### Fluid Mechanics

# Frictional Losses in Pipelines

2.16 m^{3}/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 m^{3}/h = 2.16/3600 m^{3}/sec = 0.0006 m^{3}/sec

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

Length of pipe in horizontal direction = 150 m

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

Equivalent length of fittings = 260 pipe diameters

Friction factor(f) = 0.008

Efficiency of pump(h ) = 0.6

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

Density of fluid(r) = 1 gm/cc = 1000 kg/m^{3}

Formulae:

- Bernoulli's equation
- Frictional losses per unit mass of flowing fluid
- Power required for pumping = (mass flow rate x head developed by pump)/ h
= (volumetric flow rate x pressure developed by pump)/ h

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/((p/4) x 0.04^{2}) = 0.477 m/sec

h_{f} = 2 x 0.008 x 172.4 x 0.477^{2} / 0.04 = 15.69 m^{2}/sec^{2}

frictional losses per unit weight of fluid(h) = h_{f} / g = 15.69/9.812 = 1.6 m

pump head(q) = Dz + h = 12 + 1.6 = 13.6 m

pressure developed by pump = 13.6 x 1000 x 9.812 = 133443.2 N/m^{2}

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