Turbocharger power requirements

[Smokey]

how is your driveline goining to handle this idea? the stock viper engine (505)is already double your 5.9's (245) HP rateing....and your talking 3 to 4 times what it came with stock.....just an observation:dontknow: :dontknow:

I wouldn't necessarily agree with the statement. What if your goal is 1000ft/lbs of TQ from 2000-6000RPM? Turbo is unable to spool fast enough to give 14+psi at 2000rpm and then still be able to flow well enough to continue to provide that boost (or more) at 6000RPM. If you want NO LAG then turbos are not an option and a centrifugal is out as well. If you want efficiency then a roots blower is out leaving only a twin screw. If turbos out performed charges then chargers wouldn't exist. Each has its place/purpose.

what about doin a rear mount, sequential twin? small turbo for down low, and a big one for WOT. then u get teh both worlds doesnt seem that hard to do either

 

[DevilDawg3097]

Hey Pokey, smokey Is dead on pretty much said the same thing. And the VHT you will be the first he knows to try on a gasser, he said maybe you ll have better luck:marchmellow: He said it is just like anything when it first comes out it looked good on paper and in programs but is a piece of Shite in the real world. So...

 

[pokeytemplar]

Built 48RE, upgrade the Transfer case to the Hummer H1 specs, driveshafts, halfshafts, etc...... Still figuring all of that stuff out! I am learning and building completely on my own. Darton sleeves (designing my own). Dry Sump (trying to figure out how to size it). Modify Magnum block to accept Viper heads (working on it). I just keep moving down the list of stuff that needs to be done and what I have to learn to do it. Just picked up a hydroboost setup for the Durango because the Vacuum booster was going to interfere with the V-10's valve covers.......(one more thing off of the list)

The cummins VGT's have problems with soot build up stopping the VGT from V'ing. Gasser won't have that problem. Mine would be getting the VGT to work in a non-cummins environment.

I have been toying with the idea of a quad compound. But my ideas quickly outrun my pocketbook! The twin screw would be around $10-12k for a 1800HP capable system(that is with me doing all of the work). Yes turbo's can make even more power I know but there are packaging issues. 01 Durango's have really tiny engine bays. Currently I am disassembling the fuse box so I can resize/relocate it to free up more space.

 

[WOT]

Interesting.... If there is any drive pressure though there is a cost to driving the turbo. The cost is a lot less but there is still a cost. We see a benefit from FI because our gains are more than our losses.

Which means it has a negative net cost. So to answer your question, the boost itself in the example I gave had a net cost of 0 hp since 100% of the cost due to backpressure is made up in positive downward pressure during the intake stroke.

If you are looking for how much HP is required to drive the compressor side of the turbo, that can be calculated. Otherwise, it can be set up to not cost any hp (net).

When comparing the two, the net cost of a turbo can be 0 while the net cost of a SC can never be 0.

When you say the gains are more than the losses, that is due to being able to burn more air and fuel to make up for the loss. In the example I am using, there is no cost because no additional fuel has to be burned for the system to break even.

 

[WOT]

As another example, one of the large big-rig diesel motor manufacturers at one time had a turbo on the exhuast but instead of making boost, the compressor side was removed and connected to a gear drive system going to the crank. Instead of using exhaust energy to make boost, it used it to help add power to the crank. The net result was a 20 hp increase in output, 100% of the increase coming from the exhaust energy. So, the net cost of driving the turbine side of the turbo was negative 20hp.

 

[pokeytemplar]

How do you calculate the HP required to drive the compressor side? That would help in calculating the load on the engine. In your examples above you don't quite make sense. If your examples are true then a centrifugal would have a net loss in the lower RPM's but then the net loss would be 0 at any level above it producing boost(ie 1psi). The roots and twinscrews would never have a loss as they are capable of producing boost at any RPM.

My point would be to put a wastegate/restrictor onto the exhaust of any NA engine. Dyno with it open and then dyno it with 20psi worth of restriction. You will see power down across the entire graph. This is the same thing that is happening while the drive pressure is turning the turbines. That is the "loss/cost" of driving turbo's. Belt driven systems move the losses to the crankshaft instead of the exhaust system.

In the last example it shows the 15% of efficiency that a turbo enjoys over belt driven FI solutions by utilizing wasted energy. Like I have said before this isn't a which is better I am just wanting to completely understand all of the positives and negatives of FI and turbos in this particular instance....

 

[WOT]

I think you are missing my point completely. My point is that you can have a turbo in certain rpm ranges that does not cost any horsepower to drive.

We all know that the power increases you get from either a turbo or a SC is due to the fact that you are cramming more air and fuel into the combustion chamber which allows the engine to make more power. So let me give you two hypothetical examples to make my point.

Let's say you have an engine that makes 300 hp naturatally aspirated. And we calculate that the motor should gain 200 hp if we could supply 14 psi of boost to the motor from say a huge compressed air tank (that way there is no possibility for any parasitic loss). Maybe you actually have something like this type of set up and dyno the motor. You supply 14psi of boost from the giant compressed air tank and the motor makes 500 hp. All is good.

Then you put a belt driven SC on the motor. For example purposes, let's say that at 14 psi boost, it takes 50 hp to drive the supercharger. Maybe you have an electic motor driven dyno that is capable of measuring assesory and drive train hp requirements and you put the SC on that dyno by itself and measure the hp required to drive it (some race teams have these and use them to dyno transmissions).

Now, let's say you dyno your 300 hp motor at 14 psi boost with the SC and it makes 460 hp. Since it takes 50 hp to drive the blower and the motor made 460 hp, some of the parasitic loss must have been made up by the pressurzed intake charge created by the SC actually pushing down on the pistons during the intake stroke. It's free horsepower, so you'll take it. Doing the math, the net hp cost to drive the blower is 40 hp and the SC provides a 160hp gain by allowing the engine to burn more air and fuel.

Then you take off the supercharger and put on a turbo. You carefully pick the size and A/R for the housing and run your test with 14 psi boost. You know that the motor running 14 psi boost supplied from the big compressed air tank will make 500 hp, so you should easily be able to figure out how much parasitic loss is caused by the turbo as long as you keep it exactly at 14 psi.

You dyno the engine with the turbo, and it makes 501 hp. So the turbo actually costs -1 hp to drive.

How is this possible you ask? Because the turbo gains its energy from exhaust heat and backpressure, but it gets most of the energy from the exaust heat, not the back pressure. So it can have lower backpressure on the exaust side than it does on the intake side.

 

[pokeytemplar]

Since it takes 50 hp to drive the blower and the motor made 460 hp, some of the parasitic loss must have been made up by the pressurzed intake charge created by the SC actually pushing down on the pistons during the intake stroke. It's free horsepower, so you'll take it.

I have to disagree that the compressed air "pushes" the piston down. It is impossible to move the air into the cylinder fast enough to act upon the piston. If you see any gains it would be in the increased efficiency of the combustion process.

The ONLY time a turbo costs nothing to drive is when you have spooled the turbo at max RPM and downshift. During that "spin down" time it imposes no restriction to exhaust flow because it is spinning faster than the flow and thus no "cost" to drive it.

My point is that if there is any additional back pressure then there is a loss of HP. So your turbo example really should show a 184HP gain with only 16HP used to drive it (due to its ability to utilize this "wasted" energy). If a turbine is in the exhaust path there will be a restriction because if it imposed no restriction then it would not be "driven".

 

[Smokey]

I think you are missing my point completely. My point is that you can have a turbo in certain rpm ranges that does not cost any horsepower to drive.

We all know that the power increases you get from either a turbo or a SC is due to the fact that you are cramming more air and fuel into the combustion chamber which allows the engine to make more power. So let me give you two hypothetical examples to make my point.

Let's say you have an engine that makes 300 hp naturatally aspirated. And we calculate that the motor should gain 200 hp if we could supply 14 psi of boost to the motor from say a huge compressed air tank (that way there is no possibility for any parasitic loss). Maybe you actually have something like this type of set up and dyno the motor. You supply 14psi of boost from the giant compressed air tank and the motor makes 500 hp. All is good.

Then you put a belt driven SC on the motor. For example purposes, let's say that at 14 psi boost, it takes 50 hp to drive the supercharger. Maybe you have an electic motor driven dyno that is capable of measuring assesory and drive train hp requirements and you put the SC on that dyno by itself and measure the hp required to drive it (some race teams have these and use them to dyno transmissions).

Now, let's say you dyno your 300 hp motor at 14 psi boost with the SC and it makes 460 hp. Since it takes 50 hp to drive the blower and the motor made 460 hp, some of the parasitic loss must have been made up by the pressurzed intake charge created by the SC actually pushing down on the pistons during the intake stroke. It's free horsepower, so you'll take it. Doing the math, the net hp cost to drive the blower is 40 hp and the SC provides a 160hp gain by allowing the engine to burn more air and fuel.

Then you take off the supercharger and put on a turbo. You carefully pick the size and A/R for the housing and run your test with 14 psi boost. You know that the motor running 14 psi boost supplied from the big compressed air tank will make 500 hp, so you should easily be able to figure out how much parasitic loss is caused by the turbo as long as you keep it exactly at 14 psi.

You dyno the engine with the turbo, and it makes 501 hp. So the turbo actually costs -1 hp to drive.

How is this possible you ask? Because the turbo gains its energy from exhaust heat and backpressure, but it gets most of the energy from the exaust heat, not the back pressure. So it can have lower backpressure on the exaust side than it does on the intake side.

thats a awsome example :rock:

 

[WOT]

I have to disagree that the compressed air "pushes" the piston down. It is impossible to move the air into the cylinder fast enough to act upon the piston. If you see any gains it would be in the increased efficiency of the combustion process.

OK, so you are saying that it could not possibly push down on the piston. Think of it this way: it is relieving the duty of the motor having to create vacuum to suck the air in -- tomato/tomatto. Either way you will see an increase in hp due to the reduced pumping loss on the intake side. As an example of how much power is actually required just to suck the air in, a jake brake on a large diesel can create the equavalent of 500 hp (only in reverse) of braking force by creating vacuum. Diesels do not have a throttle butterfly, so to create any off-throttle deceleration they use a device that throws the intake and exhaust valving out of time so that it creates suction inside the cylinders even though the throttle is always open.

So, without the turbo, the motor would have to suck its own air in and that takes horsepower. The turbo is using free exhaust energy to help push the pistons down (or minimize vacuum that is robbing hp).

If a turbine is in the exhaust path there will be a restriction because if it imposed no restriction then it would not be "driven".

Why won't you look at both sides of the motor? Yes, it creates a restriction, but at certain steady state conditions, it can relieve more intake restriction than it causes in exhaust restriction, thus having a net cost of 0.

thats a awsome example :rock:

Thx!

 

[Prof]

Sure glad I have a supercharger...the terminology is much more simple...boost or no boost...blows up or doesn't...

Damn that turbo stuff seems complicated!

 

[JTS VENOM PERFORMANCE]

Sure glad I have a supercharger...the terminology is much more simple...boost or no boost...blows up or doesn't...

Damn that turbo stuff seems complicated!

Prof I am seriously considering going back N/A

 

[pokeytemplar]

OK, so you are saying that it could not possibly push down on the piston. Think of it this way: it is relieving the duty of the motor having to create vacuum to suck the air in -- tomato/tomatto. Either way you will see an increase in hp due to the reduced pumping loss on the intake side. As an example of how much power is actually required just to suck the air in, a jake brake on a large diesel can create the equavalent of 500 hp (only in reverse) of braking force by creating vacuum. Diesels do not have a throttle butterfly, so to create any off-throttle deceleration they use a device that throws the intake and exhaust valving out of time so that it creates suction inside the cylinders even though the throttle is always open.

So, without the turbo, the motor would have to suck its own air in and that takes horsepower. The turbo is using free exhaust energy to help push the pistons down (or minimize vacuum that is robbing hp).

Why won't you look at both sides of the motor? Yes, it creates a restriction, but at certain steady state conditions, it can relieve more intake restriction than it causes in exhaust restriction, thus having a net cost of 0.

Yes I am saying there is no way the piston would ever be pushed by the intake charge (especially a V-10). To even come close to that possibility you would have to open the intake valves way BEFORE TDC, have more than one intake valve, and need a TREMENDOUS amount of boost (around 2,000+psi). Engines do not suck air it is pushed into the engine.

The air moves into the cylinder due to the pressure differential created by the piston moving down. Your are then limited by the intact tracts flow and the MACH index of the valves. The boost has increased the density of the air it cannot move any faster than the previous limitations allow.

Also to slow a Diesel down (This is the way my Mack's do it) from Wiki:
"An exhaust brake is a means of slowing a diesel engine by closing off the exhaust path from the engine, causing the exhaust gases to be compressed in the exhaust manifold, and in the cylinder. Since the exhaust is being compressed, and there is no fuel being applied, the engine works backwards, slowing down the vehicle. The amount of negative torque generated is usually directly proportional to the back pressure of the engine."

Oh wait what was that last sentence "The amount of negative torque generated is usually directly proportional to the back pressure of the engine." Hmmm so the Drive pressure (back pressure) creates NEGATIVE TORQUE. The power that is required to drive the turbo and make boost. The "cost" of driving the turbo.

Once again how do you calculate the cost? I never have wanted this to be a which is better thread. I will always agree that turbos are more efficient than any other type of force induction. I WILL NOT always agree that they are the best solution for force induction. There are situations where the turbo just doesn't efficiently fill the solution.

I am looking at both sides of the motor. My point is every type of FI that creates any boost has already surpassed net cost of 0.

 

[pokeytemplar]

Sure glad I have a supercharger...the terminology is much more simple...boost or no boost...blows up or doesn't...

Damn that turbo stuff seems complicated!

Turbo's are complicated because there are so many choices! Superchargers have much fewer options so it makes narrowing down a solution easier. For the 10's to make big power (1000+HP) you need at least a F1 or larger from Procharger or a w200AX or larger from Whipple. I imagine there is a Roots out there big enough as well I just haven't ever looked. With Turbo's you can go with singles, duals, quick spool/lower peak HP, slow spool/higher peak HP, etc etc..... I was looking at Garrett's site and there are around a DOZEN different turbos that would be capable of producing 1000+HP on our tens!

 

[505'sFastestViper.]

just put on a flippen turbo if its not what you want sell it and try another one no biggy!!!!!!!

 

[pokeytemplar]

just put on a flippen turbo if its not what you want sell it and try another one no biggy!!!!!!!

No biggy for someone with TONS of disposable income. I on the other hand do not. I have ONE shot to get this right. I want to make an EXTREMELY educated decision. When we are talking $8K plus for a Twin Screw or Turbos and that doesn't include all of the support equipment I won't just jump into it blindly. I am personally setting my limit at the max power I can get with pump gas (preferably without W/M). I also want the broadest, flattest TQ curve possible. This is going into a Durango. Yes it is lighter than a RC and it has AWD. The increased load of tires not spinning would help the turbo's spool faster bringing peak boost quicker in the rev range.:dontknow: The problem is this will be a mostly daily driver so one of my concerns is the power lost spooling turbos for nothing. With Twin Screws there is very little load on the engine when the bypass is open.

Just about finished up with my workshop so things will start moving considerably quicker now that I have a "clean room" to work in. The "test v-10" will be bolted in soon and I can start running all of the water/oil/PS hoses and wiring harness modifications. That is all cheap/time consuming things I can do as I save for the rest of the parts.

 

[blackviper]

Cool. Post some pics when you get started.

 

[WOT]

Yes I am saying there is no way the piston would ever be pushed by the intake charge (especially a V-10)....Engines do not suck air it is pushed into the engine.

The air moves into the cylinder due to the pressure differential created by the piston moving down...

Once again tomato/tomatto. Reducing the pumping requirement for the engine is where the 'free' power comes from. It is relieving the duty of the motor having to create vacuum to suck the air in. Either way you can have a net cost equal to or near 0 due to the reduced pumping loss on the intake side.

But hey, I am repeating myself.

Also to slow a Diesel down (This is the way my Mack's do it) from Wiki:
"An exhaust brake is a means of slowing a diesel engine by closing off the exhaust path from the engine, causing the exhaust gases to be compressed in the exhaust manifold, and in the cylinder. Since the exhaust is being compressed, and there is no fuel being applied, the engine works backwards, slowing down the vehicle. The amount of negative torque generated is usually directly proportional to the back pressure of the engine.".

I have taken apart a Jake brake off of 525HP Cat diesel out of a Peterbuilt truck, and I can tell you from personal experience, that is not how they work. They use hydraulic oil pressure to act on the valves to bleed compression at the top of the compression stroke and then close the valves at TDC thereby creating a vacuum in the cylinder during what would be the power stroke. There is no butterfly in the exhaust or any type of exhaust restriction. They may have tried exhaust throttles a long time ago, but since the late '70s, this is how engine braking has been accomplished on large diesels. Now I have seen some aftermarket exhaust brakes (butterfly in the exhaust) for small diesels, but the reports I read show that they are only minimally effective as compared to a Jake brake.

Oh wait what was that last sentence "The amount of negative torque generated is usually directly proportional to the back pressure of the engine." Hmmm so the Drive pressure (back pressure) creates NEGATIVE TORQUE. The power that is required to drive the turbo and make boost. The "cost" of driving the turbo.

Nobody has argued that back pressure doesn't cause a power rob. My point, which you just can't seem to accept, is that the reduction of pumping cost on the intake side can be made to outweigh the backpressure loss on the exhaust side for a net cost of 0.

I never have wanted this to be a which is better thread.

Who is arguing which is better? You have said this many times, yet I see noone arguing which is better.

My point is every type of FI that creates any boost has already surpassed net cost of 0.

And you would be wrong.

Think about it: if your assumption is that there has to be a net cost > 0, then how does a jet motor work? How would it be possible to have the compressor side directly attached to the turbine side with the compressor supplying all of the air that drives the turbine? How does the jet motor run if the net cost of boost has to be > 0?

I hope you take all of this is good spirit, but I promise you that I am not making this stuff up or pulling it out of my a$$.

 

[pokeytemplar]

The engine is always creating a vacuum. The free power is from the increased efficiency of the engine due to the denser air/fuel mixture in the combustion chamber. That piston is getting JERKED down by the crankshaft because another throw is being pushed by the combustion stroke. The air cannot move any faster it is just denser.

You are correct on how a Jake brake works on a Peterbuilt. I was showing you how the exhaust brake works on a Mack. The increased back pressure produces negative TQ slowing the truck down. The biggest thing you have to watch with the Mack is EGT going WAY TOO HIGH. The Mack is an exaggerated example of what back pressure does.

And I have never disagreed that a turbo doesn't create more power than it uses. I want to calculate how much power it is using! That is the purpose of the entire thread. HOW DO YOU CALCULATE THE LOSS (energy required to drive the turbo).

I am taking this all in good spirit.

 

[pokeytemplar]

Well I have made a few phone calls. I have been told that no one has calculated it or measured it. I guess if I ever want to find out I am going to need an engine dyno and $$$$$$$$$$.