Sorta useless post, how much power per PSI of boost (at least, in my case)

[Pyro]

Forget the stock turbo ratings. There is something wrong with the factory rating. A stock L28 turbo engine will do about 170 to 180 hp at the wheels with 7 psi of boost (not at the crank). I have seen stock L28 turbos make this power many times on a dyno. If they don't make that much power then there is something wrong with the engine or the tune or the air is too hot. Plus it is common for a na L28 engine to make about 125hp at the wheels.

 

A stock na engine will do 120 to 130hp at the wheels. The difference in cr from 7.4 or 8.3 or 8.8 doesn't effect the max hp as much as the mid range power and torque. So 7.4 cr makes about the same peak hp as a 8.8 cr, maybe a few hp less, but not much less.

 

So, I disagree, the formula works perfect and is used by most all turbo shops to determine how much power can be made by turbo changing an engine. Sure, there are a much of other varibles that need to be controlled, but the formula shows the max hp obtainable at a boost level starting from a set base hp. Simple Physics!!!

 

stock na L28 power=125hp

with 7 psi boost= 184hp

with 10 psi boost= 210hp

with 14 psi boost= 244hp

with 17 psi boost= 269hp

 

Quit fighting it! Science and math are your friends! hahahahaha.

 

If more than 270hp is made with a L28 at 17psi of boost then the base hp has been increased by some other means (exhaust, cam, head porting, etc.)

[Guest bastaad525]

Forget the stock turbo ratings. There is something wrong with the factory rating. A stock L28 turbo engine will do about 170 to 180 hp at the wheels with 7 psi of boost (not at the crank). I have seen stock L28 turbos make this power many times on a dyno. If they don't make that much power then there is something wrong with the engine or the tune or the air is too hot. Plus it is common for a na L28 engine to make about 125hp at the wheels.

 

A stock na engine will do 120 to 130hp at the wheels. The difference in cr from 7.4 or 8.3 or 8.8 doesn't effect the max hp as much as the mid range power and torque. So 7.4 cr makes about the same peak hp as a 8.8 cr' date=' maybe a few hp less, but not much less.

 

So, I disagree, the formula works perfect and is used by most all turbo shops to determine how much power can be made by turbo changing an engine. Sure, there are a much of other varibles that need to be controlled, but the formula shows the max hp obtainable at a boost level starting from a set base hp. Simple Physics!!!

 

stock na L28 power=125hp

with 7 psi boost= 184hp

with 10 psi boost= 210hp

with 14 psi boost= 244hp

with 17 psi boost= 269hp

 

Quit fighting it! Science and math are your friends! hahahahaha.

 

If more than 270hp is made with a L28 at 17psi of boost then the base hp has been increased by some other means (exhaust, cam, head porting, etc.)[/quote']

 

 

Alright dude I can debate this all month long if you like.

 

First off you are WAY off on the numbers you are giving for what a stock turbo or N/A will put down to the wheels.

 

i have seen a few examples of dyno sheets from bone stock 280zx turbos. They ALWAYS put down in the 140-150rwhp range at stock boost. Not 180. Don't believe me? Find Jeff P's Extreme 280zxt website, he's got documentation there of a bone stock L28ET dynoing on a Dynojet dyno and putting down about 145. If you know someone who's putting down 180rwhp stock you need to provide some scans/links/proof.

 

My 100% stock L28ET dynoed 190rwhp the first time I dynoed at 10psi. You're gonna tell me that 3 psi increase only netted me 10hp? Hardly. I've been to the dyno SEVERAL times with this motor, and have seen hp/torque jump, on average, 15hp/30ftlbs of torque every time I turned boost up 1psi. Others have seen about the same on their own dyno testing.

 

As for the N/A, my own fixed up, high compression N/A motor with big bore TB, 3-2-1 headers, mildly worked head, and a professional dyno tune only made 140rwhp. This was a rebuilt 2.9 liter motor, with mostly stock '81 ZX fuel injection. You're gonna tell me that increasing displacement, increasing compression, totally freeing up the intake and exhaust, and mild head work, ONLY gave me 10-20hp? I refuse to believe that.

 

 

 

And I replied in that other thread... I'll say it again here. How can that formula be accurate, when it doesn't even account for something as integral to a turbo motors power output as the efficiency of the turbo itself??????

 

You say you can take any motor that makes x hp, and multiple this by that and factory the amount of boost and get really close to the HP the motor will make at that boost?

 

Okay, so you mean to tell me that a bone stock 280zxt motor at 10psi on stock T3 will make the same power as that same motor on a T3/04 turbo at that same 10psi?? HELL NO!! The T3/04 will make SIGNIFICANTLY more power at the same amount of boost. Does your equation factor for that? NO!

 

 

You are right, science and math ARE your friends, and you can predict within a very close margin how much power a motor can make, IF you factor in ALL the variables. If you take a given motor, with a given turbo and know the turbos compressor map/effiiciency, know the intercoolers efficiency, know how much fuel you are supplying (a/f ratio), know how well the head flows and the cam specs, the ignition timing, the weather/temperature/barometric pressure/altitude/humity, boost level, etc. THEN, YES you could definately make a very good estimate of how much power your motor would make. Then you'd have to very accurately factor how much power your drivetrain is actually soaking up, and you could estimate wheel hp.

 

But you can't just say "take X amount of hp base and multiply by Y amount of boost and get Z horsepower" and expect that to be accurate! Sure it MIGHT be accurate in some cases but more often than not it wont be even close, especially in the case of heavily modded motors.

 

Like in that other guys thread, where he dynoed 270hp at 17psi. That was on a T3/04, and like two or three other guys in that thread who know their stuff agreed with me, that a T3/04 at 17 psi should be worth WAY more than 270rwhp. Regardless of what that useless formula has to say about it! Because we have SEEN that proven multiple times, by guys who've put down well over 300rwhp with that turbo at around those boost levels.

 

You can't argue with THAT.

[Pyro]

All I can say is.

 

You can lead a horse to water but you can't make it drink.

[proxlamus©]

my stock 280z N/A engine has a stamp in the engine bay stating 170HP to the crank...

 

 

which means... with 9psi on my N/A motor....

I SHOULD be running 272HP to the flywheel! Now account the 15% typical drivetrain loss and that means I should have 231.2 HP to the flywheel

 

Bastaad I think the only little flaw I oculd find in the first few posts, was the 280ZX turbo engine was rated at 180 HP to the CRANK!

That means at 10psi you should have 302.4hp and with the 15% drivetrain loss .. that means you should have 257.04 HP to the WHEELS

 

NOTE - this is according to Road and Track and Car and Driver

-----------------

0-60 1/4 mile Lat Braking Top Peak Peak Curb Source

(sec) (sc-mph) g (sp-ft) Spd HP/RPM TQ/RPM Wght (date)

81 ZXT (AT) 6.8 15.2 89 .78 70 196 126 180/5600 203/2800 3070 CD 5/81

81 ZXT (AT) 7.4 15.6 88 .75 60 164 129 180/5600 203/2800 2995 RT 5/81

----------------

 

This equation is somewhat farily accurate within at least 30 horsepower-ish.. LoL...

 

note.. with 14.7 for standard atmosphere is on a PERFECT day at sea level altitude at 15 degrees Celcius..

 

since I am 5,800 feet up, it is rare to get 14.7psi of standard atmosphere.. or 29.92 in as a barometer would indicate..

 

especially on a 90 degree day, the actual DENSITY altitude would be like I am actually at nearly 9,000 feet! When I am really at 5,800ft.

This is a result of the heat.. blah blah blah.. which thins the air out and would be equivelant to 9,000 ft..

 

ANYWAY.. that seems about right Bastaad... at 10psi 257rwHP thats about right... right?! LoL havent looked at ur charts yet

 

 

OK just re-read your posts.... the equation is WRONG!!! LoL.. your right Bastaad.. the equation is bogus!

[SleeperZ]

All I can say is.

 

You can lead a horse to water but you can't make it drink.

Yes, look in the mirror. The reason you canot use boost pressure as a way to calculate power is DENSITY. Take the case of adding an intercooler, and compare it to the same engine without one. You can have the same boost pressure and make more power with the intercooler. I did it. My boost controller operates off manifold pressure; I added an intercooler without changing the boost controller and gained 0.5 seconds and 10mph in the 1/4. The reason why is more air at the same pressure because the air is cooler. That's why thermal efficiency in a turbo and an intercooler is so important.

 

Take it or leave it, as you said, you can lead a horse to water...

[Pyro]

That is exactly right!

 

The equation doesn't consider air density or efficency or A/F ratio or. That is the whole point of the equation. It gives the theroitical max hp an engine will make if a certain amount of turbo boost is added.

 

The equation assumes the air temp, air density, etc is the same for the turbo engine is the same as the NA. So, in a "Perfect World", a 125 hp NA engine with 14.7 psi of boost will double it's output. Very simple.

 

Why are you guys making this so hard?

 

And don't forget, there are BIG errors in dyno numbers.

[zuhow]

So I've been following this thread and I've gotta make a couple of comments:

 

Pyros formula is a good formula, if used correctly. I think we're trying to take it beyond it's limits here though.

 

I use it frequently. It IS a really good SWAG (Scientific Wild A$$ Guess) if you dont strech it too far. Basically, it works damn good if you only project a little ways from your starting point.

 

Look at it this way, say my house was worth $100k last year and $150k this year. I can project that my house will be worth (150/100)*150=225K. probably pretty good guesstimate. The problem comes if you try to strech this 30 or 40 years out- you'll see my house is now (projected) to be worth more than the output of many nations!!!

 

That's whats going on here.

 

Bastaad, you were right on the money when you guessed a turbo w/o the turbo would put out about 100hp. If you take "the formula" and apply it in reverse (ie divide by the ratio of pressures instead of multiply) you'll come up with 98whp (at the wheels). check it out:

 

assume 145 whp for l28t bone stock

the ratio is (14.7+7)/14.7=1.48

so 145/1.48=98hp

 

still following me? good, now the problem comes when you say I'm gonna put 20 psi boost on this engine:

ratio= 14.7+20/14.7=2.36

2.36*98 (base l28T "naked" power output)=231whp

 

I agreee this number is low, as evidenced by Bastaads hard data of 240ish hp at 14psi.

 

That said, this formula is great if you say "i've got a a given engine and I want to estimate the power output if I turbo it (a little bit)"

 

(you have to assume here that you dont change anything, leave the CR, cam lift timing, etc the same)

 

You can say, which is Pyro's point- and he's totally right, that if I double the air mass flow (ie 15 psi boost, or 2x the density of normal atmosphere) I can expect (about) twice the power output.

 

Just keep in mind that you're now projecting "years" ahead like in the house example and it's bound to be off quite a bit. If you stick to "just a little bit" around your starting point, you're good to go. But it's just a SWAG, not any sort of gospel truth or theoretical limit. It's just a simple way of saying, "if I consume 20% more air I'll get about 20% more power" which is totally legimate.

 

Just remeber, all bets are off when you start talking about 'large' changes.......

 

Oh, and Bastaad, thanks for keeping my math skills sharp, you always give me something to crunch over a bit (but my employer is pissed that you're killing my productivity!)

[Pyro]

I think the problem with using the stock 280zxt engine with this equation is that there is no IC.

 

An IC is neccessary to maintain equal temperatures into the engine when using this formula.

 

I'm sure the intake temps are much higher with a t3 at 7 psi of boost with no IC as compared to the air intake temp of NA engine.

 

Use an IC on a stock 280ZX and the HP jumps 30hp. Then use the equation.

 

1 point in compression is not worth much. A stock cam at 7.4:1 versus 8.3:1 will not make much of a difference in HP. Certainly not 20 hp. Maybe 4 or 5hp, at most.

[SleeperZ]

So I've been following this thread and I've gotta make a couple of comments:

 

Pyros formula is a good formula' date=' if used correctly. I think we're trying to take it beyond it's limits here though.

 

I use it frequently. It IS a really good SWAG (Scientific Wild A$$ Guess) if you dont strech it too far. Basically, it works damn good if you only project a little ways from your starting point.

 

Look at it this way, say my house was worth $100k last year and $150k this year. I can project that my house will be worth (150/100)*150=225K. probably pretty good guesstimate. The problem comes if you try to strech this 30 or 40 years out- you'll see my house is now (projected) to be worth more than the output of many nations!!!

 

That's whats going on here.

 

Bastaad, you were right on the money when you guessed a turbo w/o the turbo would put out about 100hp. If you take "the formula" and apply it in reverse (ie divide by the ratio of pressures instead of multiply) you'll come up with 98whp (at the wheels). check it out:

 

assume 145 whp for l28t bone stock

the ratio is (14.7+7)/14.7=1.48

so 145/1.48=98hp

 

still following me? good, now the problem comes when you say I'm gonna put 20 psi boost on this engine:

ratio= 14.7+20/14.7=2.36

2.36*98 (base l28T "naked" power output)=231whp

 

I agreee this number is low, as evidenced by Bastaads hard data of 240ish hp at 14psi.

 

That said, this formula is great if you say "i've got a a given engine and I want to estimate the power output if I turbo it (a little bit)"

 

(you have to assume here that you dont change anything, leave the CR, cam lift timing, etc the same)

 

You can say, which is Pyro's point- and he's totally right, that if I double the air mass flow (ie 15 psi boost, or 2x the density of normal atmosphere) I can expect (about) twice the power output.

 

[/quote']In a sealed vessel, doubling the absolute pressure will double the density given the same temperature. But in an engine, there are many other variables at work. Boost pressure is partly made up of airflow through an orifice, which is not linear with pressure, but increases with the square of the pressure. This will factor in as volumetric efficiency so doubling the boost doesn't double airflow. That's why it's so important to drop the temperature and increase the density, in addition to removing all impediments to flow that you can.

[SleeperZ]

Now you are missing the point also.

 

I'm not saying efficency doesn't make a difference. Or course it does.

 

The formula is simple' date=' as you can tell.

 

It assumes everything is equal except for air pressure. So, the formula gives the maximum hp that a NA engine would produce at a specific boost. That means the turbo and IC are the perfect size and the ignition and fuel are just right.

 

So, a NA engine with 14.7 psi of boost would, in theory, double its output. If there is something wrong with the turbo efficency or air temp or fuel ratio then the power will be lower than the theoritical. That is all I'm saying.

 

Sure, the hp can go up at the same boost level if a better head or a bigger cam is used. But that better head and bigger cam also would increase the NA hp, so the formula remains correct.

 

And, everytime someone posts turbo dyno numbers it proves to me that this formula is a close estimate of power. But as we all know, dyno numbers can vary by a lot. 250 on one dyno could be 300 on another.[/quote']

 

I'm not sure I'm missing a point, you are talking about a theoretical system that doesn't exist. IF you had an engine with a volumetric efficiency of 100%, and IF your inlet temperature was the same as ambient, and IF your fuel mixture was constant, you are correct. Doubling the AIRFLOW would double the power. To refer to it as doubling boost pressure is misleading, and only it doubles power in a dream world.

[zuhow]

In a sealed vessel, doubling the absolute pressure will double the density given the same temperature.

 

No, this is wrong.

 

In a sealed vessel, the density is constant. INDEPENDENT of temperature and pressure. it's SEALED, so no mass goes in or out, and the volume remains constant (i assume you meant a ridig vessel) so the density, which is mass/volume is constant because the mass and volume are constant.

 

 

 

Boost pressure is partly made up of airflow through an orifice, which is not linear with pressure, but increases with the square of the pressure.

 

boost pressure has nothing whatsoever to do with flow through an orifice. flow through an orifice is an EXPANSION process

 

This will factor in as volumetric efficiency so doubling the boost doesn't double airflow. That's why it's so important to drop the temperature and increase the density, in addition to removing all impediments to flow that you can.

 

Volumetric effiency is a totally different beast. I agree that, in practice, doubling the boost pressure doesnt double the mass flow of air, but that's a fine point that isn't really important. if you double the density (however you do it) you WILL double the mass flow for a give volumetric flow rate.

 

If you're flowing 500cfm and the denisty is 0.08 lb/ft^3 and you change to a density of 0.16lb/ft^3 and keep the flow of 500cfm, you will flow twice as many lbs of air.

[Pyro]

I said from the start is was a simple formula.

 

What do you expect from 2 basic inputs in a very simple equation?

 

But, regardless of it's theroritical system that doesn't exsist, the formula still seems to work pretty well.

 

The formula is just a tool. If you want to make a 500 hp engine with 15 psi of boost then you need to start with an engine that can make at least 250 without a turbo.

[zuhow]

Sorry Sleeper Z, I cant read!

 

I see what you mean about the doubling the pressure, you're right. the sealed part confused me, I see what you mean. forget my first comment in the previous post

[SleeperZ]

Sorry Sleeper Z' date=' I cant read!

 

I see what you mean about the doubling the pressure, you're right. the sealed part confused me, I see what you mean. forget my first comment in the previous post[/quote']I gotcha. It's cool.

 

I think you need to clarify your second statement - Volumetric efficiency is never 100% because of the orifice effect, remember boost pressure is partly due to the fact there are restrictions in the intake and exhaust tracts. Forcing air through an orifice will result in a pressure drop, and doubling the pressure will only get you 70% more air through the restriction. I admit this can be minimized, but opening up a factory system like a stock ZXT, the results are dramatic. Just check out JeffP's exhaust test, factory versus 3".

[zuhow]

SleeperZ,

 

becareful on saying that pressure drop through a piping system is like flow through an orifice. Agreed that you will have a pressure drop in the intake system, however small (or large), but flow through an orifice is a very specific thing and different than the resulting pressure drop from frictional losses in the flow path.

 

In the example (of pyros formula) I dont care what the VE is, and I'm assuming that it stays the same from one boost condition to the other (I know it's not true but if the difference from one boost level to the other is 'small' it's close enough to call it even).

 

I don't follow you on "the orifice effect" can you explain to me what you mean?

 

Opening up a system doesnt make it flow better because of a lack of (or larger) orifice, it flows better because there is less turbulance in the flow stream eating up pressure as friction. You can think of it like an orifice is a very specific type of restriction, while you can always call an orifice a restriction, you can't necessarily call restriction an orifice.

 

No doubt that a better flowing engine makes more power. And that was the whole point of "the formula" it just assumes that "it's all good" to give you an idea of what you can expect. Just don't try to apply it to far away from where you're starting out, or it will waaaay off.

[SleeperZ]

I don't know how much plainer I can state it. For a given area inside a tube, you will have a certain pressure drop when you flow mass through it. Higher flows result in higher pressures, and this is described as a square function. With large areas and low flows, it appears linear, but at higher flows, the pressure builds to the square of the airflow. A doubling of airflow will result in quadrupling the backpressure. This effect is most obvious with a significant restriction in place, such as a throttle plate, but you will see smaller restrictions exhibit this at higher airflows, such as across valves, through turbines and wastegates, intercoolers and such. All of this will prevent you from making the theoretical power Pyro is referring to.

 

Maybe you can explain to me how flow through an orifice is somehow different and non-existent in an engine..

[zuhow]

Thanks SleeperZ, You've pointed out to me a completely new way of considering pipe flow. I'm willing to buy that a restriction is an orifice always now that i know what you mean by OE

 

sweet, I love this place

[Guest bastaad525]

I said from the start is was a simple formula.

 

What do you expect from 2 basic inputs in a very simple equation?

 

 

Hmmm' date=' no, what YOU said at the start was "The formula works [b']perfect[/b] because it is based on basic science."

 

So which is it? The formula works perfectly or "it's a very simple formula, what do you expect?". Seems to me you're backpedaling a bit now and finally admitting/accepting that this formula CAN be innaccurate. Yes it can get CLOSE to the right numbers, hey I admitted that in my first reply to you that it was close. It is a good estimate, sure.

 

But you are still wrong on a couple things. You saying that a bone stock L28ET can make no more than 270hp (at the crank) at 17psi REGARDLESS of the turbo is ABSOLUTELY 100% WRONG. There are definately guys here with 100% stock unimproved blocks that run that kind of boost on a T3/04 (or bigger!) and get way more than 270hp TO THE WHEELS, let alone at the crank. This isn't a formula, this is actual experiences of actual people. I was making more than 270hp at the crank at 12psi! If you're going to try to cover that by saying "oh dyno numbers are so wrong anyways" well that's a whole other can of worms, suffice it to say that too many people dyno their cars then go down to the track and run times and trap speeds that are pretty damn close to what they should run according to what the dyno said they had. You can argue that with guys who've dynoed AND been to the track (I'm not one of them).

 

Even if you talk about changing the boost level on the same turbo, that formula is off. It gives a linear rise in the amount of hp to the rise in boost. But we all know that as the boost is changed, the turbo's efficiency, and hence the amount of power it adds, will change as well. So that formula only has the ability to be accurate over a narrow spread of the possible boost range of the turbo.

 

Then you're saying "oh but an L28ET makes more power with the IC, so really we have to consider that as it's base power" well duh, of course it does. So what? Remember, this formula assumes you are starting with the HP the motor makes at ZERO boost. So, if you take away the turbo completely, the L28ET is gonna make 120-130rwhp whether it's intercooled or not!

 

It's also a FACT that if you make X amount of power at X amount of boost on the stock turbo, if you upgrade JUST the turbo and NOTHING else (well, of course you have to up fuel to support the better turbo, but NO other improvements to the engines ability to breath), and run the EXACT SAME AMOUNT OF BOOST, you WILL make more power, PERIOD! EVERYONE here will agree with that. But from what you are saying that's not possible. 17psi will be 270hp (or less) unless you improve the engine/intake/cam/exhaust somehow. But that is WRONG and has been proven as such! You can argue that with every guy on here who's upgraded to a T3/04!

 

Yes you can lead a horse to water... but if the water smells a little off don't be surprised when he doesn't want to drink it!

[Pyro]

Bastaad525,

Take it easy. I was just offering a simple formula that is based on science prinicples. But for some reason you guys don't like simple. And it does work perfectly, it is the engine that doesn't.

 

The IC is needed on the stock engine so the air temps are the same as the NA. Otherwise the formula doesn't work, the air temps NEED to be the same. Try it again with IC's installed.

 

And like I had been saying, if the output is lower than expected under boost then something is wrong. And high inlet temps is something wrong. That is why and IC is needed for the formula to work.

 

No, I didn't say a stock L28 couldn't make more than 270 at the crank with 17 psi of boost. Go back and read my old post. What I said was, an engine with a 125hp base power at the wheels would make 270 hp AT THE WHEELS, with 17 psi of boost if everything was working perfectly. If more power was made with 17 psi then the base power was higher to begin with.

 

Hey, I could care less if you want to push a bunch of buttons on your calculator and guess VE numbers. If you don't want to use the simple formula then don't. I just thought it was a useful formula and wanted to share it.

 

SleeperZ,

Fluid flow through a pipe theory is good stuff. Basic fluid dynamics, good...

However, you are not considering the mechaincal system. As you know, a pump has a much harder time pulling fluid through a pipe as compared to pushing fluid through a pipe. So, the stock intake becomes a lot less restrictive to the engine under positive pressure from the turbo as compared to the vacuum creatated by the piston. Just something else to consider.

[mobythevan]

As you know, a pump has a much harder time pulling fluid through a pipe as compared to pushing fluid through a pipe. So, the stock intake becomes a lot less restrictive to the engine under positive pressure from the turbo as compared to the vacuum creatated by the piston. Just something else to consider.

 

I don't like it when people talk about push vs pull when describing air into the engine. That works I suppose if you have a positive displacement pump, but a turbo isn't. Talking about fluid being pumped is one thing since fluid does not compress and will be postively displaced by a pump.

 

You can't turn the turbo compressor 10 revolutions and get x amount of air flow since it is not a sealed pumping device.

In the fluid example the pump "pulling" fluid through a pipe creates a vacuum on the pump side and whatever pressure is on the other end of the pipe does the pushing. When the pump is on the fluid side it does the pushing and because fluid cannot compress the pump exerts its entire force directly on the fluid. I think comparing fluid versus gas quickly breaks down since you guys have made this discussion more technical.