Exhaust gas temp's influence on turbine speed
#1
Exhaust gas temp's influence on turbine speed
Does anyone have any information, graphs, or can comment how exhaust gas temp’s in a turbocharged system may affect the turbine and impeller speed, or would temperature not be a variable because mass air flow times the pressure difference spins the turbine. Any input would be appreciated.
#2
The power that transferred from the turbine to the impeller is proportional to the temperature of the exhaust gas, the pressure ratio across the turbine, and the massflow. So yes, temperature is important.
To be more specific, the power from the turbine is the temperature change of the exhaust times massflow. The temperature change depends on the pressure ratio and turbine efficiency.
To be more specific, the power from the turbine is the temperature change of the exhaust times massflow. The temperature change depends on the pressure ratio and turbine efficiency.
#3
So if I understand you correctly, the heat energy is proportional to the pressure difference, which enables the turbine to spin. If this is true then, would the same engine with the same turbo set up require more mass airflow with a cooler exhaust temp. to produce the same amount of shaft power. Thanks for your time.
#4
Edit: I reread your post. The heat energy that is turned into shaft power is proportional to the temperature difference. But yeah, you would need more massflow with lower exhaust temp. It gets complicated if you try to think of it that way because on a given engine you can't change massflow, temperature, or pressure ratio without affecting the other two.
Last edited by P Mack; 02-09-2006 at 06:24 PM.
#5
Wt= m * cp * (Tin-Tout)
Where
Wt = Time averaged turbine power
m = time averaged exhaust MASS flowrate
cp= specific heat (kJ/kg-K) @
273K = 1.004
300K = 1.005
500K = 1.029
850K = 1.108
Tin= exhaust temp into the turbine
Tout= exhaust temp out of the turbine
"Turbochargers should be mounted as close as possible to the cylinder exhaust ports so that turbine inlet pressure, temperature, and kinetic engergy can be as high as possible" - William pulkrabek
This is why us engineering types dont like the STS system, its inefficent. Can it make power? yes.. but it looses tons of power through heatloss. At that point, just run a roots blower.. it might actually be more efficient overall..
Where
Wt = Time averaged turbine power
m = time averaged exhaust MASS flowrate
cp= specific heat (kJ/kg-K) @
273K = 1.004
300K = 1.005
500K = 1.029
850K = 1.108
Tin= exhaust temp into the turbine
Tout= exhaust temp out of the turbine
"Turbochargers should be mounted as close as possible to the cylinder exhaust ports so that turbine inlet pressure, temperature, and kinetic engergy can be as high as possible" - William pulkrabek
This is why us engineering types dont like the STS system, its inefficent. Can it make power? yes.. but it looses tons of power through heatloss. At that point, just run a roots blower.. it might actually be more efficient overall..
#6
Ok this is the one thing that I do know, thermodynamics.
Of course the more mass flow you have the more work you can get, but that’s the easy part.
so... ho=CpTo, (all of the o's are a way to express total conditions as apposed to stagnate conditions)
ho is total entropy and To is total temp, To=T+1/2V^2
delta ho is the total work that is extracted from a turbine; therefore you want a big value for delta ho.
delta ho=ho2-ho1
ho2-ho1=Cp(To2-To1)
ho2-ho1=Cp((T2+1/2v2^2)-(T1+1/2v1^2))
sorry about not having subscripts or superscripts to work with. But as you can see the energy that can be extracted out of the gas is dependent on
1. Temp difference and
2. Velocity difference between the entrance and exit of the turbine.
for a complete picture look up some h-s diagrams of a turbine where you can see the constant pressure lines
*All this is treating Cp as a constant; if the temp change is more then about 100 degrees Kelvin then it’s not good to assume that Cp is constant.
I hope that this clears everything up, I’d be happy to answer any question just let me know.
Sorry if is misspelled something I'm a Aerospace Engineer not an english major
Of course the more mass flow you have the more work you can get, but that’s the easy part.
so... ho=CpTo, (all of the o's are a way to express total conditions as apposed to stagnate conditions)
ho is total entropy and To is total temp, To=T+1/2V^2
delta ho is the total work that is extracted from a turbine; therefore you want a big value for delta ho.
delta ho=ho2-ho1
ho2-ho1=Cp(To2-To1)
ho2-ho1=Cp((T2+1/2v2^2)-(T1+1/2v1^2))
sorry about not having subscripts or superscripts to work with. But as you can see the energy that can be extracted out of the gas is dependent on
1. Temp difference and
2. Velocity difference between the entrance and exit of the turbine.
for a complete picture look up some h-s diagrams of a turbine where you can see the constant pressure lines
*All this is treating Cp as a constant; if the temp change is more then about 100 degrees Kelvin then it’s not good to assume that Cp is constant.
I hope that this clears everything up, I’d be happy to answer any question just let me know.
Sorry if is misspelled something I'm a Aerospace Engineer not an english major
#7
Originally Posted by yourtaxes=myGTO
Ok this is the one thing that I do know, thermodynamics.
Of course the more mass flow you have the more work you can get, but that’s the easy part.
so... ho=CpTo, (all of the o's are a way to express total conditions as apposed to stagnate conditions)
ho is total entropy and To is total temp, To=T+1/2V^2
delta ho is the total work that is extracted from a turbine; therefore you want a big value for delta ho.
delta ho=ho2-ho1
ho2-ho1=Cp(To2-To1)
ho2-ho1=Cp((T2+1/2v2^2)-(T1+1/2v1^2))
sorry about not having subscripts or superscripts to work with. But as you can see the energy that can be extracted out of the gas is dependent on
1. Temp difference and
2. Velocity difference between the entrance and exit of the turbine.
for a complete picture look up some h-s diagrams of a turbine where you can see the constant pressure lines
*All this is treating Cp as a constant; if the temp change is more then about 100 degrees Kelvin then it’s not good to assume that Cp is constant.
I hope that this clears everything up, I’d be happy to answer any question just let me know.
Sorry if is misspelled something I'm a Aerospace Engineer not an english major
Of course the more mass flow you have the more work you can get, but that’s the easy part.
so... ho=CpTo, (all of the o's are a way to express total conditions as apposed to stagnate conditions)
ho is total entropy and To is total temp, To=T+1/2V^2
delta ho is the total work that is extracted from a turbine; therefore you want a big value for delta ho.
delta ho=ho2-ho1
ho2-ho1=Cp(To2-To1)
ho2-ho1=Cp((T2+1/2v2^2)-(T1+1/2v1^2))
sorry about not having subscripts or superscripts to work with. But as you can see the energy that can be extracted out of the gas is dependent on
1. Temp difference and
2. Velocity difference between the entrance and exit of the turbine.
for a complete picture look up some h-s diagrams of a turbine where you can see the constant pressure lines
*All this is treating Cp as a constant; if the temp change is more then about 100 degrees Kelvin then it’s not good to assume that Cp is constant.
I hope that this clears everything up, I’d be happy to answer any question just let me know.
Sorry if is misspelled something I'm a Aerospace Engineer not an english major
Anyone want to take a stab at values to shoot for (both F/I and N/A)?
Trending Topics
#9
Originally Posted by yourtaxes=myGTO
Ok this is the one thing that I do know, thermodynamics.
Of course the more mass flow you have the more work you can get, but that’s the easy part.
so... ho=CpTo, (all of the o's are a way to express total conditions as apposed to stagnate conditions)
ho is total entropy and To is total temp, To=T+1/2V^2
delta ho is the total work that is extracted from a turbine; therefore you want a big value for delta ho.
delta ho=ho2-ho1
ho2-ho1=Cp(To2-To1)
ho2-ho1=Cp((T2+1/2v2^2)-(T1+1/2v1^2))
sorry about not having subscripts or superscripts to work with. But as you can see the energy that can be extracted out of the gas is dependent on
1. Temp difference and
2. Velocity difference between the entrance and exit of the turbine.
for a complete picture look up some h-s diagrams of a turbine where you can see the constant pressure lines
*All this is treating Cp as a constant; if the temp change is more then about 100 degrees Kelvin then it’s not good to assume that Cp is constant.
I hope that this clears everything up, I’d be happy to answer any question just let me know.
Sorry if is misspelled something I'm a Aerospace Engineer not an english major
Of course the more mass flow you have the more work you can get, but that’s the easy part.
so... ho=CpTo, (all of the o's are a way to express total conditions as apposed to stagnate conditions)
ho is total entropy and To is total temp, To=T+1/2V^2
delta ho is the total work that is extracted from a turbine; therefore you want a big value for delta ho.
delta ho=ho2-ho1
ho2-ho1=Cp(To2-To1)
ho2-ho1=Cp((T2+1/2v2^2)-(T1+1/2v1^2))
sorry about not having subscripts or superscripts to work with. But as you can see the energy that can be extracted out of the gas is dependent on
1. Temp difference and
2. Velocity difference between the entrance and exit of the turbine.
for a complete picture look up some h-s diagrams of a turbine where you can see the constant pressure lines
*All this is treating Cp as a constant; if the temp change is more then about 100 degrees Kelvin then it’s not good to assume that Cp is constant.
I hope that this clears everything up, I’d be happy to answer any question just let me know.
Sorry if is misspelled something I'm a Aerospace Engineer not an english major
You might have lost a person or two!
#10
In laymans terms (corret me if I'm wrong, its been about 4 years since thermo for me):
Heat is really just the ammount of energy in a bunch of particles. In a pile of hot air, the air molecules are moving around and such a lot more than in a pile of cold air. They have more kinetic energy.
So a pile of hot air will produce more shaft power than a pile of cold air, assuming that the volume of the system is fixed (which it is in the case of our exsaust headers) and that both "piles" have a relatively similarly amount of mass. The hot pile naturally wants to get to a cooler place more than the cool pile (1st law of thermo right?), and so it slams into the exsaust a lot with all its energy as it runs for the tailpipe.
Now in a car, your basicly going to controll egt by the amount of fuel you put in. Spark makes a big diff of course, but everyone I know goes with as much as they can as soon as they can without causing knock. Its almost a non-vairable for theoretical stuff I think.
So if you are running really lean, your going to get higher egts, and thus more exsaust energy. And so more shaft power. I love saying that.
Now what I dont understand is how some people actually run their tunes rich on purpose at low rpms to get a lower boost threshold. Is that maybe due to the mass of the fuel being added to whats being thrown at the turbine? So it violates my assumption about masses of the two piles being roughly equal?
Heat is really just the ammount of energy in a bunch of particles. In a pile of hot air, the air molecules are moving around and such a lot more than in a pile of cold air. They have more kinetic energy.
So a pile of hot air will produce more shaft power than a pile of cold air, assuming that the volume of the system is fixed (which it is in the case of our exsaust headers) and that both "piles" have a relatively similarly amount of mass. The hot pile naturally wants to get to a cooler place more than the cool pile (1st law of thermo right?), and so it slams into the exsaust a lot with all its energy as it runs for the tailpipe.
Now in a car, your basicly going to controll egt by the amount of fuel you put in. Spark makes a big diff of course, but everyone I know goes with as much as they can as soon as they can without causing knock. Its almost a non-vairable for theoretical stuff I think.
So if you are running really lean, your going to get higher egts, and thus more exsaust energy. And so more shaft power. I love saying that.
Now what I dont understand is how some people actually run their tunes rich on purpose at low rpms to get a lower boost threshold. Is that maybe due to the mass of the fuel being added to whats being thrown at the turbine? So it violates my assumption about masses of the two piles being roughly equal?
#11
If you want the turbo closest to the exhaust port as possible for the hottest air, and you want the change in the temperature between the air on the other side of the turbine and the gas coming out of the exhaust port to be as large as possible, then... in theory... wouldn't you want the intercooler as close as possible to the turbo? It just seems to me that since the main production center of the cooler air is the intercooler, the closer it is to the turbo the cooler the air will be on that side of it, thus creating a larger difference in temperatures between it and the exhaust coming out of the combustion chamber.
Does that make any sense? It makes sense to me now, but it may not 5 minutes from now.
Does that make any sense? It makes sense to me now, but it may not 5 minutes from now.
#12
Originally Posted by Sparetire
The hot pile naturally wants to get to a cooler place more than the cool pile (1st law of thermo right?)
#13
Originally Posted by ArcticZ28
If you want the turbo closest to the exhaust port as possible for the hottest air, and you want the change in the temperature between the air on the other side of the turbine and the gas coming out of the exhaust port to be as large as possible, then... in theory... wouldn't you want the intercooler as close as possible to the turbo? It just seems to me that since the main production center of the cooler air is the intercooler, the closer it is to the turbo the cooler the air will be on that side of it, thus creating a larger difference in temperatures between it and the exhaust coming out of the combustion chamber.
Does that make any sense? It makes sense to me now, but it may not 5 minutes from now.
Does that make any sense? It makes sense to me now, but it may not 5 minutes from now.
#14
Originally Posted by yourtaxes=myGTO
Ok this is the one thing that I do know, thermodynamics.
Of course the more mass flow you have the more work you can get, but that’s the easy part.
so... ho=CpTo, (all of the o's are a way to express total conditions as apposed to stagnate conditions)
ho is total entropy and To is total temp, To=T+1/2V^2
delta ho is the total work that is extracted from a turbine; therefore you want a big value for delta ho.
delta ho=ho2-ho1
ho2-ho1=Cp(To2-To1)
ho2-ho1=Cp((T2+1/2v2^2)-(T1+1/2v1^2))
sorry about not having subscripts or superscripts to work with. But as you can see the energy that can be extracted out of the gas is dependent on
1. Temp difference and
2. Velocity difference between the entrance and exit of the turbine.
for a complete picture look up some h-s diagrams of a turbine where you can see the constant pressure lines
*All this is treating Cp as a constant; if the temp change is more then about 100 degrees Kelvin then it’s not good to assume that Cp is constant.
I hope that this clears everything up, I’d be happy to answer any question just let me know.
Sorry if is misspelled something I'm a Aerospace Engineer not an english major
Of course the more mass flow you have the more work you can get, but that’s the easy part.
so... ho=CpTo, (all of the o's are a way to express total conditions as apposed to stagnate conditions)
ho is total entropy and To is total temp, To=T+1/2V^2
delta ho is the total work that is extracted from a turbine; therefore you want a big value for delta ho.
delta ho=ho2-ho1
ho2-ho1=Cp(To2-To1)
ho2-ho1=Cp((T2+1/2v2^2)-(T1+1/2v1^2))
sorry about not having subscripts or superscripts to work with. But as you can see the energy that can be extracted out of the gas is dependent on
1. Temp difference and
2. Velocity difference between the entrance and exit of the turbine.
for a complete picture look up some h-s diagrams of a turbine where you can see the constant pressure lines
*All this is treating Cp as a constant; if the temp change is more then about 100 degrees Kelvin then it’s not good to assume that Cp is constant.
I hope that this clears everything up, I’d be happy to answer any question just let me know.
Sorry if is misspelled something I'm a Aerospace Engineer not an english major
#15
Senior BSME student here...you guys did a pretty good job of explaining in terms people can understand...but until you take a bunch of thermo, heat transfer, fluid dynamics, etc, it probably won't make perfect sense to the average guy. But worth a shot nonetheless.
#16
Originally Posted by LTLHOMER
Senior BSME student here...you guys did a pretty good job of explaining in terms people can understand...but until you take a bunch of thermo, heat transfer, fluid dynamics, etc, it probably won't make perfect sense to the average guy. But worth a shot nonetheless.
#17
Let me see if I am understanding this right.
Hotter air on the inlet side of the turbo=more pressure (basic gas law)
More pressure means the air is trying to reach equilbrium faster thus it pushes the turbo with more force.
Hotter air on the inlet side of the turbo=more pressure (basic gas law)
More pressure means the air is trying to reach equilbrium faster thus it pushes the turbo with more force.
#18
Not exactly, you're missing a subtle point. Even with the same pressure, hotter air will spin the turbo harder.
Also hotter air does not necessarily equal more pressure. If you're flowing air through a pipe at a certain pressure at point A and add heat at point B, the pressure will be lower at point B.
Also hotter air does not necessarily equal more pressure. If you're flowing air through a pipe at a certain pressure at point A and add heat at point B, the pressure will be lower at point B.
#19
Originally Posted by Sparetire
In laymans terms (corret me if I'm wrong, its been about 4 years since thermo for me):
Heat is really just the ammount of energy in a bunch of particles. In a pile of hot air, the air molecules are moving around and such a lot more than in a pile of cold air. They have more kinetic energy.
So a pile of hot air will produce more shaft power than a pile of cold air, assuming that the volume of the system is fixed (which it is in the case of our exsaust headers) and that both "piles" have a relatively similarly amount of mass. The hot pile naturally wants to get to a cooler place more than the cool pile (1st law of thermo right?), and so it slams into the exsaust a lot with all its energy as it runs for the tailpipe.
Now in a car, your basicly going to controll egt by the amount of fuel you put in. Spark makes a big diff of course, but everyone I know goes with as much as they can as soon as they can without causing knock. Its almost a non-vairable for theoretical stuff I think.
So if you are running really lean, your going to get higher egts, and thus more exsaust energy. And so more shaft power. I love saying that.
Now what I dont understand is how some people actually run their tunes rich on purpose at low rpms to get a lower boost threshold. Is that maybe due to the mass of the fuel being added to whats being thrown at the turbine? So it violates my assumption about masses of the two piles being roughly equal?
Heat is really just the ammount of energy in a bunch of particles. In a pile of hot air, the air molecules are moving around and such a lot more than in a pile of cold air. They have more kinetic energy.
So a pile of hot air will produce more shaft power than a pile of cold air, assuming that the volume of the system is fixed (which it is in the case of our exsaust headers) and that both "piles" have a relatively similarly amount of mass. The hot pile naturally wants to get to a cooler place more than the cool pile (1st law of thermo right?), and so it slams into the exsaust a lot with all its energy as it runs for the tailpipe.
Now in a car, your basicly going to controll egt by the amount of fuel you put in. Spark makes a big diff of course, but everyone I know goes with as much as they can as soon as they can without causing knock. Its almost a non-vairable for theoretical stuff I think.
So if you are running really lean, your going to get higher egts, and thus more exsaust energy. And so more shaft power. I love saying that.
Now what I dont understand is how some people actually run their tunes rich on purpose at low rpms to get a lower boost threshold. Is that maybe due to the mass of the fuel being added to whats being thrown at the turbine? So it violates my assumption about masses of the two piles being roughly equal?
#20
Originally Posted by studentdriver
I think that the reduced spool time with the rich afrs is due to combustion of the excess fuel continuing as the exhaust mixture passes through the headers, but I could be wrong.