# Forced Air Diesel Fuel Burner

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#### Steamchick

##### Well-Known Member
HMEM Supporting Member
I have a notion that water tubes, where full of water, will carry maybe 250 times the mass of steam at the same velocity. (Depends on the pressure, ergo the density of the steam). But the carrying capacity of tubes for liquid is different to the vapour limit anyway, so maybe you have to consider at what point the liquid will convert to vapour, then consider the 2 lengths separately?
I am a total amateur for this, just asking so I can learn from others!
I have never seen the inovative branching before... I would have guessed a manifold into 8 tubes at the start would be simpler to manufacrture? But have never tried, so do not know?
K2

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Just an odd view, very much my opinion, so not a rule at all. But many successful products have "not the best" design in them. So, if something works at 800 turns of tube does not mean it is the best, or just the size of coil someone bought.... possibly it would be better at a properly designed length, or maybe that is the properly designed optimum length.... Who knows?
Which is why I question many things, and thy to verify them by leaning the sums....
Then I gain knowledge to design other sizes of "job".
Enjoy!
This is a curious project. Far more ambitious than I would try, with lots for me to learn about.
K2

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
I have a notion that water tubes, where full of water, will carry maybe 250 times the mass of steam at the same velocity. (Depends on the pressure, ergo the density of the steam). But the carrying capacity of tubes for liquid is different to the vapour limit anyway, so maybe you have to consider at what point the liquid will convert to vapour, then consider the 2 lengths separately?

Agreed. Calculating each section of 1, 2, 4, & 8 parallel tube lengths is the ideal approach, and of course it's also the most difficult.

One of the very innovative techniques employed in the SES boiler design to reduce tube length is seen in the routing of the tubes. Feed water flows into a single tube which coils around making several rows of tube coils at the exhaust end of the boiler, which is how all tube boilers are configured. But then the tube is routed directly into the hottest part of the boiler, in the area typically reserved for the superheat coils. At this point the tube is still filled with liquid water and therefore picks up tremendous heat; that tube is then routed back to where it left the row of initial coils and from that point forward, is laid out as any other mono-tube boiler, with the exception that the superheat section is now in a slightly different location.

This technique of placing tubing carrying liquid in the hottest part of the boiler will have dual benefits for my boiler; not only will it reduce needed tube length, but it will also cool the super hot exhaust gases down to a more favorable temperature for the R123, making it easier to avoid accidentally overheating and decomposing the Freon.

I am a total amateur for this, just asking so I can learn from others!
I have never seen the inovative branching before... I would have guessed a manifold into 8 tubes at the start would be simpler to manufacrture? But have never tried, so do not know?
K2

K2, you & I will both be learning together in this area. At the moment, my knowledge of boiler design is at a beginner level, so unless I'm incredibly lucky, whatever I build wont be the peak of efficiency. But I'm also pretty sure that whatever I build will at least work well enough to drive my turbine.

#### HMEL

##### Well-Known Member
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
You are correct boiler tubes are based on surface area required and mass flow through them. Unless forced circulation is involved the tube area is designed for a range of velocities and is a critical calculation if thermal differences are driving the flow as in many boiler tube furnaces. On rare occasion you might see a pump installed on the economizer to ensure the economizer has enough water flow for all operating conditions. It depends on the circuit.

You must consider not only the pressure drop of the working fluid in the tubes but also the pressure drops of the exhaust gases over the tube bank. In general you want nucleate boiling not flash boiling as the water is essentially cooling the tube walls.

The technique for doing this first starts with a mass and thermal balance. Then surface area needed is calculated based on heat transfer and the tubes selected based on these parameters. Sometimes based on experience a ratio is used say for instance drum volume and its surface area. Different size tubes are considered in the design when searching for best choices for performance and economics.

The fans are sized based on the pressure drop through the boiler as well as the fuel input.

The efficiency of a boiler is calculated by measuring the fuel input heating value and then measuring the mass of the flue gas and heat it contains leaving the stack.

Just thought I would throw that in as boiler efficiency is a misunderstood topic.

A well designed boiler usually takes a good engineering team even for small units.

HMEL

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker. Just exploring ideas.... on your treble-bifurcated tube arrangement: (I called it that as it divides into 2, 3 times I.E. 2 cubed = 8 - Is it that simple?). Considering Mass flow and pressure, at each junction, aside of turbulent restriction, there is "simply" a halving of pressure, as the pressure in the inlet tube to the junction sees double the area ahead of it. So "hydraulically" the pressure is halved in accordance with the increase in CSA.
So initial pressure: 600psi: 4137kPA: - from the data tables: That is off the scale: Which has a maximum at 183C of 3627kPa...
So for simplicity, to try and understand what the bifurcated arrangement is doing:
At the first split: 5.7Kg/sec of liquid enters the coil at (Say) 3600kPa.=> It gets heated (I am worried it will get over-heated?) => something that will then be halved. So the pressure will be 1800kPa (& at 141C?)+ the pressure & temperature developed by heating... (say 67kJ/Kg enthalpy: x 5.7 = 382kJ?) - So we maintain liquid at 182C and pressure of 3600kPa (or thereabouts).
So the second stage leg will have half the flow rate entering (after the first Bifurcation) = 2.85kg/sec at 182C and 3600kPa - because we have put in the heat to maintain the temperature and pressure.
So assuming this continues at each stage, where each stage is the appropriate length to get the heat from the exhaust without overheating (at which point the chemical degrades), or falling short due to cooling exhaust gases, then I am not sure at what point the liquid will turn to gas...?

Thinking a different way... The input velocity of 5.7kg/sec liquid is intended to become an exit velocity of 5.7kg/sec of gas ..
So input liquid density = 655kg/m3: x flow rate of 5.7kg/sec. = 0.0087m3/sec.
Output gas density = 449kg/m3 => flow rate = 0.0127 m3/sec. - which is only 1.5 times the inlet flow rate: But you are using 8 times the CSA of exit tubing compared to inlet tubing, so the gas velocity will be 1.5/8 = 18% of the inlet velocity. - I must have missed something to have such crazy answers? - What size are your nozzles? Maybe that should be the output CSA that I should use?
So now I have totally confused myself with the numbers!
Do you have some numbers to make sense of this, as I am obviously doing the wrong analysis!
Thanks,
K2

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Thanks HMEL, your explanation helps me a lot! - I think the comment "The technique for doing this first starts with a mass and thermal balance. Then surface area needed is calculated based on heat transfer and the tubes selected based on these parameters. Sometimes based on experience a ratio is used say for instance drum volume and its surface area. Different size tubes are considered in the design when searching for best choices for performance and economics." - is most pertinent.

I think that is what toymaker is trying to do with the "treble bifurcated" branching he has devised. Reading about his study of the SES boiler configuration, I think he plans to copy that format of tube location, but I am not sure at what point he will introduce the stages of bifurcation, nor how to work-out the tube lengths/heat flows at each stage. I think it has to come from the heat-flow and enthalpy tables, based on a model of the exhaust gas cooling curve - maybe average exhaust gas temperature at each of 4 stages in the boiler?

Unlike with steam, he has a top-limit of temperature that he must not exceed, without dangerous chemical breakdown of the gas. So I think some careful design/modelling of the heat flow is required here? He is planning on controlling the heat flow with a control that regulates the fuel input (I think?). So feedback from actual gas temperature sensors at the end of each run just before the bifurcation seem like a good idea?

Toymaker, Am I on the right track here?
K2

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
Hi Toymaker. Just exploring ideas.... on your treble-bifurcated tube arrangement: (I called it that as it divides into 2, 3 times I.E. 2 cubed = 8 - Is it that simple?). Considering Mass flow and pressure, at each junction, aside of turbulent restriction, there is "simply" a halving of pressure, as the pressure in the inlet tube to the junction sees double the area ahead of it. So "hydraulically" the pressure is halved in accordance with the increase in CSA.
So initial pressure: 600psi: 4137kPA: - from the data tables: That is off the scale: Which has a maximum at 183C of 3627kPa...

So for simplicity, to try and understand what the bifurcated arrangement is doing:
At the first split: 5.7Kg/sec of liquid enters the coil at (Say) 3600kPa.=> It gets heated (I am worried it will get over-heated?) => something that will then be halved. So the pressure will be 1800kPa (& at 141C?)+ the pressure & temperature developed by heating... (say 67kJ/Kg enthalpy: x 5.7 = 382kJ?) - So we maintain liquid at 182C and pressure of 3600kPa (or thereabouts).
So the second stage leg will have half the flow rate entering (after the first Bifurcation) = 2.85kg/sec at 182C and 3600kPa - because we have put in the heat to maintain the temperature and pressure.
So assuming this continues at each stage, where each stage is the appropriate length to get the heat from the exhaust without overheating (at which point the chemical degrades), or falling short due to cooling exhaust gases, then I am not sure at what point the liquid will turn to gas...?

Thinking a different way... The input velocity of 5.7kg/sec liquid is intended to become an exit velocity of 5.7kg/sec of gas ..
So input liquid density = 655kg/m3: x flow rate of 5.7kg/sec. = 0.0087m3/sec.
Output gas density = 449kg/m3 => flow rate = 0.0127 m3/sec. - which is only 1.5 times the inlet flow rate: But you are using 8 times the CSA of exit tubing compared to inlet tubing, so the gas velocity will be 1.5/8 = 18% of the inlet velocity. - I must have missed something to have such crazy answers? - What size are your nozzles? Maybe that should be the output CSA that I should use?
So now I have totally confused myself with the numbers!
Do you have some numbers to make sense of this, as I am obviously doing the wrong analysis!
Thanks,
K2

Before we get into numbers for my system, we need to have a brief discussion on how hydraulic systems work.

I'm 99% certain that static hydraulic pressure remains the same in all line branches, no matter how many times you split or add to the initial single input line. Example: If I seal all 8 output lines of my boiler at their exits, and set the feed pump to supply 600 psi to the single input line, then all 8 output lines would show 600 psi. You could even use different size pipe in all 8 lines and the pressure would still read 600 psi in all 8 different size lines.

Hydraulic pressures only changes when you un-cap the 8 outputs and fluid is flowing thru the boiler tubes. Pressure drops will occur over the length of tube due to flow resistance, but the pressure will not drop by 1/2 at each 1 into 2 split. And because I'm using the same diameter tube in all parallel lines, the pressure drops from flow resistance thru the tubes will all be equal. What will drop by 1/2 at each split is the rate of flow.

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
<snip>

You must consider not only the pressure drop of the working fluid in the tubes but also the pressure drops of the exhaust gases over the tube bank. In general you want nucleate boiling not flash boiling as the water is essentially cooling the tube walls.

<snip>

A well designed boiler usually takes a good engineering team even for small units.

HMEL

I understand that flowing fluid pressure drops inside a pipe the further away the fluid gets from the feed pump in a non-heated pipe, but does the pressure actually drop inside boiler tubes when the external surface of the tubes are being supplied with large amounts of heat from the burner exhaust gasses? It seems logical that as the working fluid inside the tubes is heated and expands, that the working fluid will supply it's own pressure separate from what the feed pump initially supplied, yes? In fact, if the working fluid is ever heated to a point where hot vapor pressures exceed feed pump pressure, wouldn't the working fluid begin flowing backwards into the feed pump?

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
<snip>

I think that is what toymaker is trying to do with the "treble bifurcated" branching he has devised. Reading about his study of the SES boiler configuration, I think he plans to copy that format of tube location,

Yes, I will be mostly copying the SES boiler tube routing.

but I am not sure at what point he will introduce the stages of bifurcation,

The first tube split will take place just after the single tube leaves the hottest section of the boiler, nearest the burner's exhaust.

nor how to work-out the tube lengths/heat flows at each stage. I think it has to come from the heat-flow and enthalpy tables, based on a model of the exhaust gas cooling curve - maybe average exhaust gas temperature at each of 4 stages in the boiler?

I really wish I had a Fluid Dynamics software package to play with for a few weeks. But alas, I do not, so I will make the best guesstimates I can using lots of logic combined with the limited thermodynamics knowledge I posses.

Unlike with steam, he has a top-limit of temperature that he must not exceed, without dangerous chemical breakdown of the gas. So I think some careful design/modelling of the heat flow is required here? He is planning on controlling the heat flow with a control that regulates the fuel input (I think?).

Yes, a micro controller will have complete control over fuel flow, and air flow into the burner, and feed pump pressure, and will be able to read pressures and temperatures at several locations.

So feedback from actual gas temperature sensors at the end of each run just before the bifurcation seem like a good idea?

Toymaker, Am I on the right track here?
K2

I'ld love to have all those temperature sensors inside the boiler as you've suggested, BUT, I haven't yet devised a way to keep the sensor's output wires from burning up as they will be exposed to exhaust gases.

#### HMEL

##### Well-Known Member
I understand that flowing fluid pressure drops inside a pipe the further away the fluid gets from the feed pump in a non-heated pipe, but does the pressure actually drop inside boiler tubes when the external surface of the tubes are being supplied with large amounts of heat from the burner exhaust gasses? It seems logical that as the working fluid inside the tubes is heated and expands, that the working fluid will supply it's own pressure separate from what the feed pump initially supplied, yes? In fact, if the working fluid is ever heated to a point where hot vapor pressures exceed feed pump pressure, wouldn't the working fluid begin flowing backwards into the feed pump?
No the working fluid does not supply the energy for its flow it always moves to the low pressure it sees. And this is a design function barring a stuck check valve, control valve, or a broken cap blocking a tube.

Pressure drop is key in fluid flow. Flow is always from high pressure to low pressure. Indeed under high pressure and a poor design back flow can occur. But under that condition the boiler will not work. The feed pump works against the boiler pressure and
supplies the boiler drum at a pressure higher then the operating boiler. Control valve and a check valve is installed for safety. There is also a non return valve on the boiler when the steam heads to the turbine or process.

And it is the velocity in the tube which determines the Reynolds number and heat transfer. The fluid is cooling the tube walls in the furnace and on average the difference in wall temperature and fluid is 70 to 80 degrees. For most boilers circulation is developed due to thermal differences of the fluid in the furnace walls but the wall must be designed so flow develops and heat is extracted from the furnace. You can force circulate the fluid. The water releases the steam in the drum unless you operate at critical which is a special design all by itself. There are three or four circuits which have to be considered so that the boiler works properly.

At no point in the furnace are the tubes allowed to become uncovered as the metal temperature will rise to failure.

Now for the final point. I suspect your design is working at the critical pressure and this requires a different approach as you will not have a boiler drum but the working fluid must remain a liquid until it is released to the turbine. A different can of worms. And there are a few steam boilers that operate at the critical pressure of water which if I remember right is 3206 psi. The fluid goes immediately to vapor after leaving the boiler.

HMEL

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
No the working fluid does not supply the energy for its flow

This is an old Greek steam "engine". How it works is pretty straight forward,...steam from the kettle flows into the ball and exits out of the directional nozzles, causing the ball to spin. This engine demonstrates how heat from a fire pressurizes a working fluid (steam), which then supplies the energy needed to cause the working fluid to flow through the steam riser tubes, into the ball, and out of the nozzles.

In a water tube boiler, the hot combustion gases are adding heat to the working fluid inside the water tubes creating "steam", which expands inside the tubes, thereby inducing fluid flow in addition to the flow provided by the feed pump. Even if there were no feed pump to supply pressure to force the fluid to flow, the pressure from the expanding working fluid would supply the energy to cause the fluid to flow.

it always moves to the low pressure it sees. And this is a design function barring a stuck check valve, control valve, or a broken cap blocking a tube.

Pressure drop is key in fluid flow. Flow is always from high pressure to low pressure.

Yes, of course pressure always flows from high to low pressure.

Indeed under high pressure and a poor design back flow can occur. But under that condition the boiler will not work. The feed pump works against the boiler pressure and
supplies the boiler drum at a pressure higher then the operating boiler. Control valve and a check valve is installed for safety. There is also a non return valve on the boiler when the steam heads to the turbine or process.

And it is the velocity in the tube which determines the Reynolds number and heat transfer.

Yes ! And by branching the tubes in my design, the fluid velocity in each tube is reduced by 1/2 at each branching, thereby allowing the fluid to spend more time inside the tubes and extract more heat from the combustion gasses.

The fluid is cooling the tube walls in the furnace and on average the difference in wall temperature and fluid is 70 to 80 degrees. For most boilers circulation is developed due to thermal differences of the fluid in the furnace walls but the wall must be designed so flow develops and heat is extracted from the furnace. You can force circulate the fluid. The water releases the steam in the drum unless you operate at critical which is a special design all by itself. There are three or four circuits which have to be considered so that the boiler works properly.

At no point in the furnace are the tubes allowed to become uncovered as the metal temperature will rise to failure.

Now for the final point. I suspect your design is working at the critical pressure and this requires a different approach as you will not have a boiler drum but the working fluid must remain a liquid until it is released to the turbine. A different can of worms. And there are a few steam boilers that operate at the critical pressure of water which if I remember right is 3206 psi. The fluid goes immediately to vapor after leaving the boiler.

HMEL

Yes, I believe all US nuclear subs run their steam systems at critical temperature & pressure. Interesting side bar; steam leaking from a 3000 psi pipe is invisible. So sailors are taught to look for leaks in the steam pipes by sweeping the pipe with a broom,...when the bristles on the broom are sliced off and fall to the floor, the leak has been found.

#### HMEL

##### Well-Known Member
This is an old Greek steam "engine". How it works is pretty straight forward,...steam from the kettle flows into the ball and exits out of the directional nozzles, causing the ball to spin. This engine demonstrates how heat from a fire pressurizes a working fluid (steam), which then supplies the energy needed to cause the working fluid to flow through the steam riser tubes, into the ball, and out of the nozzles.

View attachment 139667
In a water tube boiler, the hot combustion gases are adding heat to the working fluid inside the water tubes creating "steam", which expands inside the tubes, thereby inducing fluid flow in addition to the flow provided by the feed pump. Even if there were no feed pump to supply pressure to force the fluid to flow, the pressure from the expanding working fluid would supply the energy to cause the fluid to flow.

Yes, of course pressure always flows from high to low pressure.

Yes ! And by branching the tubes in my design, the fluid velocity in each tube is reduced by 1/2 at each branching, thereby allowing the fluid to spend more time inside the tubes and extract more heat from the combustion gasses.

Yes, I believe all US nuclear subs run their steam systems at critical temperature & pressure. Interesting side bar; steam leaking from a 3000 psi pipe is invisible. So sailors are taught to look for leaks in the steam pipes by sweeping the pipe with a broom,...when the bristles on the broom are sliced off and fall to the floor, the leak has been found.
Flow begins when the vessel reaches a pressure above atmospheric which is 14.7 not before. No pressure differential no flow. The pressure is a function of pv=mrt. Heat just raises the state properties of the fluid.

Each tube has to be calculated for fitting + entry losses + friction losses. Friction losses are a function of line length and the appropriate Reynolds number. So each split has a significant pressure drop and is reduced by far more then half of the original flow. These are restrictions and work like valves.

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Re: Pressure and flow: Yes Toymaker, I agree that hydraulic pressure in a fluid is constant.... if it isn't released at one end creating a pressure gradient.. and hence flow.( I think ?).
But we are talking about a "flash-tube boiler" Arrangement, where pressurised fluid is passed into a tube as a liquid, and leaves as a gas? - Or am I wrong in this?
The simple hydraulic model you explain only works for a system where the energy within the hydraulic fluid is all that exists. But as we are discussion a boiler, we are developing a system where first we raise the temperature and pressure of the liquid by heating so the liquid enthalpy increases, then a change of state occurs as we supply more energy to the liquid as it remains constant for temperature and pressure by the latent heat absorbed converts the liquid to gas, and the third stage where the gas is heated to increase the temperature and pressure further.
I was "opening the discussion" on aspects I do not really understand - the increase of cross-sectional area at each bifurcation in your design. As this discontinuity of cross-section area occurs, I think a change of state is likely, where the liquid converts some of the Liquid enthalpy into Latent heat and the liquid boils instantaneously. The liquid-gas mix then absorbs more energy in the next run, until the next bifurcation, then another "boiling phase" occurs.
The gas is performing:
P1 x V1 / T1 =>P2 x V2 / T2 at a bifurcation without any energy input, (considering volume expressed as mass flow-rate?) but where the energy is being input from hotter exhaust gases, and the fluid temperature is rising as well, the pressure must be rising - driven by the "Heat pump" action. Not a mechanical valved pump. This is how flash boilers (and conventional superheaters) work. It is the heat flow into the fluid that raises the temperature and pressure but resists "back-flow" - which I agree would naturally occur if you did NOT input any heat.
Simply: back in the 1960s, my Dad bought a gas refrigerator. My first lesson in thermodynamics as an 8 year old boy was that the pressure rises in the "gas fired boiler" coils, from cold - low pressure - liquid, to "Hot, high-pressurised gas". The Burner was the "Heat pump" developing the fluid pressure. This fluid was then cooled outside the fridge, but otherwise remained at a high pressure, until it was vented into the cooling chamber (chiller) inside the fridge. Then the gas expands into the chiller inside the coldest part of the 'fridge, and "wants heat from inside the fridge" to do so. SO in expanding as a gas, it uses the "inside heat" (pressure) until colder than the 'fridge, then sucks heat from the inside of the fridge to make up the balance. The "fridge-heated" gas then leaves at the lowest pressure point to return to the "boiler" where it is heated, and pressurised to a liquid state. I.E. the boiler introduces a pressure gradient to high pressure, temperature and volume (flow rate) then, after the external cooler has cooled the fluid at high pressure the chiller provides the pressure gradient from high to low - absorbing energy from inside the fridge to do so.
Hence, I think your system is following the similar heat flow pressure temperature diagram?
The pressure gradient while heating in your boiler uses the heat input to create higher pressure at the output than at the input point. Hence no back flow against this pressure, providing the burner does not stop!
But I am still a total amateur in thermodynamics, which is why I am learning from this thread.
K2

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
Wikipedia explains this as the "Vapor absorption cycle".
K2

#### Steamchick

##### Well-Known Member
HMEM Supporting Member
I am trying to compare this diagram to your system:
By The original uploader was Keenan Pepper at English Wikipedia. - Transferred from en.wikipedia to Commons., CC BY-SA 3.0, File:RefrigerationTS.png - Wikimedia Commons

I think if you simply re-write:
4=> 5 = expansion of liquid to liquid and gas at first phase of boiler. (post-pump). This initially uses "liquid enthalpy" (pressure) to expand the liquid into liquid and gas before the exhaust heat is added.
5=> 1 = Mid-part of boiler tube arrangement: from lowest pressure and temperature to some further mid-point where all the liquid has boiled (Latent heat enthalpy). Temperature constant as liquid boils to gas. Change of state latent heat enthalpy using the heat from the Hot exhaust gas.
1=> 2 = remaining part of your boiler tube arrangement where the gas pressure is being raised (superheating gas enthalpy)
2=> 3 = Turbine use of superheated gas: Temperature drop, pressure drop and enthalpy drop.
3=> 4 = condenser (expansion of gas causing pressure drop and temperature constant while latent heat is removed and condensate forms.)
In fact, I reckon the arrangement you have planned will combine 4 =>5 and 5 => 1 into a single curved line of 4 => 1, but this diagram has the individual steps separated for explanation. The total of 4 => 2 is the boiler part.

Does this help you? Maybe you can do a diagram to show your thinking: I.E. pressures, temperatures, enthalpy?

"I think, but I am not so sure I am right" - !!
K2

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
Re: Pressure and flow: Yes Toymaker, I agree that hydraulic pressure in a fluid is constant.... if it isn't released at one end creating a pressure gradient.. and hence flow.( I think ?).
But we are talking about a "flash-tube boiler" Arrangement, where pressurised fluid is passed into a tube as a liquid, and leaves as a gas? - Or am I wrong in this?

I think you just explained how all water tube boilers work,...a feed pump pushes a fluid into the boiler tube as a liquid and at some pressure supplied by the feed pump. Heat is supplied to the liquid filled tubes and turns the liquid into a gas.

The simple hydraulic model you explain only works for a system where the energy within the hydraulic fluid is all that exists. But as we are discussion a boiler, we are developing a system where first we raise the temperature and pressure of the liquid by heating so the liquid enthalpy increases, then a change of state occurs as we supply more energy to the liquid as it remains constant for temperature and pressure by the latent heat absorbed converts the liquid to gas, and the third stage where the gas is heated to increase the temperature and pressure further.
I was "opening the discussion" on aspects I do not really understand - the increase of cross-sectional area at each bifurcation in your design. As this discontinuity of cross-section area occurs, I think a change of state is likely, where the liquid converts some of the Liquid enthalpy into Latent heat and the liquid boils instantaneously. The liquid-gas mix then absorbs more energy in the next run, until the next bifurcation, then another "boiling phase" occurs.

I'm no expert in thermodynamics either. I believe what will occur at the 1 into 2 splits is that the mass flow rate in each of the 2 parallel tubes will be roughly 1/2 the original mass flow in the single tube.

The gas is performing:
P1 x V1 / T1 =>P2 x V2 / T2 at a bifurcation without any energy input, (considering volume expressed as mass flow-rate?) but where the energy is being input from hotter exhaust gases, and the fluid temperature is rising as well, the pressure must be rising - driven by the "Heat pump" action. Not a mechanical valved pump. This is how flash boilers (and conventional superheaters) work. It is the heat flow into the fluid that raises the temperature and pressure but resists "back-flow" - which I agree would naturally occur if you did NOT input any heat.
Simply: back in the 1960s, my Dad bought a gas refrigerator. My first lesson in thermodynamics as an 8 year old boy was that the pressure rises in the "gas fired boiler" coils, from cold - low pressure - liquid, to "Hot, high-pressurised gas". The Burner was the "Heat pump" developing the fluid pressure. This fluid was then cooled outside the fridge, but otherwise remained at a high pressure, until it was vented into the cooling chamber (chiller) inside the fridge. Then the gas expands into the chiller inside the coldest part of the 'fridge, and "wants heat from inside the fridge" to do so. SO in expanding as a gas, it uses the "inside heat" (pressure) until colder than the 'fridge, then sucks heat from the inside of the fridge to make up the balance. The "fridge-heated" gas then leaves at the lowest pressure point to return to the "boiler" where it is heated, and pressurised to a liquid state. I.E. the boiler introduces a pressure gradient to high pressure, temperature and volume (flow rate) then, after the external cooler has cooled the fluid at high pressure the chiller provides the pressure gradient from high to low - absorbing energy from inside the fridge to do so.
Hence, I think your system is following the similar heat flow pressure temperature diagram?

If you Google "Rankine cycle enthalpy diagram" you'll find lots of diagrams like this one, that show all the different points of typical steam turbine operation. This particular diagram shows a system that does not operate at the critical point.

The pressure gradient while heating in your boiler uses the heat input to create higher pressure at the output than at the input point. Hence no back flow against this pressure, providing the burner does not stop!
But I am still a total amateur in thermodynamics, which is why I am learning from this thread.
K2

All the info I've found indicates that nowhere in the system does pressure ever exceed feed pump pressure. The above temperature-Enthalpy diagram shows the feed pump raising the pressure from the condenser (P2) to the pump's output pressure, P1, and shows temperature rising in the boiler at P1 pressure.

Below is a screen grab pic from a simplistic explanation of the Rankine Cycle: Notice the graph shows Pressure on the Y axis and not temperature. Pump pressure starts low at "5" and is increased by the pump to some level at "1". Pressure does not increase further as the water is heated in the boiler from 1 to 3

Last edited:

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
I am trying to compare this diagram to your system:
By The original uploader was Keenan Pepper at English Wikipedia. - Transferred from en.wikipedia to Commons., CC BY-SA 3.0, File:RefrigerationTS.png - Wikimedia Commons

I think if you simply re-write:
4=> 5 = expansion of liquid to liquid and gas at first phase of boiler. (post-pump). This initially uses "liquid enthalpy" (pressure) to expand the liquid into liquid and gas before the exhaust heat is added.
5=> 1 = Mid-part of boiler tube arrangement: from lowest pressure and temperature to some further mid-point where all the liquid has boiled (Latent heat enthalpy). Temperature constant as liquid boils to gas. Change of state latent heat enthalpy using the heat from the Hot exhaust gas.
1=> 2 = remaining part of your boiler tube arrangement where the gas pressure is being raised (superheating gas enthalpy)
2=> 3 = Turbine use of superheated gas: Temperature drop, pressure drop and enthalpy drop.
3=> 4 = condenser (expansion of gas causing pressure drop and temperature constant while latent heat is removed and condensate forms.)
In fact, I reckon the arrangement you have planned will combine 4 =>5 and 5 => 1 into a single curved line of 4 => 1, but this diagram has the individual steps separated for explanation. The total of 4 => 2 is the boiler part.

Does this help you? Maybe you can do a diagram to show your thinking: I.E. pressures, temperatures, enthalpy?

"I think, but I am not so sure I am right" - !!
K2

I very much prefer using Rankine cycle enthalpy diagrams instead of refrigeration cycle diagrams.

#### Toymaker

##### Well-Known Member
HMEM Supporting Member
As I continue to read different studies done on Rankine engines I'm always learning new info which is helping me to further refine my overall system. A study, Solar Powered Turbine, was done in 2005 comparing two different working fluids to see which one performed best, R123 or Isobutane. The study concluded that more work was extracted from R123 when temperature and pressure approached the critical point, but was kept just below both limits. The pink dashed line in the TS diagram below indicates the optimal parameters. This is a very small change for me (accomplished in the software) and reassures me that I'm still on the right track. Unfortunately for me, the study focused on a solar heat source, which doesn't help me design a boiler.

So, my max pressure was going to be 3.67 MPa (530 psi) and max temperature was 183C.
New limits will be 340 MPa (493 psi) and 180C.

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#### Toymaker

##### Well-Known Member
HMEM Supporting Member
I am trying to compare this diagram to your system:
By The original uploader was Keenan Pepper at English Wikipedia. - Transferred from en.wikipedia to Commons., CC BY-SA 3.0, File:RefrigerationTS.png - Wikimedia Commons

I think if you simply re-write:
4=> 5 = expansion of liquid to liquid and gas at first phase of boiler. (post-pump). This initially uses "liquid enthalpy" (pressure) to expand the liquid into liquid and gas before the exhaust heat is added.
5=> 1 = Mid-part of boiler tube arrangement: from lowest pressure and temperature to some further mid-point where all the liquid has boiled (Latent heat enthalpy). Temperature constant as liquid boils to gas. Change of state latent heat enthalpy using the heat from the Hot exhaust gas.
1=> 2 = remaining part of your boiler tube arrangement where the gas pressure is being raised (superheating gas enthalpy)
2=> 3 = Turbine use of superheated gas: Temperature drop, pressure drop and enthalpy drop.
3=> 4 = condenser (expansion of gas causing pressure drop and temperature constant while latent heat is removed and condensate forms.)
In fact, I reckon the arrangement you have planned will combine 4 =>5 and 5 => 1 into a single curved line of 4 => 1, but this diagram has the individual steps separated for explanation. The total of 4 => 2 is the boiler part.

Does this help you? Maybe you can do a diagram to show your thinking: I.E. pressures, temperatures, enthalpy?

"I think, but I am not so sure I am right" - !!
K2
Just for you K2,.... I did my best to map out a Pressure-Enthalpy diagram using an actual R123 chart. Note that the vertical line from 5 to 1 should follow the constant temperature line, or perhaps lean to the right just a bit, as this line represents increased pressure from the feed pump, which adds very little heating to the working fluid. What really stuck out to me was how long R123 remains in the liquid state within the boiler, from 1 to 2. Its only from 2 to 3 that R123 is in the vapor phase. I think having a liquid inside the boiler tubes should allow for better cooling of the aluminum tubes and also quicker heat transfer into the R123.

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#### ajoeiam

##### Well-Known Member
No the working fluid does not supply the energy for its flow it always moves to the low pressure it sees. And this is a design function barring a stuck check valve, control valve, or a broken cap blocking a tube.

Pressure drop is key in fluid flow. Flow is always from high pressure to low pressure. Indeed under high pressure and a poor design back flow can occur. But under that condition the boiler will not work. The feed pump works against the boiler pressure and
supplies the boiler drum at a pressure higher then the operating boiler. Control valve and a check valve is installed for safety. There is also a non return valve on the boiler when the steam heads to the turbine or process.

And it is the velocity in the tube which determines the Reynolds number and heat transfer. The fluid is cooling the tube walls in the furnace and on average the difference in wall temperature and fluid is 70 to 80 degrees. For most boilers circulation is developed due to thermal differences of the fluid in the furnace walls but the wall must be designed so flow develops and heat is extracted from the furnace. You can force circulate the fluid. The water releases the steam in the drum unless you operate at critical which is a special design all by itself. There are three or four circuits which have to be considered so that the boiler works properly.

At no point in the furnace are the tubes allowed to become uncovered as the metal temperature will rise to failure.

Now for the final point. I suspect your design is working at the critical pressure and this requires a different approach as you will not have a boiler drum but the working fluid must remain a liquid until it is released to the turbine. A different can of worms. And there are a few steam boilers that operate at the critical pressure of water which if I remember right is 3206 psi. The fluid goes immediately to vapor after leaving the boiler.

HMEL
Trying to understand - - - -

If I want a steam boiler to produce say 1000 psi - - - does that mean my feedwater pump must also be rated at 1000 psi - - - - or not?

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