Increase efficiency of rotating shaft.

[jigar]

          swamishriji
       

I am mechanical engineering student . Here I am describing one mechanical engineering situation that is very useful in any kind of mechanical and energy consumed industries. It is not even in any applied mechanics book. I hope you will try to understand it and use it. If you find any difficulty to understand ,feel free to ask me.
we can increase efficiency of rotating shaft. we can lift mass using less energy by using simple mechanism. please take few minute to read .we can use this technology in power plants and transportation vehicles.
         The base of this theorey is force or weight is stationary and it can help to rotate pulleys . we can convert gravitational force into energy while the mass or weight is stationary.
           Click on Photos to View Figures(Energyefficiency and Energy efficiency1)

  The figure(energy efficiency) is in cross section. as shown in figure, there are two cases. in each case there are two pulleys of same diameter. each pulley is of exactly circular shape.In first case 50 k.g weight is fixed with each pulley as shown in figure. The center of each pulley is fixed. Between these two pulleys there is a stationary plate. plate will remain stationary, while rotating the pulleys because centers of pulleys are fixed. 100 k.g weight is put on this plate. The force or weight of plate is applied on these two pulleys in vertical downward direction. now we try to rotate slowly first pulley in clockwise direction and second pulley in anti clockwise direction. lubrication is provided between the contact surfaces of stationary plate and pulleys.
In second case two pulleys of same diameter. 50 k.g weight is also fixed with each pulley, but there is no stationary plate and 100 k.g weight. Now we try to rotate slowly first pulley in clockwise direction and second pulley in anti clockwise direction.
As free body diagram in first case we can rotate pulleys using less moment of force. we can lift mass using less energy.

1. In case 1) the moment necessary to rotate each pulley is

m=(-g/2cos(bita)+g1*cos(alpha))*r
where:
g is weight of plate (=100 kg)
g1 is given weight (=50 kg)
alpha and bita are angle of rotation of each pulley
r is radius of pulley
2. In case 2) weigt of plate g=0.
Figure (energy efficiency1) is in cross section. As shown in figure two pulleys and 100 k.g. plate is arranged. Pulleys are designed so that from points a and a` to points b and b`, radii of pulleys continuously decrease. Now the weight or force of plate apply on point a and a` of two pulleys in vertical downward direction. Tangential component of this force helps pulley to rotate. when point b and b` come in contact with plate, plate traveled .5 cm in vertical downward direction. The gravitational energy plate loose is mgh=.005*50 n-mtr=.25 n-mtr. But the same time work done on pulley or energy got by pulley is mgcos(thita)h=.04*50*cos(thita), where theta is 45. so energy got is .04*50*cos45=.04*50*.7=1.4 n-mtr. .04mtr is periphery of pulley from point a to b. calculation is given only for one pulley. we can lift mass using less energy.
Jigar Y. Patel
energyefficiency@zoomshare.com


Plz visit http://energyefficiency.zoomshare.com

[Gregory]

Hello Jigar,

Interesting, Your first picture...

At first sight I thought You're wrong, but after I realized why and how can it useful.
Thanks!  This is a good idea

But only works good with very low rate of CoF. And the stationary plate must be fixed only in horizontal directions to allow the rotating pulleys to take the force of the plate.
Very good, we can lift the weights for less force. I will try it for myself in a simulation to see what happens.

[jigar]

Thank you. Plz make it as soon as posible.My theory is purely depanding up on free body diagram.

[lancaIV]

US4464095,Hydraulic energy converter,Kango Iida(JPN):
only as idea,to compare !

S
  dL

[Gregory]

By the way I don't know for 100% sure, that it will work or not...
I will test it, but please be patient because I'm very busy in this field now.