## Bernoulli's Equation

Bernoulli's Equation is an expression of the Conservation of Energy for fluids. At any point, the combination of the pressure, $P$, the potential energy, $\rho g h$, and the kinetic energy, $\frac{1}{2} \rho v^2$, is the same as at any other point in an ideal fluid's path. Bernoulli's Equation equates these quantities at two points.

Bernoulli's Equation
$P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2$

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

Solved Problems for Bernoulli's Equation

 A tank of water, 4 meters deep, has a small hole in the side, 1 meter from the bottom. Assume the rate that the water lowers, at the top of the tank, is negligible. What is the velocity of the water as it exits the tank, through the small hole?

$P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2$

Both the top of the tank and the small hole are open to atmospheric pressure.

$P_1 = P_2$

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

$v_1$ is the rate that the water goes down, at the top of the tank, which is negligible.

$v_1 = 0$

$\rho g h_1 = \frac{1}{2} \rho v_2^2 + \rho g h_2$

The density cancels out. The exit velocity is independent of the density of the fluid.

$g h_1 = \frac{1}{2} v_2^2 + g h_2$

Solve for the final velocity:

$v_2 = \sqrt{ 2 g ( h_1 - h_2) } = \sqrt{ 2 (9.8 m/s^2) ( 4 m - 1 m) } = 7.7 \ m/s$

Given $h = h_1 - h_2$, the equation: $v_2 = \sqrt{ 2 g h }$ is like the equation for the velocity of an object in free fall. This is known as Torricelli's Theorem.

2nd Problem for Bernoulli's Equation

 Water flows through a pipe as shown in the illustration. $r_1 = 12 \ cm \ \ r_2 = 8 \ cm$ $h = 2.0 \ m$ At $\ r_1$, the gauge pressure is: $P_1 = 3.0 x 10^5 \ Pa$ The velocity of the water is: $v_1 = 6.0 \ m/s$ What is the gauge pressure and velocity of the fluid at $r_2$? What is the fluid flow rate throughout the pipe?

Gauge pressure and velocity
The velocity at $r_2$ is required in order to calculate the gauge pressure.
The velocity can be determined by using the Continuity Equation, discussed on the Flow Rate page:

$A_1 v_1 = A_2 v_2$

$v_2 = \large \frac{A_1 v_1}{A_2} = \frac{ \pi r_1^2 v_1}{ \pi r_2^2} = \frac{ r_1^2 v_1}{ r_2^2} = \frac{ (0.12 m)^2( 6.0 m/s) }{ (0.08 m)^2 } \normalsize = 13.5 \ m/s$

The gauge pressure can be found by using Bernoulli's Equation:

$P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2$

$h_2 = 0$

$P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2$

$P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) + \rho g h_1$
$\ \ = 3.0 x 10^5 Pa + \frac{1}{2} (1,000 kg/m^3) \left( (6.0 m/s)^2 - (13.5 m/s)^2 \right)$
$\ \ \ \ + (1,000 kg/m^3) (9.8 m/s^2) (2.0 m)$

$\ \ = 2.46 x 10^5 \ Pa$

The pressure is less than at $r_1$, at this location of higher velocity. even while the location is lower by $2.0 \ m$.

Flow rate throughout the pipe

Fluid flow rate is discussed on the Flow Rate page:

$f = A v$

Using the location at $r_1$:

$f = A v = \pi \ r_1^2 \ v_1 = \pi (0.12 m )^2 ( 6.0 m/s) = 0.27 \ m^3/s$

0.27 cubic meters, or 270 liters, of water passes any point along the pipe, per second.