Question

2. A pump is used to pump water at 60^oF from a reservoir whose surface is 4.0 ft above the centerline of the pump inlet.  The piping system from the reservoir to the pump consists of 10.5 ft of cast iron pipe with a diameter of 4.0 in and an average roughness (ε) of 0.02 inches.  There are several minor losses:  a sharp edged inlet (K – 0.5), three flanged smooth 90^o regular elbows (K – 0.3 each), and a fully open flanged globe valve (K – 6.0).  

Estimate the maximum volume flowrate (Q) that can be pumped without cavitation if the NPSH_Required is 6.5 ft  (assume P_v = 0.46 psia; P_atm = 14.7 psia).  

Assume fully turbulent flow!

Answer (Q = 596.95 gpm)

Answer 1

Ans) We know,

z_max = (P_atm - P_v) / - H_f - H_m - NPSHr .......................(1)

where, z_max = Maximum elevation of pump to avoid cavitation = 4 ft

P_atm = Atmospheric pressure = 14.7 psi or 2116.8 psf

P_v = Vapor pressure of water = 0.46 psi or 66.24 psf

H_f = Frictional head loss

H_m = Minor head loss

Friction loss, H_f = f L V² / 2 g D

For fully developed turbulent flow , f = 1/ [1.14 + 2 Log (D/e)]²

where, e is roughness of pipe = 0.02 in

=> f = 1 / [1.14 + 2 Log(4/0.02)]²

=> f = 0.030

=> H_f = 0.030(10.5) V² / (2 x 32.2 x 0.333)

=> H_f = 0.014 V²

Minor loss, H_m = K V² / 2 g

=>[0.50 + 3(0.30) + 6] V² / 2 g

=> H_m = 7.4 V² / (2 x 32.2)

   => H_m = 0.114 V²

Putting values in equation 1,

=> 4 = (2116.8 - 66.24)/62.4 - 0.014 V² - 0.114 V² - 6.5

=> 4 = 32.86 - 0.12958 V² - 6.5

=> 0.128 V² = 22.36

=> V = 13.21 ft/s

We know, flow rate (Q) = A x V

Area of pipe,A = (/4) (0.333)² = 0.08705 ft²

=> Q = 0.08705 x 13.21 = 1.15 cfs 596 gpm