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Thermodynamics... Constant volume pressure of combustion

  1. Jan 14, 2017 #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. The problem statement, all variables and given/known data

    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. Relevant 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-
     
    Last edited: Jan 14, 2017
  2. jcsd
  3. Jan 14, 2017 #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?
     
  4. Jan 14, 2017 #3
    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?
     
    Last edited: Jan 14, 2017
  5. Jan 14, 2017 #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.
     
  6. Jan 15, 2017 #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?
     
  7. Jan 15, 2017 #6
    Still trying to wrap my mind around the concept of Moles.
    0.2518309847 lb?
    12.5
    0.8818166987 lb.
    53.7137754143
    3.317310438
    4.1991271367
    16.6743863633 : 1 ???
     
  8. Jan 15, 2017 #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%
     
    Last edited: Jan 15, 2017
  9. Jan 15, 2017 #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
     
    Last edited: Jan 15, 2017
  10. Jan 15, 2017 #9
    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
     
  11. Jan 15, 2017 #10
    Interesting. I had pretty much all the same figures which I thought were in grams, so I converted them all to pounds. No need?
     
  12. Jan 15, 2017 #11
    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.
     
    Last edited: Jan 15, 2017
  13. Jan 16, 2017 #12
    Grams are fine too. But I though you wanted to work in Imperial units, since you gave the heat of combustion in BTU/lb.
     
  14. Jan 16, 2017 #13
    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?
     
  15. Jan 16, 2017 #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
     
  16. Jan 16, 2017 #15
    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-
     
  17. Jan 17, 2017 #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 in to 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.
     
  18. Jan 18, 2017 #17
    Still soaking this up. Seems a lot like plan 1.
     
  19. Jan 18, 2017 #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?
     
  20. Jan 18, 2017 #19
    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. Jan 18, 2017 #19
    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.
     
  22. Jan 18, 2017 #20
    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-
     
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