To calculate the enthalpy of air, or the change in enthalpy when the temperature changes but pressure remains at one atmosphere, given that it is never saturated and there is never evaporation and condensation, can’t one just use the specific heats of air and water vapor?
References that I turned up in a little web searching seem to make a big deal of the latent heat of vaporization in the water vapor component.
But air is just a mix of gasses, mostly N2 and O2 and H2O and CO2. There is latent heat in ALL of them. If none of them change phase during a temperature change, the latent heat is irrelevant, isn’t it? There would be no special consideration for the H2O part, right?
My typical interest is in heating an air stream that starts at typical indoor conditions including moderate relative humidity. I hope I can use simple models for the specific heats of air and water vapor, in J/(kg K). A further hope is that the different components absorb heat independently, so a mix of 10 g of H2O and 990 g of air will absorb the same heat energy in going from 20 °C to 200 °C as would those two components if heated separately. I’m not sure how to interpret an isolated component of H2O but hope it would be at the same density as it is in humid air, i. e. at a reduced pressure, and their heat absorption effects would be separable like their partial pressures are.
Thinking back to when we were going over this in my classes, what’s important about the vaporization energy of that water (which I’ll call Hv) is that it’s the energy differential between having “some air with a bit of water at the bottom” and “some air that’s slightly humid” at the same temperature and pressure; the T and P at which you are are such that both cases are possible. The other components would be in the same situation of their Hv being important only if you were at a T and P at which they have two possible physical states.
“But air is just a mix of gasses, mostly N2 and O2 and H2O and CO2. There is latent heat in ALL of them. If none of them change phase during a temperature change, the latent heat is irrelevant, isn’t it? There would be no special consideration for the H2O part, right?”
I don’t have time to think about this too much right now, but one difference I would point out between water and the other gases is that at standard conditions, water is a liquid while the others are gases.
I think one method of calculating the delta H of the water vapor would be to take it back to the water form (-heat of vaporization at T1), calculate the delta H with temperature change of the liquid water from T1 to T2, and then change it back to a vapor (+heat of vaporization at T2). I may have those pluses and minuses backwards.
Then just use Q = Mdot(H2-H1).
Q = BTU/min (required to heat the air stream)
Mdot = mass flow rate of air stream (in lbm/min)
H1 = Enthalpy (btu/lbm) from chart based on inlet conditions
H2 = Enthalpy (btu/lbm) from chart based on outlet conditions
If you’re using an electric heater multiply Q by 60 to get BTU/hr, then divide by 3412.1 to get kW. And then add another 20% as a safety factor.
Example: Heat a 130 lbm/min airstream from 34°F, 100%RH (from chart the enthalpy is 12.6 btu/lb 28.7 gr/lb moisture) to 120°F (gr/lb remains constant at 28.7, so follow the chart to the right horizontally, you will see the RH drops to about 14%, and the enthalpy is 26.1 Btu/lb).
130(26.1-12.6) = 1755… *60 min/hr = 150,300 BTU/hr = 31kw. So buy a 40 kw heater.
Hey, just remembered another quick and dirty formula for sensible heat (sorry, it’s in standard units, working for a customer who demands US units, I haven’t used metric since college).
Q = 1.08 x SCFM x delta T.
Q is in BTU/hr…
BTW, I messed up my example above… I meant to change T2 to 90°F because 120°F was off the chart. sorrrry
