# Pressure Drop Formula

Pressure drop describes the difference in the pressure between two points of a network carrying fluid. Pressure drop occurs when the frictional force caused by the resistance to flow acting on the fluid as it flows through the tube.

It has a relation between viscosity and velocity of the liquid. The main factors that determine the resistance to the liquid flow are fluid velocity through the pipe and the fluid viscosity. Pressure drop is proportional to the frictional shear forces within the pipe network.

The Pressure drop is denoted by J.

The pressure drop formula is given by

J = fLv2 / 2gD

$J\,&space;=\,&space;\frac{fLv^{2}}{2gD}$

Where,

J = pressure drop

f = friction factor

L = length of the tube

v = velocity of the fluid

g = acceleration due to gravity

D = inner diameter of the tube

Example 1

Determine the pressure drop of a fluid whose velocity is 60m/s. The length of the tube is 20m; the inner diameter is given as 0.1m, and the friction factor is 0.5.

Solution:

Given parameters are,

f = 0.5,

L = 20m,

v = 60m/s,

g = 9.8m/s,

D = 0.1m

Pressure drop formula is given by,

J = fLv2 / 2gD

J = 0.5×20×602 / 2 x 9.8 × 0.1

36000 / 1.96

= 18367.34 Pa

Example 2

Determine the pressure drop of a liquid whose velocity is 10m/s. The length of the tube is 4m; the inner diameter is given as 0.5m, and the friction factor is 0.3.

Solution:

Given parameters are,

f = 0.3,

L = 4m,

v = 10m/s,

g = 9.8m/s,

D = 0.5m

Pressure drop formula is given by,

J = fLv2 / 2gD

J = 0.3×4×100 / 2 x 9.8×0.5

120 / 9.8

= 12.24 Pa.