Question

1. A plant is located at an altitude where the atmospheric pressure head is 31 ft of water and the vapor pressure is 9.34 psi. The pump is placed 10 ft below the water tank with a total suction pipe length of 25 ft, pipe diameter 3 in., and friction coefficient 0.02. Assuming the given atmospheric pressure in the tank (where z1 = 10 ft and negligible velocity) and if the flowrate provided is 180 gpm, determine the available NPSH (ft). Neglect minor losses. ans 17.4 ft

Answer 1

Pressure equation: P = ρgh where P = pressure (N/m²), ρ = mass density of fluid (kg/m³), g = acceleration due to gravity = 9.8066 m/s², h = height of fluid column (m).

P=998.2 Kg/m3,g=9.81 M/s2 ,Ht=10Ft=3.048 Mtr.

Pressure=29847.05 N/m2.

Q=AV.

Velocity=180 Gpm*4*3(inch)83(inch)/3.14                                                            

180GPM=6.3 *10^-5 m3/s.,3inch=0.0762 mtr.

Velocity=0.01382 m/s.

Suction Head

Based on the Energy Equation - the suction head in the fluid close to the impeller^*) can be expressed as the sum of the static and velocity head

h_s = p_s / γ_liquid + v_s² / 2 g  

h_s = suction head close to the impeller (m, in)

p_s = static pressure in the fluid close to the impeller (Pa (N/m²)= 29847.05

γ_liquid = specific weight of the liquid (N/m³, lb/ft³)=9.81*1000 n/m3.

v_s = velocity of fluid (m/s)----0.01382 m/s.

g = acceleration of gravity (9.81 m/s²)

hs=((29847.05/(9.81*10^3))+(0.0762)^2/(2*9.81)).

=3.04 Mtr.

Liquids Vapor Head

h_v = p_v / γ_vapor    

h_v = vapor head (m)

p_v = vapor pressure (m)

γ_vapor = specific weight of the vapor (N/m³)

The vapor pressure in a fluid depends on the temperature. Water, our most common fluid, starts boiling at 20 oC if the absolute pressure is 2.3 kN/m2

=2.3/9.81

=0.23 Mtr.

Net Positive Suction Head - NPSH

NPSH = h_s - h_v  =3.04-0.23=2.81 Mtr=9 Ft.