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Flue gas enthalphy

  1. Mar 23, 2016 #1
    Hi

    1. The problem statement, all variables and given/known data

    I'm trying to calculate the enthalphy of flue gas leaving a gas turbine, in kJ/kg. I just want to confirm that I'm using a correct method.

    2. Relevant equations

    hfg = (VCO2⋅ hCO2 + VH2O⋅ hH2O + VN2⋅ hN2 + VO2⋅ hO2) / (VCO2 + VH2O + VN2 + VO2)

    ρfg = (p⋅ M) / (R⋅ Tfg)

    3. The attempt at a solution

    I have already calculated the total flue gas volume and the volumes of the flue gas components, and I'm fairly sure that I've got that right.

    The next step is to calculate the flue gas enthalphy using the following equation (the enthalpies are taken from a table at Tfg = 497 °C):

    hfg = (VCO2⋅ hCO2 + VH2O⋅ hH2O + VN2⋅ hN2 + VO2⋅ hO2) / (VCO2 + VH2O + VN2 + VO2)
    hfg = (0,59⋅ 991 + 0,55⋅ 776 + 5,09⋅ 653 + 0,81⋅ 688) / (0,59 + 0,55 + 5,09 + 0,81) = 695 kJ/mn3


    Then, from the ideal gas law, I get the density of the flue gas (M is an average that I've calculated, and p = 1 bar):

    ρfg = (p⋅ M) / (R⋅ Tfg)
    ρfg = (1⋅ 105⋅ 29⋅ 10-3) / (8,314⋅ 770) = 0,45 kg/m3

    And finally, to convert from kJ/mn3 to kJ/kg, I divide the calculated enthalphy with the density:

    hfg = 695 kJ/mn3 / 0,45 kg/m3 = 1532 kJ/kg


    Is my method correct? My biggest concern is whether or not m3 really cancel off mn3.

    Thanks!

    Marcus
     
    Last edited: Mar 23, 2016
  2. jcsd
  3. Mar 23, 2016 #2
    What are the units of the h's and the V's, and what do they represent? What makes you think you can get the density of the flue gas by using the molecular weight of air in ideal gas law?
     
  4. Mar 24, 2016 #3
    Thank you for your answer. I will try to make things clearer with units.

    The h represents the enthalphy. I think you are more familiar with using i. The unit in this case is ##\frac{kJ}{m_{n}^{3}}##
    The V represents the volume of each flue gas component. The unit is ##\frac{m_{n}^{3}}{m_{n}^{3}fuel}##

    ##h_{fg}=\frac{V_{CO_{2}}h_{CO_{2}}+V_{H_{2}O}h_{H_{2}O}+V_{N_{2}}h_{N_{2}}+V_{O_{2}}h_{O_{2}}}{V_{CO_{2}}+V_{H_{2}O}+V_{N_{2}}+V_{O_{2}}}=
    \frac{0,59\frac{m_{n}^{3}}{m_{n}^{3}fuel}\cdot 991\frac{kJ}{m_{n}^{3}}+0,55\frac{m_{n}^{3}}{m_{n}^{3}fuel}\cdot 776\frac{kJ}{m_{n}^{3}}+5,09\frac{m_{n}^{3}}{m_{n}^{3}fuel}\cdot 653\frac{kJ}{m_{n}^{3}}+0,81\frac{m_{n}^{3}}{m_{n}^{3}fuel}\cdot 688\frac{kJ}{m_{n}^{3}}}{0,59\frac{m_{n}^{3}}{m_{n}^{3}fuel}+0,55\frac{m_{n}^{3}}{m_{n}^{3}fuel}+5,09\frac{m_{n}^{3}}{m_{n}^{3}fuel}+0,81\frac{m_{n}^{3}}{m_{n}^{3}fuel}}=
    695\frac{kJ}{m_{n}^{3}}##

    ##\rho_{fg}=\frac{p_{fg}\cdot M_{avg}}{R\cdot T_{fg}}=\frac{10^{5}Pa\cdot 0,029 \frac{kg}{mol}}{8,314\frac{J}{mol\cdot K}\cdot 770K}=0,45\frac{kg}{m^{3}}##

    ##h_{fg}=\frac{695\frac{kJ}{m_{n}^{3}}}{0,45\frac{kg}{m^{3}}}=1532 \frac{kJ}{kg}##

    I have calculated an average molecular weight by using the volume-% of each component and the molecular weight of each component, and it happened to be the same as for air. I just assume that the gas behave like an ideal gas to make it more simple, that's the only reason for it.
     
  5. Mar 24, 2016 #4
    What is ##m_n## or ##m_n^3##? Why aren't you expressing the enthalpies on a molar basis?
     
  6. Mar 24, 2016 #5
    It's normal cubic meter.


    Ok, so I have the volume of each component of the flue gas :
    ##V_{CO_{2}}=0,59\frac{m_{n}^{3}}{m_{n}^{3}fuel}##
    ##V_{H_{2}O}=0,55\frac{m_{n}^{3}}{m_{n}^{3}fuel}##
    ##V_{N_{2}}=5,09\frac{m_{n}^{3}}{m_{n}^{3}fuel}##
    ##V_{O_{2}}=0,81\frac{m_{n}^{3}}{m_{n}^{3}fuel}##


    And thus also the volume-%, which I've read is the same as the molar-% :
    ##v_{CO_{2}}=8\%##
    ##v_{H_{2}O}=8\%##
    ##v_{N_{2}}=72\%##
    ##v_{O_{2}}=11\%##


    With the molar-% (or volume-%) known I can use the same equation as above to find the enthalphy on molar basis (see the attached file for the table from where I got the enthalpies) :

    ##h_{fg}=\frac{v_{CO_{2}}h_{CO_{2}}+v_{H_{2}O}h_{H_{2}O}+v_{N_{2}}h_{N_{2}}+v_{O_{2}}h_{O_{2}}}{v_{CO_{2}}+v_{H_{2}O}+v_{N_{2}}+v_{O_{2}}}=
    \frac{8\%\cdot 22347\frac{kJ}{kmol}+8\%\cdot 17617\frac{kJ}{kmol}+72\%\cdot 14808\frac{kJ}{kmol}+11\%\cdot 15580\frac{kJ}{kmol}}{100\%}=
    15999\frac{kJ}{kmol}##

    And then I find the average molecular weight of the flue gas with the volume-% and molecular weight of gas component :

    ##M_{avg}=v_{CO_{2}}\cdot M_{CO_{2}}+v_{H_{2}O}\cdot M_{H_{2}O}+v_{N_{2}}\cdot M_{N_{2}}+v_{O_{2}}\cdot M_{O_{2}}=8\%\cdot 44\frac{g}{mol}+8\%\cdot 18\frac{g}{mol}+72\%\cdot 28\frac{g}{mol}+11\%\cdot 32\frac{g}{mol}=29\frac{g}{mol}##

    And finally I convert the enthalphy from kJ/kmol to kJ/kg :

    ##15999\frac{kJ}{kmol}\Rightarrow 15999\frac{kJ}{kmol}\cdot \frac{1kmol}{1000mol}\cdot \frac{1}{29\frac{g}{mol}\cdot \frac{1kg}{1000g}}=552\; \frac{kJ}{kg}##


    There is a huge difference between the two enthalpies that I've calculated depending on the method used. But the last method seems better. What do you think?

    Marcus
     

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  7. Mar 24, 2016 #6
    I think your methodology in post #5 is correct. However, I am concerned that the specific enthalpies of all the gases in your table are taken to be zero at 0 C. Shouldn't the heats of formation of H2O and CO2 be included?

    Chet
     
  8. Mar 24, 2016 #7
    I don't know. But maybe you're right. I found a table and the enthalpy of formation is :

    ##h_{f_{CO_{2}}}=-394\frac{kJ}{kmol}##
    ##h_{f_{H_{2}O}}=-242\frac{kJ}{kmol}##

    Should I just subtract this number from the enthalpy that I originally used, like this :

    ##h_{fg}=\frac{8\%\cdot 22347-394\frac{kJ}{kmol}+8\%\cdot 17617-242\frac{kJ}{kmol}+72\%\cdot 14808\frac{kJ}{kmol}+11\%\cdot 15580\frac{kJ}{kmol}}{100\%}=
    15695\frac{kJ}{kmol}##

    And then convert from kJ/kmol to kJ/kg :

    ##15695\frac{kJ}{kmol}\Rightarrow 15695\frac{kJ}{kmol}\cdot \frac{1kmol}{1000mol}\cdot \frac{1}{29\frac{g}{mol}\cdot \frac{1kg}{1000g}}=541\; \frac{kJ}{kg}##

    Marcus
     
  9. Mar 24, 2016 #8
    I don't know. It isn't clear what temperature these heats of formation apply at. And, I don't know what your professor expects you to do.
     
  10. Mar 24, 2016 #9
    They apply at 25 °C and 1 atm.

    I'll ask him if it's necessary to take into consideration. The calculation of the flue gas enthalpy is just a small part of a bigger project I'm working on.

    Thank you so much for your help Chet!

    Marcus
     
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