Question

# Why is LHV used in calculation of minimum energy consumption ?

Hi all,
To set things clearly: I am a PhD student working on the process implications of electryfing the ammonia production by replacing conventional SMR with Water Electrolysis. A part of my preliminary work is to assess the difference in theoretical minimum energy consumption. To do so, I have calculated a first approximation by summing up the reactions (SMR, Water gas shift, Haber-Bosch,...) and calculating the enthalpy of the resulting "total" reaction. I have done this for the "Water Electrolysis + Haber-Bosch scenario" and validated the minimum with values from the literature.
However, for the conventional "SMR + Haber-Bosch scenario", values from the literature are different. To be more specific, here is the energy minimum calculated in the following conference paper:
(...) the theoretical minimum of energy consumption for the process itself (represented by LHV of methane) is 22.2 GJ/t NH3 (...)
So here is my question: Why use the LHV of methane (instead the enthalpy of reactions) to calculate the energy minimum ? I feel like this is incorrect as I do not take into account the synthesis of methane.
Antoine

The paper you cited also includes the combustion of methane to provide the energy needed to account for the endothermic reaction as well as bringing the reaction temperature to that where the reaction occurs (850 C). For the combustion reaction, the LHV of methane is appropriate, where as for the SMR reaction, the heat of formation is appropriate. The reactor(s) consist of tubes packed with catalyst where the SMR, WGS, and HB reactions occur. On the outside of the tubes, CH4 is burned using either air or pure O2 to provide the energy needed for the process.

This is often a point of confusion, especially in Europe and Canada. There are multiple definitions of LHV and HHV, but only one is meaningful. If the H2O in the products of combustion are in the vapor state, the pertinent quantity is LHV. If the H2O is in the liquid state, the pertinent quantity is HHV. H2O in any combustion process that doesn't take place during a blizzard at the North Pole will be in the vapor state, so HHV is a meaningless quantity on the rest of planet Earth. The confusion originally arose over bomb calorimetry results in which coal was burned in a typically spherical heavy metal container submerged in water to absorb and measure the heat released. In this case (a water bath) the H2O inside the calorimeter was in the liquid state and so the laboratory reported HHV. In order to get LHV from such a test, you would have to separate the moisture from the residue and measure it, then correct the results using the latent heat of vaporization, which took more time and cost more money, so this step was often omitted for practical considerations. We no longer do it that way so there's no scientific justification for ever using HHV. Added to this confusion was the fact that a century ago some coal suppliers variously used LHV or HHV and customers worried about being cheated. You should read up on Dmitri Mendeleev, who created the periodic table. His career was devoted to making sure people weren't cheated when buying coal. Note: If you're making calculations using ⌂Ho, ⌂So, ⌂Go, etc. (like from tables in the CRC Chem/Physics Handbook or Lange's Handbook of Chemistry) be very careful about reference conditions, as there are inconsistencies between various texts and online sources of information. NASA Glenn is the best reference for gases. If you do it right, you will get the heating values as listed. I can send you a spreadsheet to demonstrate.
By the way... If you're looking for a way to solve chemical reactions that works even in the liquid state when the fugacities and compressibilities depart from unity (ideal behavior), check this out (it's free too) http://dudleybenton.altervista.org/projects/reactions/index.html
21st Jul, 2022
Antoine Dechany
Université Catholique de Louvain - UCLouvain
First of all, thanks a lot for your detailed answer as well as the useful ressources you provided !
Now, my question is more on the "why" of using the LHV of methane to define a minimum energy consumption for the process (i.e. SMR + Haber-Bosch). Maybe this is some question of standard or something but my supervisor and I have a feeling that this is not relevant as the process does not ever witness any combustion of methane.
If you have some time, here are the exact details on my calculations.
First here are the reactions that take place:
Steam Methane Reforming (SMR):
(1) CH4(g) + H2O(g) = CO(g) + 3H2(g) ⌂Ho = 206 kJ/mol
Water Gas Shift (WGS):
(2) CO(g) + H2O(g) = CO2(g) + H2(g) ⌂Ho = -41 kJ/mol
Haber-Bosch (HB):
(3) N2(g) + 3H2(g) = 2NH3(g) ⌂Ho = -92 kJ/mol
Here are the calculations for both methods
A) the LHV-defined energy minimum (whose relevance I am questionning):
the resulting equation when adding up equation (1), (2) and (3):
3 CH4(g) + 6 H2O(g) + 4 N2(g) = 3 CO2(g) + 8 NH3(g)
As I want my units to be expressed "per ton of produced ammonia", I thus take 1000kg of NH3 as a product. This equates to 353.2 kg of methane. Multiplied by the LHV of methane (50 MJ/kg), you get 17.66 GJ/ton_NH3 (the same result as the paper I mentionned). Added to that the heat to supply to allow the SMR endothermal reaction (⌂Ho = 206 kJ/mol of methane), you would need 4.54 GJ/ton_NH3. Adding up both bold values: 22.2 GJ/ton_NH3 (the value obtained in the paper).
B) the energy minimum computed only by adding up the ⌂Ho. of each reactions:
⌂Ho_total = 3 * ⌂Ho_SMR + 3 * ⌂Ho_WGS + 4 * ⌂Ho_HB = 618 - 123 - 369 = 127 kJ/mol
Divided by the stoechiometric coefficient of NH3 (to express the units per moles of ammonia) then divided by the molar mass, I obtain 0.93 GJ/ton_NH3.
You are right. The LHV of methane isn't the right measure. Your breakdown of the process into steps and looking at each one, then the overall result is the way to go.
The paper you cited also includes the combustion of methane to provide the energy needed to account for the endothermic reaction as well as bringing the reaction temperature to that where the reaction occurs (850 C). For the combustion reaction, the LHV of methane is appropriate, where as for the SMR reaction, the heat of formation is appropriate. The reactor(s) consist of tubes packed with catalyst where the SMR, WGS, and HB reactions occur. On the outside of the tubes, CH4 is burned using either air or pure O2 to provide the energy needed for the process.
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