Newton's second law as an introduction to Betz's law

T

TomiSlav

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I am trying to derive the Betz factor from scratch and I did it but I just assumed the intial part as being right and now I have some questions about the first step. I have succesfully derived the Betz factor being 0,593 but the initial part, that includes Newton's 2nd law states that:
[math]F = ma = m\frac{\mathrm{d} v}{\mathrm{d} t}=\dot{m}\Delta v=\dot{m}(v_1{}-v_2{})=\rho Sv(v_1{}-v_2{})[/math]
I've never seen someone state that [math]m\frac{\mathrm{d} v}{\mathrm{d} t}=\dot{m}\Delta v[/math] and I don't know what it represents neither mathematically nor physically. Later on from that the maths is very simple, you just substitute from one equation to another, get the final equation, differentiate it by speed ratio, set the differentiated equation equal to zero and find the function maximum which is 0,593. I won't write the whole process here but you can find it on: This link under "Application of conservation of mass (continuity equation)".
It would be very helpful and a piece of crucial information if someone could explain the part I wrote up there to me a bit more into details.


Kind regards and thank you all in advance,


T.
 
TomiSlav said:
I've never seen someone state that mdvdt=m˙Δvm\frac{\mathrm{d} v}{\mathrm{d} t}=\dot{m}\Delta vmdtdv=m˙Δv
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I've never seen such thing either -- where did you get it from?
TomiSlav said:
It would be very helpful and a piece of crucial information if someone could explain the part I wrote up there to me a bit more into details.
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The stuff you posted makes no sense to me. Same question: where does it come from?
 
They are playing a bit fast and loose.

Here is what I believe the reasoning is for the start of the derivation. m is representing a small mass element, which I would rather call [imath]\delta m[/imath] or something. It isn't quite a differential, but it is macroscopically small.

[imath]\dot{m} \equiv \dfrac{m}{dt}[/imath]
representing a mass flow rate.

Since m is small, we may assume (with the other assumptions holding) that Newton's 2nd approximately holds. Then
[imath]F = ma = m \dfrac{dv}{dt} \approx m \dfrac{v_2 - v_1}{t_2 - t_1}[/imath]

[imath]F = \dfrac{m}{t_2 - t_1} ( v_2 - v_1) \approx \dot{m} ( v_2 - v_1 )[/imath]

There are two approximations here, one where
[imath]a \approx \dfrac{ \Delta v}{ \Delta t}[/imath]

and the other where
[imath]\dot{m} \approx \dfrac{m}{ t_2 - t_1 }[/imath]

Perhaps the two approximations together make something more palatable, but I would think that these would generate errors. Frankly, I should think that we ought to throw away Newton's 2nd for Navier-Stokes, but it could be that these particular assumptions makes it work.

-Dan

Addendum: Looking at it again, I think my preference of notation would fix a lot of confusion. Instead of a derivative for the acceleration, use
[imath]F \approx \delta m \dfrac{ \delta v}{ \delta t}[/imath]

Now all of the approximation is all in one place and the notation is less fuzzy.
 
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topsquark said:
Since m is small, we may assume (with the other assumptions holding) that Newton's 2nd approximately holds. Then
[imath]F = ma = m \dfrac{dv}{dt} \approx m \dfrac{v_2 - v_1}{t_2 - t_1}[/imath]

[imath]F = \dfrac{m}{t_2 - t_1} ( v_2 - v_1) \approx \dot{m} ( v_2 - v_1 )[/imath]
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Yeah, I think this part can hold, I just never thought of doing it that way. Thank you very much.
 
blamocur said:
Thanks. Somehow missed it the first time. And it doesn't help me much either. The derivations on this page look slightly better to me, but the typesetting there is not nice.
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Hello. The derivations on the link you provided do look slightly better, I will now look into them a bit more right now and let you know. If you're still confused about what I was exactly stuck on, I was confused why can we state that [math]m\frac{\mathrm{d} v}{\mathrm{d} t}=\dot{m}\Delta v[/math] or more detailed, why we can state that mass multiplied by the time derivative of velocity is equal to mass flow (m with a dot, time derivative of mass) multiplied by velocity change. In general, why can we state that[math]x\frac{\mathrm{d} y}{\mathrm{d} t}=\dot{x}\Delta y[/math].
Thanks in advance for all future answers.
 
TomiSlav said:
Hello. The derivations on the link you provided do look slightly better, I will now look into them a bit more right now and let you know. If you're still confused about what I was exactly stuck on, I was confused why can we state that [math]m\frac{\mathrm{d} v}{\mathrm{d} t}=\dot{m}\Delta v[/math] or more detailed, why we can state that mass multiplied by the time derivative of velocity is equal to mass flow (m with a dot, time derivative of mass) multiplied by velocity change. In general, why can we state that[math]x\frac{\mathrm{d} y}{\mathrm{d} t}=\dot{x}\Delta y[/math].
Thanks in advance for all future answers.
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I wish I could help you more. Obviously, this is a known and established phenomenon and derivation, so it's got to have a logical interpretation. Unfortunately, I've never found fluid dynamics to be much of anything less that confusing! ?

-Dan
 
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