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Joule-Thompson Intercooling?

Dialcaliper

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There are a couple problems, and you nailed one of them already.

When you compress gas, in addition to temperature changes, you are making it more dense, and vice versa when you expand it. The goal here is to increase the mass of air that reaches the cylinders. In placing a restriction after the turbo, you are essentially adding another throttle plate, and the accompanying pumping losses. In addition to allowing less air to flow through the engine, which is actually producing mechanical power, you're also reducing the flow through the turbine.

What you're effectively doing by compressing air to 25psi and then restricting it to 10psi is running the engine at part throttle, all of the time.

Turbo efficiency is expressed in terms of how "reversible" the compression is (adiabatic is the actual term for reversible compression/expansion). A 70% efficient turbo basically means that only 70% of the energy from the turbine shaft is used to compress the air, the other 30% is turned into waste heat, which ends up in the intake charge, due to the design of the turbo (if you were really clever, you might be able to find a way to put the heat elsewhere).

In other words, if you were to let the air expand unrestricted to ambient after being compressed by the turbo, you'll find that it will be *hotter* than when it started, even though it cooled during expansion. A 100% efficient compressor would end up with the air the same temperature it started at.

It's not worth pursuing unless you can gain some other advantage that offsets the loss during the pressure drop, that makes the air denser or cooler - such as an intercooler for example, or some sort of ram-air effect (intake manifold runners) that ultimately results in higher cylinder pressure.

One of the basic problems with alternate solutions like A/C coolers or peltier devices is that once you realize how massive the heat transfer has to be, a lot of portable devices, especially ones involving electricity, start to become impractical. Just look at the electric supercharger as an example.

Always remember what happens when you convert horsepower to kilowatts. (1 hp = 0.75 kW). And that a 90 amp alternator at 14 volts is only producing a peak of about 1.5 hp worth of electrical energy. And that heat transfer and power adder devices consume at least 10's of horsepower if not more.



Quoting DR1665:
Curious thought. Any thermodynamics geniuses in the house to tell me why this isn't worth exploring? (I'm sure there's math involved and math, especially on this level, scares me.)

Joule-Thompson effect basically says that, "At room temperature, all gases except hydrogen, helium and neon cool upon expansion by the Joule–Thomson process." This is similar to how your air conditioner and refrigerator work. High pressure liquid is forced through an orifice and then allowed to expand into a lower pressure environment, resulting in cooling.

Okay. So you run a 14b at 25psi. You run outlet-diameter plumbing to your intercooler, where there is a sort of merge collector at the inlet, reducing the diameter of the piping. The inlet/outlet of the cooler are larger sized, maybe 4". I dunno. The super-heated 25psi charge is forced through a smaller opening only to expand on the other side.

The charge is cooler on the other side, right? How much cooler? And how much strain does this place on the turbine and compressor? What if you ran a larger turbo? Suppose you had a 35R pushing 40psi into the IC, but only realized 10psi at the TB?

Just a curious thought. Random. I know. I should have gone to bed hours ago...

 
Last edited:

Ow Ow Ow, this thread hurts my headv /ubbthreads/images/graemlins/banghead.gif Rubegoldberg was really good at taking a simple task and complicating it beyond belief.
 

Dialcaliper

Well-known member
Joined
Jun 22, 2007
Messages
1,287
Location
Mountain View, CA
Allow me to summarize:

What you're effectively doing by compressing air to 25psi and then restricting it to 10psi is running the engine at part throttle, all of the time.
 
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