Question

6. Determine the setting of a centrifugal pump, given that flow rate = 2.4 m/min; diameter of intake pipe = 160 mm; entrance losses = one velocity head in suction pipe; required NPSH = 42 kPa; water temperature = 30°C

Answer 1

Ans) We know,

NPSHr = Pa/ $\gamma$ - Pv / $\gamma$ - z - Hs.........................(1)

where, NPSHr id required net positive suction head

Pa = Atmospheric pressure = 101.325 kPa

Pv = Vapor pressure of water at 30 degree C = 4.25 kPa

z = Elevation of pump

Hs = Head loss in suction pipe

$\gamma$ = Unit weight of water = 9.8 kN/m^3

Given, NPSHr = 42 kPa

=> NPSHr head = NPSHr/ $\gamma$ = 42 /9.81 = 4.28 m

Flow rate , Q = 2.4 m^3/min or 0.04 m^3/sec

  We know, Velocity = Q / A

Pipe area , A = ($\pi$/4)(0.16)^2= 0.02 m^2

=> Velocity,V = 0.04 / 0.02 = 2 m/s

According to question,head loss in suction pipe,Hs = V^2 / 2 g

=> Hs = 2^2 / (2 x 9.81)

=> Hs = 0.204 m

Putting values in equation 1,

' => 4.28 = (101.325/9.81) - (4.25 / 9.81) - z - 0.204

=> 4.28 = 10.328 - 0.433 - z - 0.204

=> 4.28 = 9.691 - z

=> z = 5.41 m

Hence, pump can be setted out at elevation of 5.41 m to avoid cavitation