Gasoline (p 680 kg/m3, H 0.3 mPa.s, Po 7.34 x 104 Pa) is to be transferred from

Question

Gasoline (p 680 kg/m3, H 0.3 mPa.s, Po 7.34 x 104 Pa) is to be transferred from an underground storage tank to a tanker at 0.0133 m3/s. The system is shown in the diagram below. The suction side piping is 4 inch Schedule 40 commercial steel pipe. Its total length is 15 ft. The discharge piping is 3 inch Schedule 40 commercial steel pipe and is 35 ft in length. There is one regular, flanged elbow in the suction side piping. In the discharge line, there are two regular, flanged elbows and 2 closed gate valve. Discharge pipe 40 ft Suction pipe Pump 10 ft A) Determine the pump head (in feet of fluid) required to obtain the specified flow B) Calculate the Net Positive Suction Head Available (NPSH-A), in feet of fluid. Using the centrifugal pump curve for a 3 x 4 Gould pump operating at 1750 RPM rate determine C) The appropriate pump/impeller size for this application D) The pump efficiency E) The Net Positive Suction Head Required (NPSH-R), in feet of fluid F) The motor size required, in horsepower (HP) G) Comment on the pump inlet conditions. Do you think they will allow for satisfactory pump performance? Or are you concerned that cavitation will occur inside the pump?

Explanation / Answer

Steel rougness = .1 mm

Reynold number, Re = density*velocity*diameter/viscosity

velocity = flowrate/C.S area

In suction,

velocity = .0133/(3.14/4*(4*2.54/100)^2)=1.64 m/s

=>Re=1.64*680*(4*2.54/100)/(.3*10^-3))=377681

Relative roughness of suction pipe = .1/(4*2.54*10)=.0009

By Moody's chart, friction f = .025

equivalent length of 90 degree 4 inch elbow = 10 ft

=>Suction side total length, l =15+10=25 ft = 7.5 m

Delta P in suction side because of friction = f*(.5*density*v^2*l)/pipe dia = .025*(.5*680*1.67^2*7.5)/(4*2.54/100)=1749 Pa

In discharge,

velocity = .0133/(3.14/4*(3*2.54/100)^2)=2.91 m/s

=>Re=2.91*680*(3*2.54/100)/(.3*10^-3))=502615

Relative roughness of suction pipe = .1/(3*2.54*10)=.0013

By Moody's chart, friction f = .023

equivalent length of 90 degree 3 inch elbow = 6.3 ft

equivalent length of 3 inch gate valve =30 ft

=>Discharge side total length, l =35+30+6.3 ft = 23.5 m

Delta P in discharge side because of friction = f*(.5*density*v^2*l)/pipe dia = .023*(.5*680*2.91^2*23.5)/(3*2.54/100)=20422 Pa

Total Pump head requred = (suction friction loss+Discharge friction loss pressure drop-Inlet pressure)/density/g+Height of elevation

Total Pump head requred=(1749+20422-73400)/680/10+(10+40)/3.3=7.61 m=25.11 ft

b) NPSH available = (Inlet Pressure-Friction loss Pressure drop in suction - Vapor Pressure)/density/g

Vapor pressure of gasoline not given,

Assuming Vapor pressure of gasoline = .1 bar = 10000 pa

=>NPSA available = (73400-1749-10000)/680/10*3.3=29.9 ft

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