X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from cdptpa-omtalb.mail.rr.com ([75.180.132.123] verified) by logan.com (CommuniGate Pro SMTP 5.3c2) with ESMTP id 3988382 for flyrotary@lancaironline.net; Tue, 24 Nov 2009 15:45:24 -0500 Received-SPF: pass receiver=logan.com; client-ip=75.180.132.123; envelope-from=eanderson@carolina.rr.com Received: from computername ([75.191.186.236]) by cdptpa-omta02.mail.rr.com with ESMTP id <20091124204443563.JSIC27026@cdptpa-omta02.mail.rr.com> for ; Tue, 24 Nov 2009 20:44:43 +0000 From: "Ed Anderson" To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Typical Injector Current Draw Date: Tue, 24 Nov 2009 15:44:52 -0500 MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_000_0000_01CA6D1D.1045A7C0" X-Mailer: Microsoft Office Outlook, Build 11.0.5510 Thread-Index: AcptMViNh5xGsFHAQdudiIsTEYBu7gAD+Bgg In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 Message-Id: <20091124204443563.JSIC27026@cdptpa-omta02.mail.rr.com> This is a multi-part message in MIME format. ------=_NextPart_000_0000_01CA6D1D.1045A7C0 Content-Type: multipart/alternative; boundary="----=_NextPart_001_0001_01CA6D1D.1045A7C0" ------=_NextPart_001_0001_01CA6D1D.1045A7C0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit Hi Jeff, Since your Renesis injector is the saturated type, it probably has a resistance on the order of 12 ohms (could be anywhere between around 8 - 16 ohms - would have to measure it to be certain. Given a 13.8 Volt system that would mean it would draw I = E/R = 13.8/12 = 1.5 amps when there is sufficient current to pull it open. Perhaps a bit more or a bit less. The dynamics of current changing in a magnetic field makes this a non-linear relationship - but for a back of the envelope approach lets ignore that little detail. Its unlikely the current draw would be more. This shows the current vs time for a typical saturation type injector. As you can see the current appears pretty much proportional to the amount of time the injector is open up to the point it stabilizes. Which I think would be around 1 msec (but just a guess in this case) At WOT at 6000 rpm the maximum PW would be 10 msec (because that happens to be the injection cycle for the rotary at 6000 rpm). So at that point the injector would be open/on continuously. So lets consider that the maximum power draw case - whenever the injectors are "locked Open" - you don't' want to operate there {:>). But, your one injector would be open continuously - drawing 1.5 amps. However, that would equate to the engine consuming 29.16 GPH (4 x 460 cc/min injectors) which is highly unlikely at 6000 rpm and no boost - your engine would probably bog down and flood at that flow rate at 6000 rpm. So something less than that is probably the case.. Let's say our engine will "realistically" produce 200 HP at 7500 rpm that would equate to approx 20.0 gph fuel flow using a BSFC of 0.55 to make the calculation. For 4 x 460 cc/min injectors to give you that fuel flow at 7500 rpm would require them to be open approx 5.1 msec The injection period at 7500 rpm is around 8 msec (so we have not "lock open" the injectors at this point) So the duty cycle would be 5.1/8 = 64% Therefore if one injector draws 1.5 amps when open all the time then 4 would draw 4 x 1.5 = 6 amps So with a duty cycle of 64% they would have a combined current draw of approx 6 * 0.64 = 3.825 amps at 7500 rpm. Now as the power requirement decreases the injectors do not have to stay on as long and so your long-term current draw is undoubtedly less than this - unless you run WOT all the time {:>) Now if you used smaller injectors the injectors would have to be open longer draing more current (for that 200 hp) and if larger injectors then they would be open less time drawing less power due to their lower duty cycle. Well, at least that is the way it seems to me from the back of the envelope. 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 Jeff Luckey Sent: Tuesday, November 24, 2009 1:09 PM To: Rotary motors in aircraft Subject: [FlyRotary] Typical Injector Current Draw I'm doing some electrical load analysis and need some stats for fuel injectors. What does a typical fuel injector used on a Renesis draw? I'm looking for 3 numbers: 1. If I've got an injector on the bench, what is the energized load? 2. When the injector is running in the engine, it is getting a PWM signal. What it the typical duty cycle at full throttle? This will help calculate average load over time. 3. Or - is there some rule-of-thumb number that is used for engine at full throttle? TIA -Jeff Jeff Luckey Direct Connection Systems www.DCSIT.com 949-645-8832 __________ Information from ESET NOD32 Antivirus, version of virus signature database 3267 (20080714) __________ The message was checked by ESET NOD32 Antivirus. http://www.eset.com __________ Information from ESET NOD32 Antivirus, version of virus signature database 3267 (20080714) __________ The message was checked by ESET NOD32 Antivirus. http://www.eset.com ------=_NextPart_001_0001_01CA6D1D.1045A7C0 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

Hi Jeff,

 

Since your Renesis injector is the saturated type, it probably has a resistance on the order of 12 ohms (could be anywhere = between around 8 – 16 ohms – would have to measure it to be = certain.

 

Given a 13.8 Volt system that would mean it would = draw I =3D E/R =3D 13.8/12 =3D 1.5 amps when there is sufficient current to pull it open.  Perhaps a bit more or a bit less.  The dynamics of = current changing in a magnetic field makes this a non-linear relationship = – but for a back of the envelope approach lets ignore that little = detail.  Its unlikely the current draw would be more. 

 

This shows the current vs time for a typical = saturation type injector.  As you can see the current appears pretty much = proportional to the amount of time the injector is open up to the point it = stabilizes.  Which I think would be around 1 msec (but just a guess in this = case)

 

At WOT at 6000 rpm the maximum PW would be 10 msec = (because that happens to be the injection cycle for the rotary at 6000 = rpm).  So at that point the injector would be open/on continuously.  So lets = consider that the maximum power draw case – whenever the injectors are “locked Open” – you don’t’ want to operate = there {:>).  But, your one injector would be open continuously - = drawing 1.5 amps. 

 

However, that would equate to the engine consuming = 29.16 GPH (4 x 460 cc/min injectors) which is highly unlikely at 6000 rpm and no = boost – your engine would probably bog down and flood at that flow rate = at 6000 rpm.

So something less than that is probably the = case..

 

Let’s say our engine will = “realistically” produce 200 HP at 7500 rpm that would equate to approx 20.0 gph fuel = flow using a BSFC of 0.55 to make the calculation.

 

For 4 x 460 cc/min injectors to give you that fuel flow  at 7500 rpm would require them to be open approx 5.1 = msec

The injection period at 7500 rpm is around 8 msec (so = we have not “lock open” the injectors at this = point)

 

So the duty cycle would be 5.1/8 =3D  = 64%

 

Therefore if one injector draws 1.5 amps when open = all the time then 4 would draw 4 x 1.5 =3D  6 = amps

 

So with a duty = cycle of 64% they would have  a combined current draw of approx 6 * 0.64 =3D = 3.825 amps at 7500 rpm.  Now as the power requirement decreases the = injectors do not have to stay on as long and so your long-term current draw is = undoubtedly less than this – unless you run WOT all the time = {:>)

Now if you used smaller injectors the = injectors would have to be open longer draing more current (for that 200 hp) and = if larger injectors then they would be open less time drawing less power = due to their lower duty cycle.

Well, at least that is the way it seems to = me from the back of the envelope.

 

Ed

 

Ed Anderson

Rv-6A N494BW Rotary = Powered

Matthews, NC

eanderson@carolina.rr.com=

http://www.andersonee.com

http://www.dmack.net/mazda/i= ndex.html

http://www.flyrotary.com/

http://members.cox.net/roger= sda/rotary/configs.htm#N494BW

http://www.rotaryaviation.co= m/Rotorhead%20Truth.htm

From: = Rotary motors in aircraft = [mailto:flyrotary@lancaironline.net] On Behalf Of Jeff = Luckey
Sent: Tuesday, November = 24, 2009 1:09 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] = Typical Injector Current Draw

 

I’m doing some electrical load analysis and need some stats for fuel = injectors.

 

What does a typical fuel injector used on a Renesis draw?  I’m = looking for 3 numbers:

  1. If I’ve got an injector on the = bench, what is the energized load?
  2. When the injector is running in the = engine, it is getting a PWM signal.  What it the typical duty cycle at full throttle?  This will help calculate average load over = time. 
  3. Or - is there some rule-of-thumb number = that is used for engine at full throttle?

 

TIA

 

-Jeff

 

 

Jeff Luckey

Direct Connection Systems

www.DCSIT.com<= /p>

949-645-8832

 



__________ Information from ESET NOD32 Antivirus, version of virus = signature database 3267 (20080714) __________

The message was checked by ESET NOD32 Antivirus.

http://www.eset.com



__________ Information from ESET NOD32 Antivirus, version of virus = signature database 3267 (20080714) __________

The message was checked by ESET NOD32 Antivirus.

http://www.eset.com



__________ Information from ESET NOD32 Antivirus, version of = virus signature database 3267 (20080714) __________

The message = was checked by ESET NOD32 Antivirus.

http://www.eset.com
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