V in first Equation results in the following expression:

this Equation shows that the head developed by a centrifugal pump is only a function of rpm and impeller diameter, and is not a function of specific gravity of the liquid being pumped. A pump moving a liquid up a static distance of 100 ft always has a required head of 100 ft (ignoring friction for the moment), regardless of the specific gravity of the liquid. If the liquid were water (SG = 1.0), a pressure gauge located at the pump discharge would show a pressure, using Equation 1.1, of 100 × 1.0/2.31 = 43.3 psig. If the liquid being pumped up the 100-ft height were oil, with a specific gravity of 0.8, the gauge would read 100 × 0.8/2.31 = 34.6 psig. The point of this is that the pump discharge pressure expressed in psi (or equivalent metric units such as kPa) varies with the specific gravity of the liquid, while the head expressed in feet (or meters) of liquid remains constant for liquids of different density. This is why one should always convert pressure terms into units of feet (or meters) of head when dealing with centrifugal pumps. A pump’s head–capacity curve does not require adjustment when the specific gravity of the liquid changes. On the other hand, as will be demonstrated shortly, the horsepower curve does vary with varying specific gravity.

To determine the required size of a centrifugal pump for a particular application, all the components of the system head for the system in which the pump is to operate must be added up to determine the pump total head (TH). There are four separate components of total head :