# As much steam as possible from engine exhaust gas + warm water?

S P
TL;DR Summary
How to produce as much steam as possible from exhaust gas + water in engine exhaust pipe?
Hi all,

The idea is to generate as much as possible water steam from heat of internal combustion engine exhaust gas. The so called steam generation device is a simple sloped engine exhaust pipe where water from engine cooling and hot gas is fed at the top and all the mix is exiting at the bottom.
I would like to have some basic napkin calculations for the ballpark numbers and practical cost effective idea to build it in reality.

The natural "product" in real life of such thing is not much water vapor but pretty much hot water and still hot gas (above water boiling point) in non uniform mix.

P.s. we are NOT talking about specialty marine wet exhaust systems, where the gas is bathed in cool water and exits in temperatures very close to ambient.

I would like to split the question into physics and engineering exercises.

Physics/thermodynamics:

The pretty close figures to real life of diesel engine running at some load:
Vw - volume (flow) of water per second = 0.0005 m3/s
Va - volume (flow) of air per second = 0.04 m3/s
tw - temperature of water entering the pipe = 310 K
ta - temperature of air entering the pipe = 800 K
L - length of the pipe = 2 m
R - inner radius of the pipe = 0.05 m
V - total volume of the pipe = 0.0157 m
Pipe is sloped at 30 degrees.

The consensus on various reports is that the exhaust gas from IC engine has physical properties very close to air.
At 800K, air has:
ρ - density = 0.435 kg/m3
h - specific enthalpy = 822.5 kJ/kg
s - specific entropy = 7.888 kJ/(kg·K)
Cp - specific heat at constant pressure = 1.099 kJ/(kg·K)
k - thermal conductivity = 0.0577 W/(m·K)

I have not found the "quick and dirty" list of water properties, but will update post, if necessary.

Lets assume, that heat loses from pipe external thermal radiation is 0, pressure in pipe is atmospheric or very close to atmospheric if that can be possible in real life.

From my understanding to solve this exercise I have for the beginning to know how much water is in the pipe at any time, then my knowledge stops: is there a specific heat of air and water + specific heat of the water to boil or thermal conductivity I have to calculate? There is a time constraint also, as in real life exhaust is supplied with new air and water constantly.
Maybe just assume that the ratio water : air is 5 : 400 and thats it?

Engineering:
The question is what is of the most importance: Make pipe slope as little as possible to slow water? Volume of the exhaust pipe? Somehow my guts tell me do not bother with welding some strange additional metal boxes and limit the amount of water to the just exact amount to boil it and that is.
Maybe I can have ballpark numbers how long and wide a pipe should be for the amount of water coming in for the max amount of hot water vapor?

I do not have spare engine at hand to make practical experiment :(

Thank you for your calculations and ideas.

This is my first post, and I am not much into physics/thermodynamics. Thank you all for replies.

## Answers and Replies

Mentor
Welcome to PF.

What is the application? Why do you want to maximize the production of low-pressure steam?

russ_watters
S P
Welcome to PF.

What is the application? Why do you want to maximize the production of low-pressure steam?
That is a good question. I want to get as much volume/pressure of gas (including hot vapor) as possible at the exhaust exit point with the same engine load.

Of course my assumption is that by boiling water with hot air I get more volume than from simple exhaust gas. Is that correct?

Science Advisor
Airships and lifeboats can and have condensed a greater weight of drinking water from their exhaust than the weight of fuel burned in their engines.

If circulating seawater was used to condense the exhaust to drinking water, that could preheat the engine cooling water. With evaporative cooling, that would release further water vapour for condensation to drinking water. In effect, a still to remove fresh water from the brine flowing through.

S P
Airships and lifeboats can and have condensed a greater weight of drinking water from their exhaust than the weight of fuel burned in their engines.

If circulating seawater was used to condense the exhaust to drinking water, that could preheat the engine cooling water. With evaporative cooling, that would release further water vapour for condensation to drinking water. In effect, a still to remove fresh water from the brine flowing through.
Baluncore, thanks for reply!

My use case absolutely not related to producing drinking water. I need to get as much gaseous substance volume/pressure from exhaust as I can. From engineering standpoint yes, it is the exact setup as classic marine engine - water into exhaust, but there similarities end.

Science Advisor
Of course my assumption is that by boiling water with hot air I get more volume than from simple exhaust gas. Is that correct?
Do you want pure steam without N2 or CO2, that can condense completely to water to run a steam engine, or do you want lots of wet air that you call steam?

What will you do with the product?

Mentor
I want to get as much volume/pressure of gas (including hot vapor) as possible at the exhaust exit point with the same engine load.
Since this is the exhaust pipe, the pressure atmospheric (as you said before).

Please tell us what the application is so we don't have to keep asking.

is there a specific heat of air and water + specific heat of the water to boil or thermal conductivity I have to calculate?
A 'steam table" has the properties of steam at various temperatures and pressures. But for this problem you are simply mixing two streams of different temperatures and chemicals and getting a mixture at the same temperature.

S P
Since this is the exhaust pipe, the pressure atmospheric (as you said before).

Please tell us what the application is so we don't have to keep asking.

Sorry. I do not want to disclose it, as it is not my project alone. Work is not mine alone and the opposite: do not want to be laughed at.
Do you want pure steam without N2 or CO2, that can condense completely to water to run a steam engine, or do you want lots of wet air that you call steam?

What will you do with the product?
I don't much care what is in the gas, as the exhaust gases are already there, I cannot avoid that and definitely you cannot make it more toxic by cooling it with water. If lots of wet air has MORE volume than simple mix of hot gas and hot but not boiling water. I somehow think, that empty pressure cooker will have much less pressure inside than the one with a sip of water, so the idea is plausible.

S P
A 'steam table" has the properties of steam at various temperatures and pressures. But for this problem you are simply mixing two streams of different temperatures and chemicals and getting a mixture at the same temperature.

Chemicals do not matter. Exhaust gases are very close to the ambient air, only some toxic substances make it poisonous, but from the physics/thermodynamics point - it is the same.

Putting there water will change the ratio of gases as there are already water vapor from hydrogen atoms present in fuel.

Yes, basically there is a question - how long and which diameter pipe has to be and how much water there has to flow?

Science Advisor
Yes, basically there is a question - how long and which diameter pipe has to be and how much water there has to flow?
Kind of feels like 'how much bad heat exchanger is needed to do the job of a good heat exchanger'...

russ_watters and S P
S P
Kind of feels like 'how much bad heat exchanger is needed to do the job of a good heat exchanger'...
EXACTLY!
I could add to the word "bad" also "cheap" and "already available"

Merlin3189
Science Advisor
Gold Member
I don't understand your goal: Is your objective to cool the exhaust gas or do you want to create steam for later use? Is the intent to mix them together?

Why mix the flows within the same pipe? This makes the problem much harder to analyze because there is no clear boundary and other effects come into play.

Is this water coming in a liquid stream flowing at the bottom of the pipe or vaporized like fuel in a carburetor?

If the water flows at the bottom, the relative velocity between the flows affects how much transfer there will be between the two in a very complicated manner. Imagine if one of the flows (or both) is turbulent.

Maybe you are even injecting your exhaust stream into a pipe filled with water. It is really not clear.

If you want help, you really need to clearly explain what you are trying to achieve and what are the initial conditions.

Last edited:
Bystander, berkeman, anorlunda and 2 others
Mentor
Chemicals do not matter.
Water is a chemical. What I'm saying is you have constituents input and output at certain temperatures and mass flow rates, and it's just a fairly simple mixing problem to determine the desired result from the starting conditions. Since we've been given limited information on the parameters, I'd just start by calculating how much energy you can recover by cooling the exhaust gases down to 100C and then calculating how much water that amount of energy will boil. That will establish your maximum potential flow rate of water/steam generation rate.
Yes, basically there is a question - how long and which diameter pipe has to be and how much water there has to flow?
No, the first question has to be to define the process. If you don't know what you think you can get from the process then there's no way to know what the heat exchanger should look like.
Sorry. I do not want to disclose it, as it is not my project alone. Work is not mine alone and the opposite: do not want to be laughed at.
This will severely limit the amount of help we can provide and value of the work you are doing. This isn't a place where people get laughed at for being ignorant - educating people (fixing ignorance) is the entire point of the forum. But a couple of things:

1. If you are trying to do mechanical/electrical energy recovery with a steam turbine, please know that this is already a thing that's been invented. Broadly, it is used for electricity production in many cases, and specifically it's at least been demonstrated in cars:
https://www.press.bmwgroup.com/glob...the-principle-of-the-steam-engine?language=en

2. You can't do it with atmospheric pressure steam and you can't use an open heat exchanger.

Bystander
Science Advisor
The sad thing is that, if the OP was willing to explain why he wanted to do this, someone could show him how it has been done more efficiently for the last 50 years, or explain why it is impossible.

dlgoff, russ_watters and sophiecentaur
Science Advisor
Gold Member
The sad thing is that, if the OP was willing to explain why he wanted to do this, someone could show him how it has been done more efficiently for the last 50 years, or explain why it is impossible.
This looks like the usual formula. A guy has a scheme in his mind that he feels could make him money. But he needs some free know-how but can't risk the World finding out. I reckon some of these 'schemes' are a lot older than 50 years.

S P and russ_watters
Science Advisor
Gold Member
This looks like the usual formula. A guy has a scheme in his mind that he feels could make him money. But he needs some free know-how but can't risk the World finding out. I reckon some of these 'schemes' are a lot older than 50 years.
OR, I could be more charitable and look upon it as a way of minimising the total loss of water in a 'vapour phase' cooling system. But these systems would usually use expensive de-ionised water so the vapour produced would need to be cooled externally and then returned as liquid, back to the heated part.

But why couldn't the OP have said that?

Gold Member
Or, 600C is too hot, and he wants a supply of low-medium grade heat.

Homework Helper
Gold Member
My vote goes with your original suggestion (SP). Put in no more than the amount of water that can be converted to steam at 100 C by cooling the gas to 100 C. Any more water just wastes heat in getting warmed as water, but not becoming steam.
Since you have (or want) limited space for mixing, I'd have thought, as you did, that spraying the water into the pipe would be most effective. A very fine, evenly distributed spray seems good - same sort of idea as fuel injection, fast and even mixing.

For me, the numbers don't match your plan of squirting all the cooling water through (I'm just using online calculators. Please check carefully yourself! )
It looks (to me)like you have 5 kJ/sec available from the exhaust gases. (Cooling air from 800 to 400 K)
Warming water from 310 to 374 k, then to steam, about 3 kJ/g. So you can boil about 1.7 g/sec of water.
That's a lot less than the 500 ml/sec you (sensibly) circulate through the engine. So you should simply bleed 1.7 g/sec of the hottest water to inject into the pipe.
I reckon that could give you about 2 l/sec of steam (oops, just realised that's at STP, so a bit, ?25%, more when it gets to near 400 K). Your exhaust volume was 40 l/sec, but may be a lot less (20 l/sec?) once it's cooled from 800 to 400 K.

That raises an interesting question of whether you want maximum water vapour, or maximum total volume of gas. You have to cool the exhaust gas a lot to generate this steam. If you just kept the gas hot, it would have more volume. Cooling it by half, from 800 to 400 K, halves its volume, but generates a much smaller volume of steam.

Since you're an engineer: A good solution now beats a better one at sometime or never. You won't achieve perfection, so don't worry about it. Good luck. E&OE

S P
S P
Apologies for not replying quickly,
I don't understand your goal: Is your objective to cool the exhaust gas or do you want to create steam for later use? Is the intent to mix them together?

My primary goals are 2:
Primary: to get as much VOLUME of gas state of matter. Water vapor and exhaust gas or mix of those is ok.
Secondary: exhaust gases have to be cooled to approx 150C, the lower - the better.

In my case the best would be 101 degree Celcius exhaust gas mix with water vapor, and no or little liquid state water coming out of the pipe. I can adjust water amount coming to the pipe.
Why mix the flows within the same pipe? This makes the problem much harder to analyze because there is no clear boundary and other effects come into play.

Pipe is already there. Everything else has to be built. Even if I need to switch to stainless steel pipe - that is not much of problem.

Is this water coming in a liquid stream flowing at the bottom of the pipe or vaporized like fuel in a carburetor?

At the top water is flowing into the pipe by simple low pressure pump. Like water form tap. At the bottom in ideal scenario - only gaseous substance.
You gave a good idea about Venturi principle... That may work to split water stream into small particles.
If the water flows at the bottom, the relative velocity between the flows affects how much transfer there will be between the two in a very complicated manner. Imagine if one of the flows (or both) is turbulent.

Flow patterns are definitely concept I have heard in theory, but my knowledge stops there. I only need approx ratio of the water/exhaust gas at the pipe. If the sloped pipe does speed water a lot - then I assume an overall ratio changes a little. I somehow think, that it does not matters much.
Maybe you are even injecting your exhaust stream into a pipe filled with water. It is really not clear.

I need gas volume. Gas, which can be directed somewhere - like in pipe. As much volume as I can get. I assume, that waste heat from engine is very good source to get gas from liquid.

I did not wrote 1 thing, but this I think matters little: I need approx 0.05-0.1 atmospheres of pressure above ambient. For the engine this additional back pressure is a joke.
If you want help, you really need to clearly explain what you are trying to achieve and what are the initial conditions.

Initial conditions are on the first post. My English is limited, so... it is not much I can explain more. Reason I do not want to disclose full details I wrote. Appologies

Water is a chemical. What I'm saying is you have constituents input and output at certain temperatures and mass flow rates, and it's just a fairly simple mixing problem to determine the desired result from the starting conditions. Since we've been given limited information on the parameters, I'd just start by calculating how much energy you can recover by cooling the exhaust gases down to 100C and then calculating how much water that amount of energy will boil. That will establish your maximum potential flow rate of water/steam generation rate.
Well... "Chemical" word has many meanings. Lets not get into the water definition.
Yes. You are correct - I have 2 different streams, and I need to exchange the heat between them. Mixing is allowed. Heat exchanger preferably has to be simple and cheap to produce - like exhaust pipe. In other words - how much that mixing can be done in a pipe described in the first post. We can select for calculations any length and diameter pipe.

The problem, I am not good at those calculations: go I need to calculate specific heat of air and water then additional heat required to boil the water? How I can calculate how much heat is already transferred after fixed amount of time? I need ballpark figures

russ_waters, I wrote starting parameters, which more you think are needed?

No, the first question has to be to define the process. If you don't know what you think you can get from the process then there's no way to know what the heat exchanger should look like.
I know exactly the process, just do not know the ratios and temperatures of the liquid water, water vapor and air will come out of it.

Steam turbine is the exactly opposite thing - steam in my case does not produce any mechanical work inside that tube. Or if it does - it is outside of my interest.

I am open to education. If you can show me how to calculate the heat exchange - I would be thankful if you show me how

S P
This looks like the usual formula. A guy has a scheme in his mind that he feels could make him money. But he needs some free know-how but can't risk the World finding out. I reckon some of these 'schemes' are a lot older than 50 years.

You missed Mason lodge, zombies, Game of Thrones and WORLD DOMINATION!

berkeman
S P
My vote goes with your original suggestion (SP). Put in no more than the amount of water that can be converted to steam at 100 C by cooling the gas to 100 C. Any more water just wastes heat in getting warmed as water, but not becoming steam.
Since you have (or want) limited space for mixing, I'd have thought, as you did, that spraying the water into the pipe would be most effective. A very fine, evenly distributed spray seems good - same sort of idea as fuel injection, fast and even mixing.

Thank you, Mr Merlin3189, you hit the point!
Do you suggest, that I need to split water into tiny droplets to make surface contact area between exhaust gas and water as big as possible, correct?

For me, the numbers don't match your plan of squirting all the cooling water through (I'm just using online calculators. Please check carefully yourself! )
It looks (to me)like you have 5 kJ/sec available from the exhaust gases. (Cooling air from 800 to 400 K)
Warming water from 310 to 374 k, then to steam, about 3 kJ/g. So you can boil about 1.7 g/sec of water.
That's a lot less than the 500 ml/sec you (sensibly) circulate through the engine. So you should simply bleed 1.7 g/sec of the hottest water to inject into the pipe.
I reckon that could give you about 2 l/sec of steam (oops, just realised that's at STP, so a bit, ?25%, more when it gets to near 400 K). Your exhaust volume was 40 l/sec, but may be a lot less (20 l/sec?) once it's cooled from 800 to 400 K.
That raises an interesting question of whether you want maximum water vapour, or maximum total volume of gas. You have to cool the exhaust gas a lot to generate this steam. If you just kept the gas hot, it would have more volume. Cooling it by half, from 800 to 400 K, halves its volume, but generates a much smaller volume of steam.

I assume 1 kJ/sec = 1kW, then there is a ballpark of 5-30 kJ/sec from engine depending on its load, and in the example engine I took approx numbers of 30-40kW crankshaft power diesel engine. If I preheat the supply water from engine coolant and mix with exhaust gas - there should be at least 40-50% of the crankshaft power available in the exhaust heat for my task.

~2g/s of boiled water is much less than I expected, and with cooled gas the overall volume is lost - then the my idea is not good...

Thank you very much.

I will post overall schematics of the project later.

dlgoff and Lnewqban
Mentor
Initial conditions are on the first post. My English is limited, so... it is not much I can explain more. Reason I do not want to disclose full details I wrote. Appologies

I know exactly the process, just do not know the ratios and temperatures of the liquid water, water vapor and air will come out of it.

I am open to education. If you can show me how to calculate the heat exchange - I would be thankful if you show me how
Assuming the numbers in the OP are accurate, the energy available is m * Cp * ΔT. The energy required to boil water is m * Cp * ΔT to warm it up, then m * Hvap to boil it.

That input heat is 6.1 MJ/s. That equates to about 500 gal/hour of diesel fuel (assuming a third is going out the tailpipe), which is pretty high -- is this a ship?

Anyway, at 2.26 MJ/kg to boil from 100C and 4.186 kJ/kG-C and room temperature input water, that's 2.35 kg/sec of water to steam. In your OP you said .5 kg/sec (.0005 cu m/s) of water. So your efficiency is only around 20%. That's probably realistic.

As for designing a heat exchanger, I'd start with figuring out what an exhaust pipe typically looks like for a 6 MW steam diesel engine. Where did you get 5 cm? That seems small.

Also, it's critically important how the water is injected into the pipe. If you spray it, your odds are pretty good of boiling it all off in a short time/distance. Beyond that, it's pretty difficult to do such a design analytically. Heat transfer is a bit of voodoo (it's very difficult to find the coefficients for convective heat transfer).

Last edited:
S P and jack action
S P
Assuming the numbers in the OP are accurate, the energy available is m * Cp * ΔT. The energy required to boil water is m * Cp * ΔT to warm it up, then m * Hvap to boil it.

That input heat is 6.1 MJ/s. That equates to about 500 gal/hour of diesel fuel (assuming a third is going out the tailpipe), which is pretty high -- is this a ship?

Anyway, at 2.26 MJ/kg to boil from 100C and 4.186 kJ/kG-C and room temperature input water, that's 2.35 kg/sec of water to steam. In your OP you said .5 kg/sec (.0005 cu m/s) of water. So your efficiency is only around 20%. That's probably realistic.

As for designing a heat exchanger, I'd start with figuring out what an exhaust pipe typically looks like for a 6 MW steam engine. Where did you get 5 cm? That seems small.

Also, it's critically important how the water is injected into the pipe. If you spray it, your odds are pretty good of boiling it all off in a short time/distance. Beyond that, it's pretty difficult to do such a design analytically. Heat transfer is a bit of voodoo (it's very difficult to find the coefficients for convective heat transfer).

I calculate like this:
2400 rpm of crankshaft = 40 rotations per second
diesel engine displacement is 2 L. Lets assume, that is is not turbocharged. 4 strokes.
40 rps * 2 L * 0.5 (4stroke) = 40 Liters per second on intake

I am assuming the worst case scenario, that there are just 40 Liters/s of gas on the exhaust, but in real life, as the gas is hot and expanded it is much more by volume + some additional co2, h2o from diesel fuel. For safety let it be 40 Liters/s.

In the first post I wrote
Va - volume (flow) of air per second = 0.04 m3/s, that is 40 liters/s

That is civilian engine, not a large ship... 30-50kW range at best.

Mentor
I calculate like this:
2400 rpm of crankshaft = 40 rotations per second
diesel engine displacement is 2 L. Lets assume, that is is not turbocharged. 4 strokes.
40 rps * 2 L * 0.5 (4stroke) = 40 Liters per second on intake

I am assuming the worst case scenario, that there are just 40 Liters/s of gas on the exhaust, but in real life, as the gas is hot and expanded it is much more by volume + some additional co2, h2o from diesel fuel. For safety let it be 40 Liters/s.

In the first post I wrote
Va - volume (flow) of air per second = 0.04 m3/s, that is 40 liters/s

That is civilian engine, not a large ship... 30-50kW range at best.
Well, the energy numbers don't add-up using that method, unless there's an error in my calc*. If that's the right volume of air, then the temperature is too high. Maybe a better way would be to use chemistry to find the actual fuel-air ratio and pick a fuel flow rate for a chosen output. Diesel has an energy density of 45 MJ/L and a good rule of thumb is that 1/3 of the energy goes out the tailpipe. I don't have time to do a more rigorous run through of this today...I'll try again tomorrow. The rigorous way starts with figuring out the fuel-air ratio, finds the enthalpy at the start (room temperature air and fuel) and then the end enthalpy of the cooled exhaust and then subtracts that from the heat of combustion. You don't actually need the intermediate state of post-combustion.

*Yes, one issue is the output is pressurized, not atmospheric pressure, but that just pushes the result up not down when correcting for that.

Science Advisor
Gold Member
So here's my math, if I understood you correctly:
• The exhaust gas must drop from ##800\ K## to ##374\ K##, which gives a difference of ##\Delta T_g = 426 K##;
• The specific heat capacity for air averages to ##C_{p\ g} = 1.05 \frac{kJ}{kg.K}## at those temperatures;
• You talk about volumetric flow rate, but we care about mass flow rate which we obtain by multiplying it by the density. The volumetric flow rate given is based on the air entering the engine at atmospheric pressure whose density is ##1.2 \frac{kg}{m^3}##, thus the exhaust gas mass flow rate is ##\dot{m}_g = 1.2 \times 0.04 = 0.048 \frac{kg}{s}##.
Knowing all that, we can get the heat transfer rate from the exhaust gas:
$$\dot{Q}_g = \dot{m}_g C_{p\ g} \Delta T_g = 21.47\ kW$$
On the water side, we want:
• An increase in temperature from ##310\ K## to ##373\ K##, which gives a difference of ##\Delta T_w = 63 K##;
• The specific heat capacity for water averages to ##C_{p\ w} = 4.2 \frac{kJ}{kg.K}## at those temperatures;
• The water must be evaporated and we know the heat of vaporization of water is ##h_{vap} = 2256 \frac{kJ}{kg}##;
Knowing all that, the heat transfer rate required is:
$$\dot{Q}_w = \dot{m}_w C_{p\ w} \Delta T_w + \dot{m}_w h_{vap} = \dot{m}_w (C_{p\ w} \Delta T_w + h_{vap}) = \dot{m}_w \left(2520 \frac{kJ}{kg}\right)$$
And since the heat from the gas should be equal to the heat added to the water (##\dot{Q}_g = \dot{Q}_w##):
$$\dot{m}_w = \frac{21.47\ kW}{2520 \frac{kJ}{kg}}= 0.0085 \frac{kg}{s}$$
Which, converted to a volumetric flow rate (density = ##1000\ \frac{kg}{m^3}##), gives ##0.000\ 008\ 5 \frac{m^3}{s}##.

This is about 60 times less water than you expected in the OP (##0.000\ 5 \frac{m^3}{s}##). Any more water will exit in liquid form. This also represents a maximum value because if the heat transfer is not optimized (which shouldn't be too much of a problem) or heat escape through the pipe (most likely), less water will vaporize.

If you spray the water somehow in the pipe, the design of the pipe is rather irrelevant in my opinion. 2 meters of exhaust pipe seems to be long enough for the process to complete. If you cannot spray it, you might want to let it drip on a screen set across the airflow.

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S P and russ_watters
Mentor
Thanks, I must have slipped a decimal point somewhere...