Bernoulli's Principle

Bernoulli’s Principle states that for an incompressible, non-viscous fluid in steady flow, the total mechanical energy per unit volume remains constant. This means as the speed of the fluid increases, the pressure decreases, and vice versa.

Mathematically, it is denoted as:

\[ P + \frac{1}{2} \rho v^2 + \rho g h = k \]

\(P\): Pressure
\(\rho\): Fluid density
\(v\): Fluid velocity
\(g\): Acceleration due to gravity
\(h\): Height above reference level

Each term represents:

\(P\): Pressure energy
\(\frac{1}{2} \rho v^2\): Kinetic energy per unit volume
\(\rho g h\): Gravitational potential energy per unit volume

Conditions for Validity

Incompressible fluid (density remains constant)
Non-viscous (no internal friction)
Steady flow (velocity at each point doesn’t change with time)
Along a streamline

Continuity Equation

The continuity equation states that the mass flow rate of a fluid remains constant throughout a streamline. This is expressed as:

\[ A_1 v_1 = A_2 v_2 \]

Where \(A\) is the cross-sectional area and \(v\) is the fluid velocity at that section.

Water flows through a pipe narrowing from a diameter of 10 cm to 5 cm. If the speed in the wider section is 2 m/s, what is the speed in the narrower section?

\[ A_1 v_1 = A_2 v_2 \Rightarrow \pi r_1^2 v_1 = \pi r_2^2 v_2 \]

\[ v_2 = \frac{r_1^2}{r_2^2} v_1 = \left(\frac{0.05^2}{0.025^2}\right) \cdot 2 = 8\, \text{m/s} \]

Water enters a pipe at a speed of 2 m/s and pressure of 200 kPa. At a higher point, the pipe narrows, and the velocity increases to 5 m/s. Assuming height difference of 3 m, find the pressure at the higher point.

Use: \( P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \)

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

Substitute values : \(P_1 = 200\, \text{kPa},\: \rho = 1000\:\text{kg} \text{m}^{-3},\: v_1 = 2\, \text{ms}^{-1},\: v_2 = 5\, \text{ms}^{-1}, \: (h_1 - h_2) = -3m \)

\(P_2 = 160,100\:Pa\)

Written by Thenura Dilruk