The concept of “suction energy” was developed to enhance the evaluation of the NPSH characteristics of centrifugal pumps, but I fear that it misses the mark. Suction energy is defined as the product of U1 (the peripheral velocity of the impeller eye) and S (suction specific speed). At first blush, that seems like a reasonable concept, but a mathematical analysis seems to indicate a flaw in the logic. The equation for suction energy is:


Es = U1 x S                                                               (1)


Es = suction energy

U1 = peripheral velocity of impeller eye, ft./sec.

S = suction specific speed (rpm-gpm-ft.)


The equation for S, suction specific speed, is:

S =       N√Q / (NPSHR)0.75                                       (2)



N = rotative speed of impeller, rev./min.

Q = capacity of one eye at best-efficiency-point, US gal./min.

NPSHR = net positive suction head, normally at 3 percent head drop (of first stage), feet

Using the “Classic” Relation Between NPSHR And U1

As shown in Reference 1, at no-prerotation capacity (which is at, or near, the best efficiency point—BEP), NPSHR can be expressed by the following equation:

NPSHR = U12/2g  [(K1 + K2) tan2 β1 + K2]                   (3)


g = gravitational constant, 32.2 ft./sec.2

K1 and K2 = experimental constants, established by test

β1 = inlet vane angle, normally measured at the shroud, degrees

Everything on the right side of Equation 3 can be considered constant, except for U1. We can, therefore, say that:

NPSHR = K3U12                                                        (4)

Solving for U1:

U1 = [NPSHR/K3]0.5                                                  (5)


If we plug Equations 2 and 5 into Equation 1, we get:

Es = U1 x S = [NPSHR/K3 ]0.5[ N√Q / (NPSHR)0.75]      (6)

But wait a minute. We have (NPSHR)0.5 in the numerator, multiplied by (NPSHR)0.75 in the denominator. Don’t those tend to cancel each other? Yes, they do. The resulting equation is:

Es = [ 1/K3]0.5[ N√Q / (NPSHR)0.25]                          (7)

The effect of NPSH on the characteristic has been significantly reduced, with the exponent dropping from 0.75 to 0.25, a two-thirds reduction. We’ve almost eliminated NPSH from the equation, and that should not happen. NPSH is what it is all about.

If we envision suction specific speed as a stool supported by three legs, with each leg being (NPSHR)0.25, then multiplying by U1 cuts off two of the legs.

I prefer to sit on a stool that has three legs.

Actual Speed Effect Makes the Situation Worse

Equation 4 states the classic relation between NPSHR and pump speed: the NPSHR increases as the square of the speed. That is probably correct for incipient cavitation, but numerous tests for the 3-percent-head-drop NPSHR have shown that the exponent is actually closer to 1.5.

Equation 4 can, therefore, be rewritten, for the 3 percent head drop, as follows: 

NPSHR3% = K4U11.5                                                 (8)


Solving for U1:

U1 = [NPSHR3% / K4]0.67                                         (9)


If we plug Equations 2 and 9 into Equation 1, the result is:

Es = U1 x S = [ NPSHR/K4]0.67[N√Q / (NPSHR)0.75]    (10)


Canceling the NPSHR values results in:

Es = [1/K4]0.67[N√Q / (NPSHR)0.08]                         (11)

That 0.08 exponent makes the effect of the NPSHR almost disappear. For example, if NPSHR = 10, NPSHR0.08 = 1.2. If NPSHR = 100, NPSHR0.08 = 1.45—only a 20 percent increase for a 900 percent increase in the NPSHR. (That stool is now supported by a toothpick.) 

If the 0.75 exponent were being used, a 900 percent increase in NPSHR would increase the denominator by 462 percent, which is a more appropriate impact of NPSHR on the characteristic. 

The conclusion is that suction specific speed is more efficacious than suction energy for evaluating the NPSH characteristics of centrifugal pumps.



1. Henshaw, Terry L., “Predicting NPSH for Centrifugal Pumps,”, Dec. 21, 2000.

2. Henshaw, Terry, “Stepping NPSHR to Different Speeds,” Pumps & Systems, August 2009.