Forrest Nearing

just use a regular gauge... I've had mine on since October... probably 15k miles or so. I can't remember the last time it got stuck

NoSeeUm

I have allot to learn about drive pressure and 1:1 ratios. But this what I have figured out so far.

For those that don't know exactly what Drive Ratio means and why it is important (the rest can roast me and or point out where I am mislead):
Ratio = Drive Pressure / Boost Pressure
Ratio = 40 / 35 = 1.14
Ratio = 70 / 35 = 2.0

What is more important is the actual difference in pressure between them:
DP = Drive Pressure - Boost Pressure
DP = 40 - 35 = 5
DP = 70 - 35 = 35

Explanation in a perfect world:
The energy (Hp) to drive the turbo comes from the engines exhaust gases.

At a drive ratio of 1.0 (or below) this energy comes for free. This is because even though the turbine causes exhaust pressure (the drive pressure) to go up on the exhaust stroke of the engine, the equal amout of pressure is being supplied by the boost pressure on the intake stroke of the engine. So the pressures and the forces are balanced on the intake and exhaust sides. The net load on the engine is balanced just concerning pressure and the force it exerts on the top of the pistons.

At a drive ratio of greater than 1.0 the drive pressure is higher than the intake pressure. Now because the different relative force on the intake and exhaust strokes the engine is out of balance. The forces are higher on the exhaust stroke. The engine must now use Hp to expell the exhaust gases.

The engine Hp calculation goes something like (the number for Hp Loss Factor is fictional and for example only until I can figure it out better):
Pumping Loss Hp = DP x Hp Loss Factor (caused by drive pressure)
Pumping Loss Hp = 5 x 1.5 = 8
Pumping Loss Hp = 35 x 1.5 = 53

Net Crank Hp = Total Available Crank Hp - Pumping Loss Hp
Net Crank Hp = 235 - 8 = 227
Net Crank Hp = 235 - 53 = 182

You can see how it could add up to quite a bit of Hp lost. This is made up for by adding more fuel to the engine, a higher fuel / air ratio and higher EGT's.

I don't have the exact numbers yet, but I am working on it. But as far as I have gotten, my calculations show that for example at 1000 CFM (give or take) at 2500 RPM and at 55 psi boost, my turbo set uses around 300 Hp. This is at a 1.0 drive ratio. Which is free boost, in the perfect world.

If I raise the drive ratio to 1.18 then:
Ratio = 65 / 55 = 1.18
DP = 65 - 55 = 10
Pumping Loss Hp = 10 x 1.5 = 15
Net Crank Hp = 400 - 15 = 385

If I put my HX 35 back in (and melt down my pistons as value added bonus):
Ratio = 70 / 35 = 2.0
DP = 70 - 35 = 35
Pumping Loss Hp = 35 x 1.5 = 53
Net Crank Hp = 400 - 53 = 347

Jim

NoSeeUm

FWIW if anyone cares to check my math.

Hp Comp = Hp consumed by the compressor
Hp Turb = Hp generated by the turbine (theorical)
Hp Loss = Difference between Hp Comp and Hp Turb

I know that this is not quite right, but I attempting to calculate the turbine Hp and hence the Net Crank Hp cost of drive pressure. To do this I am trying to compare the actual boost or actual Hp Comp to that of a theorical boost or theorical Hp Comp. My thinking is that if 100% of the drive pressure were converted to boost in all cases where drive ratio is greater than 1.0 and what would the difference be in the of Hp consumed by the compressor for those cases as if the ratio were equal to 1.0.

Jim

RowJ

Yes!
Not always desirable to have a negative delta... with boost higher than TIP. But for efficiency and head gasket protection at WOT... Yes!

Don't believe so. Just need some type of filtering media unless using a temporary gauge, as mentioned above.

Jim - Your getting up there with some of HOHN's threads. Need to review it several times for it to sink in.

RJ

NoSeeUm

Chase a few threads where some one has upgraded a turbo. Often they might say that EGT's are lower and they make more Hp for about the same boost.

Where does that Hp come from?

Jim

Forrest Nearing

more oxygen (denser charge) and less drive pressure

RowJ

Yup, Cooler compressed air = denser air = more O2 molecules = more hp for a given boost level.

General rule of thumb.... guys running bigger single turbos over 42-45 psi are getting "expansion' boost, not "compression" boost. The expansion boost comes from over working the compressor and heating the charged (compressed) air. Gives fun readings on the boost gauge but actually hurts HP.

RJ

NoSeeUm

Bingo!

My curiosity about drive pressure comes from my experimentation. I have an exhaust brake. Any time exhaust (drive) pressure gets to be above about 15 psi it can noticeably slow the truck down. Of course it is not linear. For example, 20 psi at 1500 RPM is not the same as 20 psi at 2000 RPM. As far as braking ability goes.

Thats why I think drive ratio is a talking point, but the difference in drive pressure over boost pressure is where the Hp is lost (or gained).

A ratio of 4.0 most would say it sucks, but:
Boost = 1
Drive = 4
DRatio = 4.0
DPressure = 3
Nada, who cares.

Which is totally different than a ratio of 1.5 where:
Boost = 40
Drive = 60
DRatio = 1.5
DPressure = 20
Now you have significant exhaust braking, contradicting everything going on over on fuel/air side of the engine that is trying to make Hp.

Now consider the Hp loss:
DPressure = 20
RPM = 2200
HpLoss = xxx

DPressure = 20
RPM = 3400
HpLoss = XXX

Jim

NoSeeUm

Yeah, thats why the Compressor Efficientcy / Density Ratio calculation is so important.

Jim

XLR8R

Well, I got the permanent TIP gauge kit installed on the '05 today at the shop...

it sure is a lot of fun to watch a NEW gauge!

I was surprised at the TIP levels I observed - 70 PSI!

Kind of puts HG issues in a whole new light, huh?

1-5-3-6-2-4

How come no one ever talks about the thermodynamic nature of the exhaust gas as it travels out of the chamber into the turbo. sure the piston pushes the exhaust out into the head, but the hot gas is still expanding as it travels down the manifold, which in my mind effectively drives the turbo. The heat of combustion is extracted by the turbine. Am I way off base here or something.

XLR8R

Hey, we (NTxDTR#8) discuss the thermodynamic properties of the exhaust pulses exiting the manifold plenum through the turbine on a regular basis!

NoSeeUm

LOL

My current theory, if you look at the problem as an individual cylinder. Any time drive is greater than boost the added pressure acts as a detriment to engine performance. Because up to then, the vast majority of the Hp needed to drive the turbo comes for free in the form of waste exhaust heat by the engine.

For my truck:
02 ISB
Displacement: 359 cu. in. (5883 cu cm)
Bore x Stroke: 4.02 x 4.72 (102.1 x 119.9)
Compression Ratio 16.3:1

Work = W = F * D
P = Pressure = Drive - Boost = 60 PSI
F = 761.54 Lbms (Cylinder Area = 12.69 square inches)
D = .39 Feet (Stroke)
W = 299.54 Ft-Lbms (What it takes to push 60 psi out of one cylinder)

There are 3 exhaust strokes per engine revolution.

So if:
RPM = 2500
W per revolution = 898.62 Ft-Lbms (299.54 * 3)
W per min = 2,246,544 Ft-Lbms per min
W per Sec = 37,442.4 Ft-Lbms per sec
1 Hp = 550 Ft-Lbms per sec

Pumping Losses = 68.08 Hp

Thats drive pressure at work. Theory valid or not?

Jim

NoSeeUm

Yeah you are right.

A guy over on NWBomber's calculated his twin set Hp using this method. The change in enthalpy calculation is over my head, but he could do it.

He came out at around 300 Hp at WOT. This is about what I came up with but I used compressor Hp by first calculating CFM.

Compressor Hp = Turbine Hp
IE:
Fan Hp = P = ( Q * p ) / ( 229 *u )
P = Power in Hp
Q = Flow Rate in CFM
p = Pressure in psi (boost)
u = Compressor Efficiency Coefficient (from the compressor map)

Jim

XLR8R

Wonder how many HP the stock Holset on my '05 is using at 3000 RPM and 75psi on the TIP gauge?