Combustion Thermodynamics 1

[macmet]

Everyone,

I have a question regarding the basic thermo of combustion.

When I finished school I went to work for a small company, designing simple combustion systems. Part of the design is a heat/mass balance. I think I understand H.M balances fairly well, but I have a question that I cannot seem to find a satisfactory answer for.

That question is, Why is it that you consider the latent heat loss associated with H20, but not with CO2? Heat of formation of CO2 doesn't seem to appear in the H/M balances. I just checked the B&W Steam book and didn't notice it in their equations either. Am I completely mistaken and overlooking something simple? Can someone point me to a good explanation of this?

Appreciate all the help I get.

 

[rmw]

Far be it from me to present myself as a guru on combustion calculations. I did some few years ago and felt like I was stumbling through it every time. My B&W is at the office so I can't look right now. So is my CE equivalent.

But, the latent heat part for water is just a thermodynamic book-keeping regarding the state change of the water present in the fuel and in the combustion air.

The heat of formation gets into chemical reactions, which to me seems different than first law book-keeping for what happens to the water present.

That is my take on your question. Now I'll try to remember to do some looking tomorrow AM.

rmw

 

[macmet]

Thanks RMW. I thought you might be able to provide some input.

I guess my confusion may come down to whether or not the latent heat is the formation of water vapor (i.e. oxidation of hydrogen in the fuel) or the phase transformation of water into steam? Meaning the hydrogen is oxidized into water, and then goes through the phase transformation into steam.

While the CO2 is a gas and always is a gas and therefore there is no phase change to account for?

Cheers.

 

[rmw]

Some of the water present is a product of combustion of the H2 in the fuel as you mention. Since that takes place in the flame zone, and asuming no condensation in the back end, that H2O doesn't change phase, but the combustion does release heat energy so heat of formation is involved. (You are clearing out cobwebs here.)

Other water is present in the combustion air as vapor depending on the relative humidity outside and it too doesn't change phase, just temperature (assuming fairly constant pressures).

The water tied up in the fuel, what ever that might be and whatever form that might be may involve a phase change. Solid fuels with liquid water in them (coal, wood, etc) have to contribute the BTU's necessary to heat the water do the latent heat phase change as well as heat the resulting vapor to combustion or furnace temperatures. Depending on the amount of water present, (some woods and lignites have high percentages of moisture in the fuel) a significant amount of the fuel is needed just to heat water.

So in your combustion calculations, (again remember my caveat about how long it has been) you have to deal with water in several ways.

CO2 on the other hand is a result of the combustion of some form of carbon in the fuel (unless, of course you have a fuel with something more than a minor amount of CO2 already present in the fuel as fired.)

rmw

 

[macmet]

RMW,

Thanks for writing back again.

Your last point explains to the point where I'm confused.

Fuel has hydrogen and carbon in it (and other things). Why do you only take into consideration the latent heat required to turn the hydrogen in steam and not the carbon into carbon dioxide gas?

Is it that CO2 is in the gas phase at room temperature and therefore goes through no phase transformations in the temperature range relevant here?

...And that the latent heat is to transform the liquid water into steam/gas and has nothing to do with the actual oxidation of hydrogen?

Any input is appreciated. I find just reading and thinking about all responses helps me figure these issues out.

 

[ccfowler]

The difference between HHV and LHV values is mainly the latent heat of water vapor condensation due to cooling combustion products to the lower termperature reference for the HHV. Other than the water vapor in the combustion products, none of the other common combustion products go through the vapor-to-liquid phase change between the two reference temperatures (for HHV vs. LHV values).

Just to add a bit of potential confusion, what about the effects of atomizing steam for firing oil? I admit that I hadn't paid any attention to this prior to this discussion, but it contributes to the mass and energy balances.

Valuable advice from a professor many years ago: First, design for graceful failure. Everything we build will eventually fail, so we must strive to avoid injuries or secondary damage when that failure occurs. Only then can practicality and economics be properly considered.

 

[macmet]

Ok, well thanks for both of your responses. I want to give this one more shot.

In biomass and solid fuels, you have a certain percentage water, so you have H2O in the fuel that has to be converted to steam. That is straight forward.

In other fuels, lets assume 100% methane. It is CH4. You have one carbon molecule and 4 hydrogen molecules. Without balancing equations, the oxidation of these two elements is,

C + 2O => CO2
2H + O => H2O

My understanding now is that heat of formation is not considered, so neither are considered.

And it is assumed that H2O is initially formed in the liquid phase, and then goes through the phase transformation into steam, and therefore you need to consider latent heat losses.

While the CO2 that forms, forms as a gas, and remains a gas, so no phase change needs to be considered.

Is that correct?

 

[ccfowler]

H2O is formed as a gas (or vapor), and some of the H2O changes to liquid form as the combustion products cool. The H2O is always in some two-phase balance at normal atmospheric temperatures and pressures.

The combustion temperatures are way too high for H2O to first form as a liquid from the reaction of two gases (CH4 and O2). At normal room temperatures and pressures, CH4, O2, H2, and CO2 are all only in single phase (gas) form. Only the H2O portion of the combustion products exist in a two-phase state, although one can argue that some small fraction of the O2 and CO2 can be considered to be dissolved within the liquid H2O phase.

Valuable advice from a professor many years ago: First, design for graceful failure. Everything we build will eventually fail, so we must strive to avoid injuries or secondary damage when that failure occurs. Only then can practicality and economics be properly considered.

 

[macmet]

So, if the combustion temperatures are too high for H2O to be in the liquid form, why do I need to consider latent heat loss at all?

I'm still confused. It makes sense to me for solid fuels for the moisture content, but not for fuels like natural gas.

 

[ccfowler]

You are getting into the issue of the difference between the HHV (higher heat value) and LHV (lower heat value) of a fuel sample.

The common practice for evaluating the heat rate or efficiency of a conventional steam power plant is to use the HHV as the reference for the fuel energy input.

Commonly, the LHV is used for evaluating the efficiency of combustion turbines.

A good case can be made for each method, presuming that everyone with an interest understands the implications of the different values for the same fuel sample.

If you are familiar with very high efficiency gas fired furnaces for home heating, you will note that the combustion products are cooled to very near room temperature with the result being a significant portion of the water vapor being condensed. The exhaust temperature is too cool for a conventional chimney to carry away the combustion products, so a powered venting arrangement is necessary.

This is a very practical example of a system that actually recovers most of the HHV of the fuel.



Valuable advice from a professor many years ago: First, design for graceful failure. Everything we build will eventually fail, so we must strive to avoid injuries or secondary damage when that failure occurs. Only then can practicality and economics be properly considered.

 

[25362]

Thermodynamically speaking, the standard heat of combustion, in kJ/mol at 25[sup]o[/sup]C and 1 atmosphere pressure for the reactants, and for the products returning to the same conditions, is based on the heats of formation, CO[sub]2[/sub] as a gas and H[sub]2[/sub]O as a liquid, as can be seen in the following equation:

[Δ][sub]c[/sub]H[sup]o[/sup]= 393.51 a + 142.915 b + [Δ][sub]f[/sub]H[sup]o[/sup](C[sub]a[/sub]H[sub]b[/sub]O[sub]c[/sub])​

The same criteria applies to other elements that may be present.

 

[macmet]

Maybe I need to take a step back here.

What I am trying to calculate is the theoretical flame temperature.

In this calcuation I take the HHV value, account for the latent heat (which leaves LHV).

I know the process of figuring this out, I guess I just do not understand why CO2 and H20 are treated differently (i.e. why is the H2O considered to go to liquid then to gas). I would think at the temperatures the process is occuring it should go straight to gas. And for fuels such as methane, the latent heat of vaporization should not be considered since liquid water should never be formed.

 

[danberry]

I think the answer you are looking for may be that water condenses at "normal" temperatures while CO2 does so only at unusually low temperatures. The 32F reference temperature for industrial combustion calculations is way above CO2 condensing temperatures, so the latent heat of CO2 can be ignored.

I suspect that if water condensation would not have interfered with bomb calorimeter measurements, the latent heat of water would also have been conveniently evacuated by selecting a higher reference temperature. No HHV, no LHV , just HV.

 

[macmet]

DanBerry,

I think what you said at the start is what I was looking to confirm. I was thinking about it this weekend when I saw some weird thing on tv using dry ice. Doesn't make a lot of sense to consider the evaporation of CO2 when at the temperature of fuel it is already in the gas phase.

If anyone else has a different view point, I would like to hear it though.

Cheers.

 

[25362]

The heat of combustion (a chemical reaction) may be calculated as the difference between the heats of formation (?[sub]f[/sub]H[sup]0[/sup]) of the products and reactants all taken at the same T,P conditions; chemists prefer 25[sup]o[/sup]C and 1 atm abs.

 

[RR79]

I completely agree with DanBerry´s first paragraph!

 

[macmet]

I agree with it too.

Dry ice at room temperature is a good example. Doesn't make a lot of sense to have to deal with latent heat of vaporization of CO2 when the CO2 formed at the fuel temperature is clearly a gas.

But, I'm still a little unclear as to why hydrogen molecules in the fuel disassociate, bond with oxygen, at these elevated temperatures, and yet you have to consider the water from room temperature. This makes me think I have to consider that hydrogen's oxidation first forms water in the liquid state and then it transforms to steam.

Am I mixing this up? I'm used to solid fuels where fuel is often wet and you need to evaporate the water. Maybe with gaseous fuels, you do not need to consider latent heat of vaporization of the hydrogen in the fuel?

Keep the comments coming. I appreciate them all.

 

[RR79]

I don´t believe H first forms water and then steam. But that is not the point!
The losses correspond to all the energy from flue gas that could not be recovered considering the initial temperature of the system (32F). If you could recover energy until the flue gas temperature reaches 32F, then you would have 0 energy loss on the flue gas (except by incomplete combustion, CO), therefore you would also have recovered the latent energy from H2O, but not yet from CO2.
As you are never going to reach the 32F, you have to consider the latent energy as a loss!

Regards,
Rogério

http://www.jlj.com.br