Re: Re: Tip nozzle paradox

[email protected] wrote:
[email protected] wrote:[email protected] wrote:

> But there is no apparent reason for that tip velocity to equal the 
> exhaust velocity "v", and I should think the ultimate tip velocity 
> would be very much higher.   
>
Sadly not.

The fuel of mass 'dm' must have been accelerated by an overall velocity 
increment of V to enter the engine (actually slightly less, but typical 
flow rates are only a few meters/second typically, so I will neglect 
it). This gives it a momentum increment of dm V.

The mass then leaves the nozzle with a velocity dm Ve. For the tip to 
have reached a maximum speed, the linear momenta balance:

dm V = dm Ve

i.e. V = Ve

(You can complicate things by considering angular momentum, but the 
result is the same- the above linear momentum relationship is for in the 
instantaneous frame of reference of the tip.)

 From an energy point of view:

Let the kinetic energy of the exhaust be 0.5 dm Ve2. But the rocket 
nozzle is not 100% efficient, with an efficiency of e. For example 0.9. 
So the overall energy of the exhaust shall be 1/e * 0.5 dm Ve2, 
including thermal energy of the exhaust.

Let the propellent energy be dm P.

Then for energy conservation (again in the instantaneous frame of 
reference of the tip.):

0.5 dm V2 + dm P =  1/e*0.5 dm Ve2

But V = Ve, so putting this in and rearranging:

V = sqr(2P e/(1-e))

So that is a closed form estimate for the final speed of the rotor. Note 
that e mostly depends on the expansion ratio of the nozzle; and in any 
case is generally pretty high. If we put it to 0.9 (90%) we get:

V = 3 sqr(2P)

The energy of a kg of propellent is generally in the multiple 
megajoules, so the rotation speed is likely to be rather excessive :-)

> The kinetic energy of the exhaust in the inertial frame should equal 
> the chemical energy of the propellant-which, at a large distance, 
> would be apparent as a positive radial velocity of propellant mass 
> increments moving outward in spiral waves.  This seems not that much 
> different from what happens in a lawn sprinkler, except that some of 
> the exhaust energy comes from chemical reactions rather than feed 
> pressure.  
>
I believe that it is the friction, air drag and viscosity losses that 
cause the water to radiate out; in the lossless case the water 
presumably would simply fall to the ground.

> --Best, Gerald
> _______________________________________________
> ERPS-list mailing list
> [email protected]
> http://lists.erps.org/mailman/listinfo/erps-list
>
>  
>