# How to estimate temperature loss of a mixture when we know the heat loss due to vaporisation of one of the component?

So I am modelling in Chemkin and want to include phase change phenomenon in my model. But there's no capability of phase change in Chemkin so I have to think of some out of the box method for this.

So I have a mixture of H2O(g), O2(g) at 1080 K and this mixture mixes with C2H5OH(l) at 298 K. Now in the real world, at this temperature C2H5OH(l) will boil and form C2H5OH(g) as ethanols B.P is 351 K. The process is happening at 10 atm. I know the latent heat of vaporization of ethanol, so how do I calculate the temperature drop of this overall mixture?

So I will know the total amount of heat loss caused by 'm' amount of ethanol(l) boiling by applying

Q=m*hvp, where hvp=latent heat of vaporisation.


One approach I can think of is, we know total enthalpy of mixture at say t=0 sec just before mixing and know the amount of heat loss, Q, so we know total enthalpy at say some t= x sec after mixture has come to equilibrium state, by applying

'total enthalpy at x sec = total enthalpy at t=0 sec - Q'.


So do I just use total enthalpy = sum of individual enthalpies of each component and try out different temperatures till I get the desired total enthalpy at x? Will it be as simple as this or there's some catch in between? Any help will be appreciated.

P.S-( Chemkin assumes every species is a ideal gas so enthalpy is only a function of temperature, neglecting effect of pressure and entropy).

Also can someone put out a link for syntax while writing chemical symbols, I searched in the help section but couldn't find anything.

• Treat ethanol liquid and ethanol vapor as separate chemical species, with different heats of formation that undergo a reversible chemical reaction with the liquid species producing the vapor species. Heats of formation of these items are available in tables. – Chet Miller Jun 22 at 19:35
• Hi Chet, thanks for replying. So I already have tried that step. I added a phase change step with the activation energy as the latent heat of vaporization, and found the pre-exponential factor by matching the output with experimental data for vaporization of that liquid species. And add heat loss to the reactor equal to the heat of the endothermic reaction. But chemkin is having convergence issues with it's solver when I add this step to the detailed mechanism of the species in the gas phase. So I am trying the approach in the question. Do you think it will work? – obiwan Jun 22 at 20:43
• I think I was too hasty when I gave you the advice that I gave. After further consideration, what I would do would be to take the total enthalpy of the liquid ethanol and the gas mixture in the initial state, and, for final state, assume that all the ethanol has vaporized while the enthalpy of the new mixture is the same as the initial. That sounds pretty much what you did. – Chet Miller Jun 23 at 3:25
• How will the enthalpy of the new mixture be same as that of the old mixture?. Shouldn't it less by the heat loss due to latent heat of vaporization? – obiwan Jun 23 at 7:56
• The increase in ethanol enthalpy by vaporization and temperature increase is equal to the enthalpy decrease of the water-oxygen mixture by temperature decrease, so the change in enthalpy of the combination is zero. – Chet Miller Jun 23 at 10:59