Calculating Injector Pulse Width
01-13-2004, 03:13 PM
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Calculating Injector Pulse Width
Last week we had a discussion about how to figure out how long we need to turn on an injector to obatain a given A/F ratio. Not many people even attempted to answer the questions and even fewer who IM'd me actually got it right.
As EFI UNIVERSITY's continued dedication to Honda-Tech members, I would like to show everyone exactly how its done.......for FREE!
Honda-tech has always been a great resource for people to gain useful knowledge about making their cars faster, and we feel it is partially our duty to help out by sharing our hard earned knowledge about one of the most complex subjects in the performance industry... fuel injection.
So, here goes...
If we know the size of a particular injector and we also know the displacement of an engine, we can readily calculate the exact amount of time {or Pulse Width} to open an injector for any amount of volumetric efficiency and engine speed.
To do this we must first determine the Volumetric Efficiency of the engine at the time we want to calculate how much fuel to use.
Essentially this is the assumption we make when we change the numbers in a base fuel table. We are assuming that a given amount of fuel added to the mixture will create the proper air/fuel ratio based on the theory that we can work backwards from fuel flow and measured A/F ratio to get volumetric efficiency.
It goes something like this:
First, lets convert our engine displacement that is given in Cubic Inches to Cubic Feet. We do this so that we can calculate the engine airflow and display it in Cubic Feet/ Minute, or CFM.
To do this, there are 1728 cubic inches in one cubic foot, so we divide the number or cubic inches the engine displaces by 1728 to get cubic feet of engine displacement.
Example: an engine that displaces 350 cu in would displace .2025 cubic feet.
Now if we multiply the engine displacement by the number of engine “cycles” we will obtain airflow numbers in Cubic Ft/min, or CFM.
To do this we divide the engine RPM by 2 because it takes two revolutions to make one engine cycle. So if our engine is operating at 6000 RPM, it will be completing 3000 cycles per minute.
On each cycle the engine displaces .2025 cubic feet of air, so 3000 times .2025 equals…
607.5 cubic feet per minute of airflow or 607.5 CFM.
*Note: it is important to remember that we are assuming that the engine is also operating at 100% VE. If our engine did not completely fill its cylinders on each cycle, it would not displace .2025 cubic feet per hour.
To calculate the air-flow for any other VE percentage, you must multiply your cubic feet of airflow number by the VE percentage you want to represent.
Example:
.2025 Cubic feet/hour x 80% VE = .162 cubic feet/hour
For Now, lets assume 100% VE though.
Okay moving on, we now know that our engine will displace 607.5 CFM at 6000 RPM.
We know that one cubic foot of air weighs .076 lbs at standard atmospheric pressure and temperature.
So, if we multiply the CFM by .076, we will get our airflow numbers in Lbs/min.
Example:
607.5 x .076 = 46.17 lbs/hr of airflow
Remember that we are looking at the total engine displacement, so we need to divide this number by the number of cylinders to find the airflow for one cylinder so that we know how long to turn on each injector.
So 46.17 lbs/min divided by 6 cylinders would be 7.695 lbs/min for one cylinder.
Now, if we know a desired A/F ratio that we want to obtain we can just divide the lbs/hr of airflow by that number to get the amount of fuel required in lbs/min.
Example:
If we wanted a 13:1 A/F ratio, we would divide 7.695 by 13
7.695 / 13 = .592 lbs/min of fuel flow to obtain a 13:1 A/F ratio
Now let’s take a look at our injector:
We have injectors that are rated in lbs/hour of fuel flow, so we need to convert them to lbs/min. To do this we simply divide by 60 because there are 60 seconds in one minute.
Example:
An injector that flows 60 lbs/hr divided by 60 seconds would flow
1 lb/min of fuel.
Now that we know the maximum amount of fuel we can get from our injector in one minute, we can move forward.
Next, lets see what the maximum amount of time is to actually hold our injector open during the engine cycle.
We know that RPM / 2 = cycles per minute
If we divide this number again by 60 seconds, we will get cycles per second.
Example:
6000 / 2 = 3000
3000 / 60 = 50 cycles per second.
Now if we divide one second by the number of cycles per second, we will find out how much time it takes for each cycle.
Example:
1 / 50 = .02 seconds per cycle
This means that the maximum amount of time we have for one engine cycle is .02 seconds, or 20 milliseconds {20ms}
So if we held our injector open for 20ms at 6000 RPM, we would have 100% duty cycle, which is the maximum amount of fuel we could pass through the injector.
Now we can divide the required amount of fuel in lbs/min by the total lbs/min of fuel available.
Example:
We discovered earlier that we would need .592 lbs/min of fuel to obtain our 13:1 A/F ratio
And then we determined that our injector was capable of flowing a total of 1 lb/min.
So we divide:
.592 / 1 = .592
This means that we need to turn on our injector for 59.2% of the total time available or a 59.2% duty cycle.
Now we can see how much actual time this represents by multiplying the actual amount of time in 100% of the cycle by the duty cycle to get an actual injector pulse width.
Example:
20 ms x .592 = .0118 OR 11.8 ms of injector pulse width!
There we did it! We now know that a 350 cubic inch engine operating at 100% VE at 6000 RPM would require 11.8 ms from a 60 lb/hr injector to obtain a 13:1 A/F ratio!
Now just for a second, lets say we calculated all this out and gave our injector a command of 11.8 ms, but in actual engine operation our A/F ratio ended up being 12:1 instead of 13:1.
This would indicate that the engine was not actually operating at 100% VE as we assumed.
Lets see how far off we were:
Divide 13 / 12 = 1.083
That means we had 8.3% too much fuel. All things being equal, we could take 100% and subtract 8.3% to realize that our actual VE was closer to 91.7%
So we could take our 11.8ms pulse width and multiply it by .917 to obtain the correct amount of fuel for 91.7% VE
Example:
11.8 x .917 = 10.82 ms of fuel required to get us back to 13:1 A/F ratio!
Enjoy! I hope this helps some of you out there!
As EFI UNIVERSITY's continued dedication to Honda-Tech members, I would like to show everyone exactly how its done.......for FREE!
Honda-tech has always been a great resource for people to gain useful knowledge about making their cars faster, and we feel it is partially our duty to help out by sharing our hard earned knowledge about one of the most complex subjects in the performance industry... fuel injection.
So, here goes...
If we know the size of a particular injector and we also know the displacement of an engine, we can readily calculate the exact amount of time {or Pulse Width} to open an injector for any amount of volumetric efficiency and engine speed.
To do this we must first determine the Volumetric Efficiency of the engine at the time we want to calculate how much fuel to use.
Essentially this is the assumption we make when we change the numbers in a base fuel table. We are assuming that a given amount of fuel added to the mixture will create the proper air/fuel ratio based on the theory that we can work backwards from fuel flow and measured A/F ratio to get volumetric efficiency.
It goes something like this:
First, lets convert our engine displacement that is given in Cubic Inches to Cubic Feet. We do this so that we can calculate the engine airflow and display it in Cubic Feet/ Minute, or CFM.
To do this, there are 1728 cubic inches in one cubic foot, so we divide the number or cubic inches the engine displaces by 1728 to get cubic feet of engine displacement.
Example: an engine that displaces 350 cu in would displace .2025 cubic feet.
Now if we multiply the engine displacement by the number of engine “cycles” we will obtain airflow numbers in Cubic Ft/min, or CFM.
To do this we divide the engine RPM by 2 because it takes two revolutions to make one engine cycle. So if our engine is operating at 6000 RPM, it will be completing 3000 cycles per minute.
On each cycle the engine displaces .2025 cubic feet of air, so 3000 times .2025 equals…
607.5 cubic feet per minute of airflow or 607.5 CFM.
*Note: it is important to remember that we are assuming that the engine is also operating at 100% VE. If our engine did not completely fill its cylinders on each cycle, it would not displace .2025 cubic feet per hour.
To calculate the air-flow for any other VE percentage, you must multiply your cubic feet of airflow number by the VE percentage you want to represent.
Example:
.2025 Cubic feet/hour x 80% VE = .162 cubic feet/hour
For Now, lets assume 100% VE though.
Okay moving on, we now know that our engine will displace 607.5 CFM at 6000 RPM.
We know that one cubic foot of air weighs .076 lbs at standard atmospheric pressure and temperature.
So, if we multiply the CFM by .076, we will get our airflow numbers in Lbs/min.
Example:
607.5 x .076 = 46.17 lbs/hr of airflow
Remember that we are looking at the total engine displacement, so we need to divide this number by the number of cylinders to find the airflow for one cylinder so that we know how long to turn on each injector.
So 46.17 lbs/min divided by 6 cylinders would be 7.695 lbs/min for one cylinder.
Now, if we know a desired A/F ratio that we want to obtain we can just divide the lbs/hr of airflow by that number to get the amount of fuel required in lbs/min.
Example:
If we wanted a 13:1 A/F ratio, we would divide 7.695 by 13
7.695 / 13 = .592 lbs/min of fuel flow to obtain a 13:1 A/F ratio
Now let’s take a look at our injector:
We have injectors that are rated in lbs/hour of fuel flow, so we need to convert them to lbs/min. To do this we simply divide by 60 because there are 60 seconds in one minute.
Example:
An injector that flows 60 lbs/hr divided by 60 seconds would flow
1 lb/min of fuel.
Now that we know the maximum amount of fuel we can get from our injector in one minute, we can move forward.
Next, lets see what the maximum amount of time is to actually hold our injector open during the engine cycle.
We know that RPM / 2 = cycles per minute
If we divide this number again by 60 seconds, we will get cycles per second.
Example:
6000 / 2 = 3000
3000 / 60 = 50 cycles per second.
Now if we divide one second by the number of cycles per second, we will find out how much time it takes for each cycle.
Example:
1 / 50 = .02 seconds per cycle
This means that the maximum amount of time we have for one engine cycle is .02 seconds, or 20 milliseconds {20ms}
So if we held our injector open for 20ms at 6000 RPM, we would have 100% duty cycle, which is the maximum amount of fuel we could pass through the injector.
Now we can divide the required amount of fuel in lbs/min by the total lbs/min of fuel available.
Example:
We discovered earlier that we would need .592 lbs/min of fuel to obtain our 13:1 A/F ratio
And then we determined that our injector was capable of flowing a total of 1 lb/min.
So we divide:
.592 / 1 = .592
This means that we need to turn on our injector for 59.2% of the total time available or a 59.2% duty cycle.
Now we can see how much actual time this represents by multiplying the actual amount of time in 100% of the cycle by the duty cycle to get an actual injector pulse width.
Example:
20 ms x .592 = .0118 OR 11.8 ms of injector pulse width!
There we did it! We now know that a 350 cubic inch engine operating at 100% VE at 6000 RPM would require 11.8 ms from a 60 lb/hr injector to obtain a 13:1 A/F ratio!
Now just for a second, lets say we calculated all this out and gave our injector a command of 11.8 ms, but in actual engine operation our A/F ratio ended up being 12:1 instead of 13:1.
This would indicate that the engine was not actually operating at 100% VE as we assumed.
Lets see how far off we were:
Divide 13 / 12 = 1.083
That means we had 8.3% too much fuel. All things being equal, we could take 100% and subtract 8.3% to realize that our actual VE was closer to 91.7%
So we could take our 11.8ms pulse width and multiply it by .917 to obtain the correct amount of fuel for 91.7% VE
Example:
11.8 x .917 = 10.82 ms of fuel required to get us back to 13:1 A/F ratio!
Enjoy! I hope this helps some of you out there!
01-13-2004, 03:17 PM
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Re: Calculating Injector Pulse Width (EFIGUY)
see, this is exactly the type of reasons I am going to take the ef101 class. everything is explain very nicely.
way to go ben
way to go ben
01-13-2004, 04:03 PM
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Re: (EFIGUY)
Now take all those calculations and impliment them into an Excel Spreadsheet and plot Duty Cycle vs RPM and IPW vs. RPM and that would be gravy. 
Good explanation for sizing injectors. I've typically sized them for targeted HP applications using a neat little program offered by RC Engineering. It works fairly well http://www.rceng.com/technical.htm .
Their is also a neat little chart to help for converting from IPW to DC and vice versa here: http://www.ecanfix.com/~mdhamilton/dutycycle1.html
Good explanation for sizing injectors. I've typically sized them for targeted HP applications using a neat little program offered by RC Engineering. It works fairly well http://www.rceng.com/technical.htm .
Their is also a neat little chart to help for converting from IPW to DC and vice versa here: http://www.ecanfix.com/~mdhamilton/dutycycle1.html
01-13-2004, 04:07 PM
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Re: Calculating Injector Pulse Width (RacerStev)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by RacerStev »</TD></TR><TR><TD CLASS="quote">What happens when we add boost? At 2 bar would we just double our pulse
width?
Steve </TD></TR></TABLE>
Ummmmm, not even close..............
width?
Steve </TD></TR></TABLE>
Ummmmm, not even close..............
01-13-2004, 04:10 PM
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Re: Calculating Injector Pulse Width (RacerStev)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by RacerStev »</TD></TR><TR><TD CLASS="quote">What happens when we add boost? At 2 bar would we just double our pulse
width?
Steve </TD></TR></TABLE>
hehe nope, i think you're a good canidate for his efi101 class
width?
Steve </TD></TR></TABLE>
hehe nope, i think you're a good canidate for his efi101 class
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01-13-2004, 04:11 PM
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Re: Calculating Injector Pulse Width (danl)
danl meant in his EXPLANATION, that doubling isnt right. you would hae to calculate the cfm with the compressed air in the cylinder and therefore follow the steps as shown by ben that will give the amount of air the required fueling and hence pulse width. i hope 
01-13-2004, 04:18 PM
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Re: Calculating Injector Pulse Width (RacerStev)
Awesome info ! I hope you don't mind I copied this thread and put it on our site http://www.team-integra.net with EFIGUY's credit of course I thought the users on team-integra could use the info.
Thanks.
I thought the users on team-integra could use the info.Thanks.
01-13-2004, 04:24 PM
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Re: Calculating Injector Pulse Width (freakie)
Steve,
The guys are right......you won't double the Injector pulse width.
This happens mostly becasue whether we have 1 psi or 30 psi of boost, the engine's VE doesn't change. The volume of the cylinder doesn't change no matter how mard we commpress the air inside of it...........
what changes is the air DENSITY...... however the change in air density doesn't change in direct proportion to manifold pressure. It also has to account for the adiabatic efficiency of the compressor. When you compress air, it gets hot, and therfore loses density..........the adiabatic efficiency of the compressor is a measurement of how much heat it imparts to the air while it is compressing it.
This is where the whole PV=NrT equation comes into play......we know the V, which is engine VE, but the P {pressure} changes the density of our volume and so does the T which is temperature.
Okay, thats about all I can give away....if you want more, you really should come to a class!
The guys are right......you won't double the Injector pulse width.
This happens mostly becasue whether we have 1 psi or 30 psi of boost, the engine's VE doesn't change. The volume of the cylinder doesn't change no matter how mard we commpress the air inside of it...........
what changes is the air DENSITY...... however the change in air density doesn't change in direct proportion to manifold pressure. It also has to account for the adiabatic efficiency of the compressor. When you compress air, it gets hot, and therfore loses density..........the adiabatic efficiency of the compressor is a measurement of how much heat it imparts to the air while it is compressing it.
This is where the whole PV=NrT equation comes into play......we know the V, which is engine VE, but the P {pressure} changes the density of our volume and so does the T which is temperature.
Okay, thats about all I can give away....if you want more, you really should come to a class!
01-13-2004, 05:24 PM
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Re: Calculating Injector Pulse Width (EFIGUY)
Ben,
In order for this to work, the fuel pressure would have to be 3 bar, correct? The flow rate of the injector is rated at 3 bar, but if you've got more/less fuel pressure than that, the amount of fuel going through the injector for any given amount of time is going to change.
Good stuff.
Sonny
01-13-2004, 05:28 PM
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Re: Calculating Injector Pulse Width (Sonny)
yep....if you change the fuel pressure, you'll need to calculate what the actual flow rate of the injector is..this is easy enough to do though!
01-13-2004, 06:12 PM
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Re: Calculating Injector Pulse Width (EFIGUY)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by EFIGUY »</TD></TR><TR><TD CLASS="quote">Steve,
This is where the whole PV=NrT equation comes into play......we know the V, which is engine VE, but the P {pressure} changes the density of our volume and so does the T which is temperature.
</TD></TR></TABLE>
Unfortunatley PV=nRT is only good for "low pressure" situations. The variables in the equasion act unlinear at higher pressures. However, it is fairly accurate at nomal atmospheric pressure (more than fairly actually). Thats just a little food for thought.
I was mortified when a chemistry professor explained this to me.
This is where the whole PV=NrT equation comes into play......we know the V, which is engine VE, but the P {pressure} changes the density of our volume and so does the T which is temperature.
</TD></TR></TABLE>
Unfortunatley PV=nRT is only good for "low pressure" situations. The variables in the equasion act unlinear at higher pressures. However, it is fairly accurate at nomal atmospheric pressure (more than fairly actually). Thats just a little food for thought.
I was mortified when a chemistry professor explained this to me.
01-13-2004, 06:15 PM
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Re: Calculating Injector Pulse Width (EFIGUY)
I guess I was being a smart ***, before. WHat I was trying to get too was the
vast differance in power between 2 engines with differant turbo setups. You could
have a few hundred hp differance depending on your cams, compression, valve
size, turbine and compressior size's, intercooled or not.
How could you really get close with the pulse width w/o trial and error?
vast differance in power between 2 engines with differant turbo setups. You could
have a few hundred hp differance depending on your cams, compression, valve
size, turbine and compressior size's, intercooled or not.
How could you really get close with the pulse width w/o trial and error?
01-13-2004, 08:43 PM
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Re: Calculating Injector Pulse Width (RacerStev)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by RacerStev »</TD></TR><TR><TD CLASS="quote">I guess I was being a smart ***, before. WHat I was trying to get too was the
vast differance in power between 2 engines with differant turbo setups. You could
have a few hundred hp differance depending on your cams, compression, valve
size, turbine and compressior size's, intercooled or not.
How could you really get close with the pulse width w/o trial and error? </TD></TR></TABLE>
Steve,
obviously, in reality we wouldn't spend all day calculating IJW's, but often I find it easy to get a good starting place......you will have to guess {based on experience} the RPM that peak torque will occur {highest VE}, but if you can do that its not a bad way to get a quick base map out of thin air.....
again, boost changes air density {weight per unit volume} not VE, so if we find the point of highest VE for N/A, we can readily calculate IJW for that point and for any amount of pressure based on compressor efficiency maps, and a few thermodynamic equations for adiabatic efficiency.
Thanks for the input though!
vast differance in power between 2 engines with differant turbo setups. You could
have a few hundred hp differance depending on your cams, compression, valve
size, turbine and compressior size's, intercooled or not.
How could you really get close with the pulse width w/o trial and error? </TD></TR></TABLE>
Steve,
obviously, in reality we wouldn't spend all day calculating IJW's, but often I find it easy to get a good starting place......you will have to guess {based on experience} the RPM that peak torque will occur {highest VE}, but if you can do that its not a bad way to get a quick base map out of thin air.....
again, boost changes air density {weight per unit volume} not VE, so if we find the point of highest VE for N/A, we can readily calculate IJW for that point and for any amount of pressure based on compressor efficiency maps, and a few thermodynamic equations for adiabatic efficiency.
Thanks for the input though!
01-13-2004, 08:55 PM
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Re: Calculating Injector Pulse Width (EFIGUY)
This is good stuff Ben. In the past, Ive calculated approx VE % on many baseline plots (not just peak, but throughout the rpm range). Each engine and setup varies, but the point at which peak occurs usually stays the same for each type of motor. I see the light for some "better baseline" mapping that might be dead on.
Im in the cen fla area. Im really thinking about taking the seminar, but its on a Thurs - Fri schedule. Will attendees get notes like these handed to them? cuz my note taking skillz sux.. can I bring tape recorder?.
Im in the cen fla area. Im really thinking about taking the seminar, but its on a Thurs - Fri schedule. Will attendees get notes like these handed to them? cuz my note taking skillz sux.. can I bring tape recorder?
.
01-13-2004, 10:03 PM
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Re: Calculating Injector Pulse Width (Quick 200k Mile Motor)
Actually, students receive a 60-70 page textbook to take home that has even more in-depth material than the class alone can provide......
I'd prefer not to have recording devices though, just as a general policy!
-Thanks
I'd prefer not to have recording devices though, just as a general policy!
-Thanks
