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algol13
4 years ago
15

The continuity equation shows that the ratio of fluid velocities within different openings is inversely proportional to the cros

s-sectional areas of those openings. Provide one example where this might be useful for finding differences in areas after measuring velocities, or in finding differences in velocities after measuring areas. Describe your example in detail making sure to talk about the different variables involved.
Physics
1 answer:
Stolb23 [73]4 years ago
7 0

Answer:

A typical example of where the continuity equation can be applied, is in finding the cross sectional area of a laminar falling stream of water from a pipe at an elevation, at the point where it hits the ground.

The speed of the water through the discharge opening of the pipe can be measured with a flow-meter, and the cross sectional opening of the pipe is known.

As the water leaves the pipe, and falls towards the ground, the water stream is accelerated under gravity, and the speed of the water stream increase. The increase in the speed of the water stream causes the cross sectional area to decrease in obedience of the continuity law and equation. If the speed of can be measured, then the area of the water stream just before the water touches the ground  can be calculated from the continuity equation.

If the speed of the water stream can not be measured, then it can be calculated using Newton' equation of motion

v^{2} = u^{2} + 2gz

v = \sqrt{u^{2} + 2gz}

where

v = the velocity of the water stream at the point where it hits the ground

u = the velocity of the water at the pipe discharge (measured with a flow meter in the pipe)

g = acceleration due to gravity

z = the elevation of the pipe above the ground.

After calculating the speed of the water stream at the point where it hits the ground, the cross sectional area of the the stream can be calculated from the continuity equation.

uA = va

where

u and v are the velocities of the water stream at the pipe discharge and at the ground respectively.

A = the cross sectional area of the pipe

a = the area of the water stream before it hits the ground.

the equation is simplified as

a = \frac{uA}{v}

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The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

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We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

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This net mass is the largest mass of cargo that the balloon can lift.

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