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)
Where:
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)
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 = U12/2g [(K1 + K2) tan2 β1 + K2] (3)
Where:
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.
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.