[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
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