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:

E_{s} = U_{1} x S (1)

Where:

E_{s}
= suction energy

U_{1}
= 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)

Where:

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 = U_{1}^{2}/2g [(K_{1} + K_{2}) tan^{2} β_{1} + K_{2}] (3)

Where:

g
= gravitational constant, 32.2 ft./sec.^{2}

K_{1} and K_{2}
= experimental constants, established by test

β_{1}
= 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 = K_{3}U_{1}^{2} (4)

Solving for U_{1}:

U_{1} = [NPSHR/K_{3}]^{0.5} (5)

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

E_{s} = U_{1} x S = [NPSHR/K_{3} ]^{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:

E_{s} = [ 1/K_{3}]^{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 U_{1} 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:

NPSHR_{3%} = K_{4}U_{1}^{1.5} (8)

Solving for U_{1}:

U_{1} = [NPSHR_{3%} / K_{4}]^{0.67} (9)

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

E_{s} = U_{1} x S = [ NPSHR/K_{4}]^{0.67}[N√Q / (NPSHR)^{0.75}] (10)

Canceling the NPSHR values results in:

E_{s} = [1/K_{4}]^{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, NPSHR^{0.08} = 1.2. If NPSHR = 100, NPSHR^{0.08} = 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.

P&S

References

1. Henshaw, Terry L., “Predicting NPSH for Centrifugal Pumps,” www.pump-zone.com, Dec. 21, 2000.

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