Forced Air Diesel Fuel Burner

[Toymaker]

Hi Toymaker. I understand why you are choosing R123 as a working fluid. The Entropy, Enthalpy and other data is included in the data sheet.
I have commitments all today, and most of the weekend, so I'll have a look at the sums next week for you. The simple aim is to produce an estimate of the surface area of coils you need to exceed to be sure you are getting enough heat from the burner exhaust gas. Your 9300sq.in. is probably OK, but I'll see what the numbers give.
As you are studying the thermodynamics, have you reached the "boiler" calculations yet? Or have you manually worked the calculations by hand (calculator, log tables, etc.) to figure out where you lost your factor of 1000? Maybe a K in kw? Or kg or similar?
K2

The YouTube course I'm watching is 52 separate episodes and is a general Thermodynamics course, not specifically focused on boilers,...I hope some of the episodes cover monotube boilers, but I don't know if they will.
However, the fourth video is an Intro to Convection and covers the main equation for determining heat transfer from a moving fluid (like combustion gases) through a solid surface such as the roof on a house or the walls of a water tube boiler.

Using the Heat Rate formula: q = hA(Ts – Ti) Note that Ts and Ti can be in either order, it's the Delta (or the difference) between the two Temperatures that is important.
where:
q is Heat Rate : Thermal energy transfer per unit time measured in Watts
h is the Convection heat transfer coefficient

A is area in square meters
Ts is temperature of solid surface in degrees C
Ti is temperature of the bulk fluid (combustion gases)

Assume the monotube is copper, desired power output is 300 Hp, and combustion gases 600 C.
Solving for A, ( surface area of the monotube):
A = q / h(Ts – Ti)
A = 223000 W / (400 W /

K) (600C – 180C)

= 1.33 square meters = 2,061 square inches

However: there are several problems with this overly simplistic equation. The equation only looks at heat transfer from the combustion gases into the tube wall, and not into the fluid inside the tube. We must also assume the combustion gases are 600 degrees throughout the entire boiler, and we know that isn't accurate; gases close to the burner exhaust will be much hotter, likely above 900 degrees, while other area inside the boiler closer to the exit will be much cooler as the exhaust gases transfer their heat into the tubes.

IMHO, To get a truly accurate idea of tubing surface area needed will require computer software capable of CFD (Computational Fluid Dynamics)

 

[Steamchick]

Fair enough. That is a true mathematical way to determine heat flow through the copper tube. And of course you can do likewise for you aluminium tube. But "empirically", as the heat flow is proportional to temperature difference (linear) I would take the mean heat flow as the average temperature difference, from exhaust gas outlet (chimney) temp minus input refrigerant temp, to exhaust gas input temp (burner outlet) minus refrigerant output temp. Actually it is the integral between those points, but being a linear function of the temp difference that is half the sum of those temp differences...
I think I have explained that, but maybe someone can correct me if I have my words or logic muxed ip!
K2
28/8 amendment:
An example to illustrate what I would do (perhaps this is wrong?)
Exhaust gas from burner, Tbo, = 900C, Exhaust output from heat exchanger, Tho, = 300C:
R123 Fluid into heat exchanger, Rei, = 50C, fluid out of heat exchanger, Reo, 180C.
Passing hottest exhaust to exchanger output of gas gives a temp difference here of Tbo - Reo = 900 - 180 = 720C, call this Tdo.
And at the outlet of exhaust gases/inlet of R123 to the heat exchanger: Tdi = 300 - 50 = 250C.
SO MEAN temp difference in exchanger is (720 +250)/2 = 970/2 = 485C.
Therefore I would plan the "boiler" (heat exchanger) to have excess surface area for a temperature difference of 485C between the External gas (Exhaust) and internal gas (R123). You already have the calculation for heat flow Exhaust to copper, you just need the same calculation from Copper to R123.
Does this help?
K2

 

[Steamchick]

On the heat flow from copper tube to internal fluid, that is by conduction, and (from what I remember from 1978 (yes, It was a calculation in my job then!), all you need is to understand the enthalpy of the gas temperature input to output, and determine the mass flow simply from that. From the mass flow, and pressure at input and output points, you can determine the velocity of the gas, which cannot be supersonic!
But memory may be imperfect after all this time.... And back then I was calculating mass and volumetric airflow of cooling air through busbar enclosures in power stations, to size the cooling plant air mover. I knew (easily) how much heat to move from the busbar thermal losses, the input and output temperatures from the air cooler, so the bit that joined it all together was the enthalpy of air at atmospheric pressure.
Does that help?
K2

 

[Toymaker]

On the heat flow from copper tube to internal fluid, that is by conduction,

Heat transfer from a solid into a fluid is almost never conduction. This slide is from the YouTube class on Heat Transfer: Conduction only occurs when the fluid is stationary. Whenever fluids are moving, Convection formulas are used to determine heat transfer.

and (from what I remember from 1978 (yes, It was a calculation in my job then!), all you need is to understand the enthalpy of the gas temperature input to output, and determine the mass flow simply from that. From the mass flow, and pressure at input and output points, you can determine the velocity of the gas, which cannot be supersonic!
But memory may be imperfect after all this time.... And back then I was calculating mass and volumetric airflow of cooling air through busbar enclosures in power stations, to size the cooling plant air mover. I knew (easily) how much heat to move from the busbar thermal losses, the input and output temperatures from the air cooler, so the bit that joined it all together was the enthalpy of air at atmospheric pressure.
Does that help?
K2

 

[Steamchick]

Thanks Toymaker. I stand corrected.
Where I was coming from is the boundary layer condition, where the layers of gas molecules very close to the surface of the solid are near stationary, so initially they transfer heat from the solid to gas (or vice-versa) by conduction. Then when the gas moves away from the surface it flows past the boundary layers and extracts the heat by a mix of conduction (contact interaction of molecules on molecules) and convection (mixing of hotter molecules moving in the mass of cooler molecules). A similar condition exists in long flue tubes of boilers, where laminar flow can develop.
Where there is turbulent gas movement adjacent to a surface then convection calculations work best - So I agree you are right for your exhaust gas around the single tube coils you are utilising.
Where you have gas forced along a pipe (or similar) where boundary layers develop, then normal "heat generated flow" of convection does not occur, or so I was taught about cooling cylindrical busbars with forced-air cooling 45 years ago.. Hence we calculated heat flow from the hot surfaces by Radiation and Conduction - using the coefficient of thermal conductivity of the atmosphere in use. Thermal resistivity of air = 2000 deg.C /W/cu.cm. used in the formula:
Pd = T /t Pi W/sq.cm., where:
Pd = heat transferred from solid surface into gas boundary layer (We used 1mm thick boundary layer as the sums were a close match to real results!).
T = Temp difference (deg.C).
t = thickness of layer (e.g. 1mm)
Pi = thermal resistivity in deg.C /W/cu.cm. (2000 deg.C /W/cu.cm. for air)
Not sure what the conditions will exist between the R123 and the pipe walls... if you are trying to calculate that heat flow. Possibly laminar flow, with a boundary layer?
The only time we used a Convection Calculation was the Still-Air case. (flat bar Busbars in enclosures without ventilation or forced air movement).
So much for education... !
K2

 

[Steamchick]

Hi Toymaker,
I just thought I would explain how I use the Du Pont tables in doing calculations of heat flow into the R123 from exhaust gas.
You will (I guess?) have worked-out the mass flow of R123 required to pump the heat (=energy) around your system. I have had a go below: But my sums don't give me answers I understand...?

I assume you will have condensed the R123 after the turbine?
If so, you will have liquid at 50C introduced to the heat exchanger:
From the Table 1 of http://www.frigoristes.fr/static/telechargement/abaques/diagramme_enthalpiqueR123.pdf
that you referred earlier: I assume you will have condensed the R123 after the turbine? (or maybe not? - which changes the equation?).
If so, you will have liquid at 50C introduced to the heat exchanger:
This needs a latent heat of evaporation to get to gas at 50C.: 161.0kJ/kg.
Then you need Enthalpy to raise it from 410.2 at 50C to 451.0 kJ/kg.at 180C. I.E. a further 40.8kJ/kg. of gas.
So total heat required is 161.0+40.8 = 201.8kJ/kg. of R123. (If I correctly understand these tables!).
Assuming you manage to get 100kW (= 100kJ/sec.) of exhaust heat into the R123 gas, that would be:
100kJ/sec of heat supplied per second, so will heat 100/201.8 = 0.496kg/sec. mass of gas. - That's 0.65 litres every second if a liquid? or 1.4l/s of gas. (=84l/min = 150cfm = a big compressor!). That's a lot of volume of gas. As a liquid, I can't get anything like that amount of water out of my taps that use 1/2" bore copper pipe and about 4 bar pressure... ?
Will your pipework and pump manage to push that much R123 through the small bore? Or will the back-pressure be so high it will remain liquid?
I am trying to understand the design as this is a big learning curve for me!
"Playing with sums" helps me understand what is going on... but I may be completely wrong, so advice and help is much appreciated.
Thanks,
K2

 

[Toymaker]

Hi Toymaker,
I just thought I would explain how I use the Du Pont tables in doing calculations of heat flow into the R123 from exhaust gas.
You will (I guess?) have worked-out the mass flow of R123 required to pump the heat (=energy) around your system. I have had a go below: But my sums don't give me answers I understand...?

I assume you will have condensed the R123 after the turbine?

Yes, this is a closed system, or re-circulating design.

If so, you will have liquid at 50C introduced to the heat exchanger:

Yes, 50C is what I had calculated as the turbine exit temperature at a pressure of 30.8psi.

From the Table 1 of http://www.frigoristes.fr/static/telechargement/abaques/diagramme_enthalpiqueR123.pdf
that you referred earlier: I assume you will have condensed the R123 after the turbine? (or maybe not? - which changes the equation?).
If so, you will have liquid at 50C introduced to the heat exchanger:

I hope to have a liquid feeding into the condenser, but I do not know what the temperature will be. You see, I'm employing another cooling technique which I haven't yet mentioned. see diagram: 3 stage turbine The large empty area just beyond the final turbine stage will contain multiple spay nozzles which will spay liquid R123 into the 50C R123 vapors as they leave the turbine. The cool spay of R123 mist will cool the 50C gases leaving the turbine,...but just how low I will be able to lower the temperature I'm not yet smart enough to determine. Which is another reason I'm watching that online course on Heat Transfer.

This needs a latent heat of evaporation to get to gas at 50C.: 161.0kJ/kg.

Why?? I don't understand what you're trying to find or show.

Then you need Enthalpy to raise it from 410.2 at 50C to 451.0 kJ/kg.at 180C. I.E. a further 40.8kJ/kg. of gas.
So total heat required is 161.0+40.8 = 201.8kJ/kg. of R123. (If I correctly understand these tables!).
Assuming you manage to get 100kW (= 100kJ/sec.) of exhaust heat into the R123 gas, that would be:
100kJ/sec of heat supplied per second, so will heat 100/201.8 = 0.496kg/sec. mass of gas. - That's 0.65 litres every second if a liquid? or 1.4l/s of gas. (=84l/min = 150cfm = a big compressor!). That's a lot of volume of gas. As a liquid, I can't get anything like that amount of water out of my taps that use 1/2" bore copper pipe and about 4 bar pressure... ?
Will your pipework and pump manage to push that much R123 through the small bore? Or will the back-pressure be so high it will remain liquid?
I am trying to understand the design as this is a big learning curve for me!
"Playing with sums" helps me understand what is going on... but I may be completely wrong, so advice and help is much appreciated.
Thanks,
K2

Maximum mass flow of the Freon, R123, through the entire system is limited by the max flow I can push through the turbine nozzle with the Freon at it's critical points.
Nozzle size: 0.000149 sqr meters
Max speed of R123 through the nozzle is the speed of sound at R123's critical temperature of 184C = 78.1 m/sec
Flow volume = (0.000149 sqr meters) x (78.1 meters/sec) = 0.0116 m3/sec (read m3 as cubic meters)
R123 Density at critical point: 449 kg/m3 (again, read m3 as cubic meters)
Therefore, mass flow rate = 0.0116 m3/sec x 449 kg/m3 = 5.21kg/sec

R123 leaving the condenser will be a liquid at 0.0007 m3/kg. Calculating total mass flow: (5.21 kg/sec) x (0.0007 m3/kg) = 0.0036 m3/sec, or 216 L/min

I plan to use a common household pressure washer as the boiler feed pump. These units typically produce 1500 to 2500 psi at flow rates of 6 to 9 liters/min. The pressure output of these pumps are regulated by the size of the spray nozzle. By enlarging the spay nozzle I can reduce the output pressure to 600 psi and increase the flow substantially. I'm hopeful I will be able to use a pressure washer as a boiler feed pump. If not, I'll look into other pumps.

 

[Steamchick]

Sounds good...
On latent heat: When you cool the gas to liquid, you then need heat - the latent heat - to get it back to a gas as it expands at the same temperature. That latent heat comes from the flames.... Essentially, you are make a "flame refrigerator" - by cooling the flames the R123 gains the heat so you can pass it on to the "expansion zone" (= turbine, condenser, etc.) and take some of the work as mechanical energy (via the turbine).
have you calculated to see it simply compressing the gas exiting the turbine will condense it? - That way the condensing pump will do the work required by the latent heat. - That's how many compressor refrigeration plants do it. But ultimately, all the power comes from the flames, so you need to consider the most efficient way (without stealing electricity from the mains to make it work)..
K2.

 

[Toymaker]

Sounds good...
On latent heat: When you cool the gas to liquid, you then need heat - the latent heat - to get it back to a gas as it expands at the same temperature. That latent heat comes from the flames.... Essentially, you are make a "flame refrigerator" - by cooling the flames the R123 gains the heat so you can pass it on to the "expansion zone" (= turbine, condenser, etc.) and take some of the work as mechanical energy (via the turbine).

OK, now I understand your logic.

have you calculated to see if simply compressing the gas exiting the turbine will condense it? - That way the condensing pump will do the work required by the latent heat. - That's how many compressor refrigeration plants do it. But ultimately, all the power comes from the flames, so you need to consider the most efficient way (without stealing electricity from the mains to make it work)..
K2.

The large expansion area which the turbine exhaust gases will enter is large enough as to allow the R123 to condense from expansion alone; spraying a cooling mist of R123 into the expansion area will ensure complete condensation. Therefore, I don't believe I will need a condensing pump. Suction from the boiler feed pump should be adequate to move the liquid R123 from the turbine's expansion chamber, into the condenser coils, which will function more like cooling coils, removing any remaining excess heat.

 

[Steamchick]

Hi Toymaker. Spell-check changed my words (again!) I typed "condensate" pump, not "condensing" pump.... but my tablet appears to have a limited dictionary and changes about 1 word in 10 for something with a different meaning. (Only my tablet, and on this site, so I don't know how to fix it! Very frustrating!).
I referred to a pump to return the condensate from the condenser output back into the boiler. When the liquid is pressurised by the feed pump, it absorbs "heat" (energy) for the difference of enthalpy from (say) 103kPa = atmospheric (Enthalpy about 227kJ/kg.) - to (say) 212.6kPa - liquid at 50C (Enthalpy = 249.2kJ/kg.) so you need 249.2-227 = 22.2kJ/kg "energy" from the pump.
THEN: When the liquid is heated to boiling point in the boiler tube, it absorbs heat from the burner gases for the difference of enthalpy from liquid to gas: I.E. the LATENT heat of 161.0kJ/kg.
Finally, the gas absorbs heat from the burner gases to raise the temperature and enthalpy from 50C / 410.2kJ/kg. to 180C / 451.0kJ/kg. : I.E. you need 451.0-410.2 = 40.8kJ/kg to raise the temperature of the gas.

To summarise: in total, you need:
Pump "energy" to pressurise condensate from atmospheric pressure to "50C" pressure of 212.61kPa: = 22.2kJ/kg
PLUS the Latent heat at 50C, gained from the burner gases: = 161.0kJ/kg.
And the energy from the burner gases to raise the temperature & pressure to 180C: = 40.8kJ/kg.
So it total, that is 22.2+161.0+40.8 = 224.0kj/kg. of energy input to the gas to get your desired output.
But as mentioned: part is from the pump (Mains powered?) and part from the burner:
The burner is supplying 161.0+40.8=201.8kJ/kg.
The burner limitation for "fuel calorific value" = 150kW, and this probably gives about 100kW to the gas. = 100kJ/second, or 6000kJ/minute:
The gas requires 201.8kJ/kg. so you can heat about 29.7kg/min of gas.
So this is comparable with your planned 5.21kg/sec = 31.26kg/min.
I think your heater is a good one?
For the pump: the energy requirement to pressurise the R123 of 22.2kJ/kg at 5.21kg/sec. translates as 115.7kJ/second = 115.7kW. (a large pressure washer! - More like a large fire engine? - Not "to scale" with the burner/boiler!).
Which suggests there is something wrong? - Can you spot what I have done - but should have done differently?
Cheers,
K2

 

[Toymaker]

Hi Toymaker.
<big snip>
For the pump: the energy requirement to pressurize the R123 of 22.2kJ/kg at 5.21kg/sec. translates as 115.7kJ/second = 115.7kW. (a large pressure washer! - More like a large fire engine? - Not "to scale" with the burner/boiler!).
Which suggests there is something wrong? - Can you spot what I have done - but should have done differently?
Cheers,
K2

My thermodynamic expertise is far too limited to check your work, but I can provide you with a few good examples.

Back in post #107 I used the R123 Physical Properties tables to determine volume flow given the 5.21 kg/sec maximum mass flow; that number worked out as 216 Liters per minute at just over 500 psi. Once I started researching pressure washers to see what was available here in Thailand I quickly discovered that even the largest pressure washer I could buy was still way too small. So, I started researching hydraulic pumps and found many good candidates. I also found this chart to be quite useful. Given that 216 l/min = 57 gpm we see from the chart that for a pump delivering 500 psi at 60gpm will need about 18 HP.

Next example - Looking at one of my favorite historical example steam automobiles: The SES Boiler I've previously mentioned is very close in size and power output to my boiler. Although details and specifications are limited, we do know the boiler and burner combined required 12 KW (16 HP) from the expander (a 4 cylinder steam engine).

So, my boiler feed pump will be a hydraulic pump with an output of just over 533 psi, lets say 600 psi, which will need about 16HP to 18HP to drive it.

Seems I'll need to build a smallish 20 HP piston engine powered by steam bled off from the boiler. That's a bigger accessory drive motor than I had anticipated earlier, but still quite doable.

 

[Toymaker]

Hi Toymaker. Spell-check changed my words (again!) I typed "condensate" pump, not "condensing" pump.... but my tablet appears to have a limited dictionary and changes about 1 word in 10 for something with a different meaning. (Only my tablet, and on this site, so I don't know how to fix it! Very frustrating!).
I referred to a pump to return the condensate from the condenser output back into the boiler. When the liquid is pressurised by the feed pump, it absorbs "heat" (energy) for the difference of enthalpy from (say) 103kPa = atmospheric (Enthalpy about 227kJ/kg.) - to (say) 212.6kPa - liquid at 50C (Enthalpy = 249.2kJ/kg.) so you need 249.2-227 = 22.2kJ/kg "energy" from the pump.
THEN: When the liquid is heated to boiling point in the boiler tube, it absorbs heat from the burner gases for the difference of enthalpy from liquid to gas: I.E. the LATENT heat of 161.0kJ/kg.
Finally, the gas absorbs heat from the burner gases to raise the temperature and enthalpy from 50C / 410.2kJ/kg. to 180C / 451.0kJ/kg. : I.E. you need 451.0-410.2 = 40.8kJ/kg to raise the temperature of the gas.

To summarise: in total, you need:
Pump "energy" to pressurise condensate from atmospheric pressure to "50C" pressure of 212.61kPa: = 22.2kJ/kg
PLUS the Latent heat at 50C, gained from the burner gases: = 161.0kJ/kg.
And the energy from the burner gases to raise the temperature & pressure to 180C: = 40.8kJ/kg.
So it total, that is 22.2+161.0+40.8 = 224.0kj/kg. of energy input to the gas to get your desired output.
But as mentioned: part is from the pump (Mains powered?) and part from the burner:
The burner is supplying 161.0+40.8=201.8kJ/kg.
The burner limitation for "fuel calorific value" = 150kW, and this probably gives about 100kW to the gas. = 100kJ/second, or 6000kJ/minute:
The gas requires 201.8kJ/kg. so you can heat about 29.7kg/min of gas.
So this is comparable with your planned 5.21kg/sec = 31.26kg/min.
I think your heater is a good one?
For the pump: the energy requirement to pressurise the R123 of 22.2kJ/kg at 5.21kg/sec. translates as 115.7kJ/second = 115.7kW. (a large pressure washer! - More like a large fire engine? - Not "to scale" with the burner/boiler!).
Which suggests there is something wrong? - Can you spot what I have done - but should have done differently?
Cheers,
K2

I took a second look at your logic and I believe I see a problem: you're only requiring the feed pump to supply 212 kPa or 31 psi. But the feed pump needs to supply at least 3675 kPa (533 psi) to keep the Freon flowing towards the turbine nozzle and not back-flowing into the feed pump.

Also, I've already demonstrated the burner's capable of doubling (and likely tripling) my original designed fuel burn and kW output, and I haven't even tried Green Twin's method of supplying pressurized fuel into the nozzle, which would likely increase the fuel burn even further. So I don't believe you need to limit the burner's output to 100 kW.

 

[Steamchick]

Thanks Toymaker. This is beginning to look like a real project in my head...
My nominal 100kW was not a burner limit, but just to give a sense of scale, but loosely, is 2/3rds of the 150kw of fuel power you are starting with (nominal value). Any boiler around 60~70% efficiency of creating "pressurised vapour from liquid" is reasonable, some are more or less efficient, but you must factor that "Efficiency" in to the calculations from "fuel power" to "vapour power". Then you must consider engine efficiency: I don't know if a steam engine as you suggest will be 40% or 80% efficient at producing shaft power from vapour power, but if you guess at 60% engine efficiency, and you have 100kW of vapour power, you seem to need 15/0.6 = 25kW of vapour power to get the shaft power you need for the compressor. - Or something like that?
I'm sure you are closer to what you want with your calculations, I'm simply getting a perspective in my head.
Thanks,
K2

 

[Toymaker]

Thanks Toymaker. This is beginning to look like a real project in my head...
My nominal 100kW was not a burner limit, but just to give a sense of scale, but loosely, is 2/3rds of the 150kw of fuel power you are starting with (nominal value). Any boiler around 60~70% efficiency of creating "pressurised vapour from liquid" is reasonable, some are more or less efficient, but you must factor that "Efficiency" in to the calculations from "fuel power" to "vapour power". Then you must consider engine efficiency: I don't know if a steam engine as you suggest will be 40% or 80% efficient at producing shaft power from vapour power, but if you guess at 60% engine efficiency, and you have 100kW of vapour power, you seem to need 15/0.6 = 25kW of vapour power to get the shaft power you need for the compressor. - Or something like that?
I'm sure you are closer to what you want with your calculations, I'm simply getting a perspective in my head.
Thanks,
K2

This entire steam turbine project is designed to be a test bed for me; a starting point to learn from. To that end, I've intentionally designed the burner with twice the capacity I need, (it's just lucky that it likely has 3 times the capacity I need). I plan to design and build a boiler that will exceed my needs by 100%; that way, in case my boiler has only 50% efficiency, I will still have 100% of the output I want. The turbine was also designed to produce 200% of the 100 HP I want. Also, to the extent possible, I have designed and constructed all the various parts to be easily replaced with upgraded parts. The nozzle in the burner can be replaced with a larger capacity nozzle. The leaf blower can be enlarged to supply more air to the burner. All the boiler tubes can be removed and replaced with a different design and/or larger, longer tubes if I need to increase boiler output. The turbine nozzle is a discrete part that can be exchanged with a larger capacity nozzle. Turbine blisks (bladed disks) can be individually replaced allowing me to try out different designs.

 

[Steamchick]

Just a word on "experience" from Boiler design books: Noting that they are referring to flue gases passing through tubes, surrounded by boiler water (at steam temperature), the length is ALWAYS limited to a maximum of 80 diameters, as after that there is insufficient heat in the flue gases to warrant any extra length. I suspect that you have the "Flash" boiler concept with your tubes, so perhaps they can be longer, as the exhaust gases start hot and cool as they pass through the heat exchanger, rather than exchanging heat at a constant temperature? - But my point is, I suspect that the "Flash boiler" guys have a concept of maximum length of tube relative to bore diameter, beyond which efficiency of heat exchange of the whole system drops off. And I also believe (from some other work I did 45 years ago?) that the bore will define a maximum flow permissible, so I guess you have based your monotube boiler on that.
I know that a guy - nickname Windy - who made a world speed record breaking flash boiler steam boat that used 3 separate chambers powered by paraffin burners - from a common pump supply - to fire 3 water-steam coils in parallel - again fed from a common water pump - because he needed the length of tube, but could not do it in a single coil... - Hence the 3 burners and 3 coils, so the tube length was within the "pass-through" length for his bore of tube. I just don't know tube lengths and bore. But I believe I used some tables/formula from a reference book in the '70s.

https://www.engineersedge.com/pipe_flow_capacity.htmhttps://michigan.swagelok.com/-/med...0 feet of,is not to exceed 5 feet per second.The Michelin Swagelok graphs remind me of what I looked at in the 1970s...
You may want to research this and consider a multi-tube arrangement to avoid choking of the single tube?
K2

 

[Toymaker]

Just a word on "experience" from Boiler design books: Noting that they are referring to flue gases passing through tubes, surrounded by boiler water (at steam temperature), the length is ALWAYS limited to a maximum of 80 diameters, as after that there is insufficient heat in the flue gases to warrant any extra length.

I'm not sure you can conclude tube lengths of a fire tube boiler should be roughly equal to tube lengths in a water tube boiler.

I suspect that you have the "Flash" boiler concept with your tubes, so perhaps they can be longer, as the exhaust gases start hot and cool as they pass through the heat exchanger, rather than exchanging heat at a constant temperature?

Automobile boilers typically have a long coil of tube placed just before the exhaust gases leave the boiler, called an "economizer", which transfers the heat from lower temp exhaust gases into the boiler tubes in order to increase fuel efficiency.

- But my point is, I suspect that the "Flash boiler" guys have a concept of maximum length of tube relative to bore diameter, beyond which efficiency of heat exchange of the whole system drops off. And I also believe (from some other work I did 45 years ago?) that the bore will define a maximum flow permissible, so I guess you have based your monotube boiler on that.

Yes, you are correct. I'm pretty sure the maximum velocity of a fluid through a tube is the speed of sound of the fluid.

The max volume flow of R123 through the boiler tubes at the feed pump outlet was determined by multiplying the speed of sound (at 30C at 4 kPa), which is 700 m/s, times the cross sectional area of the tubes ID, which is 0.0000283 m2 (read as square meters) to get 0.0198 m3/s (cubic meters/s). Our favorite R123 Physical Properties chart tells us that at 533 psi, the liquid density is at 654.7 kg/m3. So, 0.0198 m3/s x 654 kg/m3 = 12.97 kg/sec.

My turbine needs only 5.21 kg/sec maximum mass flow. Even assuming a boiler efficiency of 60% that still gives me 7.8 Kg/s or 150% of what the turbine can use at max output.

I know that a guy - nickname Windy - who made a world speed record breaking flash boiler steam boat that used 3 separate chambers powered by paraffin burners - from a common pump supply - to fire 3 water-steam coils in parallel - again fed from a common water pump - because he needed the length of tube, but could not do it in a single coil... - Hence the 3 burners and 3 coils, so the tube length was within the "pass-through" length for his bore of tube. I just don't know tube lengths and bore. But I believe I used some tables/formula from a reference book in the '70s.

https://www.engineersedge.com/pipe_flow_capacity.htmhttps://michigan.swagelok.com/-/med...0 feet of,is not to exceed 5 feet per second.The Michelin Swagelok graphs remind me of what I looked at in the 1970s...
You may want to research this and consider a multi-tube arrangement to avoid choking of the single tube?
K2

Thanks for the links, they are interesting.

 

[Steamchick]

Hi Toymaker,
I agree "I'm not sure you can conclude tube lengths of a fire tube boiler should be roughly equal to tube lengths in a water tube boiler." - but my principle was that the "experts" have historically developed practical limits on the efficiency of adding more metal, on a law of diminishing returns. So my proposal was that you check if there are limitations other than "speed of sound" that the Flash Boiler Guys use...?? It may suggest (As Windy's boiler) that a few tubes in parallel are better than the same length of tube in a single run? - But perhaps better to check now rather than find out the lessons later...?
K2

 

[ajoeiam]

Just a word on "experience" from Boiler design books: Noting that they are referring to flue gases passing through tubes, surrounded by boiler water (at steam temperature), the length is ALWAYS limited to a maximum of 80 diameters, as after that there is insufficient heat in the flue gases to warrant any extra length.

Interesting - - - wonder if that ratio is also true in a water tube boiler.
Dunno but given the large boilers are using some about 2.5" dia tubes and the runs are some 60 to 70' I don't think that ratio applies.

Thoughts please - - - re: water tube boilers specifically.

TIA

 

[Toymaker]

Hi Toymaker,
I agree "I'm not sure you can conclude tube lengths of a fire tube boiler should be roughly equal to tube lengths in a water tube boiler." - but my principle was that the "experts" have historically developed practical limits on the efficiency of adding more metal, on a law of diminishing returns. So my proposal was that you check if there are limitations other than "speed of sound" that the Flash Boiler Guys use...?? It may suggest (As Windy's boiler) that a few tubes in parallel are better than the same length of tube in a single run? - But perhaps better to check now rather than find out the lessons later...?
K2

K2, I'm a strong believer in, "no need to re-invent the wheel". If I can copy existing, working boiler designs I'm totally happy to do that. Because my end goals are to build a small, light weight boiler in the 200 HP range, I've looked mostly at newer (1960s) steam cars for inspiration and guidance. My favorite example is the SES boiler-burner, which had 75.3 square feet of tubing surface area. To duplicate that surface area I will need 900 feet of 8mm OD tubing; for now, that is my plan, because I know it works. I will continue to study and learn about the thermodynamics of heat transfer, but in the end, copying a known working design is far better than anything I, a beginner student, could possibly hope to design.

I will, however, make a major modification in tubing layout. Because I'm using a turbine and the SES car used a piston engine, my turbine needs far higher, and continuous flow rates then the SES piston engine. Which is why I chose to use parallel tubes.

I too agree there are benefits to using parallel tubes. Do you recall seeing this drawing from post #19 in my Ambitious ORC Turbine thread? It's a schematic diagram depicting how I plan to start with a single tube from the boiler feed pump, which then branches into 2 tubes, then into 4, and finally into 8 parallel tubes. The drawing should show all 8 tubes connected to the octagon shaped junction on the right side of the drawing, but I've been too lazy to draw all the proper plumbing lines.

 

[Toymaker]

Interesting - - - wonder if that ratio is also true in a water tube boiler.
Dunno but given the large boilers are using some about 2.5" dia tubes and the runs are some 60 to 70' I don't think that ratio applies.

Thoughts please - - - re: water tube boilers specifically.

TIA

This photo shows an unfinished water tube boiler designed to supply over 800 HP. The final boiler will have a 25" diameter, a 21" length, and contain 800 feet of 1/2" diameter stainless steel tube. 80 diameters of 1/2" works out to 40" or just over 3 feet.