Hi! Ken: Let's try an Arthur C. Clark's approach as in 2001 Space
Oddesy to our
little puzzle. Let's assume that we are on a hugh space station that
has been spun
up to an RPM of rotation that provides exactly one G at the outer rim of
the station.
Let us have an artificial environment with a river of water around the
station. We know
that it only takes an initial amount of thrust to spin the station up to
the RPM needed
for the G effect we need. Now we have a small meteorite strike the
station at a
tangential angle and penetrates the hull into our artificial river
around the outer edge
of the hull. Water is now jetting out into space at a tangent to the
hull and providing
thrust which in turn is causing an acceleration of our spin rate. The G
forces are multiplying rapidly now and the pressures at the tangential
hull penetration are
increasing. Now this effect is going to go exponential because it is
directly related
to the velocity squared. It will only require a short period of time
before the space
station will suffer a catastrophic centrifugal structural failure. We
are all pinned to
the floor of the station by the multiplying G forces and are helpless to
prevent the
destruction. Now try your math on this situation. Norm
Ken Carrigan wrote:
> Just had to do this.. it's been bugging me as to how this exit
> forcefrom a centrfugal nozzle head could produce overunity. So here
> issome math (sorry Engineer here) that will explain what a
> Normal"sprinker head" would achieve in power. Forgot you said this..
> butthe raduis squared does come into play.. however.. still do notsee
> where the "kick over" occurs for over unity. OK.. make believe we have
> a sprinkler (twisted L shaped on eachside) which is feed water in the
> center of it. THis sprinker is turningat a velocity of "w" and the
> length of the arms (in the L forms) are"L". The exit velocity
> realitve to the nozzles are "Ve" for exit velocity,and the area of the
> exit nozzle is "A". I think that now covers it.Assumptions have to be
> made now.. and we concider that laminerflow is occuring and no
> resistance is taken in to account (friction orheating). What we want
> to find is the power produced from thissprinkler. So..the moment
> around the z-axis.. will be the summationsof all (r xV)*p*V*A . Opps
> now "r" is radius, and "p" is density ofwater. p*V*A is called the
> mass efflus of the nozzle.. or lets say "m"for now. The moment going
> into the hub.. enterance of water... iszero cause the radius "r" is
> zero. Thus the equation reduces to onlyone sum.. or (r x V)*m. Now
> we have to take into account for therotational velocity and the
> velocity of the zozzle exit jet... in to thenormal atmosphere. Now we
> can subsitute r X V = -r*(Ve - w*r).Thus the moment applied to the
> system by the generator wouldbe -r*(Ve - W*r)*m. BUT the moment
> applied to the generator is thenegative of the moment of the generator
> on the arms. So the equationbecomes positive... no biggy! The power
> now is simple...one we have the moment.. It is the momenttime the
> rotational velocity.. (M = moment) or M*w = P for Power! For all who
> would like the in to one equation .. here...P = (w*r*Ve -
> w*w*r*r)*m ENSURE.. that proper mks units are used for example ... RPM
> hasto be changed to RPS.. etc etc.. and don;t forget 2*pi for
> rotationalvelocity! v/r Ken Carrigan PS.. NOW... WHERE IS THE
> OVERUNITY!!!Help me understand...
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Hi! Ken: Let's try an Arthur C. Clark's approach as in 2001 SpaceOddesy to our
Ken Carrigan wrote:
Just hadto do this.. it's been bugging me as to how this exit forcefroma centrfugal nozzle head could produce overunity. So here issomemath (sorry Engineer here) that will explain what a Normal"sprinkerhead" would achieve in power. Forgot you said this.. buttheraduis squared does come into play.. however.. still do notseewhere the "kick over" occurs for over unity. OK..make believe we have a sprinkler (twisted L shaped on eachside)which is feed water in the center of it. THis sprinker is turningata velocity of "w" and the length of the arms (in the L forms) are"L". The exit velocity realitve to the nozzles are "Ve" for exit velocity,andthe area of the exit nozzle is "A". I think that now covers it.Assumptionshave to be made now.. and we concider that laminerflowis occuring and no resistance is taken in to account (friction orheating). What we want to find is the power produced from thissprinkler. So..the moment around the z-axis.. will be the summationsofall (r xV)*p*V*A . Opps now "r" is radius, and "p" is density ofwater. p*V*A is called the mass efflus of the nozzle.. or lets say "m"fornow. The moment going into the hub.. enterance of water... iszerocause the radius "r" is zero. Thus the equation reduces to onlyonesum.. or (r x V)*m. Now we have to take into account for therotationalvelocity and the velocity of the zozzle exit jet... in to thenormalatmosphere. Now we can subsitute r X V = -r*(Ve - w*r).Thusthe moment applied to the system by the generator wouldbe-r*(Ve - W*r)*m. BUT the moment applied to the generator isthenegative of the moment of the generator on thearms. So the equationbecomes positive... nobiggy! The power now is simple...one we havethe moment.. It is the momenttime the rotational velocity..(M = moment) or M*w = P for Power! Forall who would like the in to one equation .. here...P= (w*r*Ve - w*w*r*r)*m ENSURE.. that proper mksunits are used for example ... RPM hasto be changedto RPS.. etc etc.. and don;t forget 2*pi for rotationalvelocity! v/rKen Carrigan PS.. NOW... WHERE IS THE OVERUNITY!!!Helpme understand...
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