Pressure Drop Formula

The pressure drop of a liquid is defined as the resistance shown by it when it is flowing through a pipe. It can be interpreted as the difference in pressure between two sites in a fluid-carrying network. It occurs due to the frictional force created by the resistance to flow acting on the fluid as it travels through the tube. Fluid velocity and fluid viscosity are two major parameters that determine the pressure drop for a liquid. It is denoted by the symbol J and its SI unit is Pascal (Pa). Its dimensional formula is [M¹ L^-1 T^-2].

Formula

J = v²fl/2Dg

where, 

J is the value of pressure drop,

v is the liquid velocity,

f is the value of friction factor, 

l is the tube length in which the liquid is flowing,

D is the tube’s inner diameter,

g is the acceleration due to gravity,

Sample Problems

Problem 1. Calculate the pressure drop of a liquid flowing in a tube of length 10 m, inner diameter 0.6 m such that its velocity is 40 m/s and friction factor is 0.2.

Solution:

We have,

l = 10, D = 0.6, v = 40 and f = 0.2

Using the formula we have,

J = v²fl/2Dg

= (40 × 40 × 0.2 × 10)/(2 x 0.6 × 9.8)

= 320/11.76

= 27.21 Pa

Problem 2. Calculate the pressure drop of a liquid flowing in a tube of length 15 m, inner diameter 1 m such that its velocity is 60 m/s and friction factor is 0.7.

Solution:

We have,

l = 15, D = 1, v = 60 and f = 0.7

Using the formula we have,

J = v²fl/2Dg

= (60 × 60 × 0.7 × 15)/(2 x 0.7 × 9.8)

= 37800/13.72

= 2755.10 Pa

Problem 3. Calculate the length of the tube if the pressure drop of a liquid is 35 Pa flowing in a tube of inner diameter of 0.5 m with a velocity of 20 m/s and the friction factor is 0.1.

Solution:

We have,

J = 35, D = 1, v = 20 and f = 0.1

Using the formula we have,

J = v²fl/2Dg

35 = (60 × 60 × 0.1 × l)/(2 x 1 × 9.8)

l = 686/360

l = 1.80 m

Problem 4. Calculate the length of the tube if the pressure drop of a liquid is 100 Pa flowing in a tube of the inner diameter of 0.1 m with a velocity of 30 m/s and the friction factor is 0.9.

Solution:

We have,

J = 100, D = 0.1, v = 30 and f = 0.9

Using the formula we have,

J = v²fl/2Dg

100 = (30 × 30 × 0.9 × l)/(2 x 0.1 × 9.8)

l = 196/810

l = 0.241 m

Problem 5. Calculate the length of the tube if the pressure drop of a liquid is 75 Pa flowing in a tube of inner diameter of 0.4 m with a velocity of 50 m/s and the friction factor is 0.2.

Solution:

We have,

J = 75, D = 0.4, v = 50 and f = 0.2

Using the formula we have,

J = v²fl/2Dg

75 = (50 × 50 × 0.2 × l)/(2 x 0.4 × 9.8)

l = 588/500

l = 1.176 m

Problem 6. Calculate the inner diameter of the tube if the pressure drop of a liquid is 42 Pa flowing in a tube of length 20 m with the velocity of 30 m/s and the friction factor is 0.6.

Solution:

We have,

J = 42, l = 20, v = 30 and f = 0.6

Using the formula we have,

J = v²fl/2Dg

42 = (30 × 30 × 0.6 × 20)/(2 x D × 9.8)

D = 10800/823.2

D = 13.11 m

Problem 7. Calculate the friction factor if the pressure drop of a liquid is 32 Pa flowing in a tube of length 5 m with a velocity of 15 m/s and diameter is 10 m.

Solution:

We have,

J = 32, l = 5, v = 15 and D = 10

Using the formula we have,

J = v²fl/2Dg

32 = (15 × 15 × f × 5)/(2 x 10 × 9.8)

f = 6272/1125

f = 5.5