Friction Loss Formula

Last Updated : 03 Feb, 2024

The flow of the fluid through a pipe is impeded by the viscous shear forces within the fluid as well as turbulence. These turbulences occur along the internal pipe wall, which is affected by the pipe material’s roughness. Friction losses are a complex function of the system geometry, and they are proportional to the fluid characteristics and flow rate. In most engineering flows, we can see that this loss is generally proportional to the square of the flow rate.

Friction Loss

Friction is defined as the amount of resistance necessary to move a body across an external surface. The friction loss, on the other hand, is linked to the movement of liquid through a pipe. As a result, the friction inside the tube causes an energy loss. As a result, it is linked to the fluid’s velocity and viscosity. Because friction loss is nothing more than energy loss, it may be calculated as h_l. Pipe friction is the word for this resistance, which is measured in meters of the fluid head. Several forms of study have been conducted in order to develop various formulae for calculating this loss.

Formula

The formula for friction loss is given as follows:

$h_l = f \times \frac {L}{D} \times \frac {v^2}{2g}$

where,

f denotes the friction factor
L denotes the pipe’s length
D is the pipe’s inner diameter
v is the liquid’s velocity
g is the constant of gravity
h_l denotes the friction lost

Sample Problems

Question 1. Compute the friction loss, if the inner diameter and length of the pipe are 1 m and 30m, and the friction factor and velocity of the liquid is 0.4 and 75m/s.

Solution:

Given: L = 30 m, D = 1 m, v = 75 m/s and f = 0.4 .

We know, g = 9.8m/s

The friction loss formula is given by: $h_l = f \times \frac {L}{D} \times \frac {v^2}{2g}$

Substituting the given values in the above formula, we obtain:

$h_l = 0.4\times \frac {30}{1} \times \frac {(75)^2}{2\times9.8}\\= 8609.69m$

Question 2. Compute the friction loss, if the inner diameter and length of the pipe are 0.4m and 10m, and friction factor and velocity of the liquid is 0.5 and 25m/s.

Solution:

Given: L = 10m, D = 0.4m, v = 25m/s and f = 0.4.

We know, g = 9.8m/s

The friction loss formula is given by: $h_l = f \times \frac {L}{D} \times \frac {v^2}{2g}$

Substituting the given values in the above formula, we obtain:

$h_l = 0.5\times \frac {10}{0.4} \times \frac {(25)^2}{2\times9.8}\\= 398.59m$

Question 3. Enlist some measures to reduce friction loss.

Solution:

Reduce interior surface roughness of the piping system
Increase pipe diameter of the piping system
Minimize length of piping system
Minimize the number of elbows, tees, valves, fittings, and other obstructions in the piping system; replace 90 degree turns with gentle bends.

Question 4. What is the total friction loss in 700 feet of three-inch hose reduced to 200 feet of 1¾-inch attack line with a one-inch straight tip?

Solution:

Figure the attack line first:

2 × 2.1²+ 2.1 × 2 × 6

= 65.52 × 2 (length)

= 131.04 psi

Next, figure the three-inch hose:

2 × 2.12 + 2.½.5

= 4.37 × 7 (length)

= 30.59 psi

Finally, add the two together: 131.04 + 30.59 = 161.63 psi.

As you can see, if the entire hose line were 900 feet of 1¾- inch hose, the friction loss would be 589.68 psi.

Question 5. Compute the friction loss, if the inner diameter and length of the pipe are 0.3m and 30m, and the friction factor and velocity of the liquid is 0.4 and 25m/s.

Solution:

Given: L = 30m, D = 0.3m, v = 25m/s and f = 0.4.

We know, g = 9.8m/s

The friction loss formula is given by: $h_l = f \times \frac {L}{D} \times \frac {v^2}{2g}$

Substituting the given values in the above formula, we obtain:

$h_l = 0.4\times \frac {30}{0.3} \times \frac {(25)^2}{2\times9.8}\\= 1275.51m$

Question 6. Compute the friction loss, if the inner diameter and length of the pipe are 0.5 m and 20m, and the friction factor and velocity of the liquid is 0.3 and 50m/s.

Solution:

Given: L = 20 m, D = 0.5 m, v = 50 m/s and f = 0.3 .

We know, g = 9.8m/s

The friction loss formula is given by: $h_l = f \times \frac {L}{D} \times \frac {v^2}{2g}$

Substituting the given values in the above formula, we obtain:

$h_l = 0.3\times \frac {20}{0.5} \times \frac {(50)^2}{2\times9.8}\\= 1530.61m$