T-(Shaped) Turbine, Which uses Centrifugal Force

[mihai.isteniuc]

Now as promise the power calculations I did. No claims here about anything. Excuse my poor english. Take your time ... this is gonna be a long post to read and digest.

Keep that in mind that I am european and will use metric system for all my calculations.

STEP 1 – What do we know?
Well we know a lot. From the document previously posted ( http://overunity.com/15318/t-shaped-turbine-which-uses-centrifugal-force/msg432132/#msg432132 ) we have find out some real data:

Water pressure: 45 psi
Flow rate: 0.375 liters/second

Assuming an efficiency of 100% the calculated turbine should produce 116 watts of mechanical power (of course in real life there isn't such thing so lets assume efficiency somewhere around 85%)

Calculated mechanical power on turbine shaft is then:

P real = P calculated X 0.85
P real = 98.6 Watts of mechanical power!!!. Close to 100 ... not bad

In a real test it is mentioned (page 45 of the document) that a portable TV set was used as load. Power requirements for the TV: 12V/60W. And the TV worked without any problems. So 60W of electrical power !!! can be expected without any problems. Great!

STEP 2 – Let's imagine ...
... now a setup for testing our machine. I choose the following physical dimensions (based somehow on my common sense and some expertise):

main pipe: 110 mm in diameter (the vertical pipe)
length of main: as short as possible to minimize friction losses between it and the water from tank.
T pipe: 40 mm in diameter (the horizontal ones)
length of T pipe: 1000 mm

STEP 3 – Basic calculations

Water pressure: 45 psi = 3.06 atmosphere (standard); ref: http://www.onlineconversion.com/pressure.htm
Flow rate: 0.375 L/sec
Speed of water for desired flow rate = (4 x flow rate) / (pipe diameter² x  Π)
Speed of water inside 40 mm pipe for desired flow rate = 0.29842 meters/second; ref: http://www.1728.org/flowrate.htm
Cross section of the pipe = Π x d² / 4 where d is diameter of the pipe
Cross section of the pipe = 3.14 x 40 x 40 / 4 = 1256 mm² = 0.001256 m²

STEP 4 – Calculating centrifugal force (at the end of 1 meter of horizontal pipe)

Fcentrifugal = m x Ω² x r
where m is the mass of rotating object in kilograms;
Ω = angular speed of rotating mass in radians/second;
r = radius of the the rotating mass;

Damn we don't know Ω. Let's make a wild guess then. I'm feeling lucky (in google words) and I say: let's rotate the damn thing with 170 rpm.

170 rpm = 17.80 rad/sec; ref: http://www.convertunits.com/from/RPM/to/radian/second

Moving mass is: volume of the water trapped inside the pipe or in other words:
m = cross section of the pipe x length of the pipe x specific density of the water
m = 0.001256 x 1 x 1000 kilograms/cubic meter = 1.256 kilograms of water

Fcentrifugal = 1.256 x 17.80 x 17.80 x 1 = 397 Newtons

STEP 5 – Calculating Coriolis force

Coriolis ca be calculated by the following formula:
Fcoriolis =-2 x m x Ω x v
Fcoriolis =-2 x 1,256 x 17.80 x 0,29842
Fcoriolis =-13.34 Newton; so Coriolis is almost 30 times lower then centrifugal. I'm tempted to ignore it.

STEP 6 – Calculating the pressure at the end of the pipe

Pressure is force per unit area. So our pressure is:
P = 397 Newtons/0.001256 m² = 316082 Newtons/m² Well that's a lot if u ask me. How much?
P =  316082 Newtons/m² = 3.12 atmosphere (again standard); ref: http://www.onlineconversion.com/pressure.htm

CONCLUSION: If we rotate a pipe full of water with 1 meter in lengh and 40 mm in diameter with 170 rpm we get around 3 atmosphere of pressure at the end of the pipe. Hey, that's great. Remember the start condition? 45 psi or 3.06 atmospheres? Then if we attach a Pelton turbine like the one designed in the document we can produce around 100 watts of mechanical power using this method. GREAT!!! ... is it or is it not???...

STEP 7 – How much power do we need to invest to achieve and maintain 170 rpm with our pipe(s)?

A. Kinetic energy (the power needed to make the system rotate from 0 rpm to 170 rpm, no water flow and no frictions). When we achieve the desired rpm if there are no losses in the system we can consider that the system will rotate from now on for ever at that rpm. The energy invested to get desired rpm can be expressed by the formula:

KE = ½  x I x Ω², where
I is momentum of inertia and Ω is angular velocity;

I can be calculated: I = ½ x m x R², where
m is total mass of the system and R is radius of the system. Will assume that total mass of the rotating system is 20 kg (this includes pipes, water trapped inside it, turbines ... all rotating elements).

I = 0.5 x 20 x 1 x 1 = 10, so
KE = 0.5 x 10 x 17.80 x 17.80 = 1584 Joules; this is the required amount of power needed to be invested to make a 20 kg mass with radius of 1 meter to rotate with 170 rpm. Of course in reality this mass is distributed around the radius of the system. Not all 20 kg are at the end of 1 meter radius, therefore the energy requirements are, I believe smaller. But what is IMPORTANT: this energy is invested once at startup, at every startup. And that's it.

B. Power to overcome LOSSES in the system (aka to maintain 170 rpm):

1. electrical;
2. friction.

1. Let's ignore them ... (for now)  we will consider 100W of mechanical power on the output of the Pelton and the mechanical power on the input required to rotate the shaft. So, mechanical vs mechanical. No more transformations and complications.

2. there are a lot of frictions and losses:
a) friction of water with the pipes;
b) friction in bearings;
c) kinetic energy to accelerate from 0 to 170 rpm, the desired flow rate (0.375 liter of water in every second). Although the mass of the water inside the pipes is constant, if we have a flow, then every new molecule of water entering the system must be accelerated from 0 to the rotating speed);
d) friction of the rotating pipes with the surrounding air;
e) other(s) ... u name it (them) if u can identify it (them) ...better yet calculate it (them).

2a) friction of water with the pipes
Hazen Williams formula  is typically used to calculate friction losses through water conveying pipe. The result represents head loss in meters at 100 meters of pipe.

Hf (m @100m) = (608,704,451 / d4.8655) x (Q / C)1.85, where
d is diameter in mm
Q is water flow in liters/min
C is roughness coefficient (typically for PVC pipes values are between 150 and 160)
Hf = (608,704,751 / 40 4.8655) x (0,375 x 60 / 155)1.85
Hf = 608,704,751 / 62,348,051 x 0.028
Hf = 0.27 → friction of water with pipes is very low at this flow speed, almost negligible.

2b) friction in bearings with good bearings this will be very low. I'm almost sure I can estimate that no more of 5W of power loss will occur in the 2 bearings.

2c) kinetic energy for 0.375 liters of water (we already used the 2 formulas)

I = ½ x m x R²
I =  ½ x 0.375 x 1 x 1
I = 0.1875

KE = ½  x I x Ω²
KE = ½  x 0.1875 x 17.80 x 17.80
KE = 29,70 Joules

2d) friction of the rotating pipes with the surrounding air = 0

Well let's put a cover on top of our machine. Seal it. Add vacuum inside the machine (extract all the air). No more power loss due to air friction. End of story here. And it is doable.

Let's make a recap now regarding power loss ... we have discovered:

1. Some low value Coriolis force will add some extra friction of water with the pipes (minor);
2. Friction of water with the pipes (minor);
3. Friction in bearings – 5 Watt;
4. Kinetic energy for 0.375 liters of water – 29.70 Watt

Total(1) = 35 Watt. I'll be generous and I'll add another 5 for friction of water inside the pipes (hope it is ok. can't imagine to be more ... after all is only 1 meter of pipe, even if the flow is turbulent inside it)
Total(2) = 40 Watt.

HUH? So we need to invest 40 Watt (mechanical) and we will obtain then 100 Watt (also mechanical)? Really?

PS. After some more consideration regarding the system the following might be true: The mechanical work is generated by centrifugal force in direct relation with the mass of water. Mass of water is 1.2 kilograms. The kinetic energy absorbed by the system (while in motion) is for accelerating the flow of water inside the pipe and that is 0.375 liters of water. In other words 1.2 liters of water generate energy, but only 0.375 liters of water absorb energy ... in every second ... hmmm


PS2. TY for taking the time to read all of this. I'll only ask: what is missing?


Mihai

[Overunityguide]

@ Mihai,

Thank you for Posting your Theoretical Calculations Online...

[Overunityguide]

In my New Video I have done Some Testing with a High Pressure / Low Flow Rate Setup. The results are Quite Promising...
In this Setup the Coriolis Effect almost looks like to be prevented. This Setup works the Best for Now... (Compared to the Tests I have done Previously)

The New Video Can be Found at: http://youtu.be/363K5AoSi0c

[telecom]

Very exciting video!
Looking forward for more results.
Regards

[Overunityguide]

Because of the promising results with my test setup during my last video, I have decided to update the theoretical self-runner design.
So for now it contains the most recent improvements. (nozzles being bend +/- 60  degrees)