Phoenix Turbine Builders Club

In-depth Turbine Analysis by Ian Main

March 2005

I looked into these turbines some time ago and I did some analyses on them and found some formulas that seem to predict them with some accuracy. I don't know if you want to call this a paper but you can present this information on your website if you wish.

Note that it is quite possible that these calculations could be off. I did them all once and haven't verified them (I have two kids after me to get back to playing with them). If you find any errors I'd be glad to hear them. I'd also like to see this spark greater discussion.

The basic premise is that a Tesla Turbine is essentially what is known as a "forced vortex". This is similar to centrifugal pumps etc. where the fluid is forced into a vortex and a difference in pressure is created.

Of course when run in reverse in 'turbine' mode, the air is forced tangentially into the input and comes out the center. However, much of the claims regarding the Tesla turbine revolve around its ability to apply backpressure and automatically balance its fluid consumption. In order to understand how much inlet pressure you should be using you need to be able to design a TT that will provide enough back pressure to offset your planned inlet pressure.

So, I did some searching in a fluid mechanics text book and found a forced vortex formula. It goes like this:

P2 - P1 = (fluid density * angular velocity ^ 2) / 2 (R2 ^ 2 - R1 ^ 2)
Where P2 and P1 are pressures, and R2 and R1 are the radius differences.

So this tells us what our change in pressure will be given the fluid density, angular velocity (which can be determined from rpm), and the internal and external radius. To give an example, let us spin a 10" TT in air at 10,000 rpm. 

We can get the fluid density from: www.sengpielaudio.com/calculator-airpressure.htm 

There are other places on the internet to get other fluid densities. So, for air at sea level at 15C we have a density of 1.23 kg/m3.

There's a handy calculator to convert RPM to radians / second (angular velocity in SI) at:
http://www.processassociates.com/process/convert/cf_vea.htm  (yes, I'm lazy.)

So, 10000 RPM is 1047 rad/s

And lastly we convert 10" and 3" (where our inner exhaust/fluid entrance is in the disk pack) to meters:
10" = 0.254m
3" = 0.0762m

So, our formula becomes:

P2 - P1 = (1.23 * 1047 ^ 2) / 2 * (0.254 ^ 2 - 0.0762 ^ 2)
P2 - P1 = 674168 * 0.0587
P2 - P1 = 39573 Pa

So we have a pressure change of about 39kPa, which is 5.6psi.

If we compare this to the results in Dr. Rice's article in the 1963 "Journal of Engineering for Power" (for verification), we see that he attained an output pressure of 28 in of water at 10,000 rpm with a similar turbine arranged as a blower. This is about 1psi. However he states in his paper that it was a very crude design so I am not too surprised as we are calculating the theoretical maximum. What this tells me is that his compressor was operating at approximately 20% efficiency (and he in fact mentions 20% efficiency relating to another compressor he built).

What it also tells us however, is that putting 100psi into the inlet of a 10" turbine and expecting it to auto regulate is ridiculous. We should be putting roughly 5psi in, probably with high flow. We know that the extra energy added will be dissipated across the disks and be made available as power, so with 5psi it should hover around roughly 10000 rpm (depending on the efficiency) and should auto regulate.

If we want to run at these much higher pressures, you can change many aspects of the turbine:

- If you change your working fluid you can get much higher pressures at much lower speeds. Water, for example, has a density of 1000kg/m3, so that would immediately boost our output pressure.
- If you vary the rotational speed it will vary the pressure at a rate of ^ 2 / 2. This is good but we all know the engineering tradeoffs associated with high RPM.
- We can increase the size of the disks and get a ^2 increase in pressure.

Of all of these, the last one seems to be the easiest. For example, a 24" turbine spinning at 10,000 rpm with a 3" input:

P2 - P1 = (1.23 * 1047 ^ 2) / 2 * (0.610 ^ 2 - 0.0762 ^ 2)
P2 - P1 = 674168 * 0.366
P2 - P1 = 246kPa

Which is about 36psi.

If you wanted to make a gas fired turbine, you'd have to know the density of the hot gasses that you are putting through it.

I do think that water would be an excellent way to test and demonstrate the Tesla turbine. It may also produce good results if used in a turbine because of the density.

For example, a 10" turbine at 3600rpm would produce a pressure change of:

P2 - P1 = (1000 * 377 ^ 2) / 2 * (0.254 ^ 2 - 0.0762 ^ 2)
P2 - P1 = 71064500 * 0.0587
P2 - P1 = 4171486Pa = 4.1MPa = 594psi

So obviously that works very well. I think this turbine would certainly auto regulate itself and would probably be too big for most applications where you only have a few PSI of head. You could get away with a much smaller turbine.

If you relate this to water pumps, we know from experience that a small centrifugal pump like the one used for home wells is usually 5" in diameter, spins at 1750rpm, and can easily produce 50+psi of water pressure (before it cuts out). It is easy to see why given these numbers.

Note that all these calculations are done with no output flow. As soon as we start taking fluid out of the outer ring of the turbine the pressure will drop.

Hope this is helpful and at least partially correct!

Ian Main

Email: ianm_AT_stemwinder_DOT_org

Calculation: speed of sound in humid air

This calculation shows the speed of sound in humid air according to Owen Cramer, "JASA, 93, p2510, 1993", with saturation vapour pressure taken from Richard S. Davis, "Metrologia, 29, p67, 1992", and a mole fraction of carbon dioxide of 0.0004.

The calculator is valid over the temperature range 0 to 30° C (273.15 - 303.15 K) and over the pressure range 75 to 102 kPa.

In the region between the air pressures 95.000 und 104.000 kPa there is no noticeable changing of the speed of sound.

temperature °Celsius
air pressure kPa
relative humidity %
CALCULATE
speed of sound m/s

Notice: The speed of sound changes with the temperature and a little bit with the humidity, but not with the air pressure.

Air Density Calculations

First, consider the ideal gas law:

(1) p · V = n · R · T

p = pressure, pascals (multiply mb by 100 to get pascals)
V = volume in m3
n = number of moles
R = gas constant
T = temperature K = °C + 273.15

Density is the number of molecules of the ideal gas in a certain volume.
In this case a molar volume, which can be mathematically expressed as:

(2) D = n / V

D = density in kg/m3
n = number of molecules
V = volume in m3

By combining the previous two equations, the expression for the density becomes:

(3)

D = density in kg/m3
p = pressure, pascals (multiply mb by 100 to get pascals)
R = gas constant = 287.05 J / (kg · K) for dry air
T = temperature K = °C + 273.15

As an example, using the standard sea level conditions of P = 101325 Pa and T = 15 °C,
the air density at sea level, can be calculated as:

D = 101325 / (287.05 · (15 + 273.15)) = 1.2250 kg/m3

This example has been derived for the dry air of the standard conditions. For real-world situations,
it is necessary to understand how the density is affected by the moisture in the air.

The density of a mixture of dry air molecules and water vapor molecules can be expressed as:

(4)

D = density in kg/m3
pd = pressure of dry air in pascals
pv = pressure of water vapor in pascals
Rd = gas constant for dry air = 287.05 J / (kg · K)
Rv = gas constant for water vapor 461.495 J / (kg · K)
T = temperature K = °C + 273.15

To determine the density of the air, it is necessary to know is the actual air pressure,
also known as absolute pressure, or station pressure, the water vapor pressure, and the temperature.

Calculation of the wavelength with frequency and temperature
Calculation of the wavelength of radio waves and acoustic waves

From http://www.sengpielaudio.com/calculator-airpressure.htm