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 (v) = 4 m/s

Equivalent Length of pipe with fittings = 2000 m

Efficiency of pump (η) = 0.7

Formula:

• Bernoulli equation:
• $$\displaystyle \frac{p_1}{\rho_1g} + \frac{v_1^2}{2g}+z_1 = \frac{p_2}{\rho_2g} + \frac{v_2^2}{2g} + z_2 + h + w - q$$
• Frictional losses:
• $$\displaystyle h_f = \frac{2fLv^2}{D}$$
• Friction factor Relation:
• $$\displaystyle f = 0.046 \text{Re}^{-0.2}$$

Calculations:

Re = Dvρ/μ

= 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 (πD2/4) x ρ

= π 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

[Index]