The Bernoulli Principle was formulated by Bernoulli, who obtained a relation between the pressure and velocity at different parts of a moving incompressible fluid. If the viscosity is negligibly small, there are no frictional forces to overcome. Hence the work done by the pressure difference per unit volume of a fluid flowing along a pipe steadily is equal to the gain in kinetic energy per unit volume plus the gain in potential energy per unit volume according to The Bernoulli Principle. Engineering in Kenya has more information.
Mathematical Definition of The Bernoulli Principle
The work done by pressure in moving a fluid through a distance = force × distance moved = pressure × area × distance moved = pressure × volume. At the beginning of the pipe where the pressure is P1, the work done per unit volume on the fluid according to The Bernoulli Principle is P1. At the other end, if the pressure is P2, then the work done per unit volume according to The Bernoulli Principle is P2.
Hence the net work done per unit volume is the difference of P2 and P1. The kinetic energy per unit volume = ½mass per unit × velocity2 = ½ρ × velocity2 where ρ is the density of the fluid. Thus if v1 and v2 are the final and initial velocities at the end and the beginning of the pipe respectively, then the kinetic energy gained per unit volume = ½ρ(v22 – v12).
Further, if h2 and h1 are the respective heights measured from a fixed level at the end and the beginning of the pipe respectively, the potential energy gained per unit volume according to The Bernoulli Principle is = mass per unit volume × g × (h2 – h1) = ρg(h2 – h1). Thus from the conservation of energy.
P1 + ½ρv12 + h1ρg = P2 + ½ρv22 + h2ρg.
P + ½ρv2 + hρg = constant where P is the pressure at any part and v is the velocity at that part. Hence for streamline motion of an incompressible non-viscous fluid, The Bernoulli Principle is proven and it states that; the sum of the pressure at any part plus the kinetic energy per unit volume plus the potential energy per unit volume is always a constant. The Bernoulli Principle shows that at points in a moving fluid where the potential energy change hρg is very small, or zero as in flows through a horizontal pipe, the pressure is low where the velocity is high; conversely, the pressure is high where the velocity is low according to The Bernoulli’s Principle.
E.g. as a numerical illustration, suppose the area of cross-section A1 of x in the figure above is 4cm2, the area A2 of y is 1cm2 and water flows past each section in laminar (streamline) flow at the rate of 400cm3s-1, then;
At x, speed v1 of water = volume per second/area = 400cm2s-1/4cm2 = 100cms-1 = 1ms-1.
At y, speed v2 of water = (400cm3s-1/1cm2)400cms-1 = 4ms-1.
Since the density of water is ρ = 1000 kgm-3, then the pressure difference P is;
P = ½ρ (v22 – v12) = ½ × 1000 (42 – 12) = 7.5 × 103 Nm-2.
P = hρg → h = P/ρg = (7.5 × 103)/ (1000×9.8) ≈ 0.77m
The pressure head is thus equivalent to 0.77m of water.
Application of The Bernoulli Principle
- A suction effect is experienced by a person standing close to the platform at a station when a fast train passes. The fast moving air between the person and the train produces a decrease in pressure and excess air pressure at the other side may push the pedestrian towards the train. This is a fatal case of The Bernoulli Principle.
- Filter pump. A filter pump has a narrow section in the middle, so that a jet of water from the tap flows faster here. This causes a drop in pressure near it and air therefore flows in from the side tube to which a vessel is connected. The air and water together are expelled through the bottom of the filter pump.
- Aerofoil lift. The curved shape of an aerofoil creates a faster flow of air over its top surface than the lower one. This is shown by the closeness of the streamlines above the aerofoil compared to those below. From The Bernoulli’s Principle, the pressure of the air below is greater than that above, and this produces the lift on the aerofoil.
E.g. water flows steadily along a horizontal pipe at a volume rate of 8 ×10-2m3s-1. The area of cross-section of the pipe is 40cm2; calculate the flow of velocity of water. Find the total pressure in the pipe if the static pressure in the horizontal pipe is 3 × 104 Pa, assuming the water is incompressible, non-viscous and its density is 1000kgm-3. What is the new flow velocity if the total pressure is 3.6 × 104Pa?
Solution: – velocity of water = volume per second/ area = (8 × 10-3)/ (40 × 10-4) = 2ms-1.
- Total pressure = static pressure + ½ρv2 = 3 × 104 + (1000 ×22)/ 2 = 3.2 × 104 Pa.
- ½ρv2 = total pressure – static pressure.
- ½ × 1000 × v2 = 3.6 × 104 – 3 × 104 = 0.6 × 104.
- V = √ [(0.6 × 104)/ 500] = 3.5ms-1.
Conclusion on The Bernoulli Principle
Probably one of the most important principles in physics, The Bernoulli Principle is applied in many areas. A streamline flow, a non-viscous fluid (liquid or gas) and the fluid being incompressible are the requirements for The Bernoulli Principle to work effectively. Eddy currents, which form when the flow is not laminar/ streamline don not work for The Bernoulli Principle.