X-Virus-Scanned: clean according to Sophos on Logan.com X-SpamCatcher-Score: 2 [X] Return-Path: Sender: To: lml@lancaironline.net Date: Sat, 24 Mar 2007 12:19:40 -0400 Message-ID: X-Original-Return-Path: Received: from elasmtp-junco.atl.sa.earthlink.net ([209.86.89.63] verified) by logan.com (CommuniGate Pro SMTP 5.1.7) with ESMTP id 1940776 for lml@lancaironline.net; Sat, 24 Mar 2007 10:57:03 -0400 Received-SPF: none receiver=logan.com; client-ip=209.86.89.63; envelope-from=rtitsworth@mindspring.com DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=dk20050327; d=mindspring.com; b=bsB2tMFZ/CbphEyv9Vbej7rbGWuQuBvJXfKCfaXx11cANjcmkGUwdLYEAZHCSr+d; h=Received:From:To:Subject:Date:Message-ID:MIME-Version:Content-Type:X-Mailer:X-MimeOLE:Thread-Index:In-Reply-To:X-ELNK-Trace:X-Originating-IP; Received: from [69.3.253.134] (helo=RDTVAIO) by elasmtp-junco.atl.sa.earthlink.net with asmtp (Exim 4.34) id 1HV7fE-0008Rf-U5 for lml@lancaironline.net; Sat, 24 Mar 2007 10:56:17 -0400 From: "rtitsworth" X-Original-To: "'Lancair Mailing List'" Subject: RE: [LML] Re: Legacy Battery Set-up X-Original-Date: Sat, 24 Mar 2007 10:55:51 -0400 X-Original-Message-ID: <000701c76e24$84683b90$86fd0345@RDTVAIO> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0008_01C76E02.FD569B90" X-Mailer: Microsoft Office Outlook 11 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 Thread-Index: AcdtuvZ72FE8w3P7QCi+ep83ouNyPAAXejSg In-Reply-To: X-ELNK-Trace: b17f11247b2ac8f0a79dc4b33984cbaa0a9da525759e26542ddfd0da269ec1f1777417d58cc91157cd0c042b896368d8350badd9bab72f9c350badd9bab72f9c X-Originating-IP: 69.3.253.134 This is a multi-part message in MIME format. ------=_NextPart_000_0008_01C76E02.FD569B90 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit Colyn, 1) My understanding is if the "B" starting battery were nearly dead and tried to draw more than 20 amps during it's re-charging cycle, the alternator's voltage drops slightly and then the battery draws less re-charge current. Sort of self-policing. I know my Cessna battery never seems to draw more than about 5 amps. Perhaps someone can confirm/enlighten/correct me. If the A side dies, and I engage the cross-tie to power stuff from the B (backup) side, the total current load will be supplied by both the B alternator and the battery (which attempts to makeup for any alternator shortfall). Thus, I could continue flight with a 20 amp total load indefinitely, or with a higher load (say 30 amps) by also slowly depleting the battery(ies) which are supplying the extra 10 amps. My general intention is to keep the "continuous" emergency/endurance equipment load safely under 20 amps, and then let the batteries kick-in (if/as needed) for intermittent stuff like flaps. Pilot heat, prop heat and de-icing, is a debate in itself which probably calls for dual full size alternators among other things. i.e. I perhaps can't continue to fly with an iced prop, but I also can't land IMC without "some" navigational avionics (perhaps a battery powered 496 handheld as a third redundancy level). Overall, say no to ice. Note, if the A side alternator fails both the A and B batteries can still be used (and left on-line at the same time to maximize reserve power). According to Bob Nuckolls (Aeroelectric connection) there is really no case where lead acid batteries drain each other, as these batteries all supply power at a voltage less than they accept it (minimum charge voltage). i.e. one battery cannot charge the other. Also note, in my simplified sketch, I have an essential/endurance bus which is supplied from both the "A" and "B" side via a diode on each. Thus, the essential bus gets power even if the A-side fails and I haven't yet activated the cross tie. Thus, the EFIS doesn't quit and have to re-boot. This also provides some additional EFIS power redundancy in the event of a simultaneous A side and cross-tie failure. Normally, my "A" side voltage regulator will be set to slightly higher than the B side (~0.1 volt). Thus, power will normally flow into the essential bus via the A side. Alternatively, I could have a manual switch on the B-side lead or a normally closed relay on the B side lead which is dis-engaged by the A side (open then A is hot). I opted for the diodes as KISS in this case and am seeking to minimize the number of switches. However, a manual switch would create some additional starting spike isolation. Perhaps a normally closed relay which is powered (dis-engaged, opened) by the starter relay. Perhaps the B side essential feed should be tied to the B-side alternator switch, since it would normally be off during starting (providing isolation). I'm also considering activating the alternators based on an oil pressure switch (one for each side as redundancy). That way the alternators com on-line automatically with oil pressure (just after engine start). Without oil pressure the alternators won't work long anyway {g}. Each could also be turned off by pulling the circuit breaker on the respective voltage regulator. I'm undecided on that, but I digress a bit. 2) If I understand your second question correctly. I have a shunt and ammeter between the battery and the bus/alternator (on both the A and B side). These are not shown on my prior sketch. On the B side it is past the starter so the "starting current" does not go though the shunt (a special high current event). Otherwise, both sides are essentially setup the same. Thus, the ammeter(s) show the current flowing into (or out of) each battery. With the alternator on, this current shown is normally very low (just any battery recharging). With the alternator off, the ammeters will show the "drain" being placed by the electrical equipment that's on at the time. This setup dose not directly show the total load on the alternator. However, it is indirectly available. If you want to know (approximately) how much total load is being placed on the alternator at any point in time, you can look at the ammeter with the alternator on (the current going into the battery) and then momentarily turn the alternator off and see how much current flows from the battery to the electrical stuff. The current flowing out from the battery to the other stuff, was previously being supplied by the alternator. So, the total alternator load is (was) the sum of the two. Note: the needle (flow) will shift from positive to negative, which just indicates current direction relative to the battery. The total current being supplied by the alternator is (was) the sum of the absolute values (disregarding the direction into/out of the battery). Some folks (manufacturers) favor putting the shunt in the alternator B lead (between the bus/battery and the alternator). That setup shows total alternator load directly, but is useless in an alternator failure scenario, when you might want to really know total current draw to preserve the remaining battery power. It also provides no mechanism for determining whether the battery is charging (health) or draining (unhealthy). Others favor putting the shunt at the head of the bus(es) i.e.before all the electrical stuff. That setup shows total equipment load directly, but provides no mechanism for determining total alternator load, nor whether the battery is charging (health) or draining (unhealthy). Thus, the std battery ammeter setup seems the best approach (as I understand it). Rick p.s. Now back to re-checking the end connections on the big ground wire (single point of failure). _____ From: Lancair Mailing List [mailto:lml@lancaironline.net] On Behalf Of colyncase on earthlink I wonder if anybody knows what happens to the 20 amp alternator if you put greater than a 20 amp load on it? Also, how do you figure out what the load is if you are draining a fully charged battery at the same time? ------=_NextPart_000_0008_01C76E02.FD569B90 Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable

Colyn,

1) My understanding is if the = “B” starting battery were nearly dead and tried to draw more than 20 amps = during it’s re-charging cycle, the alternator’s voltage drops slightly and = then the battery draws less re-charge current.  Sort of self-policing.  = I know my Cessna battery never seems to draw more than about 5 amps. =  Perhaps someone can confirm/enlighten/correct me.  =

 

If the A side dies, and I engage = the cross-tie to power stuff from the B (backup) side, the total current load will be supplied by both the B alternator and the battery (which attempts to = makeup for any alternator shortfall).  Thus, I could continue flight with a 20 = amp total load indefinitely, or with a higher load (say 30 amps) by also slowly = depleting the battery(ies) which are supplying the extra 10 amps.  My general = intention is to keep the “continuous” emergency/endurance equipment = load safely under 20 amps, and then let the batteries kick-in (if/as needed) for = intermittent stuff like flaps.  Pilot heat, prop heat and de-icing, is a debate = in itself which probably calls for dual full size alternators among other = things.  i.e. I perhaps can’t continue to fly with an iced prop, but I also = can’t land IMC without “some”  navigational avionics (perhaps = a battery powered 496 handheld as a third redundancy level).  = Overall, say no to ice.

 

Note, if the A side alternator = fails both the A and B batteries can still be used (and left on-line at the same = time to maximize reserve power).  According to Bob Nuckolls (Aeroelectric connection) there is really no case where lead acid batteries drain each = other, as these batteries all supply power at a voltage less than they accept = it (minimum charge voltage).  i.e. one battery cannot charge the = other.

 

Also note, in my simplified sketch, = I have an essential/endurance bus which is supplied from both the = “A” and “B” side via a diode on each.  Thus, the essential bus gets power even = if the A-side fails and I haven’t yet activated the cross tie.  Thus, the = EFIS doesn’t quit and have to re-boot.  This also provides some additional EFIS = power redundancy in the event of a simultaneous A side and cross-tie failure.  = Normally, my “A” side voltage regulator will be set to slightly higher = than the B side (~0.1 volt).  Thus, power will normally flow into the = essential bus via the A side.  Alternatively, I could have a manual switch on the = B-side lead or a normally closed relay on the B side lead which is dis-engaged = by the A side (open then A is hot).  I opted for the diodes as KISS in this = case and am seeking to minimize the number of switches.  However, a = manual switch would create some additional starting spike isolation.  = Perhaps a normally closed relay which is powered (dis-engaged, opened) by the = starter relay.  Perhaps the B side essential feed should be tied to the = B-side alternator switch, since it would normally be off during starting = (providing isolation). 

 

I’m also considering = activating the alternators based on an oil pressure switch (one for each side as = redundancy).  That way the alternators com on-line automatically with oil pressure = (just after engine start).  Without oil pressure the alternators = won’t work long anyway {g}.  Each could also be turned off by pulling the circuit breaker on the respective voltage regulator.  I’m = undecided on that, but I digress a bit…

 

2) If I understand your second = question correctly… I have a shunt and ammeter between the battery and the bus/alternator (on both the A and B side).  These are not shown on = my prior sketch.  On the B side it is past the starter so the = “starting current” does not go though the shunt (a special high current = event).  Otherwise, both sides are essentially setup the same.  Thus, the = ammeter(s) show the current flowing into (or out of) each battery.  With the alternator on, this current shown is normally very low (just any battery recharging).  With the alternator off, the ammeters will show the = “drain” being placed by the electrical equipment that’s on at the = time.  This setup dose not directly show the total load on the alternator.  = However, it is indirectly available.  If you want to know (approximately) = how much total load is being placed on the alternator at any point in time, you can = look at the ammeter with the alternator on (the current going into the battery) = and then momentarily turn the alternator off and see how much current flows = from the battery to the electrical stuff.  The current flowing out from = the battery to the other stuff, was previously being supplied by the alternator.  So, the total alternator load is (was) the sum of the = two.  Note: the needle (flow) = will shift from positive to negative, which just indicates current direction = relative to the battery.  The total current being supplied by the alternator is = (was) the sum of the absolute values (disregarding the direction into/out of = the battery).

 

Some folks (manufacturers) favor = putting the shunt in the alternator B lead (between the bus/battery and the = alternator).  That setup shows total alternator load directly, but is useless in an alternator failure scenario, when you might want to really know total = current draw to preserve the remaining battery power.  It also provides no mechanism for determining whether the battery is charging (health) or = draining (unhealthy).  Others favor putting the shunt at the head of the = bus(es) i.e.before all the electrical stuff.  That setup shows total equipment load = directly, but provides no mechanism for determining total alternator load, nor = whether the battery is charging (health) or draining (unhealthy).  Thus, = the std battery ammeter setup seems the best approach (as I understand = it).

 

Rick

 

p.s. Now back to re-checking the = end connections on the big ground wire (single point of = failure).

 

From: Lancair Mailing List [mailto:lml@lancaironline.net] On Behalf Of colyncase on earthlink

I wonder if anybody knows what happens to the 20 amp alternator if you put greater than a 20 amp load on = it?

 

Also, how do you figure out what the load is if you = are draining a fully charged battery at the same = time?

 

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