Fluid Flow Analysis: Bernoulli's Equation and Momentum Changes

This report analyzes fluid flow through an inclined pipe, focusing on the application of Bernoulli's equation and momentum changes. It starts with the mechanical energy balance, including pressure, height, velocity, and friction losses. The report simplifies the equation for horizontal pipes and turbulent flow, leading to the Bernoulli equation, which relates pressure and kinetic energy. The analysis then explores incompressible flow through varying pipe cross-sections, demonstrating how velocity changes affect pressure. The report derives an equation showing the relationship between velocity and pressure difference, concluding that a plot of velocity versus pressure difference should be a straight line. This analysis provides a comprehensive understanding of fluid dynamics principles.

Flow characteristics of gases
Section 2.4.2: “Momentum changes in a fluid” in Chemical Engineering vol. 1,
Coulson and Richardson, Elsevier, Oxford, 6th ed. 2013.
Consider a fluid flowing through an inclined pipe, see fig. 25.
Figure 25. Fluid flowing through an inclined pipe
The mechanical energy balance between two points (1 and 2) in the fluid (ie in Joules kg-
1) can be represented as:
P1/+ z1g + V12/21 = P2/+ z2g + V22/22 + F
Where:  is the density of the fluid in kg m-3, P is the pressure (Nm-2), z the height above
level (m), g the acceleration due to gravity (9.8 m s-2), V the average velocity (m s-1), F
losses due to friction and  a correction factor = 0.5 for laminar flow and 1.0 for
turbulent flow. Assuming a horizontal pipe (z1 = z2), turbulent flow and negligible
friction loss:
P1/+ V12/2 = P2/+ V22/2
This is the simplified Bernoulli equation, which can be represented as:
P/+ V2/2 = constant
P/is the “pressure energy” and V2/2 is the kinetic energy, both in J kg-1.
The Bernoulli equation can be expressed as:
(P2 – P1)/= (V22 - V12)/2
If a fluid is made to pass into a narrower section of pipe (i.e. from one of cross sectional
area A1 to another of area A2) at a steady mass flow rate, then the velocity of the fluid
must increase because it is squeezed. Conservation of mass i.e. mass in = mass out:

1V1A1 = 2V2A2
For incompressible flow 1 = 2 and:
V2 = V1A1/ A2
Replacing this into the Bernoulli equation above gives:
V12 = 2(P2 – P1)/{1 – (A12 /A22)
or:
V1 = constant x (P2 – P1)
Thus, a plot of V1 vs (P2 – P1) should give a straight line passing through the origin.