Turbine Inlet Pressure
09-11-2014, 05:18 PM
#16
Yes does seem strange. Its taken from the exhaust test port. I need to re-check the line from the test port to the gauge. Im thinking its leaking once it reaches 7 psi. When my boost gauge reaches full boost the crossover gauge was hitting 7 and not climbing past there. Hard to watch both gauges for obvious reasons. Need a dyno instead of back roads.
09-12-2014, 12:29 AM
#19
Rennlist Member
The turbine is exhaust driven...exhaust piping is plumbed such that exhaust gases can flow in only one direction and, as a result, drive the turbine in only one direction.
09-12-2014, 09:56 AM
#22
Race Car
Let me see if I can help clear up things a little here. Now this is how I believe stuff works. May be mistaken, but I'm pretty sure most of this is right.
An object at rest remains at rest unless acted on by a FORCE. Okay, so how does one create force? I'll stick with standard units. Pounds are units of force, and since we are all talking PSI, I'll just stay with it.
So how on Earth does exhaust, heat, whatever, create a force? Hmmm.....let's see. PSI = Pounds / square inch. Ah! We take a pressure and multiply by an area! How do we do that? The blades of a turbine wheel! So we take P / SI * SI and we get P. That's the force we put on the turbine.
The more force on teh turbine, the more you can accelerate it. The reverse happens on the impeller side. You are putting a force on the impeller, then you take the area of the blades into consideration, and the result of all of that is how much pressure you can have in the housing. This is oversimplifying it, but I'm trying to show concepts.
How do factors affect that? Well, the area of the turbine is fixed (ignoring VNT for this discussion). So we can increase the pressure hitting teh turbine to get more force through the shaft. How do we affect the pressure? There are a few ways. One, we can clean up the exhaust system so that there are fewer losses between the exhaust port and the turbine. That would include the manifold/header and turbine housing, itself. If you are familiar with A/R, the smaller the A/R, the higher the losses. So a big honkin' A/R will help (note that there are tradeoffs - this is steady state).
Aside from losses, there are other ways. Ideal gas law, PV=nRT. For just a second, assume that volume is fixed (displacement). R is a constant. That means P = F(n,T). n is the molar quantity. So you can burn more air/fuel. That means more boost or higher RPM. Or both. OR you can add heat. How do you add heat? Well, you just lose less. So when folks say heat is what drives a turbo, they are really saying that you are increasing your pressure.
So getting back to the whole pressure difference between hot side and cold side. Where those pressures are measured will have a big impact on the results. Getting back to the ideal gas law. At steady state, both sides should be in equilibrium. P1V1/(n1T1) = P2V2/(n2T2). Since we have combustion, and we add fuel after the impeller, n's are not equal. n for exhaust is about 1.36 times n for intake. T is also much, much higher. EGT's are about 1800F. Maybe you'll lose a few hundred by the time you get to the turbo, so call it 1500.....and that's a SWAG at the loss, it may be more. On the intake side, you are maybe 300 depending on boost level and efficiency. Converting each to R, 1960/760 gives another 2.6. Factoring in the 1.36 means ceteris paribus your exhaust pressure would be about 3.5 times what your intake pressure would be.
Now we'll bring V back into the equation. You get that pressure factor to drop from 3.5 down to a manageable number by increasing volume, which means increasing flow. So your exhaust side of your turbo has to flow 3.5 times better than your intake side to hit that 1:1 ratio. If your exhuast side is half what your intake side is, you are talking 7 times the flow (roughly).
I personally don't believe that is happening. I think for just about every street or amateur racing or for that matter, sub million dollar system out there, the exhaust pressure will be higher than the intake pressure. If you are measuring something drastically different, then either your measuring systems are wrong, or your losses to your gauge are much greater on the exhaust side than on the intake side.
I'll try to come back...I really gotta split.
An object at rest remains at rest unless acted on by a FORCE. Okay, so how does one create force? I'll stick with standard units. Pounds are units of force, and since we are all talking PSI, I'll just stay with it.
So how on Earth does exhaust, heat, whatever, create a force? Hmmm.....let's see. PSI = Pounds / square inch. Ah! We take a pressure and multiply by an area! How do we do that? The blades of a turbine wheel! So we take P / SI * SI and we get P. That's the force we put on the turbine.
The more force on teh turbine, the more you can accelerate it. The reverse happens on the impeller side. You are putting a force on the impeller, then you take the area of the blades into consideration, and the result of all of that is how much pressure you can have in the housing. This is oversimplifying it, but I'm trying to show concepts.
How do factors affect that? Well, the area of the turbine is fixed (ignoring VNT for this discussion). So we can increase the pressure hitting teh turbine to get more force through the shaft. How do we affect the pressure? There are a few ways. One, we can clean up the exhaust system so that there are fewer losses between the exhaust port and the turbine. That would include the manifold/header and turbine housing, itself. If you are familiar with A/R, the smaller the A/R, the higher the losses. So a big honkin' A/R will help (note that there are tradeoffs - this is steady state).
Aside from losses, there are other ways. Ideal gas law, PV=nRT. For just a second, assume that volume is fixed (displacement). R is a constant. That means P = F(n,T). n is the molar quantity. So you can burn more air/fuel. That means more boost or higher RPM. Or both. OR you can add heat. How do you add heat? Well, you just lose less. So when folks say heat is what drives a turbo, they are really saying that you are increasing your pressure.
So getting back to the whole pressure difference between hot side and cold side. Where those pressures are measured will have a big impact on the results. Getting back to the ideal gas law. At steady state, both sides should be in equilibrium. P1V1/(n1T1) = P2V2/(n2T2). Since we have combustion, and we add fuel after the impeller, n's are not equal. n for exhaust is about 1.36 times n for intake. T is also much, much higher. EGT's are about 1800F. Maybe you'll lose a few hundred by the time you get to the turbo, so call it 1500.....and that's a SWAG at the loss, it may be more. On the intake side, you are maybe 300 depending on boost level and efficiency. Converting each to R, 1960/760 gives another 2.6. Factoring in the 1.36 means ceteris paribus your exhaust pressure would be about 3.5 times what your intake pressure would be.
Now we'll bring V back into the equation. You get that pressure factor to drop from 3.5 down to a manageable number by increasing volume, which means increasing flow. So your exhaust side of your turbo has to flow 3.5 times better than your intake side to hit that 1:1 ratio. If your exhuast side is half what your intake side is, you are talking 7 times the flow (roughly).
I personally don't believe that is happening. I think for just about every street or amateur racing or for that matter, sub million dollar system out there, the exhaust pressure will be higher than the intake pressure. If you are measuring something drastically different, then either your measuring systems are wrong, or your losses to your gauge are much greater on the exhaust side than on the intake side.
I'll try to come back...I really gotta split.
09-12-2014, 11:18 AM
#23
Thanks for the much better explaination 
I posted my issue because something was not adding up with what I am seeing on the gauges. The simple answer is the pre-turbine gauge is leaking considerably somewhere.....just wanted to see if anyone had ever had a weird reading like I am getting. Unfortunately it snowed here and I cant get back out for a few days to see if I fixed my leak. No claims of a car that defies the laws of physics just posted to query thoughts from RL
I posted my issue because something was not adding up with what I am seeing on the gauges. The simple answer is the pre-turbine gauge is leaking considerably somewhere.....just wanted to see if anyone had ever had a weird reading like I am getting. Unfortunately it snowed here and I cant get back out for a few days to see if I fixed my leak. No claims of a car that defies the laws of physics just posted to query thoughts from RL
Let me see if I can help clear up things a little here. Now this is how I believe stuff works. May be mistaken, but I'm pretty sure most of this is right.
An object at rest remains at rest unless acted on by a FORCE. Okay, so how does one create force? I'll stick with standard units. Pounds are units of force, and since we are all talking PSI, I'll just stay with it.
So how on Earth does exhaust, heat, whatever, create a force? Hmmm.....let's see. PSI = Pounds / square inch. Ah! We take a pressure and multiply by an area! How do we do that? The blades of a turbine wheel! So we take P / SI * SI and we get P. That's the force we put on the turbine.
The more force on teh turbine, the more you can accelerate it. The reverse happens on the impeller side. You are putting a force on the impeller, then you take the area of the blades into consideration, and the result of all of that is how much pressure you can have in the housing. This is oversimplifying it, but I'm trying to show concepts.
How do factors affect that? Well, the area of the turbine is fixed (ignoring VNT for this discussion). So we can increase the pressure hitting teh turbine to get more force through the shaft. How do we affect the pressure? There are a few ways. One, we can clean up the exhaust system so that there are fewer losses between the exhaust port and the turbine. That would include the manifold/header and turbine housing, itself. If you are familiar with A/R, the smaller the A/R, the higher the losses. So a big honkin' A/R will help (note that there are tradeoffs - this is steady state).
Aside from losses, there are other ways. Ideal gas law, PV=nRT. For just a second, assume that volume is fixed (displacement). R is a constant. That means P = F(n,T). n is the molar quantity. So you can burn more air/fuel. That means more boost or higher RPM. Or both. OR you can add heat. How do you add heat? Well, you just lose less. So when folks say heat is what drives a turbo, they are really saying that you are increasing your pressure.
So getting back to the whole pressure difference between hot side and cold side. Where those pressures are measured will have a big impact on the results. Getting back to the ideal gas law. At steady state, both sides should be in equilibrium. P1V1/(n1T1) = P2V2/(n2T2). Since we have combustion, and we add fuel after the impeller, n's are not equal. n for exhaust is about 1.36 times n for intake. T is also much, much higher. EGT's are about 1800F. Maybe you'll lose a few hundred by the time you get to the turbo, so call it 1500.....and that's a SWAG at the loss, it may be more. On the intake side, you are maybe 300 depending on boost level and efficiency. Converting each to R, 1960/760 gives another 2.6. Factoring in the 1.36 means ceteris paribus your exhaust pressure would be about 3.5 times what your intake pressure would be.
Now we'll bring V back into the equation. You get that pressure factor to drop from 3.5 down to a manageable number by increasing volume, which means increasing flow. So your exhaust side of your turbo has to flow 3.5 times better than your intake side to hit that 1:1 ratio. If your exhuast side is half what your intake side is, you are talking 7 times the flow (roughly).
I personally don't believe that is happening. I think for just about every street or amateur racing or for that matter, sub million dollar system out there, the exhaust pressure will be higher than the intake pressure. If you are measuring something drastically different, then either your measuring systems are wrong, or your losses to your gauge are much greater on the exhaust side than on the intake side.
I'll try to come back...I really gotta split.
An object at rest remains at rest unless acted on by a FORCE. Okay, so how does one create force? I'll stick with standard units. Pounds are units of force, and since we are all talking PSI, I'll just stay with it.
So how on Earth does exhaust, heat, whatever, create a force? Hmmm.....let's see. PSI = Pounds / square inch. Ah! We take a pressure and multiply by an area! How do we do that? The blades of a turbine wheel! So we take P / SI * SI and we get P. That's the force we put on the turbine.
The more force on teh turbine, the more you can accelerate it. The reverse happens on the impeller side. You are putting a force on the impeller, then you take the area of the blades into consideration, and the result of all of that is how much pressure you can have in the housing. This is oversimplifying it, but I'm trying to show concepts.
How do factors affect that? Well, the area of the turbine is fixed (ignoring VNT for this discussion). So we can increase the pressure hitting teh turbine to get more force through the shaft. How do we affect the pressure? There are a few ways. One, we can clean up the exhaust system so that there are fewer losses between the exhaust port and the turbine. That would include the manifold/header and turbine housing, itself. If you are familiar with A/R, the smaller the A/R, the higher the losses. So a big honkin' A/R will help (note that there are tradeoffs - this is steady state).
Aside from losses, there are other ways. Ideal gas law, PV=nRT. For just a second, assume that volume is fixed (displacement). R is a constant. That means P = F(n,T). n is the molar quantity. So you can burn more air/fuel. That means more boost or higher RPM. Or both. OR you can add heat. How do you add heat? Well, you just lose less. So when folks say heat is what drives a turbo, they are really saying that you are increasing your pressure.
So getting back to the whole pressure difference between hot side and cold side. Where those pressures are measured will have a big impact on the results. Getting back to the ideal gas law. At steady state, both sides should be in equilibrium. P1V1/(n1T1) = P2V2/(n2T2). Since we have combustion, and we add fuel after the impeller, n's are not equal. n for exhaust is about 1.36 times n for intake. T is also much, much higher. EGT's are about 1800F. Maybe you'll lose a few hundred by the time you get to the turbo, so call it 1500.....and that's a SWAG at the loss, it may be more. On the intake side, you are maybe 300 depending on boost level and efficiency. Converting each to R, 1960/760 gives another 2.6. Factoring in the 1.36 means ceteris paribus your exhaust pressure would be about 3.5 times what your intake pressure would be.
Now we'll bring V back into the equation. You get that pressure factor to drop from 3.5 down to a manageable number by increasing volume, which means increasing flow. So your exhaust side of your turbo has to flow 3.5 times better than your intake side to hit that 1:1 ratio. If your exhuast side is half what your intake side is, you are talking 7 times the flow (roughly).
I personally don't believe that is happening. I think for just about every street or amateur racing or for that matter, sub million dollar system out there, the exhaust pressure will be higher than the intake pressure. If you are measuring something drastically different, then either your measuring systems are wrong, or your losses to your gauge are much greater on the exhaust side than on the intake side.
I'll try to come back...I really gotta split.
Last edited by gruhsy; 09-12-2014 at 03:05 PM.
09-12-2014, 11:21 AM
#24
Let me see if I can help clear up things a little here. Now this is how I believe stuff works. May be mistaken, but I'm pretty sure most of this is right.
An object at rest remains at rest unless acted on by a FORCE. Okay, so how does one create force? I'll stick with standard units. Pounds are units of force, and since we are all talking PSI, I'll just stay with it.
So how on Earth does exhaust, heat, whatever, create a force? Hmmm.....let's see. PSI = Pounds / square inch. Ah! We take a pressure and multiply by an area! How do we do that? The blades of a turbine wheel! So we take P / SI * SI and we get P. That's the force we put on the turbine.
The more force on teh turbine, the more you can accelerate it. The reverse happens on the impeller side. You are putting a force on the impeller, then you take the area of the blades into consideration, and the result of all of that is how much pressure you can have in the housing. This is oversimplifying it, but I'm trying to show concepts.
How do factors affect that? Well, the area of the turbine is fixed (ignoring VNT for this discussion). So we can increase the pressure hitting teh turbine to get more force through the shaft. How do we affect the pressure? There are a few ways. One, we can clean up the exhaust system so that there are fewer losses between the exhaust port and the turbine. That would include the manifold/header and turbine housing, itself. If you are familiar with A/R, the smaller the A/R, the higher the losses. So a big honkin' A/R will help (note that there are tradeoffs - this is steady state).
Aside from losses, there are other ways. Ideal gas law, PV=nRT. For just a second, assume that volume is fixed (displacement). R is a constant. That means P = F(n,T). n is the molar quantity. So you can burn more air/fuel. That means more boost or higher RPM. Or both. OR you can add heat. How do you add heat? Well, you just lose less. So when folks say heat is what drives a turbo, they are really saying that you are increasing your pressure.
So getting back to the whole pressure difference between hot side and cold side. Where those pressures are measured will have a big impact on the results. Getting back to the ideal gas law. At steady state, both sides should be in equilibrium. P1V1/(n1T1) = P2V2/(n2T2). Since we have combustion, and we add fuel after the impeller, n's are not equal. n for exhaust is about 1.36 times n for intake. T is also much, much higher. EGT's are about 1800F. Maybe you'll lose a few hundred by the time you get to the turbo, so call it 1500.....and that's a SWAG at the loss, it may be more. On the intake side, you are maybe 300 depending on boost level and efficiency. Converting each to R, 1960/760 gives another 2.6. Factoring in the 1.36 means ceteris paribus your exhaust pressure would be about 3.5 times what your intake pressure would be.
Now we'll bring V back into the equation. You get that pressure factor to drop from 3.5 down to a manageable number by increasing volume, which means increasing flow. So your exhaust side of your turbo has to flow 3.5 times better than your intake side to hit that 1:1 ratio. If your exhuast side is half what your intake side is, you are talking 7 times the flow (roughly).
I personally don't believe that is happening. I think for just about every street or amateur racing or for that matter, sub million dollar system out there, the exhaust pressure will be higher than the intake pressure. If you are measuring something drastically different, then either your measuring systems are wrong, or your losses to your gauge are much greater on the exhaust side than on the intake side.
I'll try to come back...I really gotta split.
An object at rest remains at rest unless acted on by a FORCE. Okay, so how does one create force? I'll stick with standard units. Pounds are units of force, and since we are all talking PSI, I'll just stay with it.
So how on Earth does exhaust, heat, whatever, create a force? Hmmm.....let's see. PSI = Pounds / square inch. Ah! We take a pressure and multiply by an area! How do we do that? The blades of a turbine wheel! So we take P / SI * SI and we get P. That's the force we put on the turbine.
The more force on teh turbine, the more you can accelerate it. The reverse happens on the impeller side. You are putting a force on the impeller, then you take the area of the blades into consideration, and the result of all of that is how much pressure you can have in the housing. This is oversimplifying it, but I'm trying to show concepts.
How do factors affect that? Well, the area of the turbine is fixed (ignoring VNT for this discussion). So we can increase the pressure hitting teh turbine to get more force through the shaft. How do we affect the pressure? There are a few ways. One, we can clean up the exhaust system so that there are fewer losses between the exhaust port and the turbine. That would include the manifold/header and turbine housing, itself. If you are familiar with A/R, the smaller the A/R, the higher the losses. So a big honkin' A/R will help (note that there are tradeoffs - this is steady state).
Aside from losses, there are other ways. Ideal gas law, PV=nRT. For just a second, assume that volume is fixed (displacement). R is a constant. That means P = F(n,T). n is the molar quantity. So you can burn more air/fuel. That means more boost or higher RPM. Or both. OR you can add heat. How do you add heat? Well, you just lose less. So when folks say heat is what drives a turbo, they are really saying that you are increasing your pressure.
So getting back to the whole pressure difference between hot side and cold side. Where those pressures are measured will have a big impact on the results. Getting back to the ideal gas law. At steady state, both sides should be in equilibrium. P1V1/(n1T1) = P2V2/(n2T2). Since we have combustion, and we add fuel after the impeller, n's are not equal. n for exhaust is about 1.36 times n for intake. T is also much, much higher. EGT's are about 1800F. Maybe you'll lose a few hundred by the time you get to the turbo, so call it 1500.....and that's a SWAG at the loss, it may be more. On the intake side, you are maybe 300 depending on boost level and efficiency. Converting each to R, 1960/760 gives another 2.6. Factoring in the 1.36 means ceteris paribus your exhaust pressure would be about 3.5 times what your intake pressure would be.
Now we'll bring V back into the equation. You get that pressure factor to drop from 3.5 down to a manageable number by increasing volume, which means increasing flow. So your exhaust side of your turbo has to flow 3.5 times better than your intake side to hit that 1:1 ratio. If your exhuast side is half what your intake side is, you are talking 7 times the flow (roughly).
I personally don't believe that is happening. I think for just about every street or amateur racing or for that matter, sub million dollar system out there, the exhaust pressure will be higher than the intake pressure. If you are measuring something drastically different, then either your measuring systems are wrong, or your losses to your gauge are much greater on the exhaust side than on the intake side.
I'll try to come back...I really gotta split.
09-12-2014, 11:57 AM
#25
Race Car
Though you're physics teacher would be proud, you're forgetting about the variables that allow newtons laws of motion to be true. All variables must be kept true except for the one being tested, in this case back pressure. The turbo system is not as complicated as its made up to be. If you've ever fiddled with a turbo you'd know that a turbo is not a linear motion as newtons laws imply. Turbos use a 90 degree change in motion/ flow. Heat doesn't truely turn the exhaust turbine, the flow does( which is a by-product). If it was heat alone then I'd strap a blow torch to it and be done. The flow pushes the turbine much like a small garden windmill(the plastic cheap ones you stick in the ground). Once the flow has traveled past the turbine on a stock set up it would pass through a narrow pipe and then the cat and muffler. This creates strain on the turbines ability to push the flow out of the car. Removing the cat and using a high flow muffler significantly reduces this strain therefore allowing better flow.. Better flow spins the turbine more efficiently reducing spool time and reduces back pressure. The better the flow the less back pressure you will have.
Well, your grammar teacher wouldn't be so proud. Nor would your physics teacher. And I guarantee you I've messed with more turbos for longer than you have. And the fellow referenced above who was attributed to discussing the importance of heat has been messing with turbos probably since before you were born.
Newtonian physics don't give a darn whether air turns. That would be fluid dynamics, which covers losses, which I already addressed. And your geometry teacher would be equally ashamed if he knew you were sitting here telling people things have to be at 90 degrees to put apply a force.
This may be playing with semantics a bit, but flow doesn't move the turbo. Flow is merely the result of a pressure drop. It is a pressure drop that moves the turbo, by creating a force over an area. Newton's famous law is "an object at rest will remain at rest, and an object in motion will remain in motion, unless acted on by an outside force." And said pressure drop is affected by temperature, as I showed with the ideal gas law. Which is, FWIW, physical chemistry, not physics. That's a college course, though, so maybe you didn't get that.
Again, force is what moves things. Force from fluids is created by pressure over an area. In order for said pressure to act on a device, there must be a pressure drop across it. The same pressure drop that moves the wheels in the turbo also moves air......which is what you refer to as "flow."
09-12-2014, 04:05 PM
#26
Well, your grammar teacher wouldn't be so proud. Nor would your physics teacher. And I guarantee you I've messed with more turbos for longer than you have. And the fellow referenced above who was attributed to discussing the importance of heat has been messing with turbos probably since before you were born.
Newtonian physics don't give a darn whether air turns. That would be fluid dynamics, which covers losses, which I already addressed. And your geometry teacher would be equally ashamed if he knew you were sitting here telling people things have to be at 90 degrees to put apply a force.
This may be playing with semantics a bit, but flow doesn't move the turbo. Flow is merely the result of a pressure drop. It is a pressure drop that moves the turbo, by creating a force over an area. Newton's famous law is "an object at rest will remain at rest, and an object in motion will remain in motion, unless acted on by an outside force." And said pressure drop is affected by temperature, as I showed with the ideal gas law. Which is, FWIW, physical chemistry, not physics. That's a college course, though, so maybe you didn't get that.
Again, force is what moves things. Force from fluids is created by pressure over an area. In order for said pressure to act on a device, there must be a pressure drop across it. The same pressure drop that moves the wheels in the turbo also moves air......which is what you refer to as "flow."
Last edited by mahoney944; 09-12-2014 at 04:21 PM.
09-13-2014, 07:16 AM
#29
Race Car
Machinists, engineers, and mechanics all have very different fields of expertise. You won't find me trying to tell a machinist about feed rates, cutting speeds, etc. And when you've had 3/4ths of the engines you develop professionally make the Ward's 10 Best, you can let me know. Until then, leave the engineering to the engineers.
You talking about the Borg EFR turbos? Very niece pieces. Have a friend who is an engineer working in them. Gets to use his car as a test mule for them sometimes.
