Bernoulli's Equation Derivation and Application

Table of Contents

Explore the derivation of Bernoulli’s equation, assumptions of Bernoulli’s equation, and application of Bernoulli’s equation in fluid flow operations.

Bernoulli’s equation is an important relation between Pressure energy, Potential energy, and Kinetic energy. Bernoulli’s equation is an energy balance equation.

Here, first, we will derive Bernoulli’s equation for frictionless fluid. After that we will see corrections in Bernoulli’s equation for friction, also we will derive Bernoulli’s equation for pump work.

Bernoulli’s Equation Derivation

Bernoulli’s equation for frictionless fluid can be derived based on Newton’s second law of motion for Potential Flow.

Assumptions of Bernoulli’s Equation

Steady flow: The equation assumes that the fluid flow is steady, meaning that the velocity of the fluid particles at any point in the system does not change with time.
Incompressible fluid: Bernoulli’s equation applies to incompressible fluids, where the density of the fluid remains constant throughout the flow.
Inviscid flow: The equation assumes that the fluid has no viscosity, meaning there is no internal friction or resistance to flow within the fluid.
Along a streamline: Bernoulli’s equation is valid only along a streamline, which is a line that is tangent to the velocity vector of the fluid flow at every point.
No external work: The equation assumes no external work is done on the fluid and neglects effects such as pump work or turbine work.

Derivation of Bernoulli’s Equation

Let us consider an element of length $\Delta L$ of a stream tube of the constant cross-sectional area as shown in the figure below.

Bernoulli's Equation Derivation Force Balance — Force Balance for Potential Flow

Let’s consider,
Cross-sectional area of the element = $A$
Density of the fluid = $\rho$
Velocity of the fluid at the entrance (upstream) = $u$
Velocity of the fluid at the exit (downstream) = $u +\Delta u$
Pressure at the entrance (upstream) = $P$
Pressure at the exit (downstream) = $P+\Delta P$

The forces acting on the element are as follows

The force from the upstream pressure normal to the cross-section of the tube (acting in the direction of flow) = $P\;A$

The force from the downstream pressure normal to the cross-section of the tube (acting in the opposite direction of flow) = $(P + \Delta P) A$

The force from the weight of fluid (the force of gravity acting downward) = $\rho\;A\;\Delta L \;g$

Therefore, the gravitational force acting in the opposite direction of the flow = $\rho\;A\;\Delta L \;g\;cos\theta$

The forces acting in the flow direction will be taken as positive, and the forces acting in the opposite direction will be taken as negative.

Rate of change of momentum of the fluid along the fluid element

Momentum of the fluid at the entrance (upstream) = $\dot{m}\;u$

Momentum of the fluid at the exit (downstream) = $\dot{m}\;(u + \Delta u)$

Therefore,
Rate of Change of Momentum $= \dot{m}\;(u + \Delta u) - \dot{m}\;u = \dot{m}\; \Delta u$

From Equation of Continuity $(\dot{m}=\rho\;u\;A)$

Rate of Change of Momentum $= \rho\;u\;A \; \Delta u$

According to Newton’s second law of motion

The sum of all forces acting on the element = Rate of change of momentum

$PA - (P + \Delta P) A - \rho \; A \; \Delta L \; g \;cos \theta &=& \rho \; u\;A\; \Delta u$
$PA - PA- \Delta PA - \rho \;A\; \Delta L \;g\; cos \theta &=& \rho \; u\;A \;\Delta u$
$-\Delta PA - \rho \;A\; \Delta L \;g\; cos \theta &=& \rho \;u\;A\; \Delta u$
$\therefore \Delta PA + \rho \;A\; \Delta L \;g\; cos \theta + \rho \;u\;A\; \Delta u &=& 0$

Dividing each term of Equation by $A\; \Delta L \;\rho$ , we get

$\frac{\Delta P}{\rho\Delta L} + g \; cos \theta + \frac{u \Delta u}{\Delta L} = 0$

Here, $cos\theta=\frac{\Delta Z}{\Delta L}$

Putting the value of $cos \theta$ in the above equation, we get

$\frac{1}{\rho} \frac{\Delta P}{\Delta L} + g \frac{\Delta Z}{\Delta L} + u \frac{\Delta u}{\Delta L}=0$

Writing the equation in differential form,

$\frac{1}{\rho} \frac{dP}{dL} + g \frac{dZ}{dL} + u \frac{du}{dL}=0$

(1) $\begin{equation*}\frac{dP}{\rho} + g\;dZ + d(u^2/2)}=0\end{equation*}$

Equation (1) is the Bernoulli equation in differential form.

For incompressible fluids, density is independent of pressure, integrating equation (1), we get,

(2) $\begin{equation*} \frac{P}{\rho}+g\;Z+\frac{u^2}{2} = constant \end{equation*}$

Equation (2) is the integrated form of the Bernoulli equation.

Term $\left(\frac{P}{\rho}\right)$ represents Pressure energy.

Term $\left( g\;Z\right)$ represents Potential energy.

Term $\left( \frac{u^2}{2} \right)$ represents Kinetic energy

Each term in the Bernoulli equation [Equation (2)] represents energy per unit mass of the fluid and has the units of J/kg in the SI system.

Checking the unit of each term in the Bernoulli equation,

The unit of $\frac{P}{\rho}$ : $\frac{N}{m^2}\times\frac{1}{kg/m^3} = \frac{N\;m}{kg} = \frac{J}{kg}$

The unit of $g\;Z$ : $\left (\frac{m}{s^2} \right) (m) = \left(\frac{kg}{kg} \right) \(\left (\frac{m}{s^2} \right) (m) = \left(\frac{kg\;m}{s^2}\right) \left(\frac{m}{kg}\right) = \frac{N\;m}{kg} = \frac{J}{kg}$

The unit of $\frac{u^2}{2}$ : $\frac{m^2}{s^2} = \left(\frac{kg}{kg}\right) \left(\frac{m^2}{s^2}\right) = \left(\frac{kg\;m}{s^2}\right) \left(\frac{m}{kg}\right) = \frac{N\;m}{kg} = \frac{J}{kg}$

The alternate form of Bernoulli’s formula

(3) $\begin{equation*} \frac{P}{\rho\;g}+Z+\frac{u^2}{2\;g} = constant \end{equation*}$

Each term in equation (3) represents energy per unit weight of the fluid and has length dimensions. It is called Head.

Term $\left(\frac{P}{\rho\;g}\right)$ is called Pressure Head or Static Head.

Term $Z$ is called Potential Head.

Term $\left(\frac{u^2}{2g}\right)$ is called Kinetic Head or Velocity Head.

The sum of the Pressure Head, Velocity Head, and Potential Head is the Total Head or Total energy per unit weight of the fluid.

The sum of the Pressure Head and Potential Head is called as Piezometric Head.

The Bernoulli Equation states that in a steady rotational flow of an incompressible fluid, the total energy at any point is constant.

Bernoulli Equation application between two stations (station-a and station-b)

$\frac{P_a}{\rho}+g \;Z_a+\frac{u_a^2}{2} = \frac{P_b}{\rho}+g \;Z_b+\frac{u_b^2}{2}$

Corrections in Bernoulli’s Equation

Kinetic Energy Correction

In the derivation of Bernoulli’s Equation for frictionless fluid, it is assumed that the velocity $u$ is constant over the area $A$ . But in actual practice, the velocity varies over a single cross-section and we have a velocity profile over the cross-section.

The fluid velocity is zero at the wall surface and maximum at the center of the pipe. Therefore, allowance must be made for the velocity profile of the kinetic energy term. This can be done by introducing a correction factor $\alpha$ into the kinetic energy term.

The Kinetic energy term would be written as $\left(\frac{\alpha u^2}{2}\right)$ .

For the flow of a fluid through a circular cross-section,

$\alpha =2$ for laminar flow

$\alpha = 1$ for turbulent flow.

Correction for Fluid Friction

The Bernoulli equation is derived for frictionless fluid. Therefore, it must be corrected for the existence of fluid friction whenever a boundary layer forms. Fluid friction is an irreversible conversion of mechanical energy into heat.

Thus, the quantity $\frac{P}{\rho} + \frac{u^2}{2} +g\; Z$ is not constant but always decreases in the direction of flow.

The Bernoulli equation for incompressible fluids is corrected for friction by adding a friction term on the R.H.S. of Equation.

The Bernoulli equation between stations ‘a’ and ‘b’, after making necessary corrections, in terms of energy per unit mass (J/kg) is

$frac{P_a}{\rho}+g \;Z_a+\frac{\alpha_1\u_a^2}{2} = \frac{P_b}{\rho}+g \;Z_b+\frac{\alpha_2u_b^2}{2}+h_f$

$h_f$ is the total frictional loss of energy due to friction between stations ‘a’ and ‘b’ in $J/kg$ .

The term $h_f$ indicates the friction generated per unit mass of fluid that occurs in the fluid between stations ‘a’ and ‘b’.

Application of Bernoulli’s Equation in Pump Work

A pump is installed in a flow system to increase the mechanical energy of the fluid to maintain its flow.

Assume that a pump is installed in the flow system between stations ‘a’ and ‘b’ as shown in the figure below.

Application of Bernoulli's Equation in Pump Work — Application of Bernoulli’s Equation in Pump Work

Let, $W_p$ = to the work done by the pump per unit mass of fluid.

Let, $h_{fp}$ = the total friction in the pump per unit mass of fluid (friction in bearings, seals, or stuffing box.).

The net mechanical energy delivered to the flowing fluid is the difference between the mechanical energy supplied to the pump and frictional losses within the pump.

$= W_p - h_{fp}$

But, to obtain the net mechanical energy (Net-Work) delivered to the fluid, instead of using $h_{fp}$ , a pump efficiency designated by the symbol $\eta$ is used.

It is defined as,

$W_p- h_{fp} = \eta W_p$

$\eta=\frac{W_p- h_{fp}}{W_p}$

Since η is always less than one, the mechanical energy delivered to the fluid $ηW_p$ is less than the work done by the pump. The Bernoulli equation corrected for the pump work between stations ‘a’ and ‘b’ is thus given by

$\frac{P_a}{\rho}+g \;Z_a+\frac{\alpha_au_a^2}{2}+\eta W_p = \frac{P_b}{\rho}+g \;Z_b+\frac{\alpha_bu_b^2}{2}+h_f$

FAQs

What is Bernoulli’s equation and what does it represent?

Bernoulli’s equation is an energy balance equation that relates pressure energy, potential energy, and kinetic energy in a fluid. Each term in the equation represents energy per unit mass of the fluid.

What are the assumptions of Bernoulli’s equation?

The assumptions of Bernoulli’s equation are:
Steady flow: The equation assumes that the fluid flow is steady, meaning that the velocity of the fluid particles at any point in the system does not change with time.
Incompressible fluid: Bernoulli’s equation applies to incompressible fluids, where the density of the fluid remains constant throughout the flow.
Inviscid flow: The equation assumes that the fluid has no viscosity, meaning there is no internal friction or resistance to flow within the fluid.
Along a streamline: Bernoulli’s equation is valid only along a streamline, which is a line that is tangent to the velocity vector of the fluid flow at every point.
No external work: The equation assumes no external work is done on the fluid and neglects effects such as pump work or turbine work.
These assumptions make Bernoulli’s equation applicable in certain idealized conditions and may not hold in all real-world situations.

How is Bernoulli’s equation derived?

Bernoulli’s equation can be derived from Newton’s second law of motion for Potential Flow. The derivation involves considering an element of length in a stream tube of constant cross-sectional area and balancing the forces acting on the element.

What are the corrections to Bernoulli’s equation?

There are two corrections to Bernoulli’s equation: kinetic energy correction and correction for fluid friction. The kinetic energy correction accounts for losses in kinetic energy due to the shape of the element, while the correction for fluid friction accounts for losses due to viscosity and turbulence.

How is Bernoulli’s equation applied in pump work?

Bernoulli’s equation is applicable to determine the work done by a pump. The work done by a pump is equal to the difference in energy between the inlet and outlet of the pump. Bernoulli’s equation can be used to calculate this energy difference.

What are the units of each term in Bernoulli’s equation?

Each term in Bernoulli’s equation has the units of energy per unit mass of the fluid. The pressure energy term is in units of pressure, the potential energy term is in units of length, and the kinetic energy term is in units of velocity squared. When the equation is integrated, the constant has the same units as the individual terms.

Read Also

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