**The Bernoulli’s Equation**

Daniel Bernoulli, a Swiss mathematician and physicist, studied fluid flow in 1700. An important equation expressing some properties of fluids in motion is named after him, which is known as the Bernoulli’s equation.

The Bernoulli’s equation is an application of the principle of conservation of mechanical energy to the flow of an ideal fluid. It was observed that the volume of the fluid entering a tube is the same as it emerges out of the tube. This is because the fluid is incompressible and the fluid’s density is assumed to be constant.

If we let p_{1}, v_{1}, and y_{1} be the pressure, speed, and elevation of the fluid entering the tube and p_{2}, v_{2}, and y_{2} be the pressure, speed, and elevation of the fluid going out of the tube, and applying the principle of conservation of energy to the ideal fluid flow, the relationship between the quantities mentioned can be expressed mathematically as

\(p_1 + \frac12 \rho v_1^2 + \rho g y_1 = p_2 + \frac12 \rho v_2^2 + \rho g y_2\). (1)

Generally, the expression \(\frac12 \rho v^2\) is called as the fluid’s **kinetic energy density** or the kinetic energy per unit volume. The above equation can now be rewritten as

\(p = \frac12 \rho v^2 + \rho gy = constant\) (2)

which is known as **Bernoulli’s equation**.

Bernoulli’s equation can also be applied to fluids at rest by setting v_{1} = v_{2} = 0, the equation will be expressed as

\(p_2 = p_1 + \rho g (y_1 – y_2)\). (3)

Taking y = 0, which indicates that the fluids does not change elevation as it flows, making the equation as

\(p_1 + \frac12 \rho v_1^2 = p_2 + \frac12 \rho v_2^2\). (4)

Equation (4) implies that the speed of a fluid flowing along a horizontal streamline varies inversely to the fluid’s pressure.

**Example**

Jack fetch a pail of water. As he went along the road, the pail hit a rock which makes a hole at the bottom of the pail. If the water level is 0.1 m above the hole, what is the pressure on the hole? What is the speed of the water flow in the hole?

Solution:

- We use the Bernoulli’s equation to solve the problem.

\(p_1 + \frac12 \rho v_1^2 + \rho g y_1 = p_2 + \frac12 \rho v_2^2 + \rho g y_2\)

- Since the top of the pail and the hole is both open to atmosphere, we can assume that the pressures of the fluid at the top and on the hole are both equal to atmospheric pressures. That is, \(p_1 = p_2 = p_{atm}\).

- From the relation we had in b, we can subtract p
_{1}and p_{2}out from the equation in a, and rewrite it as

\(\frac12 \rho v_1^2 + \rho g y_1 = \frac12 \rho v_2^2 + \rho g y_2\)

- At the top of the pail, the velocity of water initially is zero, thus the expression \(\frac12 \rho v_1^2\) in equation c will become zero, \(\frac12 \rho v_1^2 = 0\) and we are left with the equation

\(\rho g y_1 = \frac12 \rho v_2^2 + \rho g y_2\)

- Note that in both sides of the equation, the density, \(\rho\) is just the density of water thus, dividing both sides of equation e with \(\rho\), we have

\({\rho g y_1 \over \rho} = {\rho(\frac12 v_2^2 + g y_2) \over \rho} \implies gy_1 = \frac12 v_2^2 + g y_2\)

- Combining like terms, we can now have expression for \(v_2^2\),

\(\frac12 v_2^2 = gy_1 - gy_2\)

- Manipulating equation f, so that what is left on the left side of the equation is the expression for \(v_2^2\),

\(v_2^2 = 2g(y_1 - y_2)\)

- Squaring both sides of the equation, we now have the equation for \(v_2\),

\(v_2 = \sqrt{2g(y_1 - y_2)}\)

substituting values,

\(v_2 = \sqrt{2(9.8)(0.1 - 0)} = 1.4\;m/s\).

Thus, the speed of the water as it emerges out of the hole is **1.4 m/s**.