# Best practice no. 25

Example 2 .- Open system with liquid reservoir below pump

Problem: for the system shown in the drawing, determine net positive suction head available (NPSHA) when A. Liquid is 68°F water and reservoir is at sea level. B. Liquid is 176°F water and reservoir is at sea level. Friction head for the 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 a = 33.9 FT

h vpa = 0.78 FT

h f = 2.0 FT

PUMP CL

h s = - 10.0 FT

ATMOSPHERIC PRESSURE

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 AT SEA LEVEL

LIQUID

NPSHA = - 10 – 2 + 33.96 – 0.78 = 21.18 ft

Solution B : Given data are: h f = 2 ft y h s = - 10 ft . From table 1 at sea level , h a = 33.96 ft , and from table 2 , at temperature de 176°F , h vpa = 15.87 ft. Then:

NPSHA = - 10 – 2 + 33.96 – 15.87 = 6.09 ft

Example 3 .- Unpressurized closed system Problem: For the system shown in the drawing, determine net positive suction head available (NPSHA) when liquid is 248 °F water at sea level. Friction head for the suction piping is h f = 2 ft . Also, from drawing, static suction piping is: h s = 10 ft . Solution : The general equation for this calculation is:

h f = 2.0 FT

VAPOR PRESSURE

LIQUID

NPSHA = h s – h f + h a - h vap

h s = 10.0 FT

Given data are: h f = 2 ft and h s = - 10 ft. Because this is a closed system (without inert gas) , h a = h vpa , from table 2 , at temperature of 248°F, h vpa = 66.53 ft = h a ,

NPSHA = 8.0 FT

CL

PUMP

NPSHA = 10 – 2 + 66.53 – 66.53 = 8 ft

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