Mailing List flyrotary@lancaironline.net Message #46420
From: Ed Anderson <eanderson@carolina.rr.com>
Subject: Air/Fuel Ratio??
Date: Sun, 14 Jun 2009 19:58:24 -0400
To: 'Rotary motors in aircraft' <flyrotary@lancaironline.net>

George, not exactly certain what you mean by “calculating Air/Fuel ratio” .

 

I used the ratio of the mass of the air to the fuel – which basically relies on the approximation of 0.0765 lbs per cubic foot of air.  So you calculate your air flow in CFM then times  0.0765 to give you the air mass.  The you can take your desired A/F ratio – say 15:1  and divide 15 into your air mass and that would give you the fuel mass required to achieve that ratio. Of if you have any two of the three factors you can find the third.

 

IF you mean from a run time perspective  - how do you know you Air/Fuel ratio -  then there are expensive testing instruments and I believe some fairly accurate air/fuel ratio meters (based on the newer broad band O2 sensor)  But still several hundred dollars when I last checked.

 

The narrow band O2 sensor is much cheaper and works just fine with 100 LL for OUR use.  I generally get closer to 200 hours using 100LL before the sensor appears to lose too much sensitivity to continue to be useful.  The common notion that a few seconds running on leaded fuel will “Kill” an O2 sensor (at least the narrow band ) is simply not true – at least not for our use.  Doing that WILL apparently degrade it for its intended use in an automobile fuel system (where it needs to help the engine computer maintain a 14:1 A/F ratio), but if you just want a general indication of whether you are lean, rich or in the middle, the cheap O2 sensor works well. 

 

You can even find  some narrow band units which will read out A/F in numeric values but if using a narrow band O2 sensor, I question the accuracy of such units myself.  I suppose you could use a microprocessor and accurate analog/digital converter and if you had the “Z” curve of your O2 sensor – you might get close.

 

So basically if you know the mass of air and mass of fuel, you have your Air/Fuel ratio

 

So how to arrive at those two factors, if you know the air pressure and temperature (at normal atmospheric  values) then you essentially know the air density from which you can calculate air mass and then using your engine flow rates and with your chosen Air/Fuel ratio - calculate your fuel flow, etc.  But, frequently it’s easier to use our fuel flow (which can be measured fairly accurately)

 

So one r way to approach the problem is as follows:  You know the displacement of your engine and assuming some Ve (volumetric efficiency) (85% - 110%) you can calculate your air mass flow through the engine for any rpm.  So how to get an approximation of our volumetric efficiency (at least at WOT), its fairly simple to get close.

 

Note the ambient atmospheric pressure (manifold gauge pressure without engine running), fire up your engine and when warmed up advance it to WOT and note the atmospheric pressure inside your intake (i.e. your manifold pressure).  Lets say ambient pressure is 29.92 inches HG and lets say you are so lucky to read 29.92 “ Hg in manifold pressure - then theoretically your Ve is 100%.  But lets say your design is not perfect (few are) and your read 28.75”  then your Ve is 28.75/29.92 = 0.9608 or 96.08 % Ve –not bad, but not perfect.

 

OK calculate your volumetric flow using our old displacement formulas and as best I recall at 6000 rpm with a 13B at 100% VE = 277 CFM.  Since our intake is not perfect we take our Ve of  0.96.08Ve*277 = 266 CFM actually going through your engine.  Recalling that a cubic foot of air approx = 0.0765 lbm/Cubic Foot, we have 266 * 0.0765 = 20.36  lbsm of air per minute.

 

 Now we don’t know our Air/Fuel ratio – but we do know our fuel flow at that rpm.  Lets say its 16 gallon/hour, turning that in to lbm/min we have 16 / 60 = 0.2666 gallon/min and we know mass of gasoline is approx 6 – 6.25 lbs/gallon.  So taking 6.0 lb/gallon  we have

0.2666 * 6 = 1.6 lbm/min of fuel based on our fuel flow indication.

 

Now taking both the air mass 20.36 lbm/min and the fuel 1.6 lbm/min and we  get 20.36/1.6 = 12.725 air/fuel ratio.

 

Which is very close to the common “Best Power” ratio used by many.

 

So don’t know if this answered any of your question – but, best I could come up with {:>)

Ed

Ed Anderson

Rv-6A N494BW Rotary Powered

Matthews, NC

eanderson@carolina.rr.com

http://www.andersonee.com

http://www.dmack.net/mazda/index.html

http://www.flyrotary.com/

http://members.cox.net/rogersda/rotary/configs.htm#N494BW

http://www.rotaryaviation.com/Rotorhead%20Truth.htm


From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of George Lendich
Sent: Sunday, June 14, 2009 6:21 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Re: Wire separation?

 

Tracy or anyone,

Can you tell me what sensors you use to calculate fuel ratios.

I notice the mention of a MAP sensor (vacuum type), but that would give pressure, not density.

I notice the absence of an oxygen sensor in the case of 100LL is used - this is understood. I suppose an oxygen sensor could be used independently.

I assume RPM would be in the calculations for fuel flow, but I wonder how you actually arrive at air density? perhaps OAT is used.

Basically I wondering what the simplest air density sensor arrangement would be. I thought Oxygen sensor (not for 100LL), with a RPM and throttle position sensor.

George (down under)

 

Old time distributors and carburetors didn’t care how badly you wired the electrical system.  Digital systems are not so forgiving of poor wiring practices.  Noisy power or ground wiring is one of the most common causes of initial builder problems with computerized equipment (including ECUs). 

In addition to the wire bundling answers you got, the basic power wiring is as or even more important.  Here is a basic aircraft wiring scheme that I recommend in latest version of the installation guide.  Not detailed but somewhere to start.  Note the independent power feeds from the battery for the various systems.  This is a physical as well as electrical representation.

Tracy


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