Clem... design and equations

Ken Carrigan ( (no email) )
Sat, 12 Sep 1998 10:17:13 -0400

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Just had to do this.. it's been bugging me as to how this exit force
from a centrfugal nozzle head could produce overunity. So here is
some math (sorry Engineer here) that will explain what a Normal
"sprinker head" would achieve in power. Forgot you said this.. but
the raduis squared does come into play.. however.. still do not
see where the "kick over" occurs for over unity.

OK.. make believe we have a sprinkler (twisted L shaped on each
side) which is feed water in the center of it. THis sprinker is turning
at 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 laminer
flow is occuring and no resistance is taken in to account (friction or
heating). What we want to find is the power produced from this
sprinkler. So..the moment around the z-axis.. will be the summations
of all (r xV)*p*V*A . Opps now "r" is radius, and "p" is density of=20
water. 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... is
zero cause the radius "r" is zero. Thus the equation reduces to only
one sum.. or (r x V)*m. Now we have to take into account for the
rotational velocity and the velocity of the zozzle exit jet... in to the
normal atmosphere. Now we can subsitute r X V =3D -r*(Ve - w*r).
Thus the moment applied to the system by the generator would
be -r*(Ve - W*r)*m. BUT the moment applied to the generator is the
negative of the moment of the generator on the arms. So the equation
becomes positive... no biggy!

The power now is simple...one we have the moment.. It is the moment
time the rotational velocity.. (M =3D moment) or M*w =3D P for Power!

For all who would like the in to one equation .. here...
P =3D (w*r*Ve - w*w*r*r)*m

ENSURE.. that proper mks units are used for example ... RPM has
to be changed to RPS.. etc etc.. and don;t forget 2*pi for rotational
velocity! =20

v/r Ken Carrigan

PS.. NOW... WHERE IS THE OVERUNITY!!!
Help me understand...

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Just had to do this.. it's been =bugging me as to=20how this exit force
from a centrfugal nozzle head could =produce=20overunity.  So here is
some math (sorry Engineer here) that =will=20explain what a Normal
"sprinker head" would achieve in =power.  Forgot=20you said this.. but
the raduis squared does come into play.. however.. =still do=20not
see where the "kick over" occurs for over=20unity.
 
OK.. make believe we have a sprinkler (twisted L =shaped on=20each
side) which is feed water in the center of it.  =THis=20sprinker is turning
at a velocity of "w" and the length of the =arms (in=20the L forms) are
"L".  The exit velocity realitve to =the nozzles=20are "Ve" for exit velocity,
and the area of the exit nozzle is ="A".  I=20think that now covers it.
Assumptions have to be made now.. and we concider =that=20laminer
flow is occuring and no resistance is taken in to =account=20(friction or
heating).  What we want to find is the power =produced=20from this
sprinkler.  So..the moment around the z-axis.. =will be=20the summations
of all (r xV)*p*V*A .  Opps now "r" =is radius,=20and "p" is density of
water.  p*V*A is called the mass efflus of the =nozzle..=20or lets say "m"
for now.  The moment going into the hub.. =enterance of=20water... is
zero cause the radius "r" is zero.  =Thus the=20equation reduces to only
one sum.. or (r x V)*m.  Now we have to take =into account=20for the
rotational velocity and the velocity of the zozzle =exit jet...=20in to the
normal atmosphere.  Now we can subsitute r X V ==3D -r*(Ve -=20w*r).
Thus the moment applied to the system by the =generator=20would
be -r*(Ve - W*r)*m.   BUT the moment =applied to the=20generator is the
negative of the moment of the generator on the =arms.  So=20the equation
becomes positive... no biggy!
 
The power now is simple...one we have the moment.. =It is the=20moment
time the rotational velocity.. (M =3D moment) or M*w ==3D P =20for Power!
 
For all who would like the in to one equation ..=20here...
P =3D (w*r*Ve - w*w*r*r)*m
 
ENSURE.. that proper mks units are used for example =.... RPM=20has
to be changed to RPS.. etc etc.. and don;t forget =2*pi for=20rotational
velocity! 
 
v/r Ken Carrigan
 
PS.. NOW... WHERE IS THE OVERUNITY!!!
Help me understand...
 
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