Pressure Drop Formula, Definition, Solved Examples
Manoj Kumar17 Nov, 2023
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Pressure Drop Formula: Pressure drop is the variance in pressure observed between two locations within a fluid-carrying network. This phenomenon arises due to the frictional force generated by the resistance encountered by the fluid as it traverses through the tubing.
The relationship between viscosity and the speed of the liquid plays a crucial role in pressure drop. The primary determinants of resistance to liquid flow are the velocity of the fluid within the pipe and its viscosity. The pressure drop is directly proportional to the frictional shear forces experienced within the network of pipes.
The pressure drop, denoted as J, is calculated using the following formula:
J= f⋅ L⋅v 2 / 2⋅g⋅D
Where: J represents the pressure drop, f is the friction factor, L is the length of the tube, v is the velocity of the fluid, g is the acceleration due to gravity, and D is the inner diameter of the tube.
Pressure Drop Formula Solved Examples
Example 1: A fluid is flowing through a tube with a friction factor ( f) of 0.35. The tube has a length ( L) of 12 meters, an inner diameter ( D) of 0.08 meters, and the fluid velocity ( v) is 25 m/s. Calculate the pressure drop ( J) using the given parameters.
Solution: Given parameters: f=0.35, L=12m, v=25m/s, g=9.8m/s 2 , . D=0.08m. Using the pressure drop formula: J= f⋅L⋅v 2 / 2⋅g⋅D Substituting the values: J= 0.35⋅12⋅25 2 / 2⋅9.8⋅0.08 J= 10500/1.568 J=6691.33Pa.Example 2: Question: A liquid is flowing through a pipe with a friction factor (f) of 0.25. The pipe has a length ( L) of 18 meters, an inner diameter ( D) of 0.12 meters, and the liquid velocity ( v) is 35 m/s. Calculate the pressure drop ( J) using the given parameters.
Solution: Given parameters: f=0.25, L=18m, v=35m/s, , g=9.8m/s 2 , D=0.12m. Using the pressure drop formula: J= f⋅L⋅v 2 / 2⋅g⋅D Substituting the values: J= 0.25⋅18⋅35 2 / 2⋅9.8⋅0.12 J= 22050 / 2.352 J=9368.92Pa.Example 3: Water is flowing through a pipeline with a friction factor ( f) of 0.28. The pipeline has a length ( L) of 15 meters, an inner diameter ( D) of 0.1 meters, and the water velocity (v) is 40 m/s. Calculate the pressure drop (J) using the given parameters.
Solution: Given parameters: f=0.28, L=15m, v=40m/s, g=9.8m/s 2 , D=0.1m. Using the pressure drop formula: J= f⋅L⋅v 2 / 2⋅g⋅D Substituting the values: J= 0.28⋅15⋅40 2 / 2⋅9.8⋅0.1 J= 16800 / 1.96 J=8571.43Pa. Example 4: Calculate the pressure drop for a gas flowing through a tube with a friction factor ( f) of 0.2. The tube has a length ( L) of 8 meters, an inner diameter ( D) of 0.05 meters, and the gas velocity ( v) is 15 m/s. Solution: Given parameters: f=0.2, L=8m, v=15m/s, , g=9.8m/s 2 , D=0.05m. Using the pressure drop formula: J= f⋅L⋅v 2 / 2⋅g⋅D Substituting the values: J= 0.2⋅8⋅15 2 / 2⋅9.8⋅0.05 J= 1800 / 0.98 J=1836.73Pa. The pressure drop ( J) in a fluid-carrying system is a crucial parameter influenced by factors such as friction, length of the tube, fluid velocity, gravity, and the inner diameter of the tube. The pressure drop is calculated using the formula: J= f⋅L⋅v 2 / 2⋅g⋅D where: J is the pressure drop, f is the friction factor, L is the length of the tube, v is the velocity of the fluid, g is the acceleration due to gravity, and D is the inner diameter of the tube. Several examples have been provided to demonstrate the application of this formula in different scenarios, showcasing how varying values of friction factor, tube length, fluid velocity, and tube diameter affect the resulting pressure drop. Understanding and calculating pressure drop is essential for optimizing fluid flow in pipelines and tubes, ensuring efficient and reliable performance in various engineering and industrial applications.| Related Links | |
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Pressure Drop Formula FAQs
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