DIY Tesla Impulse Turbine

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OK, I ran the numbers on my tiny DIY water pump first using the simple equations Gemini (Google’s AI) found, and then using the step-by-step equations described in the link you provided. The results from Gemini's equations are more-or-less reasonable, but the results from the Euler’s turbomachine equations are pretty crazy.

From Gemini:
Q = π * D * B * ω / 60 = 3.1416 * 0.04878 * 0.012 * 120,000 / 60 = 3.678 cubic meters per hour
= 61.3 LPM which seems reasonable.

Using the results, 3.678 m^3/hr, and the somewhat ridiculous pressure obtained from the Eulers Turbomachine Equations, plugged into the power equation found via the internet, results in the following.

Power (kW) = Q* P * SG / 3600 = 3.678 * 480.6 / 3600 = 0.49 kW

The need for 1/2 kW to power my small pump also seems reasonable.
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Euler’s turbomachine equations:

Given:
diameters of the impeller at the inlet and outlet
r1 = 0.01374 m
r2 = 0.02439 m
Speed = 120,000 rpm
the blade angle at inlet β1 = 30°
the blade angle at outlet β2 = 20°
assume that the blade widths at inlet and outlet are: b1 = b2 = 0.012 m

First, we have to calculate the radial velocity of the flow at the outlet. From the velocity diagram, the radial velocity is equal to (we assume that the flow enters exactly normal to the impeller, so tangential component of velocity is zero):

Vr1 = u1 tan 30° = ω r1 tan 30° = 2π x (120,000/60) x 0.01374 x tan 30° = 99.6 m/s

The radial component of flow velocity determines how much the volume flow rate is entering the impeller. So when we know Vr1 at the inlet, we can determine the discharge of this pump according to the following equation. Here b1 means the blade width of the impeller at the inlet.

Q = 2π.r1.b1.Vr1 = 2π x 0.01374 x 0.012 x 99.6 = 0.1 m3/s = 6000 LPM = 100 L/s

In order to calculate the water horsepower (Pw) required, we have to determine the outlet tangential flow velocity Vt2, because it has been assumed that the inlet tangential velocity Vt1 is equal to zero.

The outlet radial flow velocity follows from conservation of Q:

Q = 2π.r2.b2.Vr2Vr2 = Q / 2π.r2.b2 = 0.1 / (2π x 0.02439 x 0.012) = 54.38 m/s

From the figure (velocity triangle) outlet blade angle, β2, can be easily represented as follows.

cot β2 = (u2 – Vt2) / Vr2

and therefore the outlet tangential flow velocity Vt2 is:

Vt2 = u2 – Vr2 . cot 20° = ω r2 – Vr2 . cot 20° = 2π x 120000/60 x 0.02439 – 54.38 x 2.75 = 306.5 – 149.5 = 156.9 m/s.

The water horsepower required is then:

Pw = ρ Q u2 Vt2 = 1000 [kg/m3] x 0.1 [m3/s] x 306.5 [m/s] x 156.9 [m/s] = 4,808,985 W

= 4,809 kW

and the pump head is:

H ≈ Pw / (ρ g Q) = 4,808,985 / (1000 x 9.81 x 0.1) = 4,902 m

= 480.6 bar

= 6970.5 psi

These results are laughable. 4,809 kW turning the 8 mm diameter stainless steel shaft seems likely to sheer the shaft, assuming the aluminum impeller didn't self destruct first, and that the Tesla turbine could actually produce that much power. It all seems most improbable.

I suspect what these numbers are telling me is that there's simply no way my tiny water pump will actually spin at 120,000 rpm.

Testing the Tesla turbine and using the water pump as a load should prove interesting :cool:
The reason it's like this is your ludicrous pump impeller tip speed from the high RPM of the tesla turbine. Honestly this is kind of the problem for most single stage turbines, they want to spin way too fast for most loads aside from turbocompressors and electric generators.

What will happen is that your pump will load the turbine down to a speed where the driving torque demand of the pump matches the output of the turbine. Obviously not ideal for turbine efficiency. You will probably find the pump outputs a lot more pressure and flow if you gear the turbine output down so it's running at a more optimal speed.

Edit: for reference, your tip speed seems to be slightly over 300 m/s. The LOX turbo pump on the F-1 engines of the Saturn V moon rocket was only running a bit over 142 m/s.
 
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The reason it's like this is your ludicrous pump impeller tip speed from the high RPM of the tesla turbine. Honestly this is kind of the problem for most single stage turbines, they want to spin way too fast for most loads aside from turbocompressors and electric generators.

What will happen is that your pump will load the turbine down to a speed where the driving torque demand of the pump matches the output of the turbine. Obviously not ideal for turbine efficiency. You will probably find the pump outputs a lot more pressure and flow if you gear the turbine output down so it's running at a more optimal speed.

Edit: for reference, your tip speed seems to be slightly over 300 m/s. The LOX turbo pump on the F-1 engines of the Saturn V moon rocket was only running a bit over 142 m/s.

I did a bit of Googling and found that covered impellors are used up to 360 m/s, and open impellors, like mine, are used up to 500 m/s.

From what I was able to find, it seems max tip speed for water pumps is limited by material strength of the impellor; 50 m/s for cast iron impellors and 110 m/s for stainless steels. I have no idea what materials manufacturers are using to reach 500 m/s; perhaps the difference is simply which safety factor is used.

I strongly suspect the Saturn V LOX pump tip speed is limited primarily by the material strength of the impellor, which is pumping a liquid with a temperature of -183 C, which makes all metals very brittle.
 
I did a bit of Googling and found that covered impellors are used up to 360 m/s, and open impellors, like mine, are used up to 500 m/s.

From what I was able to find, it seems max tip speed for water pumps is limited by material strength of the impellor; 50 m/s for cast iron impellors and 110 m/s for stainless steels. I have no idea what materials manufacturers are using to reach 500 m/s; perhaps the difference is simply which safety factor is used.

I strongly suspect the Saturn V LOX pump tip speed is limited primarily by the material strength of the impellor, which is pumping a liquid with a temperature of -183 C, which makes all metals very brittle.
That's where copper-nickle alloys come out to play.

Forgive my memory, where does the air blower go?
 
Hi Toymaker.
I have been a bit of a "Doubting Thomas" for your combined impulse/surface friction turbine, because it seemed to used conflicting rather than supporting technologies. But reading part of a paper (by some German clever bloke) There is a summary table of various experimental works of note:
CHAPTER 2. PREVIOUS RESEARCH ON TESLA TURBINES
1727107398453.png

Table 2.1: Selected performance data of previous Tesla turbines. Maximum power and maximum isentropic efficiency data is usually achieved at different operating points:
I noted that the turbine by The turbine by Steidel and Weiss [1976] is different in that the outer disk edge is not circular but has several sawtooth-shaped cut-outs.... and thought 6.7% efficiency was "not so good" - But then I read more and it seems that Lisker et.al. achieved 38% efficiency with "another rare design feature is found in the turbine by Lisker et al. [2015] who add blade inserts to the outlet region of the disk gap, essentially building a hybrid between friction and bladed turbine".
So obviously you are on the right track, Well done!
K2
P.S. file to large to paste: try searching for "Design Guidelines for Tesla Turbines Based on Comprehensive Theoretical, Numerical and Experimental Investigations. Stefan Günter Klingl, M.Eng."
 
I did a bit of Googling and found that covered impellors are used up to 360 m/s, and open impellors, like mine, are used up to 500 m/s.

From what I was able to find, it seems max tip speed for water pumps is limited by material strength of the impellor; 50 m/s for cast iron impellors and 110 m/s for stainless steels. I have no idea what materials manufacturers are using to reach 500 m/s; perhaps the difference is simply which safety factor is used.

I strongly suspect the Saturn V LOX pump tip speed is limited primarily by the material strength of the impellor, which is pumping a liquid with a temperature of -183 C, which makes all metals very brittle.
I think those higher speeds are for compressors. Gases being less dense than liquid put less stress on the impeller
 
Sorry, I thought the end goal of this was to use the whole thing to run a blower? Like a massive steam plant leaf blower?

You are correct :)

As to where the blower goes,...the air blower will be powered by one of two different steam driven turbines.

Depending on how successful my Tesla-impulse turbine turns out to be, I will either build a larger one, matched in size to use the max steam output of my boiler, or scrap the Tesla-impulse idea and use the 3-stage axial flow turbine I've already constructed. I used very standard design practices to construct the axial flow turbine, and I therefore have a high level of confidence it will work well,...BUT because it only has 3 stages, to get the most efficiency from it, I will need to use a Freon such as R123, as the working fluid, instead of water.

For me, the biggest advantage of using a Tesla turbine, instead of an axial turbine, is that a Tesla turbine can efficiently use water as the working fluid. As an added benefit, Tesla turbines can use wet steam without the corrosion issues seen in axial turbines, so, no need to add steam dryers, or water separation devices.
 
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I think those higher speeds are for compressors. Gases being less dense than liquid put less stress on the impeller

I had the same thoughts initially, that surely the higher tangential speeds were for impellors pumping gases,...but on second thought, I couldn't recall ever seeing a closed centrifugal impellor used for gases, and multiple articles cited the 360 m/s limit for closed impellors.
 
Hi Toymaker.
I have been a bit of a "Doubting Thomas" for your combined impulse/surface friction turbine, because it seemed to used conflicting rather than supporting technologies. But reading part of a paper (by some German clever bloke) There is a summary table of various experimental works of note:
CHAPTER 2. PREVIOUS RESEARCH ON TESLA TURBINES
View attachment 159860
Table 2.1: Selected performance data of previous Tesla turbines. Maximum power and maximum isentropic efficiency data is usually achieved at different operating points:
I noted that the turbine by The turbine by Steidel and Weiss [1976] is different in that the outer disk edge is not circular but has several sawtooth-shaped cut-outs.... and thought 6.7% efficiency was "not so good" - But then I read more and it seems that Lisker et.al. achieved 38% efficiency with "another rare design feature is found in the turbine by Lisker et al. [2015] who add blade inserts to the outlet region of the disk gap, essentially building a hybrid between friction and bladed turbine".
So obviously you are on the right track, Well done!
K2
P.S. file to large to paste: try searching for "Design Guidelines for Tesla Turbines Based on Comprehensive Theoretical, Numerical and Experimental Investigations. Stefan Günter Klingl, M.Eng."

Thank you much for finding this info :),....very encouraging !!

But now I don't know whether to be happy that I'm likely on the right path, or sad, that I'm not the first crazy person to think of this idea.
 
Edit: for reference, your tip speed seems to be slightly over 300 m/s. The LOX turbo pump on the F-1 engines of the Saturn V moon rocket was only running a bit over 142 m/s.
And those pumps were rated in the 200,000hp+ range
 
I had the same thoughts initially, that surely the higher tangential speeds were for impellors pumping gases,...but on second thought, I couldn't recall ever seeing a closed centrifugal impellor used for gases, and multiple articles cited the 360 m/s limit for closed impellors.
You might not have encountered a case of them, but they do exist:
Fig10.jpg

From what I've read about 150 m/s is the practical limit for liquid pumps, due to high stress on the impeller and heavy erosion (possibly due to Cavitation?) of components at higher speeds. Though that doesn't rule out some oddball designs getting away with more! For a liquid rocket pump it only has to last a few minutes, so maybe they could push the boundaries further, especially with modern materials.
 
I crunched a few more numbers for my Tesla turbine to get a better idea of how much power to expect from it.

The turbine has 6 brass nozzle arrays, each with 24 individual 0.026" diameter nozzles with divergent openings as shown in the pic below; the top nozzle array shows the divergent opening.
110119-Divergent-Nozzle-Holes-sml.jpg


The total area of all 144 individual nozzles is 0.0764 sqr inches, which is equivalent to a single 0.156" diameter nozzle.

Using this on-line calculator, at 400 psi, 2.77 LPM will flow through the nozzles.

Simple equation to find the power from a turbine:

P = m * (h1 – h2) * n

Where:
  • P = power in watts
  • m = mass flow rate of steam in kg/s
  • h1 = the enthalpy of steam at inlet in J/kg
  • h2 = the enthalpy of steam at outlet in J/kg
  • η = the efficiency of the turbine
Plugging in the numbers:
P = 0.046 kg/s * (2,803,120 j/kg – 2,675,530 j/kg) * 1
= 5,869 W = 5.869 kW = 7.87 HP

I used 100% efficiency for the calculation, which I know I wont get,... more likely something around 50%, so
2.9 kW (3.9 HP).

Those numbers seem much too small, so maybe I've done something wrong.
 
I crunched a few more numbers for my Tesla turbine to get a better idea of how much power to expect from it.

The turbine has 6 brass nozzle arrays, each with 24 individual 0.026" diameter nozzles with divergent openings as shown in the pic below; the top nozzle array shows the divergent opening.
110119-Divergent-Nozzle-Holes-sml.jpg


The total area of all 144 individual nozzles is 0.0764 sqr inches, which is equivalent to a single 0.156" diameter nozzle.

Using this on-line calculator, at 400 psi, 2.77 LPM will flow through the nozzles.

Simple equation to find the power from a turbine:

P = m * (h1 – h2) * n

Where:
  • P = power in watts
  • m = mass flow rate of steam in kg/s
  • h1 = the enthalpy of steam at inlet in J/kg
  • h2 = the enthalpy of steam at outlet in J/kg
  • η = the efficiency of the turbine
Plugging in the numbers:
P = 0.046 kg/s * (2,803,120 j/kg – 2,675,530 j/kg) * 1
= 5,869 W = 5.869 kW = 7.87 HP

I used 100% efficiency for the calculation, which I know I wont get,... more likely something around 50%, so
2.9 kW (3.9 HP).

Those numbers seem much too small, so maybe I've done something wrong.
What's your superheat temperature? I'll feed in to my spreadsheet for an equivalent deLaval turbine and see what answer we get.
 
Looking at the boiler output side, and using Boiler Horse Power, which is defined as: One BHP is the amount of energy needed to evaporate 34.5 pounds of water into steam in one hour at a temperature of 212°F.

Or: 1 BHP = 15.6 kg/hr. Since 1 liter of water weighs 1 kg, then 1 BHP = 15.6 L/hr and 0.26 LPM.

Using the flow rate from post #293: 2.77 LPM / 0.26 LPM = 10.65 BHP.

Given that one BHP is equivalent to 13.5 normal HP, then 13.5 * 10.65 = 143.8 HP.

That's a huge difference from the 7.87 HP found in post #293
 
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What's your superheat temperature? I'll feed in to my spreadsheet for an equivalent deLaval turbine and see what answer we get.
My boiler is an all copper monotube; for safety there is no superheat. steam temperatures will be limited to at or below 470 F (243 C).
At 400 psi I expect to see steam temperature of 448 F (231 C).

Will your spreadsheet work with no superheat?

Did you ever post your spreadsheet? I'm still looking for ways to get at least ballpark numbers for how much power can be extracted from a steam supply of "X" psi and "Y" flow rate.
 
My boiler is an all copper monotube; for safety there is no superheat. steam temperatures will be limited to at or below 470 F (243 C).
At 400 psi I expect to see steam temperature of 448 F (231 C).

Will your spreadsheet work with no superheat?

Did you ever post your spreadsheet? I'm still looking for ways to get at least ballpark numbers for how much power can be extracted from a steam supply of "X" psi and "Y" flow rate.
I published it here: https://www.homemodelenginemachinis...es-spreadsheets-and-a-few-observations.36226/

It works without superheat, but there are occasional issues with calculating the steam density if the steam condenses too much in the stage.

Note that my spreadsheet actually starts with your target power output and predicts the steam consumption, rather than the other way around.
 
Tesla Turbine success & failure in the span of a few minutes. I was finally able to perform a short test of my Tesla Turbine (TT) today. With the boiler running at 53% and making only 0.83 LPM dry steam, I move the 3-way valve to direct the steam into the TT,...and success,...it began to spin the motor-generator at what sounded like a good rpm. My DVM (Digital Volt Meter) was still reading resistance, so I quickly changed it to read volts,...it briefly displayed 12vdc, then quickly began to drop. I glanced at the motor-generator and noticed it was spinning, then stopping, then spinning again, as partially shown in the video.



I later found that the TT shaft was spinning inside the coupler; apparently the steam's heat had loosened the connection. Unfortunately the heat also destroyed my coupler,...the plastic spider melted!!!

Melted Coupler sml.jpg


Important to note that despite the loose & melted coupler, the TT performed flawlessly, and powered the motor-generator to produce about 3 amps at 12 volts, or about 36 watts. Also of interest, the steam pressure guage never moved, meaning I need to push a lot more steam into the TT before I will see any back pressure in the guage.

Clearly, I have some repairs to make before I can resume testing :cool:
 

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