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I’m typing this from my phone...sorry if the quality is sub par.

I have a very involved spreadsheet from a colleague of mine that was put together in 1994. The spreadsheet calculates the excess air involved in combusting landfill gas in an enclosed flare. I’m trying to step my way through it little by little.

At one point in the spreadsheet, there is a calculation for enthalpy of the inlet gas stream. For the methane portion of the inlet stream, assumed to be 50% of the LFG stream, the formula being done is very clearly:

$$MC_p\Delta{T}$$

The inlet gas stream is assumed to be $\pu{60 ^\circ F}$. The $C_p$ value for methane is assumed to be at $\pu{77 ^\circ F}$.

My question is: Why is $\Delta{T} = \pu{60-77 ^\circ F}$ and not $\pu{77-60 ^\circ F}$? My understanding is that it doesn’t matter as the change in temperature is an absolute value.

EDIT 1

The following image is the particular place in the workbook where I am having difficulty understanding what is happening. How does the inlet stream have any enthalpy at all? Nothing has happened to it yet, it hasn't undergone any reaction.

Enthalpy Calculations

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    $\begingroup$ Adding a negative value is equal to subtracting a positive value. $\endgroup$
    – Maurice
    Feb 6 at 8:31
  • $\begingroup$ I guess I’m asking why have it 60-77 as appears to 77-60. It matters because in one case you get an exothermic reaction and in the other way it’s endothermic. $\endgroup$
    – Ryan_C
    Feb 6 at 17:24
  • $\begingroup$ If you want help with this, you're going to need to provide a much more complete and precise description of your system. For instance is your "inlet stream" a mixture of 50% air and 50% methane from a landfill that you are combusting? And what is the entire system whose temperature is being increased by 17F? $\endgroup$
    – theorist
    Feb 6 at 18:14
  • $\begingroup$ 50% methane, 50% CO2 for simplicity. Combustion air is obviously oxygen and nitrogen. Don’t think I need to provide a balanced combustion reaction for everyone. The temperature increases in the entire system. System consists of CO2, CH4, O2, N2. Basic combustion reaction. $\endgroup$
    – Ryan_C
    Feb 6 at 18:19
  • $\begingroup$ Stilll doesn't make sense. If you combust a 50% air/methane mixture adiabatically, the temp increase is going to be far more than 17F. So you're not just heating up the air/methane mixture, you're heating up a bunch of stuff around it as well. That's the system to which I'm referring. $\endgroup$
    – theorist
    Feb 6 at 18:22
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I'm going to take a guess at an answer based on the OP's limited description of the set-up:

I'm going to switch terminology from my comments above. Let's let the system be the reacting mixture, and the surroundings be everything that's in thermal contact with the reacting mixture.

The way calorimetry usually works is you burn (or otherwise react) a small quantity of a substance (the system) in a large heat bath (the surroundings). You ignore the heat capacity of the system, and calculate the thermal energy released by the reaction from $q_{p, surr} = m_{surr} C_{p, surr} \Delta T_{surr}$. [I've added the "$p$" subscript to indicate you have a constant-pressure system, which is implied by the fact that you are calculating $\Delta H$, rather than $\Delta U$, from the heat flow.]

In an exothermic reaction, $\Delta T_{surr} >0 $, and thus $q_{p, surr} = \Delta H_{surr} > 0 $. But whatever heat flowed into the surroundings must have flowed out of the reacting mixture (the system), and thus $q_{p, sys} = \Delta H_{sys} = -q_{p, surr}$.

It sounds like your colleague just reversed the sign of $\Delta T$, changing it from positive to negative, to get the correct sign for the heat flow relative to the system (negative) from the direction of heat flow relative to the surroundings (positive).

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  • $\begingroup$ I just added a screenshot to my original post in an attempt to get into more of the specifics. $\endgroup$
    – Ryan_C
    Feb 7 at 14:19
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If I had to guess, I think your colleague was trying to set $\pu{77 ^\circ F}$, the value $C_p$ was for, as the "reference state." The full equation for enthalpy at a point is: $$H - H_{ref} = mC_p(T - T_{ref})$$

At your reference point, the enthalpy is $0$, so in that stream, you get: $$H - 0 = mC_p(T - 77^\text{o} F)$$

This is okay to do because enthalpy is a state function, meaning that you're always focused on the change between states (by increasing or decreasing temperature/pressure/volume), not necessarily how it was done. For instance, if you wanted to get from point $\ce{A -> B}$ and you went $\ce{A -> B or A -> C -> D -> B}$, the change in enthalpy would be the same in both scenarios.

The value that you get in that cell is the enthalpy of methane relative to the reference state.

In your energy balance calculations, the reference states typically cancel out. Here's a quick example for flow in and out of a "box" (note that since this would be a steady flow through the box, the mass flow in would equal the mass flow out, so $m_{in} = m_{out} = m$): $$(H_{in} - H_{ref}) - (H_{out} - H_{ref}) = mC_p(T_{in} - T_{ref}) - mC_p(T_{out} - T_{ref})$$ $$H_{in} - H_{ref} - H_{ref} + H_{ref} = mC_p(T_{in} - T_{out} - T_{ref} + T_{ref})$$ $$H_{in} - H_{out} = mC_p(T_{in} - T_{out})$$

You can find more information for this by googling something like "enthalpy reference state", but I hope that clears some of it up.

Update 1 based on Edit 1 Enthalpy tells you how much energy your stream has at a specific state compared to the reference. You don't need to have a reaction or something happening in that stream to get the enthalpy. The only way it would be $0$ is if it was pure methane at the same reference state, $\pu{77 ^\circ F}$ and I'm assuming $\pu{1 atm}$.

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