Equivalent Length of Fittings

Water is pumped from a reservoir to a height of 1000 m from the reservoir level, through a pipe of 15 cm I.D. at an average velocity of 4 m/s. If the pipeline along with the fittings is equivalent to 2000 m long and the overall efficiency is 70%, what is the energy required for pumping? Friction factor f = 0.046 Re-0.2.

Data:

Diameter of pipe(D) = 15 cm

Average Velocity(n) = 4 m/s

Equivalent Length of pipe with fittings = 2000 m

efficiency of pump(h) = 0.7

Formula:

  • Bernoulli equation:
  • Frictional losses:
  • Friction factor Relation:

    f = 0.046 Re-0.2

Calculations:

Re = Dnr/m

= 0.15 x 4 x 1000/0.001 = 600000

f = 0.046 x (600000)-0.2 = 0.003215

hf = 2 x 0.003215 x 2000 x 42/0.15 = 1371.73 m2/sec2

Frictional losses per unit weight (h)= hf/g = 1371.73/9.812 = 139.8 m

Power required for pumping = mass flow rate x g x h / efficiency of pump

Mass flow rate = volumetric flow rate x density

= velocity x cross sectional area x density

= 4 x (pD2/4) x r

= p x 0.152 x 1000 = 70.686 kg/sec

power required = 70.686 x 9.812 x 139.8/0.7 = 96961 watt = 138515.7 watt = 138.5 KW