PanTTher

or something to cool flow as being compressed.

What they're saying doesn't make any sense. Just because the gap between the blades and housing is reduced to nearly zero doesn't mean things are 100% efficient all of a sudden. In fact it's far from it I bet. Reducing the gap decreases (not eliminate) the tip eddy losses only. Which means the aerodynamics might be a little more efficient, hence the compressor pumps a few % more air. But other losses may increase... friction goes up if the blades rub against the coating, which results in heat loss.

Lets look at an ideal compressor case... I had to go back review my thermo. since I hadn't used it for a long time.

Lets say we can compress air in an ideal way (isentropicaly -> very gradually so not to add or take away energy).
Then this relation stands:
P/Pt = [T / Tt] ^ k /(k - 1) where k is the ratio of specific heats, k=1.4 for air. The subscript t stands for total conditions... when gas is brought to rest in an ideal way... ambient

Lets say for ****s and grins the local conditions are:
Pt = 14.7 psia, Tt = 80F + 460 = 540 Rankine on an absolute scale.
Lets say we are running 20 psig of manifold pressure, we have about 2 psig of loss across intercooler and plumbing.

Lets see what the ideal temperature rise is for compressing the air to 20 psig in the manifold, also accounting for intercooler losses.

Solving for T yields: T = Tt [P/Pt] ^ (k - 1)/k = 540 { (20 + 2 + 14.7) / 14.7 } ^ 0.285714 = 540 {1.29876} = 701 Rankine = 241F

So in an ideal situation we get a 241 - 80 = 161F rise in temperature for compressing the air 2.5 times.

The efficiency of a compressor or in this case Isentropic efficiency, is defined as the ratios of the power required to operate an ideal compressor to the power required to operate a real compressor.

Now an ideal compressor is 100% insulated (no heat energy is lost or gained) thru housing, has zero friction and or 100% mechanical efficiency, 100% aerodynamic efficiency ... in other words ideal in every perceivable way... Same is true for the turbine side... and we all know such a device doesn't exist.

From the 1st law of thermo we can say that the change in energy between two steady state operating points is equal to the difference in heat transfer into the system and work done by the system: E2 - E1 = Q - W. If we differentiate it with time we get rate change of energy.

So W/dt = Q/dt + E1/dt - E2/dt == Wdot = Qdot + mdot Cp (T1 -T2), where mdot is the mass flow rate thru compressor, Cp is Specific heat.@ Const. Pressure. Cp for air is 6006 ft^2/sec^2 R.

By substituting the pressure and temperature relationship into the energy equation we get the Isentropic or ideal power required to operate our compressor, where Qdot is zero (no external or internal losses in system).

Wdot = mdot Cp T1 {1 - [P2/P1] ^ (k-1)/K}

Lets put in some numbers and see what we get.

Say our compressor flows 30 lbm/min at the conditions we described earlier. We have to change from Lb mass to Lb force by dividing by g (32.2 ft/sec^2).

Rate change of work =
Wdot = 30 (lbm/min) 1 (min/60sec) 1/32.2 (ft/sec^2) 6006 (ft^2/sec^2 R) 540 (R) [1- 1.29876] --- Check all units
Wdot = 15046 Lb-ft/sec / 1 Hp/550 Lb-ft/sec
Wdot = 27.35 Hp

So in an ideal case it takes about 27Hp to generate 30 lbm/min of flow at 2.5 pressure ratio.

So Isentropic Compressor efficiency is then

Eff. = Wdot ideal / Wdot actual = mdot Cp T1 {1 - [P2/P1] ^ (k-1)/K} / Qdot + mdot Cp T1 {1 - [P2/P1] ^ (k-1)/K}

So what's Qdot or the energy loss due to friction in bearing, aerodynamic losses, mechanical losses, imbalance, etc.
Qdot is determined by careful measurements under controlled lab conditions. I don't have a good feel for it but lets say it takes about 20% more energy on average to overcome all the various frictions, etc to propel the turbo at that operating point under real world conditions

So Eff. = 27.35 Hp / 1.20 * 27.35 Hp = 0.83 or 83%. In retrospect 20% energy loss due to real world conditions doesn't seem much. I bet the losses are higher. So they say 100% efficient, I don't think that's possible.

ZeroClearance

Again, you are concentrating on the performance of the turbocharger. You can't lose the end result in which I have been trying to protray. Forget the turbo's for a second, Your engine will reach the maximum boost level, sooner in engine RPM... Thus, as a benefit, the turbocharger shuts down sooner, cycle times are lower, less wear and tear.. As a benifit non coated vs coated, the shaft speeds are slower, because previous to coating the engine would have to wind out to a higher RPM to make the target boost. We have data logged many hours of before and after. I'm trying to look at the before and after...

VonK :: Retired

---> but from turbulence, of which the blade eddies are only a part. The heat transfer portion is probably comparatively low, considering the short dwell time of the air in the compressor. But a compressor is a real churner, no matter how fancy the blade shapes. Lots of entropy created here.

PanTTher

and have done the analysis (CFD and FEA)and collected test data to prove it. But unfortunately the data is considered intellectual property. Either way, someone spent lots of money developing and testing turbos and they're not going to freely share it with anyone outside of the company.

J.Moss@TorqueFactory

Bottom line, that made my head hurt. But I agree.

PanTTher

<img src="http://205.219.82.43/Turbo1.jpg">
The description above the picture says John Mason did the chassis work on the car. What a coincidence! I had lunch with him on Wed.

xr4tic

not by your formulas, but by the formula I posted earlier.

I guess the question is, which formula do the turbo companies use? Probably the latter, since it gives higher numbers

xr4tic

that's pretty obvious.

I am arguing that people are saying that turbo life will increase because of reduced shaft speeds. To make such a statement should be backed by proof, and none exists.

xr4tic

Absolutely not. You make it sound like a turbo shaft speeds shoots way up, then drops down. That is not the case.

As the turbo spools up, the shaft speed quickly increases, then it hits the required boost level, the wastegate opens, sending excess exhaust gas away from the housing. Then the shaft speed stays roughly the same, slowly increasing as flow increases.

If anything, by decreasing the spool-up time, you are increase the amount of time the turbo is spending at higher shaft speeds.

You are looking at before/after results, but you are trying to infer something that can not be proven. Yes, a Z/C turbo will flow more air for a given boost level/shaft speed, but the Z/C turbo still operates the same way.

It doesn't magically reduce the shaft speed as the turbo hits full boost, that's impossible.

This is basic turbocharger theory. If you don't understand it, I suggest you start reading and looking at some compressor maps before making wild claims about shaft speeds.

PanTTher

The manufacturers use way more comprehensive theories and formulas and back it up with real test data. Then they readjust or calibrate the equations by adding correction factors based on the test data.