Frictional Losses in Pipeline

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2.16 m3/h water at 320 K is pumped through a 40 mm I.D. pipe through a length of 150 m in a horizontal direction and up through a vertical height of 12 m. In the pipe there are fittings equivalent to 260 pipe diameters. What power must be supplied to the pump if it is 60% efficient? Take the value of fanning friction factor as 0.008. Water viscosity is 0.65 cp, and density = 1 gm/cc.

Data:

Flow rate (Q) = 2.16 m3/h = 2.16/3600 m3/sec = 0.0006 m3/sec

Dia of pipe (D) = 40 mm = 0.04 m

Length of pipe in horizontal direction = 150 m

Length of pipe in vertical direction(Δz) = 12 m

Equivalent length of fittings = 260 pipe diameters

Friction factor (f) = 0.008

Efficiency of pump (η) = 0.6

Viscosity of fluid (μ) = 0.65 cp = 0.00065 kg/(m.sec)

Density of fluid (ρ) = 1 gm/cc = 1000 kg/m3

Formulae:
  1. Bernoulli's equation
    2. \(\displaystyle \frac{p_1}{\rho_1g} + \frac{v_1^2}{2g}+z_1 = \frac{p_2}{\rho_2g} + \frac{v_2^2}{2g} + z_2 + h + w - q \)
  2. Frictional losses per unit mass of flowing fluid
    3. \(\displaystyle h_f = \frac{2fLv^2}{D} \)
  3. Power required for pumping = (mass flow rate x g x head developed by pump)/η

    = (volumetric flow rate x pressure developed by pump)/η

Calculations:

Length of pipe with fittings = 150 + 12 + 260 x 0.04 m = 172.4 m

Velocity = volumetric flow rate/flow cross sectional area

= 0.0006/((π/4) x 0.042) = 0.477 m/sec

hf = 2 x 0.008 x 172.4 x 0.4772 / 0.04 = 15.69 m2/sec2

Frictional losses per unit weight of fluid (h) = hf / g = 15.69/9.812 = 1.6 m

Pump head (w) = Δz + h = 12 + 1.6 = 13.6 m

Pressure developed by pump = 13.6 x 1000 x 9.812 = 133443.2 N/m2

power required for pumping = 0.0006 x 133443.2 / 0.6 = 133.4 watt = 133.4/736 HP = 0.181 HP

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Last Modified on: 01-May-2024

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