Thermodynamics.... Constant volume pressure of combustion

In summary, the conversation is about a 42-year-old engine hobbyist seeking help on calculating and comparing the pressures that different air/fuel ratios may produce in an internal combustion engine. The person is not formally educated and would appreciate a simplified explanation. Two plans were proposed and the second one was deemed more suitable. The person also has some questions about the calculations and processes involved. The conversation also touches on writing a balanced chemical reaction equation for the reaction of different fuels with oxygen. The person shares their attempts at solving the problem and asks for clarification on their calculations.
  • #1
InquisitiveOne
36
1
Hi guys, 42 year old engine hobbyist here, not a student. I've had great luck figuring out my questions in this portion of the forum in the past and look forward to your input. Keep in mind that I'm not formally educated, so if it's possible to dumb something down a bit I'd appreciate it.1. Homework Statement

I want to calculate and compare the pressures that different air/fuel ratios may produce in an internal combustion engine if the piston were to stop at Top Dead Center and all the reactants combusted.

What pressure would a stoichiometic Air/Fuel ratio produce? What about richer or leaner ratios?

I don't need actual running engines pressures. A figure based on a static volume will be fine for my current purposes.



2. Homework Equations

In my research so far, I haven't found a direct way to calculate what I want. So, I've come up with a couple plans to use several calculative methods combined to produce the product I'm hoping for. Please review them and let me know if one is a better method than the other, or if there is another way that might be preferred for accuracy, ease of calculating, or whatever reason.

I also have some questions about the calculations/processes listed below which I'll list in #3 of this post.

PLAN 1
Step 1) Calculate Adiabatic Flame Temperature
Step 2) Using Pv = nRT, calculate the new volume per given mass
Step 3) Use calculation for Adiabatic Compression, new volume and temperature from last step, compressed to known final volume to get final pressure and temperature.

PLAN 2
Step 1) Identify a fuels BTU/lb or kJ/kg
Step 2) Calculate amount of thermal energy from the mass of fuel at stoichiometric per given mass of air.
Step 3) Average the Specific Heat Capacities of the gasses left after combustion
Step 4) Use the Isochoric Process to solve for a final temperature and pressure

Is there a better way, or is one of these two methods incomplete or otherwise not valid?
3. The attempt at a solution Questions about the proposed plans

In PLAN 1, Step 1, Calculate Adiabatic Flame Temperature...
All the examples I've been able to source for calculating flame temperature assume either stoichiometric ratios OR excess air, a lean condition. When using the calculation for a rich condition which would result in excess un-reacted fuel, how would I go about adding it into the equation?

The fuels I'd like to calculate for are gasoline (which seems to have wildly varying chemical compositions. Seems most just use Octane for calculating, C8H18), methanol, and ethanol. I'd also like to be able to account for humidity and/or water added to the combustion process.If you guys can point me in the right direction I'd appreciate it greatly.

Thank you,
-Seth-
 
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  • #2
Method 2 is more like it. You sound that you have more than just a little bit of understanding, formal training or not. Do you still remember how to write a balanced chemical reaction equation for the reaction of octane (or other hydrocarbon) with oxygen to produce water and carbon dioxide?
 
  • #3
Chestermiller said:
Method 2 is more like it. You sound that you have more than just a little bit of understanding, formal training or not. Do you still remember how to write a balanced chemical reaction equation for the reaction of octane (or other hydrocarbon) with oxygen to produce water and carbon dioxide?
I only barely grasp what I've been able to deduce in a week of trying to figure this out on my own through google searches and YouTube lectures. I'll take that as a compliment. Thank you Chestermiller.

I believe I understand roughly how to write a balanced chemical reaction equation, however I haven't actually done so yet .
I've seen that particular one so many times over the past week, I've memorized it. Let me try it out for ethanol.

C2H5OH + 3 O2 = 3 H2O + 2 CO2
And Methanol
2 CH3OH + 3 O2 = 4 H2O + 2 CO2

Thanks for prompting me to actually do that. Kinda fun.So, if I take 0.0749 lbs / 0.0340 kg of air (one cubic foot at 1 ATM and 70ºF)
A gasoline 14.7:1 ratio gives me 0.0051 lbs / 0.0023 kg of fuel
Total combined mass of 0.0800 lbs. / 0.0363kg at stoichiometric ratio
At 18552.8 BTU/lb gives available energy of 94.54 BTU / 99.75kJ to distribute to any and all post combustion gasses

So, that's steps 1 and 2 done

Step 3, Average the Specific Heat Capacities of the gasses left after combustion

For stoichiometric, I'm left with (mostly) H2O, CO2, and N2
For richer than Stoiciometric, the same plus whatever fuel isn't reacted. H2O, CO2, N2, and C8H18 in this case.

Starting with 0.0340kg of air and assuming the composition is 79% N2 and 21% O2, I get:
0.0269 kg N2
0.00714 kg O2

Given O content in the given mass of air and the percentage of total O per H2O and CO2 in the balanced chemical reaction equation of
2 C8H18 + 25 O2 = 16 CO2 + 18 H2O (36% of the oxygen atoms go into H20 and 64% into CO2), I can take the molar mass of H2O and CO2 I get the total masses of:
0.00289kg H2O
0.00354kg CO2

... Hmmm
I've tried it a couple different ways and a bunch of different times with different starting figures and final mass is close to starting mass, however always a little off.
Isn't it supposed to calc out as the same total mass, before and after combustion? Am I just losing some here and there in the calculation process?
I have 0.0363kg starting mass, air and fuel and the above post combustion figures add up to 0.03333kg, off by 2.97 grams.

Am I off somewhere?
 
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  • #4
Had a thought. I think the numbers are off because the actual ratio is slightly lower than the 14.7 rounded figure I used to calculate fuel mass from air mass. One end based on rounded figure, the other end from the actual precise stoichiometric calculation. I'll math it out tomorrow and verify the thought. Pretty sure that's why it's not adding up.
 
  • #5
Let's start over. You are moving a little too quickly.

Suppose we start with 1 lb-mole of octane. How many lb of octane is that? How many lb-moles of oxygen are required to react with the octane stoichiometrically? Now many lb of oxygen is that? How many lb-moles of nitrogen accompany the oxygen (79/21 is the molar ratio)? How many lb of nitrogen is that. What is the total number of lb of air? What is the air/fuel ratio for stoichiometric combustion of octane?
 
  • #6
Still trying to wrap my mind around the concept of Moles.
Chestermiller said:
Suppose we start with 1 lb-mole of octane. How many lb of octane is that?
0.2518309847 lb?
How many lb-moles of oxygen are required to react with the octane stoichiometrically?
12.5
Now many lb of oxygen is that?
0.8818166987 lb.
How many lb-moles of nitrogen accompany the oxygen (79/21 is the molar ratio)?
53.7137754143
How many lb of nitrogen is that?
3.317310438
What is the total number of lb of air?
4.1991271367
What is the air/fuel ratio for stoichiometric combustion of octane?
16.6743863633 : 1 ?
 
  • #7
The total mass now also balances between pre and post combustion

Post combustion
H2O ... 0.357452336 lb ... 8.03%
CO2 ... 0.7761953474 ... 17.44%
N2 ... 3.317310438 ... 74.53%
Total .. 4.4509581214 ... 100.00%
 
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  • #8
Apparently the air composition of 79/21 is based on volume. By mass it's 23.2% O2 and 76.8% N2 and other inert gasses.

Adjusting for that brings the ratios quite a bit closer to what I expected to see.
15.09:1 for Octane
8.98:1 for Ethanol
6.46:1 for Methanol
 
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  • #9
Chestermiller said:
Let's start over. You are moving a little too quickly.

Suppose we start with 1 lb-mole of octane. How many lb of octane is that? How many lb-moles of oxygen are required to react with the octane stoichiometrically? Now many lb of oxygen is that? How many lb-moles of nitrogen accompany the oxygen (79/21 is the molar ratio)? How many lb of nitrogen is that. What is the total number of lb of air? What is the air/fuel ratio for stoichiometric combustion of octane?
As you showed, the stoichiometric equation is $$C_8H_{18}+12.5\ O_2=8CO_2+9H_2O$$
1 lb mole of octane has a mass of 8(12)+18(1)=114 lb
12.5 lb-moles of oxygen are required to react with the octane stoichiometrically.
The number of lbs of oxygen in 12.5 lb-moles = (12.5)(32)=400.0
The number of lb-moles of nitrogen accompanying the oxygen is (12.5)(79)/(21)=47.02 lb-moles
The number of lbs of nitrogen in 47.02 lb-moles = (47.02)(28)=1317.4
The total number of lb of air = 400 +1317.4 = 1717.4
Stoichiometric Air/fuel ratio = (1717.4)/(114) = 15.06
 
  • #10
Chestermiller said:
As you showed, the stoichiometric equation is $$C_8H_{18}+12.5\ O_2=8CO_2+9H_2O$$
1 lb mole of octane has a mass of 8(12)+18(1)=114 lb
12.5 lb-moles of oxygen are required to react with the octane stoichiometrically.
The number of lbs of oxygen in 12.5 lb-moles = (12.5)(32)=400.0
The number of lb-moles of nitrogen accompanying the oxygen is (12.5)(79)/(21)=47.02 lb-moles
The number of lbs of nitrogen in 47.02 lb-moles = (47.02)(28)=1317.4
The total number of lb of air = 400 +1317.4 = 1717.4
Stoichiometric Air/fuel ratio = (1717.4)/(114) = 15.06
Interesting. I had pretty much all the same figures which I thought were in grams, so I converted them all to pounds. No need?
 
  • #11
Chestermiller said:
The number of lb-moles of nitrogen accompanying the oxygen is (12.5)(79)/(21)=47.02 lb-moles

Since we're doing everything by weight, wouldn't you calculate the nitrogen weight from the oxygen weight and identify the number of moles N2 by it's weight per mole?

(12.5 * 32g/mol) = 400 g
400g / (79/21) = 1504 g
1504 / 28g/mol = 54 moles of N2

Simply basing it on the mole negates the difference in the weight of the moles, doesn't it?


Wait a minute. It looks like the 79/21 ratio is based on moles or is a molar ratio, not a mass or volume ratio. I see how that works out now. 79/21 molar ratio equates to a 76.7/23.3 mass ratio.

And... I went back and reread your post, and you said exactly that. (79/21 is the molar ratio) Didn't register upstairs when I read it the first time through I guess.
 
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  • #12
InquisitiveOne said:
Interesting. I had pretty much all the same figures which I thought were in grams, so I converted them all to pounds. No need?
Grams are fine too. But I though you wanted to work in Imperial units, since you gave the heat of combustion in BTU/lb.
 
  • #13
Chestermiller said:
Grams are fine too. But I though you wanted to work in Imperial units, since you gave the heat of combustion in BTU/lb.

I really appreciate your help with this Chestermiller

Imperial is easier for me to relate to. However, SI seems easier to calculate with.

Let's work in SI.

It seems that using the mass ratio to calculate nitrogen content is slightly more accurate than using the molar ratio. If there's not a good reason to use the molar ratio, I'll continue using the mass ratio method for it's slight accuracy advantage.

I now know the mass of H2O, CO2, and N2 post combustion.
So, what's next? Average out the Specific Heat Capacities?
 
  • #14
I'm a little tied up today, so I'll have to get back with you tomorrow. When we continue, well work with grams and gram moles.

What makes you think that the mass ratio is more accurate to use than the molar ratio? You are aware that it is molar ratio that is 79:21, right?

Chet
 
  • #15
Chestermiller said:
I'm a little tied up today, so I'll have to get back with you tomorrow. When we continue, well work with grams and gram moles.

What makes you think that the mass ratio is more accurate to use than the molar ratio? You are aware that it is molar ratio that is 79:21, right?

Chet
Based on this percent of total mass composition of air breakdown, in/at engineeringtoolbox.com

% of Total Mass, Air
Oxygen .... 23.20%
Nitrogen ... 75.47%
Carbon Dioxide .. 0.046 %
Hydrogen ... ~ 0%
Argon ... 1.28 %
Neon .... 0.0012%
Helium ..... 0.00007%
Krypton .... 0.0003%
Xenon ... 0.00004%

If I take all the inert elements that aren't O2 , clump them together, and calculate them as N2 it results in a ratio of 76.8/23.2.

Taking the total mass of O2 required for stoiciometric balance, 399.99g, using the mass ratio gives 1324.09g of N2
1324.09 at 28.0134g/mole of N2 equates to 47.27g/moles of N2 to 12.5 g/moles of O2, or a slightly more accurate molar ratio of 79.1/20.9.

I'm probably being a little too picky. But shoot, it's not like I'm mathing everything out with a pencil. It's not a problem to add resolution when the computer is doing all the heavy lifting. I usually let it run out to the ten-thousandth. However, if you want to round to the nearest whole number I'll still get the point. 79:21 it is.

I'll work on this a little today and I look forward to tomorrows lesson and review of any progress I post later today.

Again, thank you for your assistance Chet.

-Seth-
 
  • #16
OK. All the numbers in post #9 are grams and g-moles instead of lb and lb-moles. Getting the final temperature and pressure of the combusted mixture at constant volume of the chamber takes a little thermodynamic care. We need to devise a thermodynamic path from the initial state to the final state (not necessarily the actual path) for which is it relatively simple to get the final temperature and pressure. For simplicity, we are going to do this assuming ideal gas behavior. Here is the path I propose:

1. Determine the change in enthalpy of the mixture of octane and oxygen (excluding nitrogen and excess air) to determine the amount of heat that would have to be removed in order for the final temperature and pressure of the products to be equal to the initial temperature and pressure of the reactants (assumed to be 25 C and 1 atm.)

2. Mix the products of the reaction at 25 C and 1 atm with the original nitrogen and excess air at 25 C and 1 atm. The change in enthalpy for this process is zero, since the heat of mixing of ideal gases is zero. So this process would be carried out at 25 C and 1 atm., and the final temperature and pressure would be 25 C and 1 atm.

3. Add back into the product mixture (at a constant pressure of 1 atm.) the heat that was removed in step 1 so that we obtain a mixture now at a higher temperature, but still at 1 atm. This will require making use of the heat capacities of the components of the final gas mixture. Since the pressure is still 1 atm., the volume is much larger than the original reaction mixture.

4. Compress the resulting mixture adiabatically and reversibly back down to the original volume of the reaction mixture to determine the final temperature and pressure.

Starting with step 1, Heats of formation of the various species of interest at 25 C and 1 atm (in kJ/mole) are as follows:

Octane (l) = -252.1
Oxygen (g) = 0
Nitrogen (g) = 0
Carbon Dioxide (g) = -393.5
Water (g) = -241.8

Therefore, the enthalpy change for the combustion of Octane will be:

$$\Delta H= (8)(-393.5)+(9)(-241.8)-(1)(-252.5)=-5072\ kJ/mole=-5.072\ MJ/mole$$
This agrees with the heat of combustion of -5075 kJ/mole at the following link: https://en.wikipedia.org/wiki/Heat_of_combustion

I think I'll stop here for now.
 
  • #17
Chestermiller said:
1. Determine the change in enthalpy of the mixture of octane and oxygen (excluding nitrogen and excess air) to determine the amount of heat that would have to be removed in order for the final temperature and pressure of the products to be equal to the initial temperature and pressure of the reactants (assumed to be 25 C and 1 atm.)

2. Mix the products of the reaction at 25 C and 1 atm with the original nitrogen and excess air at 25 C and 1 atm. The change in enthalpy for this process is zero, since the heat of mixing of ideal gases is zero. So this process would be carried out at 25 C and 1 atm., and the final temperature and pressure would be 25 C and 1 atm.

3. Add back into the product mixture (at a constant pressure of 1 atm.) the heat that was removed in step 1 so that we obtain a mixture now at a higher temperature, but still at 1 atm. This will require making use of the heat capacities of the components of the final gas mixture. Since the pressure is still 1 atm., the volume is much larger than the original reaction mixture.

4. Compress the resulting mixture adiabatically and reversibly back down to the original volume of the reaction mixture to determine the final temperature and pressure.

Starting with step 1, Heats of formation of the various species of interest at 25 C and 1 atm (in kJ/mole) are as follows:

Octane (l) = -252.1
Oxygen (g) = 0
Nitrogen (g) = 0
Carbon Dioxide (g) = -393.5
Water (g) = -241.8

Therefore, the enthalpy change for the combustion of Octane will be:

$$\Delta H= (8)(-393.5)+(9)(-241.8)-(1)(-252.5)=-5072\ kJ/mole=-5.072\ MJ/mole$$
This agrees with the heat of combustion of -5075 kJ/mole at the following link: https://en.wikipedia.org/wiki/Heat_of_combustion

I think I'll stop here for now.
Chestermiller said:
OK. All the numbers in post #9 are grams and g-moles instead of lb and lb-moles. Getting the final temperature and pressure of the combusted mixture at constant volume of the chamber takes a little thermodynamic care. We need to devise a thermodynamic path from the initial state to the final state (not necessarily the actual path) for which is it relatively simple to get the final temperature and pressure. For simplicity, we are going to do this assuming ideal gas behavior. Here is the path I propose:

1. Determine the change in enthalpy of the mixture of octane and oxygen (excluding nitrogen and excess air) to determine the amount of heat that would have to be removed in order for the final temperature and pressure of the products to be equal to the initial temperature and pressure of the reactants (assumed to be 25 C and 1 atm.)

2. Mix the products of the reaction at 25 C and 1 atm with the original nitrogen and excess air at 25 C and 1 atm. The change in enthalpy for this process is zero, since the heat of mixing of ideal gases is zero. So this process would be carried out at 25 C and 1 atm., and the final temperature and pressure would be 25 C and 1 atm.

3. Add back into the product mixture (at a constant pressure of 1 atm.) the heat that was removed in step 1 so that we obtain a mixture now at a higher temperature, but still at 1 atm. This will require making use of the heat capacities of the components of the final gas mixture. Since the pressure is still 1 atm., the volume is much larger than the original reaction mixture.

4. Compress the resulting mixture adiabatically and reversibly back down to the original volume of the reaction mixture to determine the final temperature and pressure.

Starting with step 1, Heats of formation of the various species of interest at 25 C and 1 atm (in kJ/mole) are as follows:

Octane (l) = -252.1
Oxygen (g) = 0
Nitrogen (g) = 0
Carbon Dioxide (g) = -393.5
Water (g) = -241.8

Therefore, the enthalpy change for the combustion of Octane will be:

$$\Delta H= (8)(-393.5)+(9)(-241.8)-(1)(-252.5)=-5072\ kJ/mole=-5.072\ MJ/mole$$
This agrees with the heat of combustion of -5075 kJ/mole at the following link: https://en.wikipedia.org/wiki/Heat_of_combustion

I think I'll stop here for now.
Still soaking this up. Seems a lot like plan 1.
 
  • #18
Step 2.

First let's calculate the volume of reactants we have prior to reaction. Let's pretend we have 10% excess air. Then the number of moles of oxygen and nitrogen we have prior to reaction are 12.5 x 1.1 = 13.75 and 47.02 x 1.1 = 51.72, respectively. So the total number of moles of air initially are 65.47. If we neglect the volume of the liquid octane relative to the air, then the volume of the reaction mixture is:
$$V=\frac{nRT}{P}=\frac{(65.47)(0.082)(298)}{1}=1600.\ liters=1.6\ m^3$$
This is the volume of the reaction chamber per gram mole of octane.

Now for the product mixture, prior to adding back in the heat released by the reaction. For 10% excess air in the reaction mixture, the product mixture consists of

8 moles of CO2
9 moles of water vapor
1.25 moles of oxygen
51.72 moles of nitrogen

So the total number of moles of product mixture are 69.97. So at 25 C and 1 atm., the product mixture would occupy 1710 liters = 1.710 m^3 (disregarding the fact that the water vapor would be condensed at this temperature).

Step 3

Calculate the temperature of the product mixture if the 5.072 MJ of heat (that were removed) are added back into the product mixture at 1 atm. pressure. Fortunately for us, we won't need to integrate heat capacities as a function of temperature to carry out this calculation because Moran et al, Fundamentals of Engineering Thermodynamics, give enthalpy per mole tables as a function of temperature h(T) at 1 atm. for all four species in our product mix. The enthalpy of the product mix at any temperature T is determined from the equation
$$H(T)=8h_{CO2}(T)+9h_{H2O}(T)+1.25h_{O2}(T)+51.72h_{N2}(T)$$
From Moran et al's tables, at 298 K and 1 atm., the enthalpy of the product mix is
$$H(298)=(8)(9.364)+(9)(9.904)+(1.25)(6.203)+(51.72)(8.669)=620\ kJ=0.620\ MJ$$
Therefore, at the final temperature for Step 3, we must have $$H(T)=5.072+0.620=5.692\ MJ$$

I think I'll stop here for now. Do you have access to Moran et al?
 
  • #19
InquisitiveOne said:
Still soaking this up. Seems a lot like plan 1.
Y' know, you're right. This is very much basically Plan 1. Oh well. At least well get the results without having to integrate the heat capacities using Moran.
 
  • #20
InquisitiveOne said:
Still soaking this up. Seems a lot like plan 1.
Y' know, you're right. This is very much basically Plan 1. Oh well. At least well get the results without having to integrate the heat capacities using Moran.
 
  • #21
Chestermiller said:
Y' know, you're right. This is very much basically Plan 1. Oh well. At least well get the results without having to integrate the heat capacities using Moran.

Whatever's easier. As long as the results accurately indicate a trend. I pretty much don't care how precise the product is, as long as it's repeatable per individual air/fuel ratios.

The ultimate reason behind my trying to figure this out is, engines make more horsepower with richer than stoichiometric conditions. For instance, gasoline makes peak torque/power at a ratio close to 12.5:1, when stoichiometrically all the oxygen is gone at 14.7:1. Why is that? Some say it has to do with the rate of combustion. However, I suspect that additional fuel vapor may expand at a ratio greater than the combined byproducts of combustion do alone. And given the thermal energy the extra fuel vapor takes away from the other gasses and their expansion/pressure, a balance of best possible pressure may occur at a predictable calculated ratio. Where too much additional fuel takes more thermal energy from the other gasses than the expansion of fuel vapor can make up for.

That's my hypothesis I hope to confirm or eliminate.

If you you happen to know how "max power rich" conditions are calculated, I'd certainly like to know that too. My research hasn't turned up a single lead.

That said, it would be helpful if we assumed extra fuel instead of extra air in the above examples.

Thank you Chet.

-Seth-
 
  • #22
Let's complete the case of excess air first, then go back and do fuel rich. We'll see what the calculations say.
 
  • #23
Using the tables in Moran et al, I get a temperature of the product mixture (at 1 atm.) at the end of Step 3 as 2245 K = 1972 C. Is this about what you expected? This would correspond to a volume of 1710 x 2245\298 = 12900 liters = 12.9 m^3. This is a factor of 8.06 times the volume of the original reactant mixture. So, in Step 4, the product mixture would have to be adiabatically and reversibly compressed by a factor of 8.06.

Next, we'll do Step 4 (or, in your Plan 1, Step 3).
 
  • #24
Chestermiller said:
Do you have access to Moran et al?

I don't know. What is it and how may I gain access? Nothing really relevant came up on the first page of a search.
 
  • #25
InquisitiveOne said:
I don't know. What is it and how may I gain access? Nothing really relevant came up on the first page of a search.
Fundamentals of Engineering Thermodynamics, Moran, M.J., Shapior, H.N., Boettner, D.D., and Bailey, M.B., John Wiley, 8th Edition, 2014.

Is it safe to assume that you are satisfied with what has been obtained so far? If so, we can move on to the final step. Depending on how accurate we need to get, this step could require some effort. If you are satisfied so far, I will elaborate on what the options are for carrying out step 4.
 
  • #26
Chestermiller said:
Using the tables in Moran et al, I get a temperature of the product mixture (at 1 atm.) at the end of Step 3 as 2245 K = 1972 C. Is this about what you expected? This would correspond to a volume of 1710 x 2245\298 = 12900 liters = 12.9 m^3. This is a factor of 8.06 times the volume of the original reactant mixture. So, in Step 4, the product mixture would have to be adiabatically and reversibly compressed by a factor of 8.06.

Next, we'll do Step 4 (or, in your Plan 1, Step 3).
You're ahead of me, which is fine. Once it's here, I can soak up the information at whatever rate.

I'm not sure that the final temperature is what I expected. Seems high, but if that's what it is, then that's what it is.

I'm guessing that for adiabatic compression, we'll need to average the values for K?
 
  • #27
Looks like a good book to have, although at over $100 used it's a little too spendy for my purposes.

If there's a way to do it without referencing the book, that would be my preference.
 
  • #28
InquisitiveOne said:
Looks like a good book to have, although at over $100 used it's a little too spendy for my purposes.

If there's a way to do it without referencing the book, that would be my preference.
The nice thing about the book is that they have the integrals of Cp with respect to temperature tabulated for each of our product species, up to 3200 K. These integrals are needed to get the enthalpy changes, so the tables make things very convenient. The book also has the integrals of Cp/T, which are needed in Step 4 to get the entropy changes. Of course, we can always say that we can integrate the heat capacities on our own to get the enthalpy and entropy changes, but, for that we would need analytic approximations to the heat capacities as a function of temperature over the full range of temperatures that are relevant. In any event, conceptually, the two options are equivalent.
 
  • #29
Step 4

In this step, our goal is to determine the final temperature T and pressure P obtained as a result of adiabatically and reversibly compressing the product mixture down to an 8.06x smaller volume. So we will have two equations that need to be satisfied: the volume decrease equation and the constraint that the change in entropy of the mixture is zero.

Because we are dealing with a mixture and because the temperature and pressure increase substantially, there are complexities that need to be dealt with.

1. The final pressure will land us a little bit into the non-ideal gas region. A rough estimate indicates that the final pressure will be on the order of about 15 bars (225 psia). If we want to include the non-ideal gas behavior, we will also need to choose between several methods for accounting for this both in terms of mixture rules and in terms of calculating the effect of pressure on the entropy and volume even for the pure species.

2. If we decide that it is reasonable to approximate the mixture as an ideal gas, then taking into account the mixture aspect of the behavior is not a problem (since it is well established). However, even in this case, the temperature rise is going to be substantial (final temperature on the order of 4000K), and the effect of the changes in heat capacity of the various species over the range of temperatures in Step 4 will need to be included. We will have to find heat capacity analytic approximations that are accurate over the range of 2000K to 4000K for the 4 species in the final mixture (and for octane vapor too if the reaction mixture is rich).

A very crude estimate can be made of the solution to Step 4 if we treat the mixture as ideal and if we assume some average value for the heat capacity ratio over the temperature range. For example, we might assume an average value of ##\gamma## equal to say 1.3. This will immediately give us an approximation to the final temperature and pressure.

So what are your thoughts on all this?

If you would like me to show what the equations would look like for Option 2 above, I can easily provide that.

Chet
 
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  • #30
Chestermiller said:
Step 4

In this step, our goal is to determine the final temperature T and pressure P obtained as a result of adiabatically and reversibly compressing the product mixture down to an 8.06x smaller volume. So we will have two equations that need to be satisfied: the volume decrease equation and the constraint that the change in entropy of the mixture is zero.

Because we are dealing with a mixture and because the temperature and pressure increase substantially, there are complexities that need to be dealt with.

1. The final pressure will land us a little bit into the non-ideal gas region. A rough estimate indicates that the final pressure will be on the order of about 15 bars (225 psia). If we want to include the non-ideal gas behavior, we will also need to choose between several methods for accounting for this both in terms of mixture rules and in terms of calculating the effect of pressure on the entropy and volume even for the pure species.

2. If we decide that it is reasonable to approximate the mixture as an ideal gas, then taking into account the mixture aspect of the behavior is not a problem (since it is well established). However, even in this case, the temperature rise is going to be substantial (final temperature on the order of 4000K), and the effect of the changes in heat capacity of the various species over the range of temperatures in Step 4 will need to be included. We will have to find heat capacity analytic approximations that are accurate over the range of 2000K to 4000K for the 4 species in the final mixture (and for octane vapor too if the reaction mixture is rich).

A very crude estimate can be made of the solution to Step 4 if we treat the mixture as ideal and if we assume some average value for the heat capacity ratio over the temperature range. For example, we might assume an average value of ##\gamma## equal to say 1.3. This will immediately give us an approximation to the final temperature and pressure.

So what are your thoughts on all this?

If you would like me to show what the equations would look like for Option 2 above, I can easily provide that.

Chet
Thoughts... It's new territory for me. I'm having to look up a lot of the things you're mentioning to properly understand them, so I'm not quite eye to eye with you yet. Still a day or two worth of posts behind on the learning curve.

However, I have used equations for Adiabatic Compression in the past, to calculate pressures and temperatures after turbo and super charging. Instead of assuming some average value for ##\gamma##, couldn't we calculate the actual average?

I'll take your word for it, that it's crude. As crude as it may be, do you think that doing it that way would at least indicate an air/fuel ratio that creates the most possible pressure accurately, even though the accuracy of the final temperature and pressure may questionable?

Stated another way...
If my hunch is true, that rich makes more pressure, will doing it that way indicate accurately which fuel ratio is capable of creating the best pressure, no matter what the calculations say that pressure is?

What's the best method to identify the ratio that makes the most pressure (without worrying about how accurate the calculated pressure is)?

That's all I want to calculate. If easy and crude will do that, let's do that.
 
  • #31
InquisitiveOne said:
Thoughts... It's new territory for me. I'm having to look up a lot of the things you're mentioning to properly understand them, so I'm not quite eye to eye with you yet. Still a day or two worth of posts behind on the learning curve.

However, I have used equations for Adiabatic Compression in the past, to calculate pressures and temperatures after turbo and super charging. Instead of assuming some average value for ##\gamma##, couldn't we calculate the actual average?

I'll take your word for it, that it's crude. As crude as it may be, do you think that doing it that way would at least indicate an air/fuel ratio that creates the most possible pressure accurately, even though the accuracy of the final temperature and pressure may questionable?

Stated another way...
If my hunch is true, that rich makes more pressure, will doing it that way indicate accurately which fuel ratio is capable of creating the best pressure, no matter what the calculations say that pressure is?

What's the best method to identify the ratio that makes the most pressure (without worrying about how accurate the calculated pressure is)?

That's all I want to calculate. If easy and crude will do that, let's do that.
We would have to spot check it against the more accurate calculation on a few cases to be sure. That's the only way I know of.

If we employ Option 2 (i.e., regarding the product gas as an ideal gas mixture), then we can calculate an average gamma over the range of temperatures from T3 to T4 as follows:

$$\bar{\gamma}_{3,4}=\frac{\sum{y_i}\left[\int_{T_3}^{T_4}{C_{p,i}(T)\frac{dT}{T}}\right]}{\sum{y_i}\left[\int_{T_3}^{T_4}{C_{p,i}(T)\frac{dT}{T}}\right]-R\ln{(T_4}/T_3)}=\frac{\sum{y_i}(I_i(T_4)-I_i(T_3))}{\sum{y_i}(I_i(T_4)-I_i(T_3))-R\ln{(T_4}/T_3)}\tag{1}$$where $$I_i(T)=\int_{T_{Ref}}^{T}{C_{p,i}(T')\frac{dT'}{T'}}\tag{2}$$and where ##T_{Ref}## is any specified reference temperature, and ##y_i## is the mole fraction of species i, and ##C_{p,i}(T)## is the ideal gas molar heat capacity of species i at temperature T. For convenience, values of the integral in Eqn. 2 are tabulated in Moran et al are tabulated for CO2, H2O, O2, and N2 up to temperatures of 3250 K. At higher temperatures, we need to continue the integration on our own. Eqn. 1 must be solved simultaneously with the equation: $$\frac{T_4}{T_3}=\left(\frac{V_3}{V_4}\right)^{(\bar{\gamma}_{3,4}-1)}\tag{3}$$ to get the final temperature ##T_4##. The final pressure can then be obtained from the ideal gas law.
 
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  • #32
Step 4 Continued:

There is a simple way of taking into account the non-ideality of the gas behavior of step 4, by approximating the P-V-T behavior of the gas using the van-der-Waals equation of state: $$P=\frac{nRT}{V-nb}-\frac{an^2}{V_2}$$ where the constants a and b describe the deviations from ideal gas behavior; a and b are physical properties of the individual species that are tabulated in the literature. I'm not going to go into the details, but, if the various species are described by the van der Waals equation, the results of the previous section become:$$\frac{T_4}{T_3}=\left(\frac{V_3}{V_4-nb}\right)^{(\bar{\gamma}_{3,4}-1)}\tag{3}$$ for the final temperature and $$P_4=\frac{nRT_4}{V_4-nb}-\frac{an^2}{V_4^2}\tag{2}$$ for the final pressure, with $$a=\left(\sum {y_i\sqrt{a_i}}\right)^2\tag{3}$$and $$b=\left(\sum {{y_ib_i}}\right)\tag{4}$$Eqns. 3 and 4 are empirically derived mixing rules for a combination of gases obeying the van der Waal equation of state.
 
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  • #33
Seems that you're enjoying the challenge of figuring this out too. I'm really happy to have someone so capable helping to figure this out with me. Thank you Chet. I'm still not up to speed yet and getting somewhat lost in some of the calculations. I'll post up some specific questions in the coming week.

-Seth-
 
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  • #34
I was in a similar situation as you InquisitiveOne; I'm an Electrical/Systems Engineer who has undertaken design of an HCCI engine and has had to learn all this material just as you have.

Let me start my answering your question, "The ultimate reason behind my trying to figure this out is, engines make more horsepower with richer than stoichiometric conditions. For instance, gasoline makes peak torque/power at a ratio close to 12.5:1, when stoichiometrically all the oxygen is gone at 14.7:1. Why is that?"

I like addressing this point from the perspective of HCCI in which a pre-mixed fuel/air mixture is compressed to ignition. In this approach, it's easy to imagine that each molecule of fuel is surrounded by air sufficient to combust that molecule and the mixture compressed to yield the heat necessary to initiate the reaction. In this ideal state, the optimum air/fuel ratio would be stoichiometric. The air/fuel mixture is not, however, perfectly mixed; it has local rich and lean regions. The variance in mixture homogeneity is accommodated according to objective. If the objective is maximum power, then excess fuel is provided to ensure all the air is consumed. If the objective is maximum fuel efficiency, excess air is provided to ensure all fuel is consumed.

Moving on to calculations...

The calculation of atmospheric (ambient) conditions is described at https://www.grc.nasa.gov/WWW/K-12/airplane/atmosmet.html

The fundamental calculations for Pressure (Pa), Volume (m3), and Temperature (K) are the isentropic functions given at https://en.wikipedia.org/wiki/Isentropic_process#Table_of_isentropic_relations_for_an_ideal_gas

The critical specific heat ratio (y) varies with temperature and mixture composition. The paper "https://www.researchgate.net/publication/222535464_Temperature_and_air-fuel_ratio_dependent_specific_heat_ratio_functions_for_lean_burned_and_unburned_mixture" provides a pretty good summary of the variance in y depending on temperature and mixture composition. Below is a snapshot of the key section giving the equation for y when the mixture is unburned (yu) and burned (yb) where T is temperature (K) and λ is the air-fuel ratio. Note the equaton for pure air is available on-line.

244262


Given all the above, you can create an Excel spreadsheet that starts with atmospheric conditions in the intake volume from which you can calculate mass via PV=mRT where P is in Pa, V is in m3, m (mass) is in kg, R is 286.98 and T is in K. Next, make a table of V values between initial and compression V noting that compression V/initial V is the compression ratio. If you wish, the compression V can also be calculated from a target pressure or temperature with good reason (more later).

A critical equation for calculating energy and work is Q=mCvΔT where Q is Joules, m is mass in kg, Cv is 286.98/(y-1) and ΔT is the change in temperature from one step to the next. This equation can be applied to each step in the table described in the prior paragraph to determine how much energy was added to the closed volume through compression. Conversely, it shows the amount of work done on the gas during compression. Since we're going to focus on work, we want negative values during compression, so subtract the higher temperature from the lower temperature when determining ΔT.

The Q=mCvΔT equation is also used to determine heat added by fuel during combustion. In this case, m (mass) is the kg of fuel added and Q is Joules-per-kg of the fuel according to it's Lower Heating Value (LHV) (43.4 MJ/kg for gasoline). From this, ΔT can be calculated then added to the temperature at the end of compression to find the temperature after combustion. See later notes on accuracy.

Once the combustion temperature is known, the isentropic relations and y can be used to calculate the associated rise in pressure. Using the now known Volume, Pressure, and Temperature, you can build a second table similar to the first that expands the gas. The work done during this phase is also calculated with the Q=mCvΔT equation but with positive (work output) results.

Once both tables are complete, you can calculate Work in Joules by summing the work columns from the Compression and Expansion tables. Using RPM and knowledge of your cycle (i.e. 4 vs 2 stroke), you can convert the sum to Joules per minute (J/min). Finally, you can convert J/min to HP using an on-line calculator.

Now about accuracy...

A key error source not described is friction. It can be estimated using the figure below that I compiled from two key sources, Ricardo and Heywood. The figures are for a 6 cylinder engine. To use this data, you'll have to calculate HP the way professionals do using IMEP, FMEP, and BMEP: IMEP (bar) = Work (J)/Displacement (m3), FMEP (bar) from the figure below, and BMEP (bar) = IMEP-BMEP. Once you have BMEP, the calculations at http://www.epi-eng.com/piston_engine_technology/bmep_performance_yardstick.htm and a few unit conversions can be used to calculate HP and Torque (lb-ft).

244269

The largest single error source in the model as described is heat loss which accounts for a very large part of efficiency losses. It can be approximated by assuming the cooling system holds the cylinder at around 400K (to prevent breakdown of the lubricating oil) then working through the equation at https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html using one of the heat transfer coefficients given on page 147 - 149 of the document at http://www.ni.com/pdf/manuals/NICASUM.pdf (the model being described here is so course, it really doesn't matter which one is selected). The larger issue in calculating heat loss (and another key loss, pumping loss) is the fact that it assumes knowledge of engine geometry (areas, piston speeds, etc.) which requires some pretty intense work to develop.

There is some small work done during exhaust and intake. These processes are not performed on a closed volume, so the calculation of work is different than that given (left as an exercise). Assuming properly sized valves that don't produce excess pumping loss, the primary reason for incorporating the intake and exhaust cycles is to capture their impact on heat transfer; in the prior discussion, for example, conditions at the start of compression were assumed to be ambient (typically 288K at sea level) but it's clear that contact with the 400K typical wall during intake will in fact warm the charge and affect the start condition of compression.

The method I described for calculating temperature after combustion is very coarse because it assumes all the heat of the fuel was released instantaneously. Only HCCI comes close to releasing the fuel heat so quickly (illustrated in the video below); Spark engines depend on comparatively slow flame propagation and diesel engines depend on the time required for spray injection and the kinetics of air and fuel mixing in that spray (to make matters worse, the use of crankshafts means the volume is changing during heat release, so the slower spark and diesel systems aren't actually operating using fixed volume combustion, and this impacts all the calculations). The heat release rate of the selected combustion technique combined with the ignition delay associated with octane factors heavily into timing as previously noted.

Some practical considerations...

In the Otto cycle, the peak compression temperature of the air/fuel mixture must be limited such that the bulk of heat release occurs after Top Dead Center. The timing of combustion is affected by ignition delay which is affected by octane in turn. Furthermore, the mechanical efficiency of the linear to rotary conversion system (typically a crankshaft) has some bearing on the timing of heat release. It's all quite complex.

In all cycles, peak combustion pressure must be held below around 220 bar or the lubricating oil in the cylinder and rings will auto-ignite. This phenomenon is a hot topic in design of modern turbo charged engines and, in this realm, it's called Low Speed Pre-Ignition (LSPI). You will find that this 220 bar limit sets the upper temperature limit of combustion. As a result, you will find that greater compression leaves less ΔT for generation by fuel, and this reduces power.

All modern engines use catalytic converters, and they require a minimum exhaust temperature around 533K. Ignoring heat loss, compression and expansion in an Otto cycle engine are inversely correlated, so at least 533K of heat must be added by fuel to meet this requirement.

I hope all this helps!

Rod
 

What is thermodynamics?

Thermodynamics is the branch of science that deals with the relationships between heat, energy, and work.

What is constant volume pressure of combustion?

Constant volume pressure of combustion is the pressure that occurs during a chemical reaction at a constant volume.

How is constant volume pressure of combustion measured?

Constant volume pressure of combustion is typically measured using a bomb calorimeter, which is a device that allows for the controlled combustion of a substance at a constant volume.

What factors affect constant volume pressure of combustion?

The factors that affect constant volume pressure of combustion include the type of fuel, the amount of oxygen present, and the temperature and pressure of the environment.

What are the practical applications of understanding constant volume pressure of combustion?

Understanding constant volume pressure of combustion is important in various fields, such as energy production, engine design, and industrial processes. It allows for the optimization of combustion reactions and the development of more efficient and environmentally friendly technologies.

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