A pump is put into service at the coast where the barometric pressure is 760 mm Hg. The conditions of service are : Flow rate 0,08 m2/s, suction lift 3,5 metres, suction pipe friction loss 0,9 metres, water temperature 65°C, water velocity 4 m/s. Under these conditions of service, the pump requires an NPSH of 2,1 metres. Assuming the density of water as 980,6 kg/m3, establish whether it will operate satisfactorily.

NPSH isa measure of how likely the fluid at the suction end of the pump is to experience cavitation. We always need NPSH to be above a given treshold required by the pump.

The definition of NPSH is:

[tex]NPSH= \frac{P_{atm}-P_{v}}{\rho \, g}-h-h_f[/tex]

Where:

[tex]P_{atm}[/tex] and [tex]P_{v}[/tex] are atmospheric and vapour pressure correspondingly.
[tex]h[/tex] is suction lift and [tex]h_f[/tex] is friction loss in meters.
[tex]\rho[/tex] is the fluid's density
[tex]g[/tex] is gravity, at sea level taken as [tex]9.81 \, m/s^2[/tex]

Extra Data: Water's vapour pressure at 65°C

[tex]P_v =25.022 \, kPa[/tex]

We can now calculate the NPSH

[tex]NPSH= \frac{101325-25022}{980.6\times9.81}\, m-3.5\, m- 0.9 \, m\\NPSH= 7.932\, m-3.5\, m- 0.9 \, m\\NPSH=3.532 \, m[/tex]

We can see that our NPSH is greater than the required NPSH:[tex]NPSH>NPSH_{req} = 2.1 \, m[/tex]

Thus our pump will operate without cavitation.