Air conditioning the charge air. (math content)

I'm working on this idea for my challenge car. I am planning to have a big mid-engine shoved into a small car where it doesn't belong. I would like to put a turbo and intercooler on it just to run up the score, and there is not much real estate or airflow where an inter cooler needs to be for plumbing reasons.

Then I had a brilliant (or not) idea - Just run the charge pipe to a box housing the A/C evaporator. The A/C system should be capable of removing something on the order of 15k BTU/hr of heat from the charge air, right? I have no idea how this compares to the performance of a traditional intercooler.

It seems I have searched for "evaporator in the charge pipe" I have come up with forum posts (not here) where someone suggests doing this and is then flamed for even suggesting such a retarded idea. In my mind it seems like it should work, so I thought I would kill some time when I should be working and do some calculations.

The motor I'm considering turbo'ing is a 281 cu in (4.6L) V8. Time for some math. Please check my math. I suck at math.

Volume flow into the motor at average racing speed is: 4000 RPM X 1 intake stroke per 4 strokes = 1000 intake strokes per min (i.s./min) (1000i.s./min)X(281cu in/i.s.) = 281000 cu in/min = 4683 cu in/sec = 162.6 cu ft / min

Here's where I start making stuff up: A reasonable cooler inlet charge temperature at 8lbs of boost is 175F? Is this reasonable?

if yes then according to: http://www.engineeringtoolbox.com/air-temperature-pressure-density-d_771.html

my charge air coming in is around .095 lb/cu ft, and specific heat capacity of 1.022kJ/kgK (please pardon my mixture of units).

Since I can't remember a lick of the Thermo classes I took, I just plugged this stuff into this calculator for "Thermal Energy":http://www.pipeflowcalculations.com/heater/

and tweaked the outlet temp until the Power output was very nearly 15k BTU/h. And the delta T was 66.3F. This seems like a pretty good temp drop - certainly comparable to a big air-to-air intercooler, right?

How much HP does the "A/C intercooler" add?

n(lbs/min)= P(psia) x V(cu.ft./min) x 29/(10.73 x T(deg R)), a formula I stole here:

http://www.gnttype.org/techarea/turbo/turboflow.html

n (lbs/min)= 22.7 psia x 162.6 cfm x 29/(10.73 x 634.67 deg R) = 15.72 lb of air per minute no i/c n (lbs/min)= 22.7 psia x 162.6 cfm x 29/(10.73 x 568.38 deg R) = 17.55 lb of air per minute w/ i/c

for a difference of 17.55lb/min-15.72lb/min = 1.83lb/min air at stoic, that's (1.83lb/min air) / 13.7 = 0.1336lb/min fuel

fuel is 125k BTU/gallons, 7.29lb/gallon, so 125k BTU / 7.29lb = 17.15kBTU/lb 17.15k BTU/lb X 0.1336 lb/min = (2.29k BTU/min)/[(42.2BTU/min)/HP]= 54HP (ideal)

Crappy old Ford motor is probably 60% thermal efficiency, 54*.60 = 32.5 BHP

And my input was 15k BTU/hr from an AC unit that too is probably 70% efficient, thus costing 21.4k BTU/hr=357 BTU/min/42.2= 8.46 HP

Net HP gain: 32.5-8.46 = 24HP

Now that SEEMS worth the extra 20lbs in AC crap. And I get to use the stock accessory belt.

However, I can't believe I'm the first person to have thought of this, and if it worked, it seems like it would be commonplace. What am I missing?

I believe the 90 (or there abouts) Lotus Esprit used a setup like this. I could be wrong though.

Ok never mind. I'm wrong. Sorry

Some older BMWs ran the fuel line through the a/c plenum to keep it cool, but that was more for preventing vapor lock than for performance.

I didn't read through all your math, but I'm going to suggest you have over estimated the quantity of cooling you can get. If the air volume through the AC evaperator was sufficient to feed an engine, those guys that hook the heater fan up to the top of the carb would be making a lot more power.

Lisa, in this house, we obey the laws of thermodynamics!

Assumption was that a big Lincoln Continental cooling system can produce 15kBTU/h... which seems believable considering it can keep the cavernous interior of a Continental to 60 degrees when its 100+ outside.

And these people over at Yahoo think it could even be more: http://answers.yahoo.com/question/index?qid=20110607175226AAiktT1

any room for air to water intercooling?

In reply to mad_machine:

perhaps room, but very little budget- this is a GRM Challenger. AC stuff came 'free' with the parts car.

Well, this setup will certainly cool the incoming air a little. Here's some issues with the calculations though.

1. Your air flow rate is way too low: "4000 RPM X 1 intake stroke per 4 strokes" there are 2 strokes per revolution and you forgot to put that in the equation so that will increase your air flow rate x2. Also, these calculations assume 100% volumetric efficiency for a NA engine. You'll have more flow if you run boost. It'll be about 50% more air at 8 psi. So you need to increase your air flow rate by 150%. This give you a flow rate of 406.5 CFM.

2. Since your air flow rate is now much higher, your temperature drop through the A/C evaporator will decrease by the same proportion. So now you only get a 26.5 degree drop.

3. Your charge temperatures post-compression and pre-intercooler are too low. If you assume a compressor efficiency of 0.70 (probably reasonable for an older automotive turbocharger), and you want 8 psi with a 90 degree F ambient temp, then your charge temp will be 194 F rather than 175.

4. Your thermal efficiency for the engine is way too high. It'll be around 30-35% especially on a race tune. I'm assuming you're defining that efficiency as EDIT: (Power at Crankshaft)/(Power Available in Fuel Burned). Edit: rule of thumb is that about 1/3 of energy available in fuel goes to crank power, 1/3 goes to the exhaust heat, 1/3 goes to the coolant and oil coolers (if installed).

5. In that same equation, it looks like you're probably using the higher-heating-value for the fuel to get 125k BTU/gal. Probably better to use lover-heating-value (LHV) which is closer to 115k BTU/gal.

6. The A/C unit produces 15k BTU/hr cooling, and most older automotive AC systems probably have a COP around 2.0; therefore it requires 7.5k BTU/hr to run - that's 5.9 hp.

Overall, you've got the equations right, just some of the assumed values aren't great. I think what you'll find is that there just isn't enough heat being rejected from an A/C system to cool an intake charge with any reasonable amount of boost. Basically, you'll have a lighter and far cooler intake with a well designed and sized air-air intercooler.

If you want to use the A/C evaporator, consider running chilled water though it and have a reservoir in the car. Remember, at the challenge you only need 30 seconds of cooling at a time.

BTW, sweet to see someone actually running some numbers on a project like this - props.

JohnyHachi6 wrote: If you want to use the A/C evaporator, consider running chilled water though it and have a reservoir in the car. Remember, at the challenge you only need 30 seconds of cooling at a time.

If you are going to do that.. consider dryice

1. 2 strokes per rotation. Oops. That was a biggie.

Fixing errors: (2000 i.s./min)X(281cu in/i.s.) = 562000 cu in/min = 9366 cu in/sec = 325.2 cu ft / min

A reasonable cooler inlet charge temperature at 8lbs of boost is 194F

according to: http://www.engineeringtoolbox.com/air-temperature-pressure-density-d_771.html

my charge air coming in is around .091 lb/cu ft, and specific heat capacity of 1.009kJ/kgK (please pardon my mixture of units).

and tweaked the outlet temp until the Power output was very nearly 15k BTU/h. And the delta T was 33.5

How much HP does the "A/C intercooler" add?

n(lbs/min)= P(psia) x V(cu.ft./min) x 29/(10.73 x T(deg R)), a formula I stole here:

http://www.gnttype.org/techarea/turbo/turboflow.html

n (lbs/min)= 22.7 psia x 325.2 cfm x 29/(10.73 x 653.67 deg R) = 30.52 lb of air per minute no i/c n (lbs/min)= 22.7 psia x 325.2 cfm x 29/(10.73 x 620.17 deg R) = 32.17 lb of air per minute w/ i/c

for a difference of 32.17lb/min-30.52lb/min =1.65lb/min air at stoic, that's (1.65lb/min air) / 13.7 = 0.120lb/min fuel

fuel is 115k BTU/gallons, 7.29lb/gallon, so 115k BTU / 7.29lb = 15.77kBTU/lb 15.77k BTU/lb X 0.120 lb/min = (1.89k BTU/min)/[(42.2BTU/min)/HP]= 44.84HP (ideal)

Crappy old Ford motor is probably 30% thermal efficiency, 44.84*.30 = 13.45 BHP

Net HP gain: 13.45-5.9 = 7.55HP less attractive, but still positive...

Yeah, that's right about where I'd expect. A small gain, but still a gain.

7.55HP gain on a 350 or so motor. 2.1% yay

Yeah... air-to-air is prolly better.

I need it to work at steady state as well, as this is going to be a Chumpcar as well.

1st law of physics... conservation of energy.

You will spend more HP driving the compressor than you will get back cooling the intake charge.

Period.

No way around it.

Here's the thing though.

Most systems shut the compressor down at WOT.

You'll still be sucking cold air through the box, it just won't be cold indefinately.

Might work ok for short runs but not anything long-term.

Shawn

In reply to curtis73:

Conservation of energy really doesn't apply like that to this problem. It's not like the A/C compressor is helping to spin the engine, it's allowing the engine to burn more fuel and thus make more power.

If conservation of energy applied like you wrote, then how would a supercharger make additional power? Wouldn't you "spend more hp" driving the supercharger than you got back from compressing the charge?

It feels like there will be too many losses but I really hope you try it, that's the only way to know for sure and I'd be curious as all hell to see what happens.

In reply to JohnyHachi6:

I agree.

curtis73 wrote: 1st law of physics... conservation of energy. You will spend more HP driving the compressor than you will get back cooling the intake charge. Period. No way around it.

Bristol did it in the 70s. Ford did it on one of the Lightnings.

It doesn't violate any laws of thermodynamics. You're not getting more power directly from the A/C's working fluid (which WOULD violate laws of thermodynamics), what you are doing is enabling the engine to do its job better.

If you want to look at it from a thermodynamics perspective, intercooling is throwing power away, since it reduces the temperature delta between the combustion chamber and the water jackets. This statement, like yours, ignores that we need to keep the combustion event itself happy, which requires a cooler charge.

I have pondered this sort of a setup breifly, on a previous project. Never really got as far as sitting down for calculations, just a thought experiment. I did run across one potential issue that you have not covered. It was suggested to me that the AC working fluid (R134A or R12) may be very EXPLOSIVE in the combustion process. More so than gasoline/ethanol mix which burns. I did not spend much time trying to veryif, but it would be worth checking, since any AC leak in the intercooler section would be ingested to the combustion side. Again, no verification, but something to consider/research...

In reply to stafford1500:

Oh, that's interesting. R-134 is a halogenated hydrocarbon, and hydrocarbons do burn... I hadn't thought about that either.

Way back in the Dark Ages someone (I think it was Grumpy Jenkins) built an intake which had a 'box' through which he ran R12 refrigerant which of course required a compressor, condenser, hoses etc. I do not recall the numbers, but IIRC it did make more than it used, at least on the dyno. Due to the complexity (mostly due to sealing problems with the box) he dropped that idea, cut the lid off the box and filled the intake 'valley' with dry ice.

Be careful, burned R-134 is deadly poison