Latent Heat of Vaporization

[davisb]

I have a solution at 447 psig and 100 deg. F. I want to find the latent heat of vaporization using Aspen Plus (an old version). I input the feed stream at pressure and temp stated above and flashed it across a "Flash2" box setting the output parameters of 100 deg F and a vapor fraction of 0.0001. From the output, I took the difference in enthalpies of the feed stream and outlet vapor stream to get the latent heat. Is this correct?

When I do the same and hold the pressure constant, my output temperature of the vapor stream is around 800 deg F and I get a totally different latent heat (around 1200 Btu/lb less).

From the definition latent heat, I know that it is the amount of heat neccessary to vaporize 1 unit mass at the same temp. So, I want to say that the first way I did it is correct. Just need some confirmation, please.

DavisB

 

[dcasto]

I'd do a 2 flashs on the material. The first is bubble point, the second is dew point. The difference in ethalpy is the latent heat.

 

[Latexman]

If you need [Δ]H_vap at 100^o F and the corresponding pressure, use the first value.

If you need [Δ]H_vap at 800^o F and 447 psig, use the second value.

Good luck,
Latexman

 

[MortenA]

dcasto

It depends on what you need it for.

If its for sizing a PSV for the firecase then davisb's method is correct. Your method would give the teh "average" dHvap since the first fraction that boils of is much lighter than the last.

Best regards

Morten

 

[MortenA]

I usually do it like this:

I take a copy of the inlet stream, and sets the pressure to the PSV relief pressure (SP+21%)

Then i add a heater and use a small value for heat added (in comparisson to the flow - just make sure that is a measurable but not large part of the stream that evaporates).

Then dH vap is equal to the energy added / the mass that evaporates.

Best regards

Morten