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

h_{f} = 2 x 0.003215 x 2000 x 4^{2}/0.15 = 1371.73 m^{2}/sec^{2}

Frictional losses per unit weight (h)= h_{f}/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 (πD^{2}/4) x ρ

= π x 0.15^{2} 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**

Last Modified on: 01-May-2024

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