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Old 01-18-2012, 04:12 PM   #18
grambles423
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Some Drivetrain Loss Fun

Quote:
Originally Posted by Banshee View Post
sorry... I meant a constant amount of power lost regardless of how much horsepower or torque the engine makes. Example: a constant 25hp lost whether the engine is making 200 or 300 hp. But I suppose that point is moot because more power usually means higher temps which would increase drivetrain loss.

I really should stay away from these technical questions...
No no, I see what you mean. I think I have a fair answer for you.

In regards to the question, both instances you described are roughly the same thing, except one is a percentage function of the power provide which would increase with power increases and the other is just a constant 25HP off.

200 HP at assumed 10% loss = 180WHP
150 HP at assumed 10% loss = 135WHP
200 HP at 25HP constant assumed loss = 175WHP (12.5%)
150 HP at 25HP constant assumed loss = 125WHP (17%)

However, lets get technical with it:

What makes the power? The force pushing down on the piston which turns the crank. This force is generated by a number of things (Cylinder Pressure, Temperature, Fuel Heating value, Bore, Stroke, etc. etc.)

The engine power is roughly described, in theory, like this. Taken from another thread I did this in:

Quote:
Brake Horsepower (Watts) = N*Displacement(in cubic Meters)* (RPM/2)* nc*nm*nth*nv*Qhv*(F/A)*Air Density

N = Number of cylinders
Displacement = Bore*Stroke*pi/4
RPM/2 = Must be placed in Revs/Second = RPM/2 * (1/60 {s/min})
nc = Combustion efficiency = about .98-.99 on newer cars (can be calculated further, but these estimations are fine)
nm = mechanical efficiency = Function of RPM/heat/etc range of about .8-.93 (can be calculated further, but these estimations are fine)
nv = volumetric efficiency (what percentage is your TB open?)= 100% at WOT (Possibly a little more depending on the flow. Most I've seen is 108%)
nth = thermal efficiency = 0.8*(1-[compression ratio^(-.35)])
Qhv = Heating value of the fuel = 43,000,000 J/kg

Air Density = Pressure Entering / (287.2 J/kg-K * Intake Temperature)
Pressure Entering = Atmospheric Pressure * Pressure Ratio (This is a function of RPM) - J/m^3
Now that you have the power the engine makes you can determine how much force can be placed on the crankshaft to provide the neccessary torque at the flywheel. This is a function of material properties and mechanical efficiencies which decrease/increase throughout the rev band due to fluids heating up and metal expanding when hot.

Nevertheless........what an automaker displays as their "specific output" is what it has been measured at on the engine dyno and calculated through various equations and testing.

NOW...onto Driveline loss

Now that you know what your engine makes how do you quantify it through the wheels?

Easy: Power = Difference in Kinetic Energy over the amount of time it takes

Easiest way to do that is to dyno it in a controlled enviorment. Normally, this "constant" weight and constant air temp and etc. etc. is propelled by your wheels. This is a great indication of the power through your wheels...but it doesnt really coorelate to what your engine is making because of the weight of the driveshafts, gears, wheels, tires, hubs, etc. etc.

You could do a velocity run with a certain scan rate in VAGCOM but that accounts for drag forces which will make your power output significantly lower.

So how can you relate wheel HP to crank then? Well....you can theorhetically calculate it, but that leaves room for error (Plus it gets really confusing quick. Fluids dynamics + thermodynamics + variating degrees of freedom, frictional factors, stress analysis, plastic yielding effects with respect to temp), or you can do a very long process:

1) Dyno your engine
2) Using the same dyno, dyno your engine + Trans (1 for every gear since the mass changes for different gears
3) Using the same dyno, dyno your engine + trans + driveshafts
4) Using the same dyno, dyno your E + T + DS + Hubs
5) Using same dyno, dyno your E + T + DS + Hub + Wheels

This will take various fixtures to mount the dyno to...but would be one of the most accurate ways to relate your efficiencies together and finally provide a coorelation between brake and wheel horesepower.

Whats easier? Assuming constant percentage difference calculated from factory dyno and vendor dyno? or the process described above? Remember, no dyno is the same. All calibrations are different for various altitudes and temperatures and etc. etc.

1/4 Mile times are a good indication of power. Its as simple as how long did it take to propel this 3000lb vehicle in a 1/4 mile and what was its final speed?
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