In this lecture the following are introduced:

•The Venturi tube
• The Pitot tube
• The Prandtl tube
• Aerofoils
• Archimedes' principle

Applications of Bernoulli's Equation and the Equation of Continuity

The equation of continuity says that when flow area decreases then the fluid speed increases. Bernoulli's equation says that when the speed of a fluid increases the pressure drops. Thus a decrease in flow area means a decrease in pressure.

Venturi Meter Application

The Venturi meter measures the flow speed from the pressure drop at a constriction in a pipe.

Example
A pipeline carries water with density 1000 kg.m^-3. The difference in pressure betweeen the main pipeline and the throat of a Venturi meter is 9 kPa. The areas of the main pipe and the throat are 0.04m² and 0.02m².
(i) Use Bernoulli's equation, to show that the speed of the water in the main pipe is given by:

.

Where Δp is the pressure difference, ρ is the density, and v_th is the velocity of water in the throat.

(ii) Apply the equation of continuity to show that the speed of the water in the main pipe is also given by:

(iii) Hence, find the speed of water in the main pipe.

answer (i)
answer (ii)
answer (iii)

Note that the pressure difference is measured by the difference in the heights of fluid in the columns and expressed in Pascal. Because this theory does not include frictional effects the calculated value is too large (by up to 10%).

The Pitot Tube Application

When you put a right angled glass tube into a flowing stream, the liquid in the tube rises above the surface of the stream. Henri Pitot (1695 - 1771) first used this application in 1732 to measure the speed of the Seine River at Paris and he developed what is now known as the Pitot tube.

The flow divides at point 2 which is the entrance to the tube, producing a stagnation point. The speed of the fluid at point 2 and inside the tube is thus zero (v₂ = 0). The speed of the fluid at point 1 is v₁, and the pressure here (which has to be a fair distance away) is then lower than at point 2 because the speed is not zero.

In what follows, the density of the flowing fluid is ρ, and the density of the manometer fluid (in red) is ρ'. Heights are measured from the base line AB, where the pressures P_A and P_B are equal. The manometer fluid stands a height h above the base line, and the tube entrance is a height y₁ above that. In Bernoulli's equation, points 1 and 2 are at the same height, so the heights cancel and v₂ = 0, this gives:

The approximation is fine when using Mercury because its density is much greater than commonly used fluids.

The Prandtl Tube Application

The Prandtl tube is a modification of the Pitot tube, using two co-axial cylinders. The outer cylinder has a series of holes around it. It is also called a Pitot-Static tube. The same maths applies as above.

The Aerofoil Application

Part of the reason why an aeroplanes fly is due to the aerofoil shape of their wings. The air moves more quickly over the upper surface than the lower surface thus creating a net upward pressure. That this is not the whole story is obvious because aeroplanes can fly upside down!

If the angle of attack is too great then turbulence is generated (as shown below right), Bernoulli's equation fails to hold, and the wing stalls.

  

Archimedes and Hiero's crown

When a gold crown was made for Hiero, he asked that Archimedes determine whether it was pure gold or mixed with silver. When he was meditating on the problem in the public baths, he allegedly hit on the solution, shouted 'ευρηκα, and dashed home naked.

The (alleged) solution was to put a weight of gold equal to the crown, and known to be pure, into a bowl which was filled with water to the brim. Then the gold would be removed and the king's crown put in, in its place. An alloy of lighter silver would increase the volume of the crown and cause the bowl to overflow. The goldsmith was (allegedly) beheaded when the crown was found to be not pure gold.

Chris Rorres, at Drexel University, criticised this solution, because:
• it doesn't use Archimedes' principle, (the apparent loss in weight equals the weight of fluid displaced) and
• could not have been done with the precision of their instruments.

His solution to Archimedes and Hiero's crown

Archimedes' principle by considering pressures

Take a mass with constant cross-sectional area, floating partially submerged in water.
For equilibrium, the weight and force of the air pressure downwards, are balanced by the upward force from the water pressure.
Since it is floating, it has lost all of its weight. There is then an upward force balancing weight and air pressure. The upward bouyancy force is called the Archimedean upthrust.

The Archimedian upthrust (or buoyancy) of a partially or fully submerged object is equal to the weight of fluid displaced by the object.

Example 1:
An object of mass 0.5 kg is made from material of density 4000 kg.m^-3 and suspended by a string so that it is totally immersed in a liquid of density 1500 kg.m^-3. Find the tension in the string

 

Example 2:
An iceberg of density 917 kg.m^-3 and regular cross-sectional area floats in salt water of density 1030 kg.m^-3. Find the fraction that is submerged

 

Example 3:
A frog in a hemispherical cockleshell finds that she just floats without sinking in a pea-green sea of density 1350 kg.m^-3. Given that the cockleshell has a radius of 60mm and has negligible mass, find the mass of the frog.

Example 4:
The supertanker Globtik London has a mass of 2.2x10⁸ kg when empty and can carry 0.5x10⁶ m³ of oil with density 875 kg m^-3. The density of sea water can be taken to be 1020 kg m^-3. Assume that the shape is a rectangular prism 380 m long, 60 m wide, and 40 m high. Find how deep the hull is submerged in the water.

Example 5
Sea water has a density 1025 kg.m^-3. A submarine has a mass of 22.5x10⁶kg when floating in sea water so that 10% of its volume is above the water. Find the mass of water that must be taken into its tanks so that it can fully submerge.

on the surface:
weight of submarine plus empty tank =weight of water displaced
22.5 x 10⁶ x g = (0.9 x volume of sub) x 1025 x g
divide by g
22.5 x 10⁶ = 0.9 x volume of sub x 1025
volume of sub x 1025 = 22.5 x 10⁶ /0.9 = 25 x 10⁶ kg

under water:
weight of submarine plus water in tank =weight of water displaced
(22.5 x 10⁶ + mass of water) x g = volume of sub x 1025 x g
divide by g
22.5 x 10⁶ + mass of water = 25 x 10⁶
mass of water = (25 - 22.5) x 10⁶
                      = 2.5 x 10⁶ kg

Example 6:
A person can walk on water (density 1000 kg.m^-3 ) if they wear very big shoes. A pair of shoes are shaped like a rectangular boxes. Each shoe weighs 2 kg and is 200 mm wide and 300 mm deep. If a person of mass 80 kg wearing these shoes, slides on water so that the top of the shoes are level with the surface of the water, find the length of shoes needed.

weight of water displaced = weight of person + shoes
2 x (length x 0.2 x 0.3) x 1000 x 9.8 = (80 + 2 x 2) x 9.8
divide by g
120 x length = 84
length = 0.7m

Summarising:

Bernoulli's equation can be applied to many situations where there is streamline flow.
e.g. the Venturi tube, the Pitot and Prandtl tubes, and aerofoils.
Archimedes' principle says that the apparent loss in weight of a body partially or totally immersed in a fluid is equal to the weight of fluid displaced.