Thermo has been awhile for me as well.

Does this explaination make sense? I am trying to clear the actual micro/macro thermodynamic details at the atomic scale in my mind.

The enthalphy of the exhaust going through the scroll should remain roughly the same; however, we are converting some of the temperature component of said enthalphy into an increased velocity component.

This results in a temp decrease and increase in kinetic energy (KE = 1/2mV^2).

The exhaust's KE is then transfered to the turbine and the gases exit at basically the same temp before the turbine (after the scroll) but without much of its KE.

I just don't see how temperature (on an atomic level) can power the turbine directly. I know the average velocity of the atoms/molecules increases with temperature, but the directions are random, which I would think would make it impossible for the atoms/molecule's KE to be passed to the turbine in a baised way capable of providing motion.


Thoughts?

 

Ok. . . did some rough Thermo calc's today concerning exhaust energy applied to the turbine wheel. It works like this:

Power applied to the turbine = mass flow rate X change in enthalpy (a function of temperature) + change in kinetic energy. . . the question being: "How much kinetic energy does the exhaust apply to the turbine versus enthalpy?"

I had already calculated in the past that the power required to spin a compressor is roughly 60 hp, making a few assumptions around power level, etc. . .

So. . .

KE = 1/2 m dot * delta V^2

Assumptions
1. 2/3 of the air injested by the engine goes through the turbine, while 1/3 goes through the wastegate.
2. 91 lb/min is the exhaust rate.
3. Gas flow rate through the turbine is 2/3 of sonic, or "choke" velocity.
4. The turbine blades are traveling 1/2 the gas velocity, since that's how most turbines are designed to operate.
5. . . . so the change in velocity is half of the inlet velocity.

After running through the calc's, I get ~6 hp for the Kinetic energy, that is the energy applied to the turbine wheel due solely to momentum of high velocity gas molecules impacting the blades.

Basically, about 10% of the energy applied to the turbine comes from kinetic energy, while the other 90% comes from enthalpy. Enthalpy in a gas is related to temperature ONLY. However, when you extract enthalpy (thus temperature) from the gas, the pressure is forced to drop also.

Mike

 

A. How did you get 60hp to spin the compressor?
B. What are your thoughts on the STS rear mount turbos if 90% of the energy needed to drive a turbine comes from heat?

 

I'd say 60 hp is as good a starting point as any. It's about what a moderate centrifugal supercharger requires. The centrifugal has some amount of mechanical loss through the gearbox, but it's within the range of the assumption he started with. The diffence between the turbo and the centrifugal is that the centrifugal has to take it from the crank while the turbo extracts it from the exhaust.

I've never measured it myself, but I've read that at full tilt boogie exhaust temps drop by about 400 degrees in the turbine housing.

The STS turbos use a smaller-than-you'd-expect turbine housing and aren't as efficient as front-mounted turbos. But, they're better than no turbo at all!

 

Nice post Mike, that got my wheels turning!! I'll have to admit, those aren't the results I would have expected. One question, assumption #3...that's an average velocity, right?

 

a. Like I said, I calculated that over a year ago making some assumptions around air flow and boost pressure. It's pretty simple, even if you factor in compressor efficiency.

b. My thoughts on STS? Since alot of heat is lost in the exhaust, they cause alot more pressure drop (backpressure) than conventional. Furthermore, they don't spool up if the exhaust is cold.

Russel, yes assumption 3 was that the average velocity at the smallest passage in the turbine is 2/3 choke flow.

 

There are some seriously f-ed up explanations in this post. First, the 1000hp vehicle at 6000 rpm vs. the 1000 hp vehicle at 7million rpm will accelerate at exactly the same rate at that instant in their powerband, assuming everything else is equal.

I'd like someone to describe the mechanics (the physics) behind the transfer of enthalpy into kinetic motion. When you say that 6hp comes from kinetic energy, I think you're missing the point. Heat/enthalpy causes the kinetic energy. Heat is kinetic energy. That kinetic energy is focused in a given direction by canalizing the kinetic energy. We force the pressurized particles in one direction (towards the turbine). The force imparted on the turbine is a result of the change in kinetic energy of all of those particles.

Of course you will lose heat as the particles move through the turbine. This happens for two reasons: First, some of the kinetic energy that formerly was "heat" is transfered from the particles to the turbine as they pass through the turbine. Second, and more importantly, after the turbine there is a pressure drop. The pressure drop obviously causes a corresponding temperature drop.

Note that the heat that is physically transferred into the material of the turbine wheel does nothing to increase the power or spin of the turbine.

Chris

 

Mike, you're asking the wrong question in: "How much kinetic energy does the exhaust apply to the turbine versus enthalpy?"

Kinetic energy is part of enthalpy. I hate using formulas rather than concepts. Formulas are rarely necessary, but since "enthalpy" in and of itself is nothing but a contrived catch-all with the creation of it's formula, bringing the formula up is a necessity when discussing enthalpy. H=E+pV. "E," the internal energy, is comprised of Kinetic energy, and certain potential energies.

Regardless, the useful energy derived by a turbine is entirely a result of the tranfer of kinetic energy from molecules/particles to the turbine wheel. That kinetic energy was originally caused by the heating of the molecules in the combustion chamber.