Apply the mass continuity and Bernoulli’s equations to flow through pipes

This section covers the development and utilisation of the mass continuity equation and Bernoulli’s equation as applied to the steady flow of incompressible fluids (liquids) through pipes.

Mass continuity equation

Consider sections 1 and 2 in the tapered pipe shown in figure below which is full of steadily flowing fluid. At Section 1, the cross-sectional area is A₁, and the velocity of the fluid is C₁; at section 2, the area and velocity are A₂ and C₂ respectively.

Continuity Equation

© A.Henderson, UHI

Steady flow conditions prevail when the rate at which the mass of fluid entering the pipe at datum position 1 is the same as the rate at which it leaves at datum 2; i.e., the mass flow rate is constant.

Now,

Volume of fluid passing section 1 per unit time = volume of fluid passing section 2 per unit time

$A_1{C}_1=A_2{C}_2$ 

(Because volume per unit time is equal to cross sectional area time velocity)

$\text{∴}\text{ Volumetric flow rate }V=A_1{C}_1=A_2{C}_2$ 

Volumetric flow rates for fluids can be converted to mass flow rates by introducing density into the equation. In outcome 1, density was defined as follows:

$\text{Density (ρ)}=\frac{\text{MASS}}{\text{VOLUME}}=\frac{m}{V}$ 

This means we can calculate mass flow rate using:

$\dot{m}\rho \times V=\rho \times AC$ 

This equation, $\dot{m}=\rho AC$ is known as the mass continuity equation 

Where:

$\dot{m}$ is the mass flow rate in kilograms per second (kgs^-1)

$\rho$ is the density of the fluid in kilograms per cubic metre (kgm^-3)

$A$ is the cross-sectional area of the pipe in metres squared (m²)

$C$ is the velocity of the fluid in metres per second (ms^-1)

Branched piping systems

In many pipework systems, a single pipeline will split into two or more branches as shown in figure below:

© A.Henderson, UHI

Like the tapered pipe example already dealt with, steady flow conditions will apply when the volumetric and mass flow rates passing Section 1 equal the combined total of volume and mass flow rates passing sections 2 and 3, i.e.:

${\dot{V}}_1={\dot{V}}_2+{\dot{V}}_3$ 

This means we can also say:

$\dot{V}=A_1{C}_1=A_2{C}_2+A_3{C}_3$ 

And

$\dot{m}=\rho A_1{C}_1=\rho (A_2{C}_2+A_3{C}_3)$ 

In our studies of incompressible fluid flow along pipes, the density of the fluid is assumed to remain constant throughout a process.

Energy of a flowing fluid

In outcome 3 we determined that the total energy of a mass, m, of flowing fluid had four components.

Potential Energy $mgZ\text{ (J)}$ 

Kinetic Energy $\frac{m{C}^2}{2}\text{ (J)}$ 

Internal Energy $\text{U}\text{ (J)}$ 

Flow Energy $pV\text{ (J)}$ 

Thus, in specific terms for a mass of 1 kg, the total energy is:

$gZ+\frac{C^2}{2}+u+pv$ 

where u is the specific internal energy of the fluid and v is the specific volume of the fluid.

The specific internal energy term, u, depends wholly on fluid temperature. In most hydrodynamic situations the change in fluid temperature is very small, so the internal energy term has little significance and can be neglected.

The specific volume v of a fluid in m³ kg^-1 is the reciprocal of density,

i.e., $v=\frac{1}{\rho }$ 

Hence, by substitution, the above expression can be modified to give specific energy of moving fluid as:

$gZ+\frac{C^2}{2}+\frac{p}{\rho }$ 

Further dividing throughout by g, gives:

$Z+\frac{C^2}{2g}+\frac{p}{\rho g}(J{kg}^{-1}\times s^2{m}^{-1}$$=Nm{kg}^{-1}\times s^2{m}^{-1}$ $={kgms}^{-2}\times {mkg}^{-1}\times s^2{m}^{-1})=m$ 

All three terms now have the dimension of length (metres)

$\frac{p}{\rho g}$ is the pressure head of a fluid at pressure p and the sum of all three terms is called the total head. IE:

$\text{Total head }=Z+\frac{C^2}{2g}+\frac{p}{\rho g}$ 

Bernoulli’s Equation

Consider unit mass of fluid flowing at a steady rate through the system/pipeline shown in the figure below.

© A.Henderson, UHI

Applying the Principle of Conservation of Energy, we can see that the specific energy of the fluid entering the system must be the same as the specific energy of the fluid leaving the system:

Specific Potential Energy  +  Specific Kinetic Energy At Entry  +  Specific Flow Energy  =  Specific Potential Energy + Specific Kinetic Energy At Exit  +  Specific Flow Energy

$g{Z}_1+\frac{C_1^2}{2}+\frac{p_1}{\rho \rho }=g{Z}_2+\frac{C_2^2}{2}+\frac{p_2}{\rho \rho }$ 

Which also equates to a constant measured in Joules per kilogram (Jkg^-1).

If we divide both sides of the equation by g, the acceleration due to gravity, we get:

$Z_1+\frac{C_1^2}{2}+\frac{p_1}{\rho g}=Z_2+\frac{C_2^2}{2}+\frac{p_2}{\rho }$ 

Which again equates to constant, this time measured in metres.

This equation is known as Bernoulli’s Equation.

Each quantity in the Bernoulli Equation is measured in terms of head of liquid – that is the height of liquid above some given datum. The unit of measure for each quantity is therefore the metre (m).

Frictional Resistance to Flow (loss of energy)

Bernoulli’s equation assumes there to be no frictional resistance to the flow of an incompressible fluid/liquid through a system/pipeline. In practical applications however, frictional resistance to flow is always present and reduces the available energy in the fluid at exit from the system. This means we should modify the equation above to be:

Specific energy in the fluid entering the system  =  Specific energy in the fluid leaving the system  +  Specific energy to overcome frictional resistance

Let Z_F­=Frcitional resistance “Head” (note, friction does not have a “head” as such, but it easier to consider it in these terms)

Then, we can restate Bernoulli’s equation as:

$Z_1+\frac{C_1^2}{2}+\frac{p_1}{\rho g}=Z_2+\frac{C_2^2}{2}+\frac{p_2}{\rho }+Z_F$ 

As before, each term represents Head of Liquid in units of m (metres):

$Z$  is the potential head

$\frac{C^2}{2g}$  is the kinetic head

$\frac{p}{\rho g}$  is the pressure head