jump to navigation

Volumetric Efficiency 2011/04/15

Posted by Michael in 2JZduino.
trackback

Early on in developing 2JZduino I made an assumption that the Injector Pulse length for each combustion cycle was exclusively a function of Manifold Pressure. The reasoning was that independent of all other variables the volumetric efficiency of the air intake system would remain mostly constant and a certain Manifold Pressure correlates strongly to the volume of air in the combustion chamber.

This is grossly incorrect. In hindsight it’s obvious that the engine replaces air at varying efficiencies (which explains things like why there is such a strong torque curve in a 4-stroke engine), but I was mistaken in the magnitude of the effect. And being so large, the effect of VE has a significant impact on fuel trims (which were previously assuming VE ~= 90%), upwards of 400% under some engine conditions.

To go about resolving this I logged a fair amount of engine data intentionally exercising a range of engine speeds and manifold air pressures so that I could calculate volumetric efficiency in post processing. The datalogger object in the 2JZduino code measures Engine Hz, MAP, AFR (banks 1 & 2), and Injector Pulse Length every 167ms. With this information I have measurements for the amount of air in the combustion chamber (calculated from AFR and Injected Fuel amount), and quantity of air for 100% efficiency (manifold pressure). The actual volumetric efficiency for each data-point is then calculated as the ratio of the two. The calculation looks like this…

2JZduino datalogger Measurements:
Hz = instantaneous engine speed
MAP = instantaneous manifold air (absolute) pressure, kPa
AFR = Air Fuel Ratio (latent)
T_inj = injector pulse length for previous combustion cycle, ms

Constants and Knowns:
Q_inj = 3.32 g/s = fuel injector flow rate
Inj_Latency = 1 ms = opening time of the injector solenoid
V_cyl = 0.5L = cylinder volume (swept)
CR = 10.5 = engine compression ratio
d_Air = 1.25 g/L = approximate air density
P_atm = 101 kPa = atmospheric pressure

Solving for the combustion chamber air mass that corresponds to 100% volumetric efficiency:
V_chamber = 0.5 + 0.5/(10.5-1) = 0.553L = total volume of combustion chamber
M_air_100% = 0.553 * d_Air = 691mg = air mass for 100% VE and MAP = P_atm

Solving for the instantaneous air mass in the combustion chamber for the instantaneous measurements: e.g. for Hz = 38, MAP = 66 (absolute, T_inj = 5.0, AFR = 14.1
M_fuel = (T_inj – Inj_Latency) * Q_inj = 13.3 mg
M_air = M_fuel * AFR = 186.5 mg

Solving for measured VE:
VE = M_air / (M_air_100% * MAP/P_atm) = 186.5 / (691 * 66/101) = 41.3%

Using these calculations for every logged data point I was then able in post processing to begin separating/sorting/analyzing the relationship between engine speed, manifold pressure, and volumetric efficiency. I eventually arrived at the following formula empirically that calculates volumetric efficiency (percent) as a function of engine speed and manifold pressure…

Eq. #1)
VE(Hz, MAP) = 12 + (Hz * MAP)/320 + MAP/2.67 – Hz/16 + 100/Hz
…constrained as VE > 20% and VE > 90%

This relationship was then used to calculate and predict (rather than measure) the volumetric efficiency for each data point in the datalog, and then further predict the pulse length that would be required for the measured AFR. Finally the error is calculated between actual injector pulse length and predicted injector pulse length. Continuing from the example above the calculation is as follows…
VE(38, 66) = 44.8%
M’_fuel = (V_chamber * d_Air * MAP/P_atm)/AFR * VE(Hz, MAP) = 14.4 mg
T’_inj = Inj_Latency + M’_fuel/Q_inj = 5.3 ms
Pulse Length Error = 5.3 – 5.0 = 0.3 ms

As verification to the empirical VE formula, the graph below shows select portions of the data-logs that present the measured and calculated (using Eq. #1) injector pulse lengths alongside the error between them. For reference the corresponding engine MAP is also shown.

Clearly, the errors between calculated and measured injector pulse lengths are held mostly to less than 1ms. Exceptions to this occur primarily under conditions where injector pulse-lengths change rapidly (caused by rapid changes in throttle position). In these instances the response times of sensors and even the 2ZJduino are the likely sources of error.

This particular calculation for VE (Eq. #1) shows to be a rather good approximation. Now with an understood of volumetric efficiency for my particular 2JZ-GE, next steps are to apply the calculated VE values to the fuel trim map in 2JZduino. More on this in a future post.

For reference, below is a map of measured Volumetric Efficiency of my 2JZduino for Engine Hz vs. Manifold Air Pressure…

Advertisements

Comments»

1. Mark_F - 2011/08/12

This is fantastic information and excellent testing. What might you do for a living?

email me at /\/\feldhake@NgOmSaPiAlm.com

Michael - 2011/08/13

Thanks Mark. I graduated mechanical engineering from University of Waterloo. I now work within the high-tech sector in the region.

2. Jared - 2013/09/29

would 2hz not be 120 rpm? If so, how did you get readings that low for the VE? Also, what is the max VE of your 2jzge? I am not an engineer so… Thank you, Jared

Michael - 2013/09/30

Yes, that’s 120RPM. I did *not* get (reliable) readings that low. That’s the calculated VE table after coming up with the equations for the model. The max. VE measured while the throttle is wide open with the method described above is ~70%. (I”m assuming it’s WOT that you’re interested in).

3. Soumya Sen - 2015/10/29

Hey, this is a wonderful method to calculate Volumetric efficiency. I would like to know a few things here –
1. Is this empirical formula applicable for general cars? What changes are to be made to use it for other cars?
2. At rpm 1443, imap = 88.23529 I get VE = 109%. Is this possible? The data is from Honda Brio.

Michael - 2015/10/29

1) No. This is applicable first and specifically to *my* engine configuration (with aftermarket intake and exhaust). Every engine is different.
2) A VE > 100% *is* possible in theory, but it would only exist for a narrow set of operating conditions. And of course for a forced induction setup VE > 100%. If you’re using my empirical equation to figure VE = 109% this is a misapplication of the eq. (see #1).
Hope this helps.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: