What is pump head? and how is it different from the difference in elevation between the suction and delivery reservoir? Also why must the kinetic energy of the fluid leaving the pump must be least? I mean if it leaves with more velocity then it can go farther up to the delivery reservoir. The energy equation we write is

$$P/\rho(g) + v_2/2g + z_1 = P_2/\rho(g) + v_2(2)/2g + z_2 + Head$$

• Where does it say that "the kinetic energy of the fluid leaving the pump must be least?" – Chet Miller Sep 6 '16 at 12:14
• So that the difference in pressure is higher and the head developed is higher – Stealth Sep 6 '16 at 13:09
• Your expression for the Bernoulli equation is incorrect for a pump. The pump contributes (on its own) an increase in head, over and above the head change given by the Bernoulli equation as you have written it. – Chet Miller Sep 6 '16 at 15:15
• Yea...I did some more research on HEAD and found my mistake – Stealth Sep 7 '16 at 5:12

The head of a pump is a measure of how big of a pressure difference that pump can generate. I am not sure what the historical or practical reason for it is, but head is expressed as the height of a water column. The pressure $p$ required for such column with head $h$ can be calculated with,

$$p=\rho\,g\,h,$$

where $\rho$ is the density of water and $g$ the acceleration due to gravity.

HEAD is another expression of pressure. 1 pound per sq. inch pressure is equal to 2.31 feet of head. That is, for each 2.31 feet high a column of water is at 60 degrees F, it exerts a force of 1 pound per sq. inch on its base. It is derived as follows:

One ft^3 of water at 60 degrees F weighs 62.4 pounds. Since the base of the ft^3 consist of 144 in^2, the force or weight on each in^2 is 0.433 pounds. We can now write a simple equation to determine how high the column must be to exert one pound force on one sq. inch.

Equation: HEAD H = 1 / 0.433 = 2.31 ft.

The above is only true for water at 60 F. For other liquids at different temperatures mult. the above by the liquid's specific gravity.

• Moderator - in my opinion, this question has received the best answers that it is going to receive. Should it keep periodically popping up for review? – David White Jan 11 at 3:37

The pump head of a centrifugal pump is more subtle than just the pressure difference that the pump produces. Most centrifugal pumps are driven by AC electric motors. Those motors produce a constant rpm (e.g, 1800 rpm or 3600 rpm) due to the frequency of the supply current and the construction details of the motor. Several physical phenomena occur as a result of this:

1) Since the pump is usually directly coupled to the motor, the pump impeller turns at the same rpm as the motor.

2) Motor horsepower requirements are determined by the flow rate through the pump AND the density of the pumped fluid. If a motor/pump combination is designed to pump something like gasoline and water is pumped instead, the higher density of the water will require that the motor draws more current in order to keep the pump impeller spinning at a constant rpm. In addition, for a constant impeller speed, the outlet pressure of the pumped fluid is directly related to the fluid density, with higher density fluids producing a higher outlet pressure.

3) For a constant impeller speed, the velocity of the liquid exiting the impeller is constant. This means that if the pump outlet was pointed directly up, with no attached piping, the liquid flow coming out of the pump would rise to a height that depends on its velocity exiting the pump. This height is known as the pump head.

4) The pump head is a function of the impeller speed, and it will be the same for all pumped fluids (in principle), while the pump discharge pressure will be a function of impeller speed AND fluid density. Thus, for the same pump head, different fluids will produce different pump discharge pressures and motor power requirements.

It is convenient to specify pump characteristics that will not vary with the pumped fluid, and pump head is one of those characteristics.