I was hoping someone could check my math/educate me'¦ I'm trying to calculate how much "intercooler" effect fuel has on an engine'¦

Excuse me for doing everything in metric, but I'm trying to follow examples in a text book to help me, and they don't believe in the engish system'¦

What I did was calculate the mass of air and fuel within the cylinder on one stroke at 100% VE. I calculated that at a 9:1 A/F ratio (True A/F, not on a gas scale) that I would have .001375kg of air, and .000152kg of fuel. This was done using predicted turbo outlet temperatures of ~230 degree's F (384 degree's K) a 366" motor, @ 6000rpm 100%VE, and ~15psi (73% compressor efficency).

I used the following formula to balance the fuel/air temps to achieve the equilibreum, and see what the true Air temp drop was from the fuel. I assumed the energy transfer into the fuel would end up below the vaporization point of E85 (which it does).

Mass of Air x Specific Heat of Air (at 384 degree's K) x (Turbo outlet temp - Final Temp) = Mass of Fuel x Specific Heat of Fuel x (Final Temp-Fuel Temp) =

Or to simplify

ma x cp (air) x (Ta-T) = mf x cp (fuel) x (T - Tf) Solve for T

.001375kg (1.011 kJ/kg-K) (384-T) = .000152kg (2.44 kJ/kg-K) (T-303)

Solving for T puts me at 367 degree's K, or to the punchline a converted drop of 17 degree's F.

With a starting point of 230 degree's F, and a drop of 17 degree's F, (running it through my turbo calculator rather then do the intercooler efficency drops), it looks like it provides about a 20% intercooler effect.

Now this is only true if ALL of the energy can transfer before it hits the back of the intake valve. I'm not sure if there is enough time to do that or not? I'm sort of surprised that it is so low though? I expected to see a much bigger temp drop then that.

Thoughts from any of you guys that hang out here clearly smarter then me?

Am I doing something wrong, or just expecting too much?

Obviously the fuel ratio of 9:1 is stoich, so it would be better as we added more fuel, but I'm guessing not by that much. It sort of makes me think though that every time I do a non-intercooled calculation to figure out mass flows, that I REALLY shouldn't use 0% for the intercooling value at all. The fuel will do at least 20%, plus who knows how much more heat transfer into the rest of the cold side pipes/intake/etc...

 

Your temp drop is extremely low because of Your assumption that none of the fuel vaporizes. That portion of the function will likely need to include time and/or surface area (droplet size), depending on any new assumptions.

Regards,

H. Kurt Betton

 

I definately understand the phase change of the fuel to vapor would help a BUNCH to cool things... I get that part.

But why would the fuel vaporize if the temp is under the fuel boiling point?

Thinking out loud I suppose a non-homogenous mix could transfer more heat to a lean area and cause it to vaporize? But isn't the mixture fairly homogeneous as it comes out of the venturi's?

Also, if the fuel vaporizes it doesn't it expand to take up the volume where air (oxygen) would usually be? Isn't that a bad thing?

 

Ready for a thesis? I'll try to keep it short.

Q1: How is phase change occuring below boiling point?
Phase change occurs as a result of evaporation, as well as boiling. Evaporation occurs mostly at the surface of a liquid (most times below boiling point), whereas boiling occurs throughout the entire mass of the liquid (hence the bubbling). Proof of evaporation: Taking a bath w/o drying off. The residual water evaporates but we never get to 212degF body or surface temps. You will also note the cool (cold?) feeling afterwards. Latent heat of vaporization is at play here and, for the most part, ignores the boiling point of the liquid.

As You can probably imagine, the larger the surface area:mass ratio (think small droplets vs. one pot full, both of the same total volume), the larger the proportion of liquid that evaporates compared to the proportion that boils. (Can You see this calculation getting more and more complex?)

For simplicity, I would assume the mixture is homogenous, & the temperature distribution in the air stream is equal. In reality they're not, but that's more about engine & carb design, than liquid-vapor intercooling.


Q2: Doesn't the vaporized fuel take up useful air volume?
Yes, but not much more than the liquid fuel would have taken up. Think of the fuel vapor as infinitely small liquid droplets that take up the same volume (ignoring thermal volumetric effects) as a single droplet of fuel.


Out of curiosity, is this for a particular application, or an academic exercise (both of which are perfectly acceptable)? I can see this being complex enough to make a Master's Thesis out of, or simplified to just get a rough idea of IAT's. Depends on what You're trying to get out of it.

In more specific terms, after You calculate the temp drop, what do You plan to do with it?

Regards,

Kurt

 

Of course I was ready for a thesis... You don't start a thread like this without hoping for one...

Q1- That makes perfect sense now. If I think about the system from an energy standpoint rather then a "temperature" stand point, on the micro scale it makes better sense. Local small particles can gain enough energy from the surrounding area to cause the phase change. More surface/less volume (finer droplettes) expidite this process. So the calculation would need to take into account the droplette size, as well as the temperature gradient away from the wall, etc... much to complicated to do by hand. I think I got it now.

Q2- I suppose that is true if the air (oxygen) can then infuse the area between the now infinantely smaller droplettes. Mass stays the same, so ultimately the volume doesn't change much?

Q3 "Out of curiosity..." - hmm... that's a tough question to answer... Well the thought incepted in my head as to try and figure out how much power is gained from E85 over gas just from the latent heat of vaporization difference. Colder charge = more density = more power. But then it turned into the fact that most HP calculators under predict on non-IC carb'd motors. I was then wondering if I could calculate the "intercooler efficency" of a given fuel if I knew the mass of that fuel, and the temperatures at hand. If I had an effective "intercooler efficency" (because I know the number really shouldn't be zero as I always input), then I could get a better power number estimate.

I am actually just about to graduate with my BSME (it only took me 14 years... )... so I have a "rusty" engineering background to pull from (I had Thermo over 8 years ago). So it's somewhere between an academic exercise and the desire to just understand how stuff works...

In the end, I'm working on finishing my turbo project up right now (actually as my senior project), and I know the thing can make more power then the short block can handle so none of this really matters. Just trying to educate myself... that's my ultimate goal.

I sort of figured the caclulations would be too difficult and too "assumption filled" to be all that useful. But I thought it might just shed some light on how it works.

 

Sounds like You have the hang of it now. If You wanted to get ridiculously complex You could consider the 'external' influences on intake air temp & density, since they affect phase change (and thereby delta temp...) on a microscopic level. A few of the 'external' influences that come to mind are:

• Thermal transfer from the intake & heads back into the airstream
• Mechanical effects (such as turbulence & air vibration) on phase change
• Ambient air humidity
• etc

A good followup exercise would be to start laying out an equation with ALL the variables, and the sub-equations required to calculate some of the variables. In the process You can start plugging in some numbers and play with them within a realistic range to see their effects on the final temperature. There is usually a hidden gem that gives the most improvement for the least effort - giving You something to focus on during Your next build...

Regards,
Kurt

P.S. Don't let anybody give You flak about the 14yr degree (not even Yourself!). It's easier & quicker to recite a formula than it is to understand it. The current college philosophy encourages students to recite. As long as You don't subscribe to this philosophy You'll be in good shape for the long run. EVERYTHING that is artificially inflated will eventually deflate to its true levels.

 

I would also say that EVERYONE has a different 'path' to travel when it comes to education. I put my way through school (attending several different universities) and eventually graduated in my field(s) of study.
It took longer than 4 years but a bit less than your effort. That being said, I have worked with a variety of people over the years who could do some things very well and other things were more difficult for them.

I have 34 years at Ford Engineering as a Senior Engineer and still have to open my old books and refresh my mind on the equations.

This Forum is all about helping people solve difficult problems in an easy manner, it is not about egos.

I would like to be one of the first to congratulate you on your upcoming BSME degree.

Tom Vaught

 

Thats cool. I have a feeling you took allot of breaks because life got in the way? I.e. Financial issues, kids, jobs... Not everyone can just hit a degree running full speed. By the time I realized I actually wanted to pursue an engineering degree I'd already had a real career that paid descent enuff to get me a house and a life that I couldn't just up and ditch for dorm life, top ramen and student loans. Hell my work will even pay for school but it doesn't mean I'm dumb enuff to work 40-50hrs a week and think I can take diff equations and not wanna kill myself every waking hour. Congrats on finishing