Mohan Sundar / EV & Engineering

Mastering Bernoulli’s Equation

If you have ever wondered how a massive steel airplane stays suspended in thin air, how a simple perfume atomizer sprays a fine mist, or how a race car stays glued to the track at $200\text{ mph}$ , you are looking at the real-world consequences of Bernoulli’s Equation.
Named after the Swiss mathematician and physicist Daniel Bernoulli, who published it in his book Hydrodynamica in 1738, this equation is the cornerstone of fluid dynamics. It relates pressure, velocity, and elevation along a continuous stream of flowing fluid. This comprehensive guide breaks down the physical importance of Bernoulli's principle, walks through its mathematical derivation, and applies it to practical engineering problems.

diagram showing Bernoulli’s equation and fluid flow through a curved pipe with pressure, velocity, elevation, and cross-sectional area labels for fluid mechanics education.

the Bernoulli equation can be written as:

$P + \frac{1}{2} ρ v^{2} + ρ g h = co n s t an t$

here:

P = Pressure energy per unit volume (Pa)
ρ = Density of the fluid (kg/m³)
v = Velocity of the fluid (m/s)
g = Acceleration due to gravity (m/s²)
h = Height above a reference point (m)

According to Bernoulli’s theorem, the work done by fluid can be written as,

dW = F1dx1 - F2dx2

dW = p1A1dx1 - p2A2dx2

dW = p1dv - p2dv = (p1 - p2)dv

As known, the work done on the fluid was due to the conservation of change in gravitational potential energy and change in kinetic energy. The change in the fluid's kinetic energy is given by:

$d K = \frac{1}{2} m_{2}^{2} v_{2}^{2} - \frac{1}{2} m_{1}^{2} v_{1}^{2}$

$d K = \frac{1}{2} ρ d v (v_{2}^{2} - v_{1}^{2})$

Change in the potential energy can be written as:

$d U = m_{2} g y_{2} - m_{1} g y_{1} = ρ g d v (y_{2} - y_{1})$

Now, according to the theorem, the energy of the system is the same as the summation of the change in the kinetic fluid's energy and the change in the fluid's potential energy. This can be mathematically written as:

$d W = d K + d U$

Putting values by the above equations:

$(p 1 - p 2) d v = \frac{1}{2} ρ d v (v_{2}^{2} - v_{1}^{2}) + ρ g d v (y_{2} - y_{1})$

$p 1 - p 2 = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) + ρ g (y_{2} - y_{1})$

By rearranging the equation, we will get the Bernoulli's equation:

$p_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g y_{1} = p_{2} + \frac{1}{2} v_{2}^{2} + ρ g y_{2}$

$p + \frac{1}{2} ρ v + ρ g y = co n s t an t$

Components of Bernoulli’s Equation

Pressure Head: The pressure head is the height of a fluid column that corresponds to the pressure at a given point in the fluid. It is defined as:

Pressure Head = $\frac{P}{ρ g}$

Velocity Head: The velocity head is the height equivalent to the kinetic energy per unit weight of the fluid. It is calculated as:

Velocity Head = $\frac{v ^{2}}{2 g}$

Potential Head: The potential head is the height of the fluid column and essentially is due to the gravitational potential energy. It is given as:

Potential head = y

The head in fluid mechanics is defined as the height of a fluid column which represents certain energy in the fluid system. That is, fluid pressure, velocity, or potential can be expressed conveniently in terms of energy and hence can be compared relatively more easily across fluid systems under practical applications, such as pipes, pumps, and turbines.

Bernoulli’s Equation with Head terms:

Using all the three head terms Bernoulli’s Equation can be re-written as:

$\frac{P}{ρ g} + \frac{v ^{2}}{2 g} + y = co n s t an t$

4.0Continuity equation

The continuity equation is based on the principle of mass conservation. For an incompressible, steady-flowing fluid, the rate at which mass enters a pipe must be equal to the rate at which mass exits. The equation is:

A1v1 = A2v2

Here:

A1 and A2 are the cross-sectional areas at two points in the flow.
v1 and v2 are the velocities of the fluid at those points.
3. Real-Life Applications
Aerodynamic Lift (Airplane Wings)
An aircraft wing (airfoil) is intentionally engineered with a curved upper surface and a flatter lower surface. As air hits the leading edge, the stream passing over the curved top is forced to travel faster than the air passing underneath. According to Bernoulli’s principle, this high-velocity air on top drops the static pressure above the wing, while the slower air underneath maintains a higher pressure. This pressure difference creates an upward mechanical force known as lift.
Venturi Tubes and Carburetors
A Venturi tube is a pipeline configuration that features a constricted throat section. When a fluid passes through this narrow throat, it must accelerate to maintain mass conservation. This velocity surge causes a massive localized pressure drop. In older automotive carburetors, this low-pressure throat acts like a vacuum, drawing raw fuel into the fast-moving air stream to create an optimized combustion mixture without needing a mechanical computer injector.
Pitot Tubes (Speed Indicators)
Pitot tubes are the physical sensory probes mounted onto aircraft noses or race car bodywork. They feature an open hole facing directly into the oncoming wind. As air enters the tube, it is forced to come to a complete stop ( $v = 0$ ). This point of zero velocity is called the stagnation point. By comparing this high "stagnation pressure" against the surrounding ambient "static pressure" via Bernoulli’s equation, the flight computer can dynamically calculate the exact airspeed of the vehicle.
Solved Examples
Problem 1: Water is flowing from a tank through a pipe. The pressure at the bottom of the tank (point 1) is 150,000 Pa, and the height is 10 m. Find the pressure at the top of the tank (point 2), where the height is 0 m. The velocity at the bottom of the tank is 1 m/s, and at the top of the tank, the velocity is negligible.
Solution: Given that: P1 = 150000Pa, h1=10m, v1 = 1m/s, h2 = 0m, v2 = 0m/s
Using Bernoulli's equation between points 1 and 2
$p_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g y_{1} = p_{2} + \frac{1}{2} v 22 + ρ g y_{2}$
$150000 + \frac{1}{2} (1000) (1)^{2} + (1000) (9.8) (10) = P_{2} + \frac{1}{2} (1000) (0)^{2} + (1000) (9.8) (0)$
$P_{2} = 150000 + 500 + 98000 = 2, 48, 500$

Problem 2: An aeroplane wing is designed such that the air above the wing moves faster than the air below the wing. The velocity of the air above the wing is 200 m/s, and the velocity of the air below the wing is 150 m/s. Given that the air density is 1.225 kg/m3, calculate the difference in pressure between the upper and lower surfaces of the wing that produces lift.
Solution: Let the pressure below the wing be P1 and the pressure above the wing be P2.
It is given that the velocity above the wing is v1 = 200m/s and the velocity below the wing v2 = 150m/s, so here we apply Bernoulli’s equation between the two points:
$p_{1} + \frac{1}{2} ρ v_{2}^{2} = p_{2} + \frac{1}{2} v_{1}^{2}$
$P_{1} - P_{2} = \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2})$
$P_{1} - P_{2} = \frac{1}{2} \times 1.225 \times (20 0^{2} - 15 0^{2})$
$P_{1} - P_{2} = \frac{1}{2} \times 1.225 \times (40000 - 22500)$
$P_{1} - P_{2} = \frac{1}{2} \times 1.225 \times 17500$
$P_{1} - P_{2} = 10718.75 P a$
Hence, the pressure difference between the upper and lower wing is 10718.75Pa.

Problem 3: Water flows vertically through a pipe. The water velocity at a point 4 m above the ground is 2 m/s, and the pressure at this point is 250 kPa. Calculate the velocity of water 6 m below this point, where the pressure is 300 kPa.
Given:
Height h1 = 4m on point 1.
Height h2 = –6m on point 2.
v1 = 2m/s, =1000kg/m3, g = 9.8m/s2,
P1 = 250kPa = 250000Pa
P2 = 300kPa = 300000Pa
Solution: By applying Bernaulli’s equation:
$p_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = p_{2} + \frac{1}{2} v_{2}^{2} + ρ g h^{2}$
$\frac{1}{2} ρ v_{2}^{2} + = p_{2} - p_{1} + \frac{1}{2} v_{1}^{2} + ρ g (h_{2} - h_{1})$
$v_{2} + = \frac{2}{ρ} (p_{2} - p_{1} + \frac{1}{2} v_{1}^{2} + ρ g (h_{2} - h_{1}))$
$v_{2} = \frac{2}{1000} (250000 - 300000 + \frac{1}{2} 1000 \times 2^{2} + 10009.8 (4 - (- 6)))$
$v_{2} = \frac{2}{1000} \times 50000$
$v_{2} = 100$
$v_{2} = 10 m / s$