How Can Bernoulli's Equation Be Applied to Flight?

[adamlep14]

Homework Statement

i need to make a sample problem for a physics project using Bernoulli's equation, the real life situation it must be applied to is flight

Homework Equations

P+.5pv^2+pgy = constant

The Attempt at a Solution

this physics is too involved for my understanding. I've looked in countless textbooks and on many webpages for a sample to base mine on but to no avail.

help much appreciated!

 

[da_willem]

This is a double post, you shouldn't do that! But the physics of the Bernouilli equation is not very involved at all. You probagbly know that there is such a thing as energy (like that of moving boedies or heated objects etc.) and it is conserved. Now Bernoullis equation says that in the absence of friction the energy of a fluid can be approximated by three relevant terms:

P+.5\rho v^2+\rho gy = constant

The 2nd term is the kinetic (motion) energy of the fluid moving with a velocity v; you might know the expression .5 m v^2 for a single object with mass m. In this expression it is divided by the volume to yield the density \rho.

The third expression is the gravitational energy, i.e. the 'potential energy' an object has in a gravitational field. You might know the expression mgh, which is the same when you divide it again by the volume. When you release such an object this energy decreases and the body starts moving, so the kinetic energy increases. Assuming the first term Bernoullis equation implies that the increase in kinetic energy equals the decrease in gravitational 'potential' energy to yield a sum that remains constant.

The first term indicates the heat energy of the fluid (again divided by the volume to yield a description per volume) in terms of the pressure of the fluid.

Now pick a pressure P, a density \rho and a velocity (g=9,8m/s^2). Now Bernouillis law says that at a different point in space (in absence of friction) the sum of the three energies should be the same. E.g. it allows you to calculate the fluids velocity when at the other point all other quantities are known.

Good luck!

 

[l1nkh0gthr0b]

I am having an issue reconciling the units in the Bernoulli equation I find on the internet:

V^2/2 + P/p = k

V^2/2 units => ft^2/sec^2

P/p units => ft

It would appear that the V^2/2 kinetic term needs to be devided by g (ft/sec^2) but I find that in none of the equations as written, so it leaves a gap in my understanding. What am I missing here?

 

[Redbelly98]

P/p units => ft

This is not true.

What are you using for units of P and p? They refer to pressure and density, respectively.

 

[l1nkh0gthr0b]

units of P are #/ft^2;
units of p are #/ft^3;
then units of P/p are ft.

 

[Redbelly98]

Not quite.

For P, we use pounds-force.
For p, we use pounds-mass.

They are different units.

 

[l1nkh0gthr0b]

Yes, and the two are related by (g/gc), which makes them interchangable on the Earth's surface. For simplicity I omitted the conversion.