Supercharger does it affect gas?
#2
CTS-V Import Slayer
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Depends on how you drive it as with any other mod. Punch it a lot...your milage will suffer. If you are asking if you punch it a lot will it burn more fuel than if you had punched it without the supercharger....yes it will use more fuel. Drive it normal and I am sure that you will find the milage loss is minimal. There will be a slight loss either way though....magnified by the amt of time on the accelerator.
#3
Instructor
The supercharger will cost you some milage just because its attached to the engine. It drains horsepower all the time and only is a net plus when you step on it. Contact Roadrage on this forum and he can most likly tell you exactly how much.
#5
Yes and no...A supercharger does require power from the engine to turn the blower. This is true. However most positive displacement superchargers have by-pass valves which reduce the effort of the blower at part throttle. An Eaton hybrid roots blower doesn't compress air very well in it's housing, but rather paddles it from inlet to outlet. So there isn't much compression occuring when you are off the throttle and whatever is being moved is being by-passed anyway.
So it's a matter of how much energy is required to to turn the supercharger at part throttle conditions. Now when you go WOT, all things go out the window. First off all, the Air Fuel ratio gets rich quick (or it better). The extra fuel is required to match the extra air flow and even more fuel is used to cool the air charge mixture and combustion chambers.
How much does the blower change fuel economy at steady cruise speeds? Hard to say, but I doubt it will be more than 1 mpg. On my '95 Eaton supercharged 4.6L V-8 T-Bird, I noticed MORE fuel economy loss when switching to a colder thermostat then when adding the blower.
After all, the hotter an engine runs closer to its adiabatic efficiency, the more power it makes. Run an engine colder than adiabatic and fuel economy will suffer.
A-Train
So it's a matter of how much energy is required to to turn the supercharger at part throttle conditions. Now when you go WOT, all things go out the window. First off all, the Air Fuel ratio gets rich quick (or it better). The extra fuel is required to match the extra air flow and even more fuel is used to cool the air charge mixture and combustion chambers.
How much does the blower change fuel economy at steady cruise speeds? Hard to say, but I doubt it will be more than 1 mpg. On my '95 Eaton supercharged 4.6L V-8 T-Bird, I noticed MORE fuel economy loss when switching to a colder thermostat then when adding the blower.
After all, the hotter an engine runs closer to its adiabatic efficiency, the more power it makes. Run an engine colder than adiabatic and fuel economy will suffer.
A-Train
#6
tehLEGOman
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Lets just say i'm making more visits to the gas station with the supercharger installed.
Averaging 21 mpg in the city.
I'm about to travel 500 miles to Florida..i'll let you know the average highway mpg after the trip.
Averaging 21 mpg in the city.
I'm about to travel 500 miles to Florida..i'll let you know the average highway mpg after the trip.
#7
tehLEGOman
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Heres the return trip computer readout.
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#10
tehLEGOman
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It was reading 35 MPG until i had a little encounter with a bike.
What are people gettin on the naturally aspirated TLs?
What are people gettin on the naturally aspirated TLs?
#13
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Originally Posted by becker800
I usually average 24 mpg here in PA.
I did average 29 mpg on a trip to NY in the summer/fall time.
Winter Ive been only averaging 23 about
I did average 29 mpg on a trip to NY in the summer/fall time.
Winter Ive been only averaging 23 about
I find that it will be up around 28-29, but if I run to the store or somewhere with a lot of stops and starts, It'll drop quick! Definitely my love of the upper rev range driving this!
#14
WTF? When our TL was brand spanking new we took it to Cape May and drove 70 mph with the cruise control on, the best we saw was 29 mpg. How is it that you're getting 34 AVERAGE with a blower being dragged on the accessories?
Unless you had to upgrade the duel injectors with the supercharger. Then that makes sense. The ECM takes the injector pulse width over time and calculates fuel economy. If you swapped to a larger injector then the PCM calculates the fuel economy incorrectly.
A-Train
Unless you had to upgrade the duel injectors with the supercharger. Then that makes sense. The ECM takes the injector pulse width over time and calculates fuel economy. If you swapped to a larger injector then the PCM calculates the fuel economy incorrectly.
A-Train
#16
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Originally Posted by Ol Blue Eyes
does anybody else get better gas mileage going 80mph vs. 70mph.
sorry i'm jacking the thread for a sec
sorry i'm jacking the thread for a sec
BTW, mine is a 5AT
#17
I'm actually getting good mileage in my 5AT with the blower. 23 in mixed highway and city driving (and that is me beating on it fairly often). Highway only is in the 26-27 area. So I've lost only a few MPG with the blower and I would attribute that to flogging it more...not the blower itself.
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Originally Posted by Atrain
Unless you had to upgrade the duel injectors with the supercharger. Then that makes sense. The ECM takes the injector pulse width over time and calculates fuel economy. If you swapped to a larger injector then the PCM calculates the fuel economy incorrectly.
#19
This is scary...
Wid-TL,
Speaking of the Ford trucks in the Mythbusters episode:
The Ford EEC-V computers calculate fuel delivery like this.
The EEC-V knows the injector size, Mass Air Flow transfer function and desired Air Fuel ratio.
The Mass Air meter (MAF) is attached to the airbox and houses two resistors that are exposed to the incoming airflow. The EEC-V applies voltage to the resistor and it heats up. The incoming air cools the resistor off and the EEC-V applies more voltage to try and maintain the temperature. This will continue on and on until the volatage reaches 5V. Once the voltage reaches 5V, you have pegged the MAF. This hard to do and requires VE (volumetric efficiency) of 100% or more.
The EEC-V then knows the cfm or airflow on the inlet side. It takes theoretical airflow and divides it by actual airflow and gets VE. Now the sick part.
The EEC is looking for an Air Fuel ratio which is called COMMANDED A/F. Let's say that at WOT (wide open throttle) the EEC wants an A/F ratio of 12.0:1. In case anyone is lost here. A/F ratio is the ratio of air to fuel. The chemically correct formula for gasoline is 14.64:1 so you want to run an A/F ratio of 14.64:1 (some say 14.7:1, that's fine). That means 14.64 parts air to 1 part fuel. This is called Stoich.
Now you have a reading from the MAF to the EEC and it takes the voltage and applies it to the transfer function and gets lb./min of airflow. Let's say this truck with a 4.6L V-8 will see 25 lb./min of airflow at WOT.
The EEC-V takes 25 lb/min. and divides it by 12.0:1 A/F ratio and gets 2.083 lb./min of fuel required. Fuel Injectors for Fords are rated lb./hr. So we need to multiply by 60 to get lb./hr.
25 divided by 12 = 2.0833 lb./min.
2.083 X 60 = 124.9 lb./hr. of fuel required
At this point we have the total fuel required to maintain the 12.0:1 A/F ratio at WOT. So how much is needed at each fuel injector? The Ford trucks with the V-8 have 8 fuel injectors so we divide by 8.
124.9 divided by 8 fuel injectors = 15.6 lb./hr. of fuel required at each fuel injector
The stock Ford 4.6L SOHC 2V fuel injector is 19 lb./hr. at 39.5 psi. 15.6 lb./hr. into 19 lb./hr. = 0.82% duty cycle.
This how the EEC-IV and EEC-V calculate fuel delivery, trust me. So the EEC takes the duty cycle over time to calculate average fuel economy. If you changed fuel injectors and did not reflash the EEC, but rather cheated through a MAF upgrade or fmu change then the calculated fuel economy is wrong.
Just to set the record straight I have no idea how the TL calculates the average fuel economy. I also thought about how he got 34 mpg on the screen. Running at 70 mph downhill in nuetral can do wonderful things for fuel economy.
A-Train
Speaking of the Ford trucks in the Mythbusters episode:
The Ford EEC-V computers calculate fuel delivery like this.
The EEC-V knows the injector size, Mass Air Flow transfer function and desired Air Fuel ratio.
The Mass Air meter (MAF) is attached to the airbox and houses two resistors that are exposed to the incoming airflow. The EEC-V applies voltage to the resistor and it heats up. The incoming air cools the resistor off and the EEC-V applies more voltage to try and maintain the temperature. This will continue on and on until the volatage reaches 5V. Once the voltage reaches 5V, you have pegged the MAF. This hard to do and requires VE (volumetric efficiency) of 100% or more.
The EEC-V then knows the cfm or airflow on the inlet side. It takes theoretical airflow and divides it by actual airflow and gets VE. Now the sick part.
The EEC is looking for an Air Fuel ratio which is called COMMANDED A/F. Let's say that at WOT (wide open throttle) the EEC wants an A/F ratio of 12.0:1. In case anyone is lost here. A/F ratio is the ratio of air to fuel. The chemically correct formula for gasoline is 14.64:1 so you want to run an A/F ratio of 14.64:1 (some say 14.7:1, that's fine). That means 14.64 parts air to 1 part fuel. This is called Stoich.
Now you have a reading from the MAF to the EEC and it takes the voltage and applies it to the transfer function and gets lb./min of airflow. Let's say this truck with a 4.6L V-8 will see 25 lb./min of airflow at WOT.
The EEC-V takes 25 lb/min. and divides it by 12.0:1 A/F ratio and gets 2.083 lb./min of fuel required. Fuel Injectors for Fords are rated lb./hr. So we need to multiply by 60 to get lb./hr.
25 divided by 12 = 2.0833 lb./min.
2.083 X 60 = 124.9 lb./hr. of fuel required
At this point we have the total fuel required to maintain the 12.0:1 A/F ratio at WOT. So how much is needed at each fuel injector? The Ford trucks with the V-8 have 8 fuel injectors so we divide by 8.
124.9 divided by 8 fuel injectors = 15.6 lb./hr. of fuel required at each fuel injector
The stock Ford 4.6L SOHC 2V fuel injector is 19 lb./hr. at 39.5 psi. 15.6 lb./hr. into 19 lb./hr. = 0.82% duty cycle.
This how the EEC-IV and EEC-V calculate fuel delivery, trust me. So the EEC takes the duty cycle over time to calculate average fuel economy. If you changed fuel injectors and did not reflash the EEC, but rather cheated through a MAF upgrade or fmu change then the calculated fuel economy is wrong.
Just to set the record straight I have no idea how the TL calculates the average fuel economy. I also thought about how he got 34 mpg on the screen. Running at 70 mph downhill in nuetral can do wonderful things for fuel economy.
A-Train
#20
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Originally Posted by Atrain
Wid-TL,
Speaking of the Ford trucks in the Mythbusters episode:
The Ford EEC-V computers calculate fuel delivery like this.
The EEC-V knows the injector size, Mass Air Flow transfer function and desired Air Fuel ratio.
The Mass Air meter (MAF) is attached to the airbox and houses two resistors that are exposed to the incoming airflow. The EEC-V applies voltage to the resistor and it heats up. The incoming air cools the resistor off and the EEC-V applies more voltage to try and maintain the temperature. This will continue on and on until the volatage reaches 5V. Once the voltage reaches 5V, you have pegged the MAF. This hard to do and requires VE (volumetric efficiency) of 100% or more.
The EEC-V then knows the cfm or airflow on the inlet side. It takes theoretical airflow and divides it by actual airflow and gets VE. Now the sick part.
The EEC is looking for an Air Fuel ratio which is called COMMANDED A/F. Let's say that at WOT (wide open throttle) the EEC wants an A/F ratio of 12.0:1. In case anyone is lost here. A/F ratio is the ratio of air to fuel. The chemically correct formula for gasoline is 14.64:1 so you want to run an A/F ratio of 14.64:1 (some say 14.7:1, that's fine). That means 14.64 parts air to 1 part fuel. This is called Stoich.
Now you have a reading from the MAF to the EEC and it takes the voltage and applies it to the transfer function and gets lb./min of airflow. Let's say this truck with a 4.6L V-8 will see 25 lb./min of airflow at WOT.
The EEC-V takes 25 lb/min. and divides it by 12.0:1 A/F ratio and gets 2.083 lb./min of fuel required. Fuel Injectors for Fords are rated lb./hr. So we need to multiply by 60 to get lb./hr.
25 divided by 12 = 2.0833 lb./min.
2.083 X 60 = 124.9 lb./hr. of fuel required
At this point we have the total fuel required to maintain the 12.0:1 A/F ratio at WOT. So how much is needed at each fuel injector? The Ford trucks with the V-8 have 8 fuel injectors so we divide by 8.
124.9 divided by 8 fuel injectors = 15.6 lb./hr. of fuel required at each fuel injector
The stock Ford 4.6L SOHC 2V fuel injector is 19 lb./hr. at 39.5 psi. 15.6 lb./hr. into 19 lb./hr. = 0.82% duty cycle.
This how the EEC-IV and EEC-V calculate fuel delivery, trust me. So the EEC takes the duty cycle over time to calculate average fuel economy. If you changed fuel injectors and did not reflash the EEC, but rather cheated through a MAF upgrade or fmu change then the calculated fuel economy is wrong.
Just to set the record straight I have no idea how the TL calculates the average fuel economy. I also thought about how he got 34 mpg on the screen. Running at 70 mph downhill in nuetral can do wonderful things for fuel economy.
A-Train
Speaking of the Ford trucks in the Mythbusters episode:
The Ford EEC-V computers calculate fuel delivery like this.
The EEC-V knows the injector size, Mass Air Flow transfer function and desired Air Fuel ratio.
The Mass Air meter (MAF) is attached to the airbox and houses two resistors that are exposed to the incoming airflow. The EEC-V applies voltage to the resistor and it heats up. The incoming air cools the resistor off and the EEC-V applies more voltage to try and maintain the temperature. This will continue on and on until the volatage reaches 5V. Once the voltage reaches 5V, you have pegged the MAF. This hard to do and requires VE (volumetric efficiency) of 100% or more.
The EEC-V then knows the cfm or airflow on the inlet side. It takes theoretical airflow and divides it by actual airflow and gets VE. Now the sick part.
The EEC is looking for an Air Fuel ratio which is called COMMANDED A/F. Let's say that at WOT (wide open throttle) the EEC wants an A/F ratio of 12.0:1. In case anyone is lost here. A/F ratio is the ratio of air to fuel. The chemically correct formula for gasoline is 14.64:1 so you want to run an A/F ratio of 14.64:1 (some say 14.7:1, that's fine). That means 14.64 parts air to 1 part fuel. This is called Stoich.
Now you have a reading from the MAF to the EEC and it takes the voltage and applies it to the transfer function and gets lb./min of airflow. Let's say this truck with a 4.6L V-8 will see 25 lb./min of airflow at WOT.
The EEC-V takes 25 lb/min. and divides it by 12.0:1 A/F ratio and gets 2.083 lb./min of fuel required. Fuel Injectors for Fords are rated lb./hr. So we need to multiply by 60 to get lb./hr.
25 divided by 12 = 2.0833 lb./min.
2.083 X 60 = 124.9 lb./hr. of fuel required
At this point we have the total fuel required to maintain the 12.0:1 A/F ratio at WOT. So how much is needed at each fuel injector? The Ford trucks with the V-8 have 8 fuel injectors so we divide by 8.
124.9 divided by 8 fuel injectors = 15.6 lb./hr. of fuel required at each fuel injector
The stock Ford 4.6L SOHC 2V fuel injector is 19 lb./hr. at 39.5 psi. 15.6 lb./hr. into 19 lb./hr. = 0.82% duty cycle.
This how the EEC-IV and EEC-V calculate fuel delivery, trust me. So the EEC takes the duty cycle over time to calculate average fuel economy. If you changed fuel injectors and did not reflash the EEC, but rather cheated through a MAF upgrade or fmu change then the calculated fuel economy is wrong.
Just to set the record straight I have no idea how the TL calculates the average fuel economy. I also thought about how he got 34 mpg on the screen. Running at 70 mph downhill in nuetral can do wonderful things for fuel economy.
A-Train
Just kidding, that's some incredible info in your post. Thanks for taking the time to write it up!!
#21
tehLEGOman
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I'm running a single outlet exhaust setup.
I should have taken the picture while it was reading 35.
I should have taken the picture while it was reading 35.
#22
tehLEGOman
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Originally Posted by Atrain
Wid-TL,
Speaking of the Ford trucks in the Mythbusters episode:
The Ford EEC-V computers calculate fuel delivery like this.
The EEC-V knows the injector size, Mass Air Flow transfer function and desired Air Fuel ratio.
The Mass Air meter (MAF) is attached to the airbox and houses two resistors that are exposed to the incoming airflow. The EEC-V applies voltage to the resistor and it heats up. The incoming air cools the resistor off and the EEC-V applies more voltage to try and maintain the temperature. This will continue on and on until the volatage reaches 5V. Once the voltage reaches 5V, you have pegged the MAF. This hard to do and requires VE (volumetric efficiency) of 100% or more.
The EEC-V then knows the cfm or airflow on the inlet side. It takes theoretical airflow and divides it by actual airflow and gets VE. Now the sick part.
The EEC is looking for an Air Fuel ratio which is called COMMANDED A/F. Let's say that at WOT (wide open throttle) the EEC wants an A/F ratio of 12.0:1. In case anyone is lost here. A/F ratio is the ratio of air to fuel. The chemically correct formula for gasoline is 14.64:1 so you want to run an A/F ratio of 14.64:1 (some say 14.7:1, that's fine). That means 14.64 parts air to 1 part fuel. This is called Stoich.
Now you have a reading from the MAF to the EEC and it takes the voltage and applies it to the transfer function and gets lb./min of airflow. Let's say this truck with a 4.6L V-8 will see 25 lb./min of airflow at WOT.
The EEC-V takes 25 lb/min. and divides it by 12.0:1 A/F ratio and gets 2.083 lb./min of fuel required. Fuel Injectors for Fords are rated lb./hr. So we need to multiply by 60 to get lb./hr.
25 divided by 12 = 2.0833 lb./min.
2.083 X 60 = 124.9 lb./hr. of fuel required
At this point we have the total fuel required to maintain the 12.0:1 A/F ratio at WOT. So how much is needed at each fuel injector? The Ford trucks with the V-8 have 8 fuel injectors so we divide by 8.
124.9 divided by 8 fuel injectors = 15.6 lb./hr. of fuel required at each fuel injector
The stock Ford 4.6L SOHC 2V fuel injector is 19 lb./hr. at 39.5 psi. 15.6 lb./hr. into 19 lb./hr. = 0.82% duty cycle.
This how the EEC-IV and EEC-V calculate fuel delivery, trust me. So the EEC takes the duty cycle over time to calculate average fuel economy. If you changed fuel injectors and did not reflash the EEC, but rather cheated through a MAF upgrade or fmu change then the calculated fuel economy is wrong.
Just to set the record straight I have no idea how the TL calculates the average fuel economy. I also thought about how he got 34 mpg on the screen. Running at 70 mph downhill in nuetral can do wonderful things for fuel economy.
A-Train
Speaking of the Ford trucks in the Mythbusters episode:
The Ford EEC-V computers calculate fuel delivery like this.
The EEC-V knows the injector size, Mass Air Flow transfer function and desired Air Fuel ratio.
The Mass Air meter (MAF) is attached to the airbox and houses two resistors that are exposed to the incoming airflow. The EEC-V applies voltage to the resistor and it heats up. The incoming air cools the resistor off and the EEC-V applies more voltage to try and maintain the temperature. This will continue on and on until the volatage reaches 5V. Once the voltage reaches 5V, you have pegged the MAF. This hard to do and requires VE (volumetric efficiency) of 100% or more.
The EEC-V then knows the cfm or airflow on the inlet side. It takes theoretical airflow and divides it by actual airflow and gets VE. Now the sick part.
The EEC is looking for an Air Fuel ratio which is called COMMANDED A/F. Let's say that at WOT (wide open throttle) the EEC wants an A/F ratio of 12.0:1. In case anyone is lost here. A/F ratio is the ratio of air to fuel. The chemically correct formula for gasoline is 14.64:1 so you want to run an A/F ratio of 14.64:1 (some say 14.7:1, that's fine). That means 14.64 parts air to 1 part fuel. This is called Stoich.
Now you have a reading from the MAF to the EEC and it takes the voltage and applies it to the transfer function and gets lb./min of airflow. Let's say this truck with a 4.6L V-8 will see 25 lb./min of airflow at WOT.
The EEC-V takes 25 lb/min. and divides it by 12.0:1 A/F ratio and gets 2.083 lb./min of fuel required. Fuel Injectors for Fords are rated lb./hr. So we need to multiply by 60 to get lb./hr.
25 divided by 12 = 2.0833 lb./min.
2.083 X 60 = 124.9 lb./hr. of fuel required
At this point we have the total fuel required to maintain the 12.0:1 A/F ratio at WOT. So how much is needed at each fuel injector? The Ford trucks with the V-8 have 8 fuel injectors so we divide by 8.
124.9 divided by 8 fuel injectors = 15.6 lb./hr. of fuel required at each fuel injector
The stock Ford 4.6L SOHC 2V fuel injector is 19 lb./hr. at 39.5 psi. 15.6 lb./hr. into 19 lb./hr. = 0.82% duty cycle.
This how the EEC-IV and EEC-V calculate fuel delivery, trust me. So the EEC takes the duty cycle over time to calculate average fuel economy. If you changed fuel injectors and did not reflash the EEC, but rather cheated through a MAF upgrade or fmu change then the calculated fuel economy is wrong.
Just to set the record straight I have no idea how the TL calculates the average fuel economy. I also thought about how he got 34 mpg on the screen. Running at 70 mph downhill in nuetral can do wonderful things for fuel economy.
A-Train
The contour of the land in Florida is relatively flat which might have an effect on the MPG readout.
The kit did not supply upgraded fuel injectors.
The sunauto voltage regulator might be part of the reason why i was able to come up with this number.
Am I the only one getting ~34 MPG on the highway? If so this is strange.
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