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Author Topic: Update on the Implosion motor project.  (Read 1070 times)

Offline TommeyLReed

  • Hero Member
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  • Posts: 511
Update on the Implosion motor project.
« on: April 22, 2021, 04:34:34 PM »
Hello and happy New Year to all,
 I've been really busy working on this project for many years now. With many unsatisfied out come, make me look at this project in a different way. Theories without building a prototype is and will always be foolish thinking. This is why I believe many people are just wasting time of others to comment.
Being a honest person won't make me stand out like a rock star, but my experience in building and testing stuff will always have a bigger impact then any theories out there.
I really enjoy experimenting and building stuff, this Implosion motor project also known as," Clem Engine"; is very interesting,challenging and fun to build. I have learn so much working with this theory of creating a type of a water spout vortex motor where it keeps me energized like a young man.

I believe testing and data is very important. It also gives you a better understand of what's really going on.
This centrifugal impeller design I'm working on is the main reason I keep improving in my build.https://www.youtube.com/watch?v=WUY_T4xiLtM&t=24s
This deal with fluid mechanics and centrifugal forces. When things rotate odd things start to change flow and pressure of fluid.
Some interesting math is involve.
Centrifugal force : Fc=mv^2/rfluid jet thrust : 1.57*psi*dia*diaflow rate: .408*gpm/dia^2

This impeller is 18" diameter with 8 jets, size will change thrust/gpm and pressure needed.
For example:

jet size of 0.125 at a pressure of 500 psi with produce a thrust of 1.57*500*0.125*0.125 =12.26lb/force.using 2 jets, a total force generated is 2x12.26=24.52lb/force.
This seems like a lot of force, and it is. But, without flow we wouldn't have a clue of how much energy is needed to generate that pressure and velocity
velocity: \/32+32*(500psi/0.43) = 192.98ft/sec
18" impeller at peak thrust would rotate it at at:(18"*pi)/12 = 4,71 ft/circumfrence192.98*60=11,578.8 ft/min11,578.8/4.71 = 2,458.34 rpm's
at 2,458.34 rpm's no power is being produced, in fact at 100% this can't be calculated without a load.
At peak rpm's thrust is about nothing due to the fact rotation is moving the same speed, if not a little lower then thrust velocity due air and friction drag.
Now, what if we put it at a 50% load?What does this do?First of all now the rpm's is at 1,229.17 and thrust is now 1/2 of total peak 24.52lb/thrust.
To calculate hp we must first find ft/lb of torque.
18"/2 =9" or .75 foot.(24.52/2)*.75 =9.195ft/lb1,229x9.195/5252 = 2.15hp

Now we need to find power input needed to produce psi and gpm of fluid.
0.408* gpm/dia^20.408* ?/(0.125*2)^2...... GPM?0.408/ (192.98/2)=\/236.4950x(0.25^2) = 14.8709375 gpm.408*(14.7809375/.25^2)= 96.72ft/sec



(14.7809375*500 psi)/1714 = 4.31 hp needed to make 2.15 hp.
But my experiments shows far less hp is needed due to centrifugal force being added, and should show OverUnity!

centrifugal calculated.http://www.calctool.org/CALC/phys/newtonian/centrifugal
18" x 1.5" in height has a area of (9"^2Xpi)x1.5") = 0.22 cu/ft or about 1.65 gal of hydraulic fluid or 7.2lb*1.65 = 11.88lb
centrifugal force generated at 1229 rpm's with a mass fluid of 11.88lb would generate 4587 pound of outward fluid force.
In fact my data shows as the rotate increase pressure decrease.