AFR's and Lambda
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On The Tree
Joined: Mar 2014
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From: Lakeville, MN
AFR's and Lambda Let's say for the sake of argument you're hitting .8 lambda with gasoline. How would you go about calculating what your lambda would be with a fuel change to say E85 given you're injecting the same amount of fuel? Do you suppose cross multiplication is applicable?
Maybe:
(1/(1+14.7))/0.8 = (1/(1+9.76))/X
This gives you 1.167, which is maybe in the ballpark? You'll obviously go lean if you don't add any fuel.
The math sort of works, but I'm not sure I buy it.
Edit: Also I'm under the impression E85 is denser then gasoline? Surely that needs to be taken into account?
Maybe:
(1/(1+14.7))/0.8 = 1.04(1/(1+9.76))/X
This gives you 1.214, which definitely isn't right the added density should make it richer when compared to the original equation?
I'll get there eventually, maybe...
Edit 2: I think this looks better. 0.8 Lambda is 11.76:1 on gasoline, that equates to 7.84% gasoline by weight. E85 is 1.04 times the density of gasoline and if the same volume were injected you would be at 8.15% E86 by weight. This is an AFR of 11.27:1 and a lambda of 1.155?
0.8 * 14.7 = 11.76
(1/(1 + 11.76)) = .0784
.0784 * 1.04 = .0815
(1/.0815) - 1 = 11.27
11.27 / 9.76 = 1.155
Maybe:
(1/(1+14.7))/0.8 = (1/(1+9.76))/X
This gives you 1.167, which is maybe in the ballpark? You'll obviously go lean if you don't add any fuel.
The math sort of works, but I'm not sure I buy it.
Edit: Also I'm under the impression E85 is denser then gasoline? Surely that needs to be taken into account?
Maybe:
(1/(1+14.7))/0.8 = 1.04(1/(1+9.76))/X
This gives you 1.214, which definitely isn't right the added density should make it richer when compared to the original equation?
I'll get there eventually, maybe...
Edit 2: I think this looks better. 0.8 Lambda is 11.76:1 on gasoline, that equates to 7.84% gasoline by weight. E85 is 1.04 times the density of gasoline and if the same volume were injected you would be at 8.15% E86 by weight. This is an AFR of 11.27:1 and a lambda of 1.155?
0.8 * 14.7 = 11.76
(1/(1 + 11.76)) = .0784
.0784 * 1.04 = .0815
(1/.0815) - 1 = 11.27
11.27 / 9.76 = 1.155
Last edited by InfrareV; Apr 28, 2014 at 08:26 PM.
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Joined: Apr 2002
Posts: 6,080
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From: So.Cal.
The EQR or Lambda stays almost the same... the Stoich AFR changes (there is a parameter for this in your calibration)...
e.g. so if your PE commands EQR 1.165 then AFR = 9.76/1.165 = 8.38
you then measure using wideband as suggested and make adjustment;
you then have to watch the injector duty cycles to make sure the injectors still have room.
e.g. so if your PE commands EQR 1.165 then AFR = 9.76/1.165 = 8.38
you then measure using wideband as suggested and make adjustment;
you then have to watch the injector duty cycles to make sure the injectors still have room.
