EFI and everything else you want to calculate
15-Jan-2004, 11:32 AM
#1
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Join Date: Jan 2003
Location: on a race track
Posts: 17,846
EFI and everything else you want to calculate
ok followers of Theoldone.com
have a look:
http://www.theoldone.com/forum/topic.asp?TOPIC_ID=14918
I found it to be very interesting
have a look:
http://www.theoldone.com/forum/topic.asp?TOPIC_ID=14918
I found it to be very interesting
15-Jan-2004, 12:08 PM
#2
hehe, for the benefit of those not signed up for the goodness of the old one:
www.efi101.com
this is a reproduced article... if there is a problem with me posting this, would the author of the article please PM me.
thank you!!
Here's a little article I wrote for the EFI 101 textbook.
Hope it helps a few people out! Enjoy.
Calculating the necessary Injector Pulse Width for a known injector size and engine displacement.
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.
Here’s how it works:
1} 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!
-Ben
www.efi101.com
this is a reproduced article... if there is a problem with me posting this, would the author of the article please PM me.
thank you!!Here's a little article I wrote for the EFI 101 textbook.
Hope it helps a few people out! Enjoy.
Calculating the necessary Injector Pulse Width for a known injector size and engine displacement.
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.
Here’s how it works:
1} 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!
-Ben
30-Jan-2004, 12:33 AM
#3
Registered User
Join Date: Feb 2003
Location: In a tree
Posts: 320
meh.
check out www.efi101.com
and see if you can make it to a school session.
itll be the best money spent if your doing a build up.and are wanting to tune or to learn more about it
check out www.efi101.com
and see if you can make it to a school session.
itll be the best money spent if your doing a build up.and are wanting to tune or to learn more about it
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