# Ambitious ORC Turbine

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#### Toymaker

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker, just some clarification of my understanding of De Laval nozzles.
1. I agree the total CSA of the combined set of nozzles (throat CSA) determines the total flow, subject to pressure difference across the throat.
2. I do not know how to design the converging nozzle.
3. I have read about the diverging nozzle with gas mixtures accelerating to sonic velocities. BUT the sonic limit (speed of sound) varies as the pressure and density of fluid (gas mixture). In rocketry, the diverging nozzle is different for an atmospheric and vacuum environment, and some nozzles have variable bells to optimise the shape as the rocket passes through the thinning atmosphere with altitude. I think that the nozzle can choke the output (velocity) if the wrong shape.
As you are sending gas at the orifice from the pressure of the liquid, to a space beyond the nozzle at "some other pressure", I think the divergent nozzle shape is critical, both to the pressures, the density of gas at the end of the nozzle, and also the speed of sound for the gas at the pressure within the chamber at the end of the nozzle.I think that if the nozzle is too wide, you won't develop the velocity of gas that would be achieve with a correct nozzle. I also think that if the nozzle is too long it will choke the flow and reduce performance of the gas stream, both mass and velocity.
Please research this as I am not an expert, just a novice at the early stages of learning these subjects. I am quite likely to be wrong.....
K2

From what I've learned about nozzles, choking is not usually a big problem. Induced turbulent flow is the biggest concern, which happens when the gases are allowed to expand too quickly, but as long as the nozzle expansion angle stays somewhere close to 15°, there shouldn't be a problem. Here are couple of images of real existing ORC nozzles. The color images were generated using CFD (Computational Fluid Dynamics) software; notice how the upper portion of the nozzle is not fully complete, but instead relies on the fluid flow from the adjacent nozzle to help shape the flow. The convergent-divergent shape is also obvious, and exists only in 2 dimensions as these nozzles are not round holes.
Also notice how thin the nozzle material becomes at the nozzle exit; when you actually machine these nozzles you will find that compromises will need to be made as the designed material thickness becomes less than 0.010", which from a practical standpoint, wont work.

In a multi-stage axial flow design, you should also keep the expansion angle of the entire turbine housing close to 15° so as to avoid turbulent flow through the blades and stators.

Also, in the 2 dimensional plan view, notice how both the second stage blades and stators (black & white drwg above) roughly form convergent-divergent nozzles, the entrance to all the blades and stators is larger then the exit, forcing the gases to accelerate through the blade row.

Finally, here's my nozzle and blading design. Ignore purple lines as they're only construction lines from my CAD drawing. The gap between adjacent nozzles is 1.5mm. Notice the nozzles convergent-divergent geometry.

Added one final pic,...this is my actual nozzle. It's not easy to see, but if you focus on the root area you can see the convergent-divergent geometry.

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#### Toymaker

##### Well-Known Member
HMEM Supporting Member
I agree with Steamchick, Heat transfer depends on the working fluid and the velocity in the tube. It also depends on residence time of the combustion gases and it gets a bit more complicated when radiation is involved not to mention consideration of the approach temps. The examples you cited deal with the latent heat of vaporization in open heaters and are not applicable.

Yes, I too agree with Steamchick's assertions concerning heat transfer. The purpose of my example was only show that so long as the heat transfer from the aluminum into the contained fluid is adequate that the fluid's heat absorption will prevent the aluminum from rising higher then the contained fluid. The process by which the fluid is absorbing the heat out of the aluminum, thereby keeping it's temperature from rising too high, is not important for the point I'm trying to make. Whether the heat transfer is due to latent heat of vaporization, or fluid flow rate through the tube, my point is that the temperature of the aluminum tube will be only slightly higher than the fluid it contains. If you read post #35 you'll see how I plan to demonstrate this.

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#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker. Post #41 is just brilliant! Explained a lot of what I was trying to understand. I am convinced it will work!
K2

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker.
I feel I disagree (without technical backing) about "wet steam" versus "dry steam". But that is from a real lack of technical data. Should you use Steam at a pressure of (say) 6 barg. (90psi) = 166C, and superheat to (say) 300C. before expanding through the turbine to (say) 20psi, before entering a condenser, and returning to the boiler coils via a pump, I would guess the steam is still dry? > 126C.?

I know that wet steam occurs when saturated steam and condensate water molecules are mixed. And that sometimes, heat loss in piping causes some of the saturated steam to condense and create a steam/water mix. But quite honestly, I still don't fully understand how steam could possibly condense and form liquid droplets inside a pipe filled with steam. Why those little suspended water droplet fail to re-vaporize baffles me.

It appears to me that your design, for whatever reason, has few blades at each stage.
<snip>
Your "large slot" (few blades) design appears to me to be losing some power transfer at each stage by the low number of blades? (compared to pictures of conventional "parson's turbines"?). I.E. the large gaps between power blades means the "jets" of gas from the fixed blades "waste" a lot of flow before the next blade comes along?

You're absolutely right that my design has far fewer blades than a typical turbine,...and I would love to have more blades,... but with the small overall size of this turbine the diameter of the blisk at the blade root was also very small and I'm limited to how many blades I can squeeze onto the blisk.

Below is a short video of an earlier turbine I built and have not yet tested. The OD of the larger rotor is a bit over 7", more than double the OD of my current turbine. The larger size gave me enough room to fit in a more standard number of blades. The potential power of the turbine below scares the crap out of me, which is why I decided to put this one the shelf until I have far more experience with steam turbines

#### Henry K

##### Well-Known Member
R123 is one of the refrigerants that has negative impacts on ozone and CO2. Check Wikipedia or other sources. As a hobby, I would not want to contribute these problems even in a trivial way.

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
R123 is one of the refrigerants that has negative impacts on ozone and CO2. Check Wikipedia or other sources. As a hobby, I would not want to contribute these problems even in a trivial way.

This single graph from an R123 Environmental Study should tell you everything you need to know about the environmental impact of R123. The impact to both Global Warming and to Ozone Depletion are very nearly zero.
It's only the poor wording of the Montreal Protocol that may end up banning R123.

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker, Maybe I can fumble my way through an explanation re: "I still don't fully understand how steam could possibly condense and form liquid droplets inside a pipe filled with steam.":- As I see it... If fine spray from the rapid boiling in the boiler (bubbles bursting) is not separated from the steam then some un-boiled water will pass down the steam pipe at the same temperature as the steam - but not the same phase. I.E. these droplets will need extra latent heat to vapourise... but the steam pipe actually "extracts" a little heat as the steam expands to accelerate down the steam pipe, so there is no "spare heat" to vapourise these droplets. In fact, without extra heat more steam condenses to power the wet steam down the pipe. So water in 2 states hits the turbine. (No good!). The "Obvious" countermeasure, is to have some extra coils in the flue gases to add extra heat to the wet steam, to provide the extra latent heat needed to dry the steam, the elevate the temperature further, so the steam is carrying a lot more energy into the engine. (superheating). That is, so the steam reaching the engine is hotter and still at a higher pressure than the "condensing pressure" (equal or less than the boiler pressure). Flash boilers (like yours = coils in flames/hot exhaust) do this "somewhere" along the pipework in the flames/hot exhaust gases..... Except, with your high pressure (523psi) and temperature controlled pipework (at 183C), you plan to have liquid right up to the nozzles: AT which point, the pressure of the liquid (523psi) and heat therein will expand/cool to a state of lower pressure, that cannot sustain the liquid state and will use some of the energy (temperature) to provide the latent heat to convert the liquid to gas at the lower pressure/same temperature. The worry, is that there won't be the gas pressure at the narrowest point of the nozzles, to make the expansion "sonic". But whatever energy that remains for expansion of the gas - after the latent heat extracted from the pressure/temperature has vapourised the liquid - will start at the lower pressure and then accelerate the gas into the turbine.
Your plan: "I will try to operate the boiler in the super critical pressure-temperature region for R123, meaning that I need to keep the pressure at or a little above 526 psi (3627 kPa ), and the temperature at or a little below 183°C, which will keep the R123 in a liquid state. My goal is keep the working fluid, R123, in it's liquid stated until it reaches the De Laval nozzles (or convergent-divergent nozzles). As the R123 liquid passes through the nozzles it will flash into a vapor at very high velocity as it impacts the first row of turbine blades."
I think the "trick" you will be performing, is in controlling the pumping of the fluid with enough flow to raise the pressure (against the venting at the nozzles) and keep the temperature at the hottest point of the boiler coils to 183C or below.
Just trying to work this out (My confused brain will probably get this wrong? - But a clever person will correct me!):
Enthalpy of 1kg of R123 at 183C:
Liquid: 422.6Kj/Kg. Latent heat to vapourise to gas = 18.9 kj/kg.
Therefore to change state from liquid to gas without addition of external heat, the gas only has Enthalpy of 422.6-18.9 = 403.7 Kj/Kg.... Thus the temperature of the gas will be at 39C: having taken the heat from the liquid to vapourise to gas. - See table: Thermodynamic Properties of HCFC-123, SI units (frigoristes.fr)
I.E. the gas at 39C will start to expand and accelerate down the expansion nozzle... So then you are in the realms of determining what pressure you can attain at the end of the turbine, so you can see what energy you can extract from the gas (in the perfect 100% efficient case). However, life has its little kick-backs, in that at each stage the gas is expanding (and cooling): but only "half" the stages are dynamic, and extract some heat /velocity from the gas into kinetic energy of the turbine wheel (and some is conducted away to the shaft). At the fixed blades, the heat within the gas is extracted as heat conducted away to the shaft/body, and some to accelerate the gas through the blades with the change of momentum and friction heating the blades in the process. 39C is about the temperature of the human body, and usually a room would be considered to be at 20~25C. So that leaves Enthalpy of 394.9Kj/Kg. at 25C.: you can extract a total of 8.6Kj/Kg. through the turbine. If the efficiency is 25% (a random guess?)- this then becomes 2.15Kj/Kg. converted to shaft energy for the pumps, generator or whatever you will put onto the shaft as the load.
So how many Kgs of gas will you pump per second?
What does this all mean?
K2

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#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Maybe I was wrong with my thermodynamic calculation...?
Your plan: "I will try to operate the boiler in the super critical pressure-temperature region for R123, meaning that I need to keep the pressure at or a little above 526 psi (3627 kPa ), and the temperature at or a little below 183°C, which will keep the R123 in a liquid state. My goal is keep the working fluid, R123, in it's liquid stated until it reaches the De Laval nozzles (or convergent-divergent nozzles). As the R123 liquid passes through the nozzles it will flash into a vapor".
Simply:
• Liquid R123 at 183C = pressure 3627kPa, Enthalpy = 422.6Kj/Kg.
• Gaseous R123 with Enthalpy = 422.7Kj/Kg is at 71C and pressure = 388kPa. I.E. JUST to make the physical change of state from liquid to gas. - Without any other losses to heat the intake metalwork of the turbine. (More later!). So your starting point cannot be a gas pressure above 388kPa (abs).
• If the turbine gas exit (The point just after the last moving blade) is at 20C, then the gas will be at a pressure of 75kPa (abs) with Enthalpy of 392Kj/Kg. I.E. It will have converted ~30Kj/Kg of Enthalpy into kinetic energy (gas and turbine motion) and heat (friction of gas against fixed and moving blades and walls of housing, bearings, conduction to cooler surfaces, etc.). This is the aspect that determines efficiency.
• For momentum exchange, I understand mathematically you cannot have more than 50% of the momentum transferred from moving fluid to moving component. But considering the thermal losses (Perhaps this will be well lagged?) I guess the losses could be between 20 and 50% of the input heat (Enthalpy of gas). so you could easily have efficiency of 50% x 50% = 25% - or even less? So for the power you need, you only have maybe 25% of 30Kj/Kg = 7.5Kj/Kg. - so use this to convert to how many Kgs of gas per second you need to pump through the system. (Then working backwards, you can determine how much heat is needed to achieve this?).
• You can also work out the condenser size, based on post turbine condensing of the gas at (or below) 75kPa (abs). into liquid.
• But there is a catch... 75kPa is sub-atmospheric pressure. So must be generated by the suction side of the pump that converts the gas back to liquid. So you need to do the sums to convert the gas back to liquid considering Enthalpy... temperature and pressure....
• More stuff: The liquid has t=o leave the heat source and get to the nozzle point before expanding the gas... but there will be energy (Fluid Enthaply) lost during that phase, buy conduction of heat to pipes - and then lost - and from friction (losses) of the fluid moving in the pipes. Also, there must be some (small) pressure drop to move the fluid along the pipes. So the Pressure at the nozzle (liquid) will be lower as a result, and the starting Enthalpy I have used will be "a perfect number" - not a real one. When you have a flow in Kgs per second, you can do calcs on the pipework to understand the pressure drop in the pipework, to get a better starting Enthalpy.
Sorry to be a damp squib, but I begin to doubt that R123 will work the way you expected? Can you comment? (I may be wrong - as usual! - I am learning thermodynamics from scratch on this one!)
Of course, using steam tables, or for any other medium, you can do the same calculations...
K2

#### HMEL

##### Well-Known Member
Before I start winding the boiler's aluminum tubes I plan to first test a theory I have. I will run a single, short length of aluminum tube, bent into a "U" shape, and place it directly in the hottest part of the burner flames. Using only water supplied from my house, I will run a continuous flow through the tube. I'm betting the tube will not melt under those conditions. Next step will be to slow the water flow rate down to a point at which steam is coming out of the open end; again I'm betting that the tube will remain undamaged. Final step, place an orifice onto the open end of the tube which restricts steam flowing out of the tube. Monitor the tube & steam temperatures using a non-contact thermometer and allow those temperatures to reach 400°F. Again, I'm betting the tubing remains undamaged even when exposed to the burner's hottest exhaust gases. As this point, I will be confident that my aluminum tube boiler will work as designed.

After the boiler is physically finished, the first few tests will be using water, not R123.

Safety relief valves on the boiler will vent the excess vapors into the condenser, not out into the open air. Only if condenser pressures get too high will R123 be vented into the air.

If you'll go back to the first post, you will find a schematic block diagram showing all the pressures and temperatures and RPM sensor which the FADEC (aka on-board computer) will be monitoring and controlling.

Since I'm not designing this engine with any pre-determined output power levels, I really don't have a need to worry about various pressure losses,...Whatever power I get will make me happy.
You should know that aluminum yield strength vs temperature of most aluminum alloys downward descent of what many call the knee starts at 200 degrees C.(392F) At a temperature of 275C the material has lost over 50% of its strength. It is for this reason that many codes do not allow aluminum to be placed in a fired system. These numbers come from certified laboratories and testing facilities. The test you are designing will be performed at atmosphere or 24.7 psia far below what you intend to operate. At 400C degrees there will be little or no strength. Because of the way you have split the manifold for the heat exchanger the flow will decrease by half as it passes through one of the splitters. In the design of a good safe heat exchanger you need to stay above that knee. Your burner design will easily give you a bulk temperature of over 1600 degree F. You should redesign your test to replicate your operating conditions.

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
You should know that aluminum yield strength vs temperature of most aluminum alloys downward descent of what many call the knee starts at 200 degrees C.(392F) At a temperature of 275C the material has lost over 50% of its strength. It is for this reason that many codes do not allow aluminum to be placed in a fired system. These numbers come from certified laboratories and testing facilities.

Aluminum's 200°C knee is the main reason I will be keeping boiler temperature below 200°C. Looking at the temperature to yield strength graph below, my never exceed temperature of 184°C seems reasonable, yes?

The test you are designing will be performed at atmosphere or 24.7 psia far below what you intend to operate. At 400C degrees there will be little or no strength. Because of the way you have split the manifold for the heat exchanger the flow will decrease by half as it passes through one of the splitters. In the design of a good safe heat exchanger you need to stay above that knee. Your burner design will easily give you a bulk temperature of over 1600 degree F. You should redesign your test to replicate your operating conditions.

Part of my working career included testing various parts of airplanes, and our moto was test-like-you-fly, So I like your idea of a slight redesign of my test,...it shouldn't be too difficult to place a valve and a pressure gauge onto the open end of the boiler tube to restrict flow and increase pressure up operational levels.

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker, Maybe I can fumble my way through an explanation re: "I still don't fully understand how steam could possibly condense and form liquid droplets inside a pipe filled with steam.":- As I see it... If fine spray from the rapid boiling in the boiler (bubbles bursting) is not separated from the steam then some un-boiled water will pass down the steam pipe at the same temperature as the steam - but not the same phase. I.E. these droplets will need extra latent heat to vapourise... but the steam pipe actually "extracts" a little heat as the steam expands to accelerate down the steam pipe, so there is no "spare heat" to vapourise these droplets. In fact, without extra heat more steam condenses to power the wet steam down the pipe. So water in 2 states hits the turbine. (No good!). The "Obvious" countermeasure, is to have some extra coils in the flue gases to add extra heat to the wet steam, to provide the extra latent heat needed to dry the steam, the elevate the temperature further, so the steam is carrying a lot more energy into the engine. (superheating). That is, so the steam reaching the engine is hotter and still at a higher pressure than the "condensing pressure" (equal or less than the boiler pressure).

Thank you,...This is one of the best explanations on wet steam I've read.

Your plan: "I will try to operate the boiler in the super critical pressure-temperature region for R123, meaning that I need to keep the pressure at or a little above 526 psi (3627 kPa ), and the temperature at or a little below 183°C, which will keep the R123 in a liquid state. My goal is keep the working fluid, R123, in it's liquid stated until it reaches the De Laval nozzles (or convergent-divergent nozzles). As the R123 liquid passes through the nozzles it will flash into a vapor at very high velocity as it impacts the first row of turbine blades."
I think the "trick" you will be performing, is in controlling the pumping of the fluid with enough flow to raise the pressure (against the venting at the nozzles) and keep the temperature at the hottest point of the boiler coils to 183C or below.

My current design uses a pressure washer, driven by a DC motor for the boiler feed pump, which allows the FADEC to control motor speed and thereby output pressure and flow rate. The FADEC also has control of both the air flow into the burner and fuel flow. By monitoring pressure & temperature of the working fluid at the boilers output and turbine RPM, the FADEC will be able to regulate all three of these parameters by controlling burner output levels and feed pump pressure & flow. This is all pretty standard process control. My biggest concern is avoiding feedback induced oscillations.

Just trying to work this out (My confused brain will probably get this wrong? - But a clever person will correct me!):
Enthalpy of 1kg of R123 at 183C:
Liquid: 422.6Kj/Kg. Latent heat to vapourise to gas = 18.9 kj/kg.
Therefore to change state from liquid to gas without addition of external heat, the gas only has Enthalpy of 422.6-18.9 = 403.7 Kj/Kg.... Thus the temperature of the gas will be at 39C: having taken the heat from the liquid to vapourise to gas. - See table: Thermodynamic Properties of HCFC-123, SI units (frigoristes.fr)
<snip>
K2

I don't understand how you arrived at this conclusion, "Thus the temperature of the gas will be at 39C"

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
<snip>
• Gaseous R123 with Enthalpy = 422.7Kj/Kg is at 71C and pressure = 388kPa. I.E. JUST to make the physical change of state from liquid to gas. -
<snip>

Where did you get these values? I'm not following your logic.

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker, maybe my logic is at fault? I would not be surprised as I am not experienced in thermodynamics. But the tables in the link I posted are much like steam tables. My logic goes like this in post #47.
• The Enthalpy (the absolute energy contained within, or needed to get to this state...) of liquid R123 at 183C is given in the table. = 422.6Kj/Kg.
• So I considered that to get the Enthalpy (energy of latent heat of vaporisation) to vaporise the liquid to gas I should deduct this value from the enthalpy of the liquid. 422.6 - 18.9 = 403.7 Kj/Kg.
• Thus the remaining enthalpy (energy within the gas) = 403.7Kj/Kg., would define the gas at a temperature and with the related lower pressure as per the table. Therefore I took that value of enthalpy and from the table for Gas determined the related temperature as 39C.
• So in reality, the liquid at 183C has cooled to gas at 39C. With related pressure drop. All the heat required to expend the liquid into gas has come form within the liquid, as it is an adiabatic expansion to create the gas from the liquid. I.E. This is how a refrigerator works to get the cold gas for keeping your foodstuff cold, and ice cream frozen.
• Maybe I am in error by subtracting the latent heat from the enthalpy of the liquid? A proper refrigeration Engineer (NOT a clever and we'll trained fitter who knows how to mend a fridge!) will do these calculations as his job so can teach us all, perhaps? That's why I reconsidered in post#48 by not subtracting the latent heat of vaporisation from the enthalpy of the liquid. And my confusion demonstrates why this needs expert advice, so we can all learn!
I am sorry I am unclear about some of this, but I find text books blow my mind. So I learn by trying to solve problems, so I can teach myself. However, I do need an expert to say "Good, the right answer", or "Wrong, it should be _______!".
The calculations for steam are basically the same, but use different tables. And I am not an expert with interpreting those either!
Cheers,
K2

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker, maybe my logic is at fault? I would not be surprised as I am not experienced in thermodynamics. But the tables in the link I posted are much like steam tables. My logic goes like this in post #47.
• The Enthalpy (the absolute energy contained within, or needed to get to this state...) of liquid R123 at 183C is given in the table. = 422.6Kj/Kg.
• So I considered that to get the Enthalpy (energy of latent heat of vaporisation) to vaporise the liquid to gas I should deduct this value from the enthalpy of the liquid. 422.6 - 18.9 = 403.7 Kj/Kg.
• Thus the remaining enthalpy (energy within the gas) = 403.7Kj/Kg., would define the gas at a temperature and with the related lower pressure as per the table. Therefore I took that value of enthalpy and from the table for Gas determined the related temperature as 39C.
• So in reality, the liquid at 183C has cooled to gas at 39C. With related pressure drop. All the heat required to expend the liquid into gas has come form within the liquid, as it is an adiabatic expansion to create the gas from the liquid. I.E. This is how a refrigerator works to get the cold gas for keeping your foodstuff cold, and ice cream frozen.
• Maybe I am in error by subtracting the latent heat from the enthalpy of the liquid? A proper refrigeration Engineer (NOT a clever and we'll trained fitter who knows how to mend a fridge!) will do these calculations as his job so can teach us all, perhaps? That's why I reconsidered in post#48 by not subtracting the latent heat of vaporisation from the enthalpy of the liquid. And my confusion demonstrates why this needs expert advice, so we can all learn!
I am sorry I am unclear about some of this, but I find text books blow my mind. So I learn by trying to solve problems, so I can teach myself. However, I do need an expert to say "Good, the right answer", or "Wrong, it should be _______!".
The calculations for steam are basically the same, but use different tables. And I am not an expert with interpreting those either!
Cheers,
K2

I'm not an expert in thermodynamics either,...I'm just an old retired electronic engineer whom likes to play with turbines.
Hopefully an expert will chime in and add some much needed clarity.

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
My brother has given me some advice. I THINK, my later post #48 is correct, so the gas after expansion is at 71C.
K2

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
My brother has given me some advice. I THINK, my later post #48 is correct, so the gas after expansion is at 71C.
K2

Remember the definition of "latent heat", is the heat required to convert a solid into a liquid or vapor, or a liquid into a vapor, without change of temperature. Table 1 of the the Thermodynamics Properties chart shows us that at 183°C R123 can be either a liquid or a vapor, the only difference is 18.9 kJ/kg of latent heat. Even the pressure remains the same in both states. My point being that I don't believe you can change the temperature of a gas or a liquid by only changing it's latent heat.

Whether I have a liquid or a vapor on the input side of the nozzle, the biggest nozzle related changes to the working fluid will come as it passes through the nozzle and expands towards the first row of blades; which will be governed by the ideal gas law, PV = nRT . At the nozzle's most restricted point the total area is 0.22 sqr". At the face of the first row of blades, the area is 1.185 sqr" giving a 5.38 increase in area and volume. Using Table 1: at 183°C the vapor volume is 0.0022. We now know the vapor will expand 5.38 times as it reaches the first blade row, so 0.0022 x 5.38 = 0.0118. Again from Table 1, a volume of 0.0118 corresponds to a temperature of 126°C or 127°C.

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker, I agree with the PV=RT aspect of expansion from nozzle to blades...
But I think you mis-understand the latent heat bit - If I am right, which may be not so, if I misunderstand it!
But Here goes. (See also post #48?). N.B. Pressures are absolute, Temperatures degrees C.
The Enthalpy of the liquid is the "energy within" the liquid R123 at 183C and pressure = 3627kPa. So, Hf liquid = 422.6Kj/Kg.
The conversion from liquid to gaseous state is ADIABATIC: I.E. NO ENERGY flows in from outside.
So to achieve the gaseous state needs an energy (from somewhere...) of 18.9Kj/Kg. Being ADIABATIC, the only place this can come from is the energy stored within the pressure and temperature of the liquid: I.E. Hg liquid = E Gaseous: Hence The GASEOUS state has to be at a lower temperature and pressure - I.E. relating to the Enthalpy of 422.6Kj/Kg. - because there is no energy input of extracted (Enthalpy is constant at an Adiabatic change).
So referring to the tables of GAS Enthalpy of 422.6Kj/Kg. I found the temperature and pressure of the gas after conversion of the physical state = 71C and 388kPa.

Then the PV = RT from nozzle pressure and temperature of the gas is used as P/T = R/V.. or T/P = V/R, or V = RT/P, or however you wish to calculate what happens to the gas after the conversion of physical state.
It is 46 years since I did any Thermodynamics, except for 3 or 4 boiler calculations, so I am a "beginner", and likely to be wrong... Maybe a refrigeration Engineer or thermodynamics teacher can decide who is right, as I am still not 100% sure! - But it seems to be logical (to me) that this is what it should be...

K2

#### Attachments

• R123 Saturation Enthalpy etc.pdf
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#### Toymaker

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker, I agree with the PV=RT aspect of expansion from nozzle to blades...
But I think you mis-understand the latent heat bit - If I am right, which may be not so, if I misunderstand it!
But Here goes. (See also post #48?). N.B. Pressures are absolute, Temperatures degrees C.
The Enthalpy of the liquid is the "energy within" the liquid R123 at 183C and pressure = 3627kPa. So, Hf liquid = 422.6Kj/Kg.
The conversion from liquid to gaseous state is ADIABATIC: I.E. NO ENERGY flows in from outside.
So to achieve the gaseous state needs an energy (from somewhere...) of 18.9Kj/Kg. Being ADIABATIC, the only place this can come from is the energy stored within the pressure and temperature of the liquid: I.E. Hg liquid = E Gaseous: Hence The GASEOUS state has to be at a lower temperature and pressure - I.E. relating to the Enthalpy of 422.6Kj/Kg. - because there is no energy input of extracted (Enthalpy is constant at an Adiabatic change).
So referring to the tables of GAS Enthalpy of 422.6Kj/Kg. I found the temperature and pressure of the gas after conversion of the physical state = 71C and 388kPa.

Then the PV = RT from nozzle pressure and temperature of the gas is used as P/T = R/V.. or T/P = V/R, or V = RT/P, or however you wish to calculate what happens to the gas after the conversion of physical state.
It is 46 years since I did any Thermodynamics, except for 3 or 4 boiler calculations, so I am a "beginner", and likely to be wrong... Maybe a refrigeration Engineer or thermodynamics teacher can decide who is right, as I am still not 100% sure! - But it seems to be logical (to me) that this is what it should be...

K2

Hi Steamchick, If you stop using the supercritical approach of having the working fluid remain a liquid right up to the nozzle exit, and instead do the calculations for a vapor state at the nozzle input and output, then you end up with the calculations I made in post #56 using PV = nRT and yielding 126°C at a pressure of 1352kPa.

Your calculations which start with the working fluid in a liquid state, result in a much lower temperature and pressure just transitioning from liquid to vapor and without even expanding through the nozzle.

We cant both be right,...so my money is on PV = nRT yielding the correct answer.

#### solarenergyadventures

##### Active Member
Hi guys! I've got to jump in here regarding PV=nRT. You cannot use PV=nRT when dealing with a Vapor according to my thermo book. This is why we use the Steam Tables. There is no simple mathematical expression for pressure, volume, and temperature when dealing with vapors. The vapor temperature must be so high above the critical point that the substance takes on the properties of a perfect gas. ONLY Then is PV=nRT valid. The steam tables are derived empirically from extensive laboratory testing of the substance in question, so the enthalpies listed in the tables are actually real-world results of the steam at various temperatures and pressures.

Regarding nozzle choking in a DeLaval nozzle. You must have choked flow at the narrowest part of the nozzle, otherwise the gas will not accelerate to supersonic velocities.

I'm also working on ORC engines, so I've done a fair bit of thermo research.
Cheers M8s

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
Hi guys! I've got to jump in here regarding PV=nRT. You cannot use PV=nRT when dealing with a Vapor according to my thermo book. This is why we use the Steam Tables. There is no simple mathematical expression for pressure, volume, and temperature when dealing with vapors. The vapor temperature must be so high above the critical point that the substance takes on the properties of a perfect gas. ONLY Then is PV=nRT valid. The steam tables are derived empirically from extensive laboratory testing of the substance in question, so the enthalpies listed in the tables are actually real-world results of the steam at various temperatures and pressures.

Regarding nozzle choking in a DeLaval nozzle. You must have choked flow at the narrowest part of the nozzle, otherwise the gas will not accelerate to supersonic velocities.

I'm also working on ORC engines, so I've done a fair bit of thermo research.
Cheers M8s

Glad you jumped in Solarenergyadventures,...always nice to get input from someone with actual experience with ORC.

Some 50 years ago I was taught that the Ideal Gas Law, PV=nRT, only works for an ideal gas, which doesn't actually exist in the real world,...but that PV=nRT will get you pretty close; I've never had a reason to doubt that,...until today.

So, looking a little closer at the thermo tables Steamchick and I have both been using, under the heading, Equations, in the DuPont Suva paper on the Thermodynamic Properties of HCFC123 it's stated that the Modified Benedict-Webb-Rubin (MBWR) equation of state was used to calculate the tables of thermodynamic properties,...PV=nRT was not used.

Since I was using the DuPont Suva table to get the values I referenced in posts #56 & #58, I was wrong to state that those values were derived from PV=nRT.

But, I'm still left with the question: am I using the DuPont Suva table correctly? I determined the gas volume expanded 5.38 times as it passes through the turbine nozzle. At the most restricted part of the nozzle, Using Table 1, I find at 183°C the gas volume is 0.0022. The gas will expand 5.38 times as it passes through the nozzle, so 0.0022 x 5.38 = 0.0118. Again from Table 1, the new gas volume of 0.0118 corresponds to a temperature of 126°C or 127°C. Therefore, the nozzle will have a temperature drop from 183°C to 126°C.

If you've used similar thermo charts in your work, I'ld appreciate your feedback.

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