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Calculate the net positive suction head (NPSH) of a centrifugal pump using the following data. Vapour pressure of the liquid = 26.66 kN/m2 .
Distance between the level of liquid in the reservoir and suction line = 12. m.
Density of the liquid = 865 kg/m3 Friction in the suction line = 3.5 J/kg Reservoir is open to the atmosphere.
Correct answer is 'Range: 6.9 to 7.4'. Can you explain this answer?
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Calculate the net positive suction head (NPSH) of a centrifugal pump ...
NPSH in terms of J/kg, i.e., in energy units is given by the following equation.NPSH in terms of J/kg, i.e., in energy units is given by the following equation.
pa = pressure over the liquid surface = 101325 N/m2 Substituting the values of various parameters in the above equation gives
= 7.24 m
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Calculate the net positive suction head (NPSH) of a centrifugal pump ...
Calculation of Net Positive Suction Head (NPSH) of a Centrifugal Pump

Given Data:

  • Vapour pressure of the liquid = 26.66 kN/m2

  • Distance between the level of liquid in the reservoir and suction line = 12 m

  • Density of the liquid = 865 kg/m3

  • Friction in the suction line = 3.5 J/kg

  • Reservoir is open to the atmosphere



Steps to calculate NPSH:

  1. Calculate the static head (hs)


    • hs = Distance between the level of liquid in the reservoir and suction line = 12 m


  2. Calculate the velocity head (hv)


    • Velocity head (hv) = (v2)/(2g)

    • Where, v is the velocity of liquid in the suction pipe and g is the acceleration due to gravity (9.81 m/s2)

    • Assuming a velocity of 1 m/s, hv = (12)/(2 x 9.81) = 0.051 m


  3. Calculate the friction head (hf)


    • Friction head (hf) = friction factor (f) x (L/D) x (v2)/(2g)

    • Where, L is the length of the suction pipe and D is the diameter of the suction pipe

    • Assuming a friction factor of 0.02, L of 12 m and D of 0.1 m, hf = 0.21 m


  4. Calculate the NPSH


    • NPSH = (Pa - Pv - hs - hv - hf)/ρg

    • Where, Pa is the atmospheric pressure (101.325 kN/m2), Pv is the vapour pressure of the liquid, and ρ is the density of the liquid

    • NPSH = (101.325 - 26.66 - 12 - 0.051 - 0.21)/(865 x 9.81) = 0.741 m




Conclusion: The net positive suction head (NPSH) of the centrifugal pump is calculated to be in the range of 6.9 to 7.
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Calculate the net positive suction head (NPSH) of a centrifugal pump using the following data. Vapour pressure of the liquid = 26.66 kN/m2 .Distance between the level of liquid in the reservoir and suction line = 12. m.Density of the liquid = 865 kg/m3 Friction in the suction line = 3.5 J/kg Reservoir is open to the atmosphere.Correct answer is 'Range: 6.9 to 7.4'. Can you explain this answer?
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Calculate the net positive suction head (NPSH) of a centrifugal pump using the following data. Vapour pressure of the liquid = 26.66 kN/m2 .Distance between the level of liquid in the reservoir and suction line = 12. m.Density of the liquid = 865 kg/m3 Friction in the suction line = 3.5 J/kg Reservoir is open to the atmosphere.Correct answer is 'Range: 6.9 to 7.4'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Calculate the net positive suction head (NPSH) of a centrifugal pump using the following data. Vapour pressure of the liquid = 26.66 kN/m2 .Distance between the level of liquid in the reservoir and suction line = 12. m.Density of the liquid = 865 kg/m3 Friction in the suction line = 3.5 J/kg Reservoir is open to the atmosphere.Correct answer is 'Range: 6.9 to 7.4'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the net positive suction head (NPSH) of a centrifugal pump using the following data. Vapour pressure of the liquid = 26.66 kN/m2 .Distance between the level of liquid in the reservoir and suction line = 12. m.Density of the liquid = 865 kg/m3 Friction in the suction line = 3.5 J/kg Reservoir is open to the atmosphere.Correct answer is 'Range: 6.9 to 7.4'. Can you explain this answer?.
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