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I am doing an extended essay on Terminal Velocity and I need the derivation for the drag force equation: 1/2*C*A*P*v^2

Can you please post the derivation for me?

Thanks a lot.

Tsering

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I am doing an extended essay on Terminal Velocity and I need the derivation for the drag force equation: 1/2*C*A*P*v^2

Can you please post the derivation for me?

Thanks a lot.

Tsering

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Clausius2

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The coefficient of Drag is defined as:

[tex] C_D=\frac{P}{1/2 \rho U^2} [/tex]

Of course, the left term is COUPLED with the flow around the body considered. I mean, to obtain Cd, you'll have to solve the field of Pressures P around the body. And believe me, that's not a simple task.

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Andrew Mason

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The drag force equation is a constructive theory based on the experimental evidence that drag force is proportional to the square of the speed, the air density and the effective drag surface area. The drag coefficient is simply the proportionality constant that relates drag force to these factors for a given object, so there is no way to 'derive' it other than by experiment.tnorkhangpa said:Hi Guys,

I am doing an extended essay on Terminal Velocity and I need the derivation for the drag force equation: 1/2*C*A*P*v^2

AM

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But there is also the theory of the linear dependency of drag force on the velocity of the falling object.

Can you explain that bit as well?

Tsering

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Clausius2

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There is neither linear dependence on velocity, nor cuadratic dependence on velocity. In fact, the drag coefficient is not constant at all. The dependence of drag force on velocity can be much more complex. In usual industrial problems the drag coefficient depends on non dimensional parameters which are a function of the geometry, velocity, flow physical properties and time.

However, most of physics books do not tell nothing about this complexity, but if you advance in the study of aerodynamics you will hear again about what I have just told you.

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Andrew Mason

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For small velocities, the drag force is roughly proportional to velocity. Look up Stokes law. But, as Clausius says, it is just a rough approximation.tnorkhangpa said:

But there is also the theory of the linear dependency of drag force on the velocity of the falling object.

Can you explain that bit as well?

Tsering

If dF/dv = constant, k, then [itex]dF/dt = dF/dv * dv/dt = ka = Fk/m[/itex]. So F must be an exponential function of time ([itex]F \propto v \propto c + e^{kt/m}[/itex])

Once the drag force reaches the weight of the object, there can be no further acceleration and the object falls at constant speed - terminal velocity, which is the asymptote of the velocity-time graph.

AM

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Clausius2

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Could you write here the derivation?. I don't believe that. I know there is a way to relate Drag force and free stream velocity in the case of two dimensional incompressible and steady flow, being the proportional scalar a function of the velocity profile downstream.SergejVictorov said:

In industrial problems sure it doesn't work. There is only a few simplified cases where Cd can be derived as I posted above.

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Let's start with the kinetic energy of the fluid in front of the body:

[tex] E=\frac{1}{2}m v^2 [/tex]

The mass is given by the area times an arbitrary length times the density:

[tex] m=A s \rho [/tex]

[tex] E=\frac{1}{2}A s \rho v^2 [/tex]

Since work or energy is also:

[tex] E=F s \rightarrow F=\frac{E}{s} [/tex]

It follows that:

[tex] F=\frac{1}{2}A \rho v^2 [/tex]

It's even easier if you take the dynamic pressure,

you can simply multiply by the frontal area to get the force.

[tex] p_{dyn}=\frac{1}{2}\rho v^2 [/tex]

You plug in a coefficient because you don't want all of the force to act on the body.

[tex] F_{drag}=p_{dyn} A C_D = \frac{1}{2}\rho A C_D v^2 [/tex]

As stated before, this is only a very rough approximation. It is for turbulent flows past a free body with high Reynolds numbers. It uses the fact that the fluid must be displaced.

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Clausius2

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SergejVictorov said:

Let's start with the kinetic energy of the fluid in front of the body:

[tex] E=\frac{1}{2}m v^2 [/tex]

The mass is given by the area times an arbitrary length times the density:

[tex] m=A s \rho [/tex]

[tex] E=\frac{1}{2}A s \rho v^2 [/tex]

Since work or energy is also:

[tex] E=F s \rightarrow F=\frac{E}{s} [/tex]

It follows that:

[tex] F=\frac{1}{2}A \rho v^2 [/tex]

It's even easier if you take the dynamic pressure,

you can simply multiply by the frontal area to get the force.

[tex] p_{dyn}=\frac{1}{2}\rho v^2 [/tex]

You plug in a coefficient because you don't want all of the force to act on the body.

[tex] F_{drag}=p_{dyn} A C_D = \frac{1}{2}\rho A C_D v^2 [/tex]

As stated before, this is only a very rough approximation. It is for turbulent flows past a free body with high Reynolds numbers. It uses the fact that the fluid must be displaced.

:surprised :surprised And you call this a DERIVATION??????

If some PF mentor read this, please for the sake of the forum delete that post.

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FredGarvin

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[tex] D = \int dF_{x} = \int p cos \theta dA + \int \tau_{w} sin \theta dA [/tex]

Where:

[tex] \theta [/tex] is a function of the geometry of the object

[tex] \tau_{w}[/tex] is the wall shear stress distribution

[tex]p[/tex] is the normal stresses due to pressure distribution

It essentially boils down to two P/A components added together. The tough part is in describing the distributions.

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Daniel.

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arildno

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We can even make a graph out of this!

However, as Clausius2 has pointed out, the functional dependency of the coefficient (in, say, the velocity) is, in general impossible to gain from analytical considerations.

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Clausius2

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I'm not a professor. But if you don't know nothing about geometry and ask me what is the shape of the earth, I'd never say "plane". I'd say: "you only should know the shape is a bit more complex", without giving a mathematical definition of an sphere.dextercioby said:particlemechanics...What formulas /what derivation would u use...?

Daniel.

Once Einstein said that a good physicist must know how to explain his achievements to every people no matter what previous knowledge about the stuff they have. Sometimes I have to admit I am unable to do that, because I do not agree with Einstein (this is the unique thing I do not agree with him by the way ) and because there are some things which cannot be explained on an easy way. If you try to reduce complex subjects to easy ones, surely you will find yourself lying.

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Daniel.

P.S.Einstein knew what he was talking about...

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arildno

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As examples:

In the Stokes flow regime (Re<<1) for a sphere, the friction coefficient goes as the inverse velocity (giving the familiar linear force/velocity relationship), whereas the drag crisis in turbulent flow represents an upper limit of the regime where the friction coefficient is roughly constant.

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arildno

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Clausius2

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Some elementary proofs are false, due to its "elementarity".dextercioby said:

Daniel.

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I've said this before:Cocky attitude and bad grammar don't go well together.Clausius2 said:Some elementary proofs are false, due to its "elementarity".

Daniel.

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