DIY Tesla Impulse Turbine

[Steamchick]

Thanks Toymaker. Sounds good! I am very interested in your turbine, as the one I have have been trying (made by the late Chairman of my local model engineering club) is a bit unknown, and I have hesitated dismantling it as I understand he carefully assembled it for good balance. Very similar in size to yours.
From what you are doing, I feel more confident about giving mine a lot more steam and heat!
Thanks!
K2

 

[Nerd1000]

Do you have an equation to calculate flow rate through a centrifugal water pump which includes impeller blade height or thickness?

Being that a centrifugal pump is not a positive displacement machine, the flow rate will be set by many variables, most notably the amount of restriction on the flow at the inlet and outlet. You can get an idea of the amount of pressure the impeller can generate at a given flow rate by using velocity triangles and the Euler pump and turbine equation.

 

[Toymaker]

Thank you Gemini (Google's AI), for finding the flow rate equation I've been looking for.

The flow rate of a centrifugal pump impeller can be determined using the following equation:

Q = π * D * B * ω / 60

where:

• Q is the flow rate (in cubic meters per hour)
• π is a mathematical constant approximately equal to 3.14159
• D is the impeller diameter (in meters)
• B is the impeller width (in meters)
• ω is the angular velocity of the impeller (in revolutions per minute)
• 60 is a constant to convert from revolutions per minute to revolutions per hour

Once Q is known, the following equation can be used to find power needed to drive the pump.

Power (kW) = Q* P * SG / 3600
where:

• Q is the flow rate (in cubic meters per hour)
• P is pressure in Bar
• SG is Fluid's Specific Gravity (water = 1)

 

[Steamchick]

This seems to be straightforward and simple:
I think it should also apply to a Tesla pump? - using the plate gaps to be the equivalent of "B"?
Outside my experience, so maybe this is a stupid question? But on various types of rotary oil pumps (on cars, motorcycle engines, etc.) the clearances between rotors and housings are critical to prevent "flow-back" and loss of pressure &/or flow. E.g.

But I guess that is because they work towards their limit of pressure, and not so much flow. Are these "displacement pumps" rather than "inertial pumps"?
What is the corollary for leak-back on a centrifugal pump? Side rotor clearance I guess, and is there a problem with clearance around the outer diameter of the rotor? I can imagine that fluid leaking at the sides could travel from the outer "higher speed (and pressure?) zone" to the inner "lower speed zone" causing loss of performance from re-cycling of fluid, which is minimised by close fitting of parts. But does it matter if the housing around the outer diameter is close fitting or not? - Maybe not as on the outside there is only fluid at pressure and flow for the outlet, and no inlet to leak back to?
Sorry, I can't get my head around what is going on, to work it out.
Maybe it is something like:- the pump can deliver a maximum flow based on the equation above, but pressure is based on how the flow is restricted downstream - or not - at anything below a maximum pressure determined by the rotor outer diameter speed?
I have a feeling that max pressure is at zero flow, and max flow is at zero pressure. but we work somewhere between those points in real life.
Can anyone teach me what is really happening? - or point me to a website so I can learn a bit? (I have been otherwise engaged so not had time yet to study this properly). And I'm feeling a bit "thick" today... so ignore me if I am too far off track.
Thanks,
K2

 

[Nerd1000]

This seems to be straightforward and simple:
I think it should also apply to a Tesla pump? - using the plate gaps to be the equivalent of "B"?
Outside my experience, so maybe this is a stupid question? But on various types of rotary oil pumps (on cars, motorcycle engines, etc.) the clearances between rotors and housings are critical to prevent "flow-back" and loss of pressure &/or flow. E.g.
View attachment 159822View attachment 159823View attachment 159824 But I guess that is because they work towards their limit of pressure, and not so much flow. Are these "displacement pumps" rather than "inertial pumps"?
What is the corollary for leak-back on a centrifugal pump? Side rotor clearance I guess, and is there a problem with clearance around the outer diameter of the rotor? I can imagine that fluid leaking at the sides could travel from the outer "higher speed (and pressure?) zone" to the inner "lower speed zone" causing loss of performance from re-cycling of fluid, which is minimised by close fitting of parts. But does it matter if the housing around the outer diameter is close fitting or not? - Maybe not as on the outside there is only fluid at pressure and flow for the outlet, and no inlet to leak back to?
Sorry, I can't get my head around what is going on, to work it out.
Maybe it is something like:- the pump can deliver a maximum flow based on the equation above, but pressure is based on how the flow is restricted downstream - or not - at anything below a maximum pressure determined by the rotor outer diameter speed?
I have a feeling that max pressure is at zero flow, and max flow is at zero pressure. but we work somewhere between those points in real life.
Can anyone teach me what is really happening? - or point me to a website so I can learn a bit? (I have been otherwise engaged so not had time yet to study this properly). And I'm feeling a bit "thick" today... so ignore me if I am too far off track.
Thanks,
K2

This covers it pretty well.

https://www.nuclear-power.com/nucle...trifugal-pumps/eulers-turbomachine-equations/

 

[Toymaker]

This covers it pretty well.

https://www.nuclear-power.com/nucle...trifugal-pumps/eulers-turbomachine-equations/

Thanks for the link,...but can you tell me what ρ is?

I'm assuming it's the mass density of water, 1000kg/m^3, but it's not defined in the linked text.

 

[Toymaker]

This seems to be straightforward and simple:
I think it should also apply to a Tesla pump? - using the plate gaps to be the equivalent of "B"?
Outside my experience, so maybe this is a stupid question? But on various types of rotary oil pumps (on cars, motorcycle engines, etc.) the clearances between rotors and housings are critical to prevent "flow-back" and loss of pressure &/or flow. E.g.
View attachment 159822View attachment 159823View attachment 159824 But I guess that is because they work towards their limit of pressure, and not so much flow. Are these "displacement pumps" rather than "inertial pumps"?
What is the corollary for leak-back on a centrifugal pump? Side rotor clearance I guess, and is there a problem with clearance around the outer diameter of the rotor? I can imagine that fluid leaking at the sides could travel from the outer "higher speed (and pressure?) zone" to the inner "lower speed zone" causing loss of performance from re-cycling of fluid, which is minimised by close fitting of parts. But does it matter if the housing around the outer diameter is close fitting or not? - Maybe not as on the outside there is only fluid at pressure and flow for the outlet, and no inlet to leak back to?
Sorry, I can't get my head around what is going on, to work it out.
Maybe it is something like:- the pump can deliver a maximum flow based on the equation above, but pressure is based on how the flow is restricted downstream - or not - at anything below a maximum pressure determined by the rotor outer diameter speed?
I have a feeling that max pressure is at zero flow, and max flow is at zero pressure. but we work somewhere between those points in real life.
Can anyone teach me what is really happening? - or point me to a website so I can learn a bit? (I have been otherwise engaged so not had time yet to study this properly). And I'm feeling a bit "thick" today... so ignore me if I am too far off track.
Thanks,
K2

All the pumps you've mentioned are positive displacement,... beyond that, I'm not qualified to answer your other questions. However, there must be a few good reasons we don't see these pumps used in home-use pressure washers; sealing issues, expensive to hold needed tolerances, work great with oils, but not water. The correct answers are above my pay grade.

 

[Steamchick]

Thanks Toymaker.
You have a pay grade?
I'm a humble "Worth-less Git". - Or so I was told many times in the past...

Thanks for the feedback.
K2
P.S. Wictionary quotes: "Git" - in case it isn't known to non-UK persons - is a term of (mild?) insult denoting an unpleasant, silly, incompetent, annoying, senile, elderly or childish person.[1]. The phrase "grumpy old git", denoting a cantankerous old man, is used with particular frequency.
- So perhaps they were correct? - And I simple understood it to be someone who was an inconvenience, as in "please GO AWAY!" (Seen in Western movies when I was young, as "Git" - or "Skedaddle". an abbreviation of "You are a nuisance, please go away". E.g. "Git Along Little Dogies" - Classic Old Western Movie.

 

[HMEL]

This seems to be straightforward and simple:
I think it should also apply to a Tesla pump? - using the plate gaps to be the equivalent of "B"?
Outside my experience, so maybe this is a stupid question? But on various types of rotary oil pumps (on cars, motorcycle engines, etc.) the clearances between rotors and housings are critical to prevent "flow-back" and loss of pressure &/or flow. E.g.
View attachment 159822View attachment 159823View attachment 159824 But I guess that is because they work towards their limit of pressure, and not so much flow. Are these "displacement pumps" rather than "inertial pumps"?
What is the corollary for leak-back on a centrifugal pump? Side rotor clearance I guess, and is there a problem with clearance around the outer diameter of the rotor? I can imagine that fluid leaking at the sides could travel from the outer "higher speed (and pressure?) zone" to the inner "lower speed zone" causing loss of performance from re-cycling of fluid, which is minimised by close fitting of parts. But does it matter if the housing around the outer diameter is close fitting or not? - Maybe not as on the outside there is only fluid at pressure and flow for the outlet, and no inlet to leak back to?
Sorry, I can't get my head around what is going on, to work it out.
Maybe it is something like:- the pump can deliver a maximum flow based on the equation above, but pressure is based on how the flow is restricted downstream - or not - at anything below a maximum pressure determined by the rotor outer diameter speed?
I have a feeling that max pressure is at zero flow, and max flow is at zero pressure. but we work somewhere between those points in real life.
Can anyone teach me what is really happening? - or point me to a website so I can learn a bit? (I have been otherwise engaged so not had time yet to study this properly). And I'm feeling a bit "thick" today... so ignore me if I am too far off track.
Thanks,

This seems to be straightforward and simple:
I think it should also apply to a Tesla pump? - using the plate gaps to be the equivalent of "B"?
Outside my experience, so maybe this is a stupid question? But on various types of rotary oil pumps (on cars, motorcycle engines, etc.) the clearances between rotors and housings are critical to prevent "flow-back" and loss of pressure &/or flow. E.g.
View attachment 159822View attachment 159823View attachment 159824 But I guess that is because they work towards their limit of pressure, and not so much flow. Are these "displacement pumps" rather than "inertial pumps"?
What is the corollary for leak-back on a centrifugal pump? Side rotor clearance I guess, and is there a problem with clearance around the outer diameter of the rotor? I can imagine that fluid leaking at the sides could travel from the outer "higher speed (and pressure?) zone" to the inner "lower speed zone" causing loss of performance from re-cycling of fluid, which is minimised by close fitting of parts. But does it matter if the housing around the outer diameter is close fitting or not? - Maybe not as on the outside there is only fluid at pressure and flow for the outlet, and no inlet to leak back to?
Sorry, I can't get my head around what is going on, to work it out.
Maybe it is something like:- the pump can deliver a maximum flow based on the equation above, but pressure is based on how the flow is restricted downstream - or not - at anything below a maximum pressure determined by the rotor outer diameter speed?
I have a feeling that max pressure is at zero flow, and max flow is at zero pressure. but we work somewhere between those points in real life.
Can anyone teach me what is really happening? - or point me to a website so I can learn a bit? (I have been otherwise engaged so not had time yet to study this properly). And I'm feeling a bit "thick" today... so ignore me if I am too far off track.
Thanks,
K2

k2 There are three types of pumps you are asking a question about. They all require a clearance tolerance.

The oil pumps are displacement pumps. Each gear revolves in opposite direction so the space between the gears is capturing the fluid and delivering it to the exit chamber. The contact between the gears is the seal for the rotating gear as well as the housing clearance. They build fairly high pressure and many of these pumps require pressure relief and re circulation for control. Rpm or pressure control valves maintain the developed pressure. For engines the oil is constant volume but the volume varies with rpm as well so the pressure should rise with rpm. If the oil gets to hot the viscosity changes will cause the oil pressure to drop. So understanding what the fluid will do is as important as understanding the pump.

A Tesla pump works on a different principle. It uses the shear properties of the fluid on the disk. And although the tolerance is not as critical the spacing between the rotors are critical. Theoretically the fluid interacts with the disk wall. Fluid pressure is a result of the rpm as well as the fluid viscosity. They work better for high viscosity fluids.

Conventional pumps operate based on volume and rpm. They have been around for many years but the analysis is not always so straight forward. If you ever have a chance to corner a manufacture of these pumps they will probably tell you they test the pumps in house to generate the curves and they are as much experimental as analytical. The side walls of these pumps are specially constructed and often they come supplied with bars that are fixed to the housing to close the sidewall clearance. I used to call these wizzer bars but the nomenclature is different depending on the pump manufacture. They can also be constructed with tight side wall clearances

There are other types of volumetric pumps such as a tapered spiral, archemidies spiral, and probably a few I forgot.

In all these systems the fluid properties are critical Cavitation occurs if the fluid is not at the correct temperature and pressure. For this reason boiler feedpumps often have more then one impeller. Each impeller raising the pressure in stages. For instance a 500 psig system might have six to eight impellers mounted on the same shaft.

The above descriptions are very simple explanations. Pumps are used so often we sometimes forget how complex they really can be.

 

[Steamchick]

Thanks HMEL,
Things are a lot clearer now.
And Thanks Nerd1000 - that link gives me a lot to work with - and it seems pretty straightforward! (Unlike some things!).
It also appears that the centrifugal pump methodology could be applied to a Tesla turbine, if angles of flow are guessed appropriately? But I want to try and find how Tesla worked it out.
K2

 

[Steamchick]

I found this:
https://www.researchgate.net/public...cal_Numerical_and_Experimental_Investigations

This is where I fell asleep - as my brain does not understand the maths.
Extract from:

Design Guidelines for Tesla Turbines Based on Comprehensive Theoretical, Numerical and Experimental Investigations​

• March 2024

• Advisor: Michael Pfitzner, Stefan Lecheler

Authors:

Stefan Klingl

• Universität der Bundeswehr München

I flew the white flag for this one...

PREVIOUS RESEARCH ON TESLA TURBINES
Because of the annular shape of the fluid domain in a Tesla turbine rotor, it is useful to work in a cylindrical coordinate system, using the coordinate vector r = (r, φ, z) and the velocity vector u = (ur, uφ, uz). Note that the centrifugal and Coriolis terms are still included, so that uφ describes the circumferential velocity relative to the disk surface. The system rotates with the angular velocity ω around the z-Axis, thus
Ω = (0, 0, ω).∂ur∂t +1ρ∂p∂r + ur∂ur∂r +uφr∂ur∂φ −u2φr+ uz∂ur∂z − 2ωuφ − rω2
(2.3a)+ ν urr2−1r∂ur∂r −∂2ur∂r2−1r2∂2ur∂φ2+2r2∂uφ∂φ −∂2ur∂z2!= 0∂uφ∂t +1ρr∂p∂φ + ur∂uφ∂r +uφr∂uφ∂φ +uφurr+ uz∂uφ∂z + 2ωur
(2.3b)+ ν uφr2−1r∂uφ∂r −∂2uφ∂r2−1r2∂2uφ∂φ2−2r2∂ur∂φ −∂2uφ∂z2!= 0∂uz∂t +1ρ∂p∂z + ur∂uz∂r +uφr∂uz∂φ + uz∂uz∂z
(2.3c)+ ν −1r∂uz∂r −∂2uz∂r2−1r2∂2uz∂φ2−∂2uz∂z2!= 0urr+∂ur∂r +1r∂uφ∂φ +∂uz∂z = 0
(2.4)The corresponding boundary conditions are inlet and outlet boundaries at the outer and inner radius r1 and r2 respectively and periodicity in circumferential direction. The walls are fixed at z = ±s and thus are rotating along with the reference frame. There is no standardized method for nondimensionalization and naming of non-dimensional parameters when analysing Tesla turbines.
(etc.)
I didn't know there was such a word as "nondimensionalization".. !
Perhaps I should leave it to clever people?
K2

 

[Nerd1000]

Thanks for the link,...but can you tell me what ρ is?

I'm assuming it's the mass density of water, 1000kg/m^3, but it's not defined in the linked text.

Yes it is fluid density

 

[wce4]

Thanks Toymaker.
You have a pay grade?
I'm a humble "Worth-less Git". - Or so I was told many times in the past...

Thanks for the feedback.
K2
P.S. Wictionary quotes: "Git" - in case it isn't known to non-UK persons - is a term of (mild?) insult denoting an unpleasant, silly, incompetent, annoying, senile, elderly or childish person.[1]. The phrase "grumpy old git", denoting a cantankerous old man, is used with particular frequency.
- So perhaps they were correct? - And I simple understood it to be someone who was an inconvenience, as in "please GO AWAY!" (Seen in Western movies when I was young, as "Git" - or "Skedaddle". an abbreviation of "You are a nuisance, please go away". E.g. "Git Along Little Dogies" - Classic Old Western MIovie.

In my opinion, you don't fall under the category of Git, but rather a Philomath type person.
Although your replies may be a little lengthy at times, you are definitely a philomathic.

 

[Steamchick]

Thanks Wce, I just looked that one up. I'll take it as a compliment - Thankyou!
But I don't mind being called a "stupid old git" - or anything. I grew up being called all things with the surname Chicken, and sometimes people are right when they call me something. Who am I to judge the opinions of others? So I use that to try and improve (rarely succeed though!).
I was taught at a young age that you can't teach an idiot to be clever, but sometimes an idiot will present the cleverest idea, because their brains work things out slightly differently. So if I am an idiot, then sometimes I shall be right. (I probably won't know when I am right and when wrong, but carry on regardless!). Good enough for me. So I try to never write anyone off, if I don't think they are right, I just explain what I think is right as an alternative for them to decide.
Except when it comes to "industrial safety" or the Law, as those both have simple "musts" and "Must-nots". It comes from a lesson my Dad taught me when I had my fingers in the electrical socket, and was about to switch "ON" when he caught me and told me there are some things you NEVER do. - "Don't even think about it, some things are simple "musts or must-nots".
I guess I was only 4 or 5 years old...? - but I remember his words well. (Dementure brain?). Despite everything I have reached my 3 score years and 10 quoted in the Bible as Man's lifetime. - Hence "Old" is not an insult to me.
Cheers,
Ken
(P.S. that was another long-winded tale... - Sorry!)

 

[Toymaker]

This covers it pretty well.

https://www.nuclear-power.com/nucle...trifugal-pumps/eulers-turbomachine-equations/

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.
____________________________________________________________________________________

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.Vr2 ⇒ Vr2 = 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

 

[The Ignoble Troll]

All the pumps you've mentioned are positive displacement,... beyond that, I'm not qualified to answer your other questions. However, there must be a few good reasons we don't see these pumps used in home-use pressure washers; sealing issues, expensive to hold needed tolerances, work great with oils, but not water. The correct answers are above my pay grade.

All the pumps listed above require ultra high pressure lubrication. Lubrication beyond the point that fluid film lubrication breaks down. That's the roll of zinc or calcium packages in hydraulics oil.

Pumping water, they would be basically unlubricated.

 

[The Ignoble Troll]

I have a feeling that max pressure is at zero flow, and max flow is at zero pressure. but we work somewhere between those points in real life.
Can anyone teach me what is really happening? - or point me to a website so I can learn a bit? (I have been otherwise engaged so not had time yet to study this properly). And I'm feeling a bit "thick" today... so ignore me if I am too far off track.
Thanks,
K2

For centrifugal pumps, this is true.

Deadheading them causes max pressure.

 

[Steamchick]

Thanks Troll.
Well done Toymaker.
K2

 

[wce4]

I'll take it as a compliment - Thankyou!

It was meant to be a compliment and you are welcome.

 

[The Ignoble Troll]

Thanks Troll.
Well done Toymaker.
K2

I should mention that with certain centrifugal actions, dead heading also causes less loading on the prime mover, because it is maintaining static pressure and not moving anything. So peak pressure and peak power input share a weird relationship.

You can feel this effect by covering and uncovering the output of a hand cranked blacksmith blower.