Bernoulli's Principle - Definition, Principle, Application, Limitations, FAQs

What is Bernoulli’s Theorem?

Bernoulli’s principle states that:

Bernoulli's Principle states that for an incompressible, non-viscous fluid in steady flow, the total energy of the fluid (the sum of pressure energy, kinetic energy, and potential energy per unit volume) remains constant along a streamline.

Bernoulli’s Principle Formula

The formula for Bernoulli’s principle is given as follows:

$
P+\frac{1}{2} \rho v^2+\rho g h=\text { constant }
$

where,

• $P=$ Pressure of the fluid
• $\rho=$ Density of the fluid
• $v=$ Speed of the fluid
• $h=$ Height above the reference level

According to Bernoulli's principle, an increase in a fluid's speed is characterized by a reduction in static pressure or a decrease in the fluid's potential energy.
The principle was first stated in 1738 by Daniel Bernoulli in his book Hydrodynamic. Although Bernoulli deduced that pressure decreases as flow speed increases, it was Leonhard Euler who first put Bernoulli equation in its current form in 1752. The theory only applies to isentropic flows, in which the impacts of irreversible processes such as turbulence and non-adiabatic processes such as heat radiation are minor and may be ignored.

The assumption of conservation of energy can be used to derive Bernoulli's principle formula. In a continuous influx, the aggregate of all sources of energy in a fluid anywhere along the flow path is the same across all points along that flow path. For this to happen, the sum of kinetic energy, potential energy, and internal energy must remain constant.

Bernoulli's principle can be applied to a variety of fluid flows, yielding a number of different Bernoulli's equations. For incompressible flows, Bernoulli's equation in its simplest version is valid. Most liquids and gases, for example, move with a low Mach number. At increasing Mach numbers, more complex forms can be applied to compressible flows.

Principle of Conservation of Energy

Since energy cannot be generated or destroyed, it can be converted into another form. The principle of energy conservation is what this is characterized as. This principle is the basis for Bernoulli’s theorem. According to the principle of conservation of energy, the total energy at any point remains constant. There are three different forms of energy.

• Potential energy
• Pressure energy
• Kinetic energy

Potential Energy

The energy a fluid possesses due to its position above or below a datum line:

$
\text { Potential Energy }=W \cdot h=m g h=m g Z
$

Where:

• $h$ or $Z$ : Height of the fluid particle above the datum line (in meters)
• $m$ : Mass of the fluid (in kg )
• $W=m g$ : Weight of the fluid (in N )
• $g$ : Acceleration due to gravity (in $\mathrm{m} / \mathrm{s}^2$ )

Potential Energy per unit mass:

$
\frac{\text { Potential Energy }}{m}=g Z \quad(\text { in } \mathrm{J} / \mathrm{kg})
$

Pressure Energy

The energy a fluid possesses due to its pressure:
Pressure Energy $=P \cdot U$
Where:

• $P$ : Pressure of the fluid (in $\mathrm{N} / \mathrm{m}^2$ )
• $U$ : Specific volume of the fluid (in $\mathrm{m}^3 / \mathrm{kg}$ )

If $U=1$, then:

$
\text { Pressure Energy }=P \quad(\text { in } \mathrm{J} / \mathrm{kg})
$

Kinetic Energy

The energy of the fluid by virtue of its velocity is called kinetic energy.

$
\text { Kinetic Energy }=\frac{1}{2} m V^2
$

Where $m$ is the mass of the fluid $(\mathrm{kg})$ and $V$ is the velocity of the fluid $(\mathrm{m} / \mathrm{s})$.
Kinetic Energy per unit mass (specific energy):

$
\frac{\text { Kinetic Energy }}{m}=\frac{V^2}{2}
$

Total Energy of the Fluid (according to the principle of conservation of energy):
Total Energy = Potential Energy + Kinetic Energy + Pressure Energy

Substituting each term:

$
\text { Total Energy }=g Z+\frac{V^2}{2}+\frac{P}{\rho}
$

Dividing the total energy by $g$ to express in terms of total head:

$
\text { Total Head or Total Energy }=Z+\frac{V^2}{2 g}+\frac{P}{\rho g}
$

Alternatively, using specific weight $(w=\rho g)$ :

Total Head or Total Energy $=Z+\frac{V^2}{2 g}+\frac{P}{w}$
Bernoulli's Equation:
Since the total head remains constant:

$
Z_1+\frac{V_1^2}{2 g}+\frac{P_1}{w}=Z_2+\frac{V_2^2}{2 g}+\frac{P_2}{w}
$

Where:

• $Z_1, Z_2$ : Elevation head at points 1 and 2
• $\frac{V_1^2}{2 g}, \frac{V_2^2}{2 g}$ : Velocity head at points 1 and 2
• $\frac{P_1}{w}, \frac{P_2}{w}$ : Pressure head at points 1 and 2

This is the complete expression for Bernoulli’s equation.

Application of Bernoulli’s Theorem

• Aeroplane Wings (Lift of an Aircraft): Air moves faster over the curved upper surface of the wing than below it, creating lower pressure on top and higher pressure below, which lifts the plane.
• Atomiser or Spray: In perfumes or insect sprays, fast-moving air from the nozzle reduces pressure and draws liquid upward to form a mist.
• Chimney: Wind blowing across the top of a chimney creates low pressure, helping smoke rise easily.
• Venturi Meter: It is used to measure the speed or flow rate of fluids in pipes.
• Spinning Ball (Magnus Effect): The curved airflow around a spinning ball creates pressure difference, making it swing or curve in motion.

Limitations of Bernoulli’s Theorem

• Applies only to ideal fluids: It is valid only for fluids that are non-viscous (no friction) and incompressible.
• Steady flow condition: The theorem works only when the fluid flow is steady, not when it changes with time.
• No energy loss considered: It assumes no loss of energy due to heat, friction, or turbulence, which is not true in real situations.
• Applies along a streamline only: It is valid only along a single streamline, not for the entire fluid.
• Cannot be applied to fluids with high viscosity: For fluids like oil or honey, Bernoulli’s theorem gives inaccurate results because of internal friction.

Recommended Topic Video

Solved Examples Based on Bernoulli's Theorem

Example 1: When an air bubble of radius $r$ rises from the bottom to the surface of a lake, its radius becomes $\frac{5 r}{4}$. Taking the atmospheric pressure to be equal to the pressure of a 10 m height of the water column, calculate the approximate depth (in meters) of the lake. Assume the temperature remains constant, and ignore surface tension.

1) 11.2

2) 8.7

3) 9.5

4) 10.5

Solution

- At the bottom surface:

$
P_1=P_a+\rho g h
$

Where:
$P_a$ : Atmospheric pressure
$\rho$ : Density of the fluid
$g$ : Acceleration due to gravity
$h$ : Height of the fluid column
At the upper surface:

$
P_2=P_a
$

Using the continuity equation $P_1 V_1=P_2 V_2$ :

$
P_1 \cdot \frac{4}{3} \pi r^3=P_2 \cdot \frac{4}{3} \pi \cdot 12564 \cdot r^3
$

Simplify:

$
P_1=12564 \cdot P_2
$

Thus,

$
\frac{P_1}{P_2}=12564
$

Substituting the Pressures:
Substitute $P_1=P_a+\rho g h$ and $P_2=P_a$ :

$
\frac{P_a+\rho g h}{P_a}=12564
$

$
1+\frac{\rho g h}{P_a}=12564
$

Rearranging:

$
\frac{\rho g h}{P_a}=12564-1=6164
$

Solve for $h$ :

$
\rho g h=6164 \cdot P_a
$

Assume $P_a=10 \mathrm{~N} / \mathrm{m}^2$ :

$
\begin{aligned}
& \rho g \cdot h=6164 \cdot 10 \\
& h=\frac{6164}{\rho g}=9.5 \mathrm{~m}
\end{aligned}
$

Hence, the answer is the option (3).

Example 2: Water from a tap emerges vertically downwards with an initial speed of $1.0 \mathrm{~m} / \mathrm{s}$. The cross-sectional area of the tap is $10^{-4} \mathrm{~m}^2$. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. Find the cross-sectional area $\left(A_2\right)$ of the stream at 0.15 m below the tap, expressed as $x \times 10^{-5} \mathrm{~m}^2$. What is the value of $x$ ?
(Take $g=10 \mathrm{~m} / \mathrm{s}^2$ ).

1) 2

2) 5

3) 0.5

4) 0.2

Solution:

Using Bernoulli's theorem we get

$
V_2^2-V_1^2=2 g h
$

Rearranging to solve for $V_2$ :

$
V_2=\sqrt{V_1^2+2 g h}
$

Substituting the values:

$
\begin{gathered}
h=0.15 \mathrm{~m}, V_1=1 \mathrm{~m} / \mathrm{s}, g=9.8 \mathrm{~m} / \mathrm{s}^2 \\
V_2=\sqrt{1^2+2(9.8)(0.15)} \\
V_2=\sqrt{1+2.94}=\sqrt{3.94} \approx 2 \mathrm{~m} / \mathrm{s}
\end{gathered}
$

$\begin{gathered}\left(10^{-4}\right)(1)=A_2(2) \\ A_2=\frac{10^{-4}}{2}=0.5 \times 10^{-4}=5 \times 10^{-5} \mathrm{~m}^2\end{gathered}$

Hence, the answer is the option (2).

Example 3: Bernoulli's principle is based on

1) Conservation of momentum

2) Conservation of mass

3) Conservation of energy

4) Both (1) and (3)

Solution:

Bernoulli's Principle

The total energy (Pressure energy, Potential energy, and Kinetic energy ) per unit volume or mass of an incompressible and nonviscous fluid in steady flow through a pipe remains constant.

wherein

The proper expression for Bernoulli's equation is:

$
P+\rho g h+\frac{1}{2} \rho v^2=\mathrm{constant}
$

Where:

• $P$ : Pressure energy
• $\rho g h$ : Potential energy per unit volume
• $\frac{1}{2} \rho v^2$ : Kinetic energy per unit volume

So each term represents energy. So it is the conservation of energy.

Hence, the answer is the option (3).

Example 4: According to Bernoulli's equation

$
\frac{P}{\rho g}+h+\frac{v^2}{2 g}=\text { constant }
$ant A+B+C

The terms A, B, and C are generally called

1) Gravitational head, pressure head and Velocity head

2) Gravity, gravitational head and velocity head.

3) Pressure head, gravitational head and Velocity head.

4) Gravity, pressure head and velocity head

Solution:

Bernoulli's theorem for unit mass

$
\frac{P}{\rho g}+h+\frac{v^2}{2 g}=\text { constant }
$

Where:

• $\frac{P}{\rho g}$ : Pressure head
• $\frac{v^2}{2 g}$ : Velocity head
• $h$ : Gravitational head (elevation head)

Hence, the answer is the option (3).

Example 5: Water enters a house through a pipe with an inlet diameter of 2.0 cm at an absolute pressure of 4.0 × 10⁵ Pa (about 4 atm). A 1.0 cm diameter pipe leads to the second-floor bathroom 5.0 m above. When the flow speed at the inlet pipe is 1.5 m/s, what will be the flow speed, pressure and volume flow rate in the bathroom respectively?

1. $6 \mathrm{~m} / \mathrm{s}, 6.6 \times 10^5 \mathrm{~Pa}, 4.7 \times 10^{-4} \mathrm{~m}^3 / \mathrm{s}$
2. $3 \mathrm{~m} / \mathrm{s}, 3.3 \times 10^5 \mathrm{~Pa}, 5.7 \times 10^{-4} \mathrm{~m}^3 / \mathrm{s}$
3. $4 \mathrm{~m} / \mathrm{s}, 4 \times 10^5 \mathrm{~Pa}, 3.2 \times 10^{-4} \mathrm{~m}^3 / \mathrm{s}$
4. $6 \mathrm{~m} / \mathrm{s}, 3.3 \times 10^5 \mathrm{~Pa}, 4.7 \times 10^{-4} \mathrm{~m}^3 / \mathrm{s}$

Solution:

Let points 1 and 2 be at the inlet pipe and at the bathroom, then from the continuity equation,

Step 1: Continuity Equation
From the continuity equation:

$
A_1 v_1=A_2 v_2
$

Solving for $v_2$ :

$
v_2=6.0 \mathrm{~m} / \mathrm{s}
$

Step 2: Bernoulli's Equation
Applying Bernoulli's equation between the inlet ( $y_1=0$ ) and the bathroom ( $y_2=5.0 \mathrm{~m}$ ):

$
P_2=P_1-\frac{1}{2} \rho\left(v_2^2-v_1^2\right)-\rho g\left(y_2-y_1\right)
$

Substitute the given values to find $P_2$ :

$
P_2=3.3 \times 10^5 \mathrm{~Pa}
$

The volume flow rate $(Q)$ is given by:

$
Q=A_2 v_2=A_1 v_1
$

Substitute $A_2=\frac{\pi}{4}(0.01)^2$ and $v_2=6.0 \mathrm{~m} / \mathrm{s}$ :

$
\begin{gathered}
Q=\frac{\pi}{4}(0.01)^2 \cdot 6 \\
Q=4.7 \times 10^{-4} \mathrm{~m}^3 / \mathrm{s}
\end{gathered}
$

Hence, the answer is the option (4).