Turbine work

[AnArKey]

This is a theoretical question. Please don't tell me "Don't do this," I know.

The actual "work" done by a turbine is determined by the pressure ratio from the inlet to the outlet, and the mass flow through it right?

In looking at Turbine flow maps from Garrett, I think another principle is a given turbine will only flow a certain amount of mass flow though it. As more "work" is needed by the compressor, not much more mass flow travels through the turbine, rather the pressure ratio goes up, providing the extra power.

So if you had a REALLY small turbine, say a small T3 turbine wheel and .36AR, and had it connected to a big compressor, say a T66 wheel. To maintain enough "work" to keep the compressor going, it would still be able to do it, BUT it would require a insane amount of backpressure in the inlet.

So a turbine designed to flow only say 6lb/min of mass flow, was connected to a compressor flowing 72lb/min of air at 29psi (3:1 pressure ratio), should require (give or take) the following at the turbine:

72/6=12
3*12=36

36:1 pressure ratio

36*14.7=529psi

529psi at turbine inlet
6lb/min mass flow

Right?

Obviously, you can see why it's a bad idea to have a way too small turbine, the backpressure would kill the engine. But what if it was 6 lb/min of CO2 gas at 529psi? See where I'm going? CO2 powered turbo.

 

[Ormand]

Maybe??

I think the trade-off in turbine sizing is bigger turbine, slower spool up but lower back pressure. So going to a small "exhaust powered" turbine would give "instant" spool up, but result in high exhaust back pressure, and thus less on the top end. The CO2 idea might let you use the small turbine, to get quick spool, but you wouldn't mind the back pressure, because of the the high CO2 pressure. It would use a lot of CO2, though. Not too bad for drag races, but maybe a problem for the street. Getting control of it would be the hard part. You would need a valve to control CO2 flow, and somehow get it to open on throttle position/engine load. Not necessarily control with the chip, but that might be the easy way.??

 

[trading t/a]

You could just max out the co2 pressure and use a bypass molded into the turbine housing. It would waste some co2 but prolly would be cheaper than designing a system to meter airflow vs engine load. Another plus would be not needing an insane IC cause there woul be no heat transfer from the turbine just the natural thermal increases when air is compressed

 

[AnArKey]

6 lb/min usage is acceptable. That's over 3min of full boost time per 20lb cylinder of CO2. That's double the amount of usage time the methanol injection I was running, which I didn't bleed dry even on long trips. 700RWHP means when you use the power, you can't use it for very long until you are pushing 150MPH. If this thing works, I'd probably have more like 800RWHP after I switched to NA cams and exhaust, and had a little overlap in the cams to blow all the exhaust out of the cylinders.

The same type of valve that I'm going to use to meter propane as my primary fuel, I can use another to meter the boost. My engine computer (AEM EMS), even has a full 3D wastegate map, and additional feedback controls to get the boost correct. So it would be totally closed loop, no wasted/vented CO2.

From the theory, the smaller the turbine, the better, up to the point where I'll need more backpressure than either the CO2 can generate, or the turbine can handle. REAL cold CO2 and massive backpressure, think the turbine blades may shatter?

I really need a test rig for this.

_______________________

I was able to find a equation for the potential energy of a compressed gas: W=P*delta(V)

That means the work goes up as the sqaure of backpressure, rather than in a linear fashion, since the change in volume, not counting for temp changes, is the same as the pressure ratio (3 bar gas explands 3:1 as it transitions to 1bar gas).

This translates I think into only needing perhaps 100psi of pressure at the inlet of the super small turbine to drive the large compressor. Not so concerned about transing turbine blades at that pressure.

I also realized, I can continue to use the CO2 after that point. It will be high volume, low pressure, VERY cold gas. Why not use run a intercooler with it? If you buy a Spearco liquid/air intercooler CORE only, it has no end tanks. Air flows short way front to back, and water is supposed to flow end to end. Cap the ends that normally have small water inlets, instead with a 2.5" inlet. Run the cold air out of the exhaust side of the turbine, through the "inside" of the core. It will be like -50F going through.

It may turn out feasible to run it the other way, liquid CO2 injected into "inside" of liquid intercooler core, expanding to 100-120psi, ultra cold gas, exiting out of intercooler, into the turbine, expanding to atomospheric pressure, then exhausted out the bottom. That would keep the cold where we want it, and end up with warmer gas for the turbo, which is a good thing. Just a matter of if the core can handle 120psi internally without bursting. If it bursted in use, it probably wouldn't shed any parts, rather just dump CO2 into the engine, which would shut it down instantly without damage.

 

[JDEstill]

Here's my take:

- compressor, 72% efficiency, moving 72 lb/min at a pressure ratio of 3.0 requires 120 hp of power to drive it.

- the turbine has to therefore extract 120 hp plus a little more for mechanical losses

- now you can figure up the flow and pressure ratio for the turbine required to deliver 120-125 hp.

- CO2 at 500 psig and 100F, dropping it down to 5 psig through the turbine; 72% turbine efficiency - you need 138 lb/min of CO2 to get your 120 hp. The CO2 leaving will be at about -100F.

Basically, you've got to supply a lot of power to the compressor. When you look at the turbine curves in the Garrett catalog, don't forget that the flow numbers are *corrected*, not actual. Take into account the temperature and pressure of the gas going through the turbine, and your answer will change a good bit.

Exhaust is good in that it has a lot of energy that is easily to extract due to it's high temperature. High pressure CO2 also has a lot of energy, but you'll still need a lot of it to supply the power that you need. Can you image taking all the exhaust that the engine puts out over a quarter mile run and putting it in a bottle? You'll end up with a pretty big bottle

John

 

[mgmshar]

What JDE said is correct, although I didn't take the time to double-check his math. Let me make this more complicated. For an adiabatic (meaning no heat transfer), isentropic (meaning 100% efficient) compression process, the equation for "power" (assuming air is an ideal gas) is:

POWER = { mdot * k * R * T1 * [1 - (P2/P1)^((k-1)/k)] } / { 1 - k }

(wow!)

where,

POWER = shaft horsepower required, in kJ/sec.
mdot = mass flow rate of air (in kg/sec)
k = ratio of specific heats for air (around 1.395 for air in the temperature ranges we're talking about)
R = the universal gas constant for air (0.287 kJ/kg-K)
T1 = the air inlet temperature to the compressor (in degrees K)
P2/P1 = the pressure ratio for the turbo outlet to inlet (in absolute, not gauge, pressures)

When you're all done, you have the 100%-efficiency power required to compress the inlet air. If you then devide by the compressor efficiency, you get the actual power required.

Here's an example, from my car: about 300 grams/sec (0.3 kg/sec) mass air flow, 20 psi of boost, 60 degree F air coming into the turbo. Let's say the compressor efficiency is around 70%.

P2/P1 = (20+14.7psi) / (14.7 psi) = 2.36 (remember, absolute pressures, not gauge pressures)
T1 = 60 deg. F = 289 Kelvin (K)

POWER = { 0.3kg/s * 1.395 * 0.287kJ/kg-K * 289K * [1-(2.36^0.283)] } / { 1 - 1.395 }

POWER = 24.17 kJ/sec, if the compressor is 100% efficient. For a 70% efficient compressor:

ACTUAL POWER = POWER / EFFICIENCY = 24.17kJ/sec / 0.7 = 34.53 kJ/sec, which converts to 46.3hp (1 hp = 0.746kJ/sec). That's for my typical low-12/high-11 second car. For faster cars, the power required to drive the compressor will go up dramatically, since "mdot" will be a lot higher.

See, it's easy! (Kids, don't try this at home).

You can use the exact same equations to estimate how much CO2 ("mdot") you would need to produce the 34.53kJ/sec that the compressor needs. For CO2, your values for "k" will be around 1.26, "R" will be 0.189 kJ/kg, and your "T1" will be whatever temperature the CO2 is leaving the bottle at. You will have to guess what your "P2/P1" and your turbine efficiency are.

One other thing - metals can do some goofy things at very low temperatures. When you start injecting CO2 at very low temperatures into a turbine that's designed for high temperatures, internal metal stresses and clearances are going to be a lot different than what they were designed for. It wouldn't surprise me if turbing housings and blades started to crack at those ultra-low temperatures. Something to consider.

Good Luck,

 

[AnArKey]

Equations disgree completely. Those equations make pressure and volume change across the turbine the opposite of the data I found. Your math suggest 10x the pressure only produces ~4x the work, is this right? Seems counter intuitive.

 

[JDEstill]

Derek -

It may seem counterintuitive, but that's the way it is. Pressure ratio - plot the function for the pressure ratio term: 1 - (P2/P1)^(k-1/k). Pressures in absolute terms of course, with P2 as the turbine outlet and P1 as the turbine inlet. You will see that it isn't a straight line, and that you can't make any general statements about what happens when you double the inlet pressure.

You will see that there is a diminshing returns thing going on. At first an increase in turbine inlet pressure has a big effect - going from 10 to 20 psi at the inlet will more than double the amount of power extracted. But as the inlet pressure keeps getting higher, the power that can be extracted keeps going up in smaller and smaller steps. When you start getting up to the 300-500 psi range, doubling the inlet pressure might only give you 20% more power.

Also remember pressure is not the only variable. Other important things are:
- molar gas flow
- inlet temperature

Molar flow - to get the same power out of a turbine, holding temperatures and pressures constant, you have to supply ~2.44 X more lbs of CO2 than you would of steam, because CO2 (mw = 44) is so much heavier than steam (mw = 18). In other words, with the same turbine inlet pressure and temperature, you could replace 100 lb/min of CO2 with 41 lb/min of steam and get the same amount of power.

Inlet temperature - power extracted from a given flow at a given pressure ratio will go up by the factor of (T2 + 460) / (T1 + 460) So if you raise the inlet temperature from 500 F to 700 F the power you can extract will go up by (700+460)/(500+460) = 20.8%

John