Bernoulli’s Equation

When a fluid speeds up, its kinetic energy increases. That extra kinetic energy comes from work done by pressure differences and gravity. Bernoulli’s equation is the principle of conservation of energy applied to a moving, ideal fluid.

When fluid flows from a wider pipe into a narrower pipe, it speeds up to maintain a constant flow rate. To accelerate that fluid, a net force must push it forward, meaning the pressure behind the fluid in the wide section is higher than the pressure ahead of it in the narrow section. Higher speed corresponds to lower pressure. Additionally, pumping a fluid upward requires work against gravity, causing pressure to drop with increasing elevation.

Derivation of Bernoulli's Equation from Energy Conservation

Consider a small volume of fluid $dV$ with mass $dm = \rho dV$ moving along a pipe from point 1 (height $h_1$ , speed $v_1$ , pressure $p_1$ ) to point 2 (height $h_2$ , speed $v_2$ , pressure $p_2$ ).

The net work done on this fluid element by surrounding pressure forces is:

dW = (p_1 - p_2) dV

By the work-energy theorem, this work equals the change in kinetic energy plus the change in gravitational potential energy ( $dW = dK + dU$ ):

(p_1 - p_2) dV = \frac{1}{2} (\rho dV) (v_2^2 - v_1^2) + (\rho dV) g (h_2 - h_1)

Dividing the entire equation by $dV$ removes volume and yields energy per unit volume:

p_1 - p_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2 + \rho g h_2 - \rho g h_1

Rearranging all point 1 terms to the left and point 2 terms to the right gives Bernoulli's equation:

p_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = p_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2

(1)

Because points 1 and 2 are arbitrary choices along a path, this total sum remains constant along any streamline:

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

(2)

Each term in Bernoulli's equation represents an energy density with units of Joules per cubic meter, which simplifies directly to Pascals ( $\text{J/m}^3 = \text{N/m}^2 = \text{Pa}$ ). The term $p$ is static pressure, $\frac{1}{2} \rho v^2$ is dynamic pressure (kinetic energy per unit volume), and $\rho g h$ is hydrostatic pressure density (potential energy per unit volume).

Special Cases

If the fluid is static ( $v_1 = v_2 = 0$ ), the kinetic energy terms drop out, leaving $p_1 + \rho g h_1 = p_2 + \rho g h_2$ . Setting $h_1 - h_2 = h$ reproduces the hydrostatic pressure formula $p_2 = p_1 + \rho g h$ .

If fluid flows horizontally without changing elevation ( $h_1 = h_2$ ), potential energy terms cancel, leaving Bernoulli's principle:

p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2

(3)

This equation shows that as fluid speed increases, static pressure decreases. If $v_2 > v_1$ , then $p_2$ must be less than $p_1$ for the sum to remain equal.

Applications

Entrainment occurs when a fluid moves rapidly through a narrow opening, dropping its pressure below atmospheric pressure and drawing surrounding fluid into the stream. Atomizers, carburetors, aspirators, and Bunsen burners rely on this effect.

Velocity measurement devices like Pitot-Prandtl tubes measure aircraft speed by bringing oncoming fluid to a complete stop at an opening ( $v_1 = 0$ ). Comparing this stagnation pressure $p_1$ to static pressure $p_2$ from a side opening allows calculation of airspeed via $v \propto \sqrt{p_1 - p_2}$ .

Example 1

Water ( $\rho = 1000 \text{ kg/m}^3$ ) flows horizontally through a pipe. At point 1, the water speed is $1.96 \text{ m/s}$ and the absolute pressure is $4.24 \cdot 10^5 \text{ Pa}$ . The pipe then narrows, increasing the water speed at point 2 to $25.5 \text{ m/s}$ . Find the absolute pressure at point 2.

Hint Click to unblur

Since the pipe is horizontal ( $h_1 = h_2$ ), use Bernoulli's principle (Equation 3) to solve for $p_2$ .

Solution Click to unblur

Start with Bernoulli's principle for level flow (Equation 3):

p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2

Isolate $p_2$ :

p_2 = p_1 + \frac{1}{2} \rho (v_1^2 - v_2^2)

Substitute the known values:

p_2 = 4.24 \cdot 10^5 \text{ Pa} + \frac{1}{2} (1000 \text{ kg/m}^3) \left[ (1.96 \text{ m/s})^2 - (25.5 \text{ m/s})^2 \right]

p_2 = 4.24 \cdot 10^5 \text{ Pa} + 500 \cdot (3.84 - 650.25) \text{ Pa}

p_2 = 4.24 \cdot 10^5 \text{ Pa} - 3.23 \cdot 10^5 \text{ Pa} = 1.01 \cdot 10^5 \text{ Pa}

The pressure at the narrow section drops to atmospheric pressure ( $1.01 \cdot 10^5 \text{ Pa}$ ) due to the sharp increase in speed.