Best practice no. 25

Where:

P g = gauge reading, psi

P a = atmospheric pressure, psi

P vpa = absolute vapor pressure of liquid pumping temperature, psi.

V = velocity in suction line, ft/sec.

g = acceleration resulting from gravity (constant at 32,17 ft/sec/sec)

Y = difference in elevation between pump centerline and gauge, ft. (If gauge is above centerline. Y is a positive value, if below Y is negative).

Typical calculations for various pumping system arrangements are provided in the accompanying examples.

Example 1 .- Open system with liquid reservoir above pump

Problem: Fort eh system shown in the drawing, determine net positive suction available (NPSHA) when A. liquid is 68°F water and reservoir is at sea level. B. liquid is 68°F water and reservoir is 6000 ft above sea level. C. liquid is 60°F gasoline and reservoir is at sea level. Friction head for suction piping is: h f = 2 ft . Also, from drawing, static suction head is h s = 10 ft . Solution A : The general equation for these calculations is:

PRESSURE (ENERGY) AT SURFACE OF LIQUID

h vpa = 0.78 FT

h f = 2.0 FT

h a = 33.9 FT

NPSHA

NPSHA = h s – h f + h a - h vap

Given data are: h f = 2 ft and h s = 10 ft from table 1, at sea level, h a = 33,96 ft , and from table 2 , at temperature of 68°F, h vpa = 0.78 ft. Then,

ATMOSPHERIC PRESSURE

ATMOSPHERIC PRESSURE AT SEA LEVEL

LIQUID

NPSHA = 10 – 2 + 33.96 – 0.78 = 41.18 ft

Solution B : Given data are: h f = 2 ft and h s = 10 ft , from table 1 at 6000 ft above sea level, h a = 27.3 ft and from table 2 at temperature of 68°F, h vpa = 0.78 ft . Then:

h s = 10.0 FT

CL PUMP

NPSHA = 10 – 2 + 27.3 – 0.78 = 34.52 ft

Solution C : Given data are: h f = 2 ft and h s = 10 ft , from table 1 at sea level, h a = 33.96 ft. Absolute vapor pressure of gasoline at 60°F cab be obtained from standard vapor pressure tables and is: h vpa =7.7 ft. Accordingly:

NPSHA = 10 – 2 + 33.96 – 7.7 = 34.26 ft