Friction Loss Formula

Friction Loss

Friction is the resistance required for moving a body through an external surface. But the friction loss is connected to the flow of liquid through a pipe. In other sense, it is a kind of energy loss because of the friction inside the tube. It is intimately related to the velocity and viscosity of the fluid. Friction loss can be articulated as hl as friction loss is nothing but the energy or head loss. Friction loss formula is articulated as,

\(\begin{array}{l}h_1 = f\times \frac{L}{D}\times \frac{v^2}{2g}\end{array} \)


the friction factor is f
the length of pipe is L
the inner diameter of the pipe is D
the velocity of the liquid is v
the gravitational constant  is g

the friction loss is hl

Solved Examples

Answered problems of friction loss are stated below.

Problem 1: Compute the friction loss, if the inner diameter and length of the pipe are 0.3m and 30m, respectively. It is also given that the friction factor and velocity of the liquid is 0.4 and 25m/s?


Given that,
Length of the pipe, L = 30m; internal diameter of the pipe, D = 0.3m;

velocity of the liquid, v = 25m/s;

friction factor, f = 0.4

g = 9.8m/s

The friction loss formula is,

\(\begin{array}{l}h_1 = f\times \frac{L}{D}\times \frac{v^2}{2g}\end{array} \)
\(\begin{array}{l}h_1 = 0.4\times \frac{30}{0.3}\times \frac{25^2}{2\times 9.8}=1275.51\ m\end{array} \)

Problem 2: Compute the friction loss if the friction factor is 0.3 and velocity of the flow is 50m/s. Given numerics are the length of the pipe 20m, inner diameter 0.5m and gravitational constant 9.8m/s.

Known factors are the velocity of flow is 50m/s, length of the pipe 20m, inner diameter 0.5m, friction factor 0.3 and gravitational constant 9.8m/s,

The friction loss formula is articulated as,

\(\begin{array}{l}h_1 = f\times \frac{L}{D}\times \frac{v^2}{2g}\end{array} \)
\(\begin{array}{l}h_1 = 0.3\times \frac{20}{0.5}\times \frac{50^2}{2\times 9.8}=1530.61\ m\end{array} \)

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