DIY_EFI Digest Sunday, 7 February 1999 Volume 04 : Number 095 In this issue: Re: fusible link questions RE: fusible link questions Re: Digital: fuel pressure gage; boost gage RE: fusible link questions Re: fusible link questions Re: fusible link questions Re: fusible link questions Re: fusible link questions Re: fusible link questions Re: fusible link questions Re: Digital: fuel pressure gage; boost gage See the end of the digest for information on subscribing to the DIY_EFI or DIY_EFI-Digest mailing lists. ---------------------------------------------------------------------- From: "Clarence L.Snyder" Date: Sat, 06 Feb 1999 21:40:40 -0500 Subject: Re: fusible link questions David A. Cooley wrote: > > At 12:45 PM 2/6/99 -0800, you wrote: > > >> > > >Ford used 2 alternators on some Lincolns; one was 110 volts and used for > >the rear window heater. What did they change in the alternator to do this? > > > > Just the wiring... > You can buy a box that hooks up to the alternator on most cars that > provides 110VAC for running power tools remotely, while still charging the > battery. > > =========================================================== > David Cooley N5XMT Internet: N5XMT@xxx.net > Packet: N5XMT@xxx. Member #7068 > I am Pentium of Borg...division is futile...you will be approximated. > =========================================================== You get 110 VDC only, and cannot charge the battery at the same time - sorry. To get 110 VAC and still charge the battery you need an inverter. I've got one in my van for my computer, and one I can put in the van for power tool use - 1500 VA so it REALLY makes the alternator work at full load. ------------------------------ From: Don.F.Broadus@xxx.com Date: Sat, 6 Feb 1999 21:17:57 -0600 Subject: RE: fusible link questions You are correct that the speed of the moon roof motor would vary with engine speed, but that wouldn't be much of a concern with a moon roof. I don't think the frequency at 3600 RPM would be 60 HZ. I'm not sure of the exact number of poles in the stator It might be 12 and the rotor had a different number like 7. I would venture to say the AC frequency would be in the vacinity of 100 to 400 HZ. that would vary with engine speed. Take Care Don > -----Original Message----- > From: bearbvd@xxx.net] > Sent: Saturday, February 06, 1999 7:40 PM > To: diy_efi@xxx.edu > Subject: RE: fusible link questions > > >I have worked on a few of the 2 alternator Fords (1977era) . The shop > manual > >shows one alternator for the moon roof and windshield de-icer. And one > >alternator for the 12 volt systems. The one alternator is indeed 110 > volts > >but at 400 cycles 3 phase AC. It looks like all they did is leave off the > >diodes and full field the rotor. The people I talked to said they seldom > >failed. The windshield had 3 heating elements wired in a star > configuration > >and the moon roof was a 3 phase motor star wound. To close the moon roof > 2 > >phases were > >flopped just like a regular 3 phase motor. Don > > Makes perfect sence, except that the frequency of the AC produced would > have varied with engine speed. Alternators are wound like a two pole > motor, > so 3600 ALTERNATOR rpm would have given 60 cycle AC. > > Regards, Greg > > > > > >> -----Original Message----- > >> From: David A. Cooley [SMTP:n5xmt@xxx.net] > >> Sent: Saturday, February 06, 1999 3:21 PM > >> To: diy_efi@xxx.edu > >> Subject: Re: fusible link questions > >> > >> At 12:45 PM 2/6/99 -0800, you wrote: > >> > >> >> > > >> >Ford used 2 alternators on some Lincolns; one was 110 volts and used > for > >> >the rear window heater. What did they change in the alternator to do > >> this? > >> > > >> > >> Just the wiring... > >> You can buy a box that hooks up to the alternator on most cars that > >> provides 110VAC for running power tools remotely, while still charging > the > >> battery. > >> > >> =========================================================== > >> David Cooley N5XMT Internet: N5XMT@xxx.net > >> Packet: N5XMT@xxx. Member #7068 > >> I am Pentium of Borg...division is futile...you will be > >> approximated. > >> =========================================================== > ------------------------------ From: JRECPA@xxx.com Date: Sat, 6 Feb 1999 22:25:03 EST Subject: Re: Digital: fuel pressure gage; boost gage In a message dated 2/6/99 7:46:19 PM US Mountain Standard Time, Don.F.Broadus@xxx.com writes: << Autometer has an isolated system that uses a rubber diaphragm to isolate the gas liquid. A gauge on the other side of the diaphragm is responding to air pressure. So I would see how much the isolator is and find a pressure transducer and a volt meter. Don >> I have an 1998 PAW catalog and they list the isolator for $39.95 part #ato-5280 rated to 200psi working pressure. James ------------------------------ From: bearbvd@xxx.net (Greg Hermann) Date: Sat, 6 Feb 1999 21:15:49 -0700 Subject: RE: fusible link questions >You are correct that the speed of the moon roof motor would vary with engine >speed, but that wouldn't be much of a concern with a moon roof. I don't >think the frequency at 3600 RPM would be 60 HZ. I'm not sure of the exact >number of poles in the stator It might >be 12 and the rotor had a different number like 7. I would venture to say >the AC frequency would be in the vacinity of 100 to 400 >HZ. that would vary with engine speed. Take Care > Don Thought I said ALTENATOR RPM--with a typical pulley ratio that would give you 60 HZ at about 1000 ENGINE rpm--- Most alternators I have seen have six coils in the stator--for a three phase machine, coils/#of phases give the number of poles, so 2 poles it would be, and 3600 rpm on the alternator=60 Hz. If you got 12 coils, thazza 4 pole machine, and you would git 60 Hz @ 1800 machine rpm, ad infinitum. Regards, Greg > > > >> -----Original Message----- >> From: bearbvd@xxx.net] >> Sent: Saturday, February 06, 1999 7:40 PM >> To: diy_efi@xxx.edu >> Subject: RE: fusible link questions >> >> >I have worked on a few of the 2 alternator Fords (1977era) . The shop >> manual >> >shows one alternator for the moon roof and windshield de-icer. And one >> >alternator for the 12 volt systems. The one alternator is indeed 110 >> volts >> >but at 400 cycles 3 phase AC. It looks like all they did is leave off the >> >diodes and full field the rotor. The people I talked to said they seldom >> >failed. The windshield had 3 heating elements wired in a star >> configuration >> >and the moon roof was a 3 phase motor star wound. To close the moon roof >> 2 >> >phases were >> >flopped just like a regular 3 phase motor. Don >> >> Makes perfect sence, except that the frequency of the AC produced would >> have varied with engine speed. Alternators are wound like a two pole >> motor, >> so 3600 ALTERNATOR rpm would have given 60 cycle AC. >> >> Regards, Greg >> > >> > >> >> -----Original Message----- >> >> From: David A. Cooley [SMTP:n5xmt@xxx.net] >> >> Sent: Saturday, February 06, 1999 3:21 PM >> >> To: diy_efi@xxx.edu >> >> Subject: Re: fusible link questions >> >> >> >> At 12:45 PM 2/6/99 -0800, you wrote: >> >> >> >> >> > >> >> >Ford used 2 alternators on some Lincolns; one was 110 volts and used >> for >> >> >the rear window heater. What did they change in the alternator to do >> >> this? >> >> > >> >> >> >> Just the wiring... >> >> You can buy a box that hooks up to the alternator on most cars that >> >> provides 110VAC for running power tools remotely, while still charging >> the >> >> battery. >> >> >> >> =========================================================== >> >> David Cooley N5XMT Internet: N5XMT@xxx.net >> >> Packet: N5XMT@xxx. Member #7068 >> >> I am Pentium of Borg...division is futile...you will be >> >> approximated. >> >> =========================================================== >> ------------------------------ From: Jim Davies Date: Sat, 6 Feb 1999 20:56:45 -0800 (PST) Subject: Re: fusible link questions On Sat, 6 Feb 1999, David A. Cooley wrote: > >Ford used 2 alternators on some Lincolns; one was 110 volts and used for > >the rear window heater. What did they change in the alternator to do this? > > > > Just the wiring... > You can buy a box that hooks up to the alternator on most cars that > provides 110VAC for running power tools remotely, while still charging the > battery. > Yes, I have one kicking around somewhere. I havent stripped the 110 volt ford alternator, so I wonder what it was like inside. ------------------------------ From: "Tom Parker" Date: 07 Feb 99 18:15:54 +1200 Subject: Re: fusible link questions John Hess wrote: >Actually, you haven't really considered that: >while P=IE (the basic formula) >E=IR, therefore substituting IR for E: >P=I*IR or I^2R >I=E/R, therefore substituting E/R for I >P=E*E/R or E^2/R, etc ------------------------------ From: "Tom Parker" Date: 07 Feb 99 18:19:12 +1200 Subject: Re: fusible link questions Clarence L.Snyder wrote: >> The only extra electrical load on the alternator will be in the extra >> current flowing in the field windings. The field windings will have to >> dissapate heat due to the P = I^2 R equation, so if you double your field >> winding current, you will have to dissapate 4 times the heat in them. >Mabee I didn't explain it 100%, but I THINK that's what I said. Double >the power output of the alternator = 4 X the heat to dissipate, due to >I^2 R losses. > (there are also extra >> mechanical loads, you will stress the bearings and the shaft of the >> alternator if you ask it to draw such a huge amount of power out of the >> engine). >> >> If you double the voltage output of the alternator main windings, the heat >> dissapated in the main windings will remain the same, as the heat generated >> there is due to the current through them only! >Correct - stator current will remain constant, so I^2 R losses will >remain constant. Stator heating will not be the problem. - >However, if more POWER is required at the same voltage, then both field >and stator heating increase at the same "square of increase" rate. What is the current in and voltage across the field windings necessary to generate the power proposed? I'm willing to bet that it isn't that huge. Certainly the full power output proposed by the original poster will not be dissapated in the field windings. I've no idea what current is normally presant in the field windings, but I wouldn't expect it to be large, and I would expect that it could be increased significantly before you have problems with overheating. The other issue noone has mentioned is the voltage output, can the diodes and other components survive the high voltage? - -- Tom Parker - tparker@xxx.nz - http://www.geocities.com/MotorCity/Track/8381/ ------------------------------ From: Raymond C Drouillard Date: Sat, 6 Feb 1999 01:25:31 -0500 Subject: Re: fusible link questions Actually, I have considered all of that stuff. I have done those derivations many times. Read again what I wrote about it. What you are missing is the difference between a source and a straight resistance. You can measure the voltage across a resister and the current through it, multiply them together, and you'll know how much power is being dissapated, and how much heat is being generated. Let's do a thought experiment. Let's say that you have a battery and attach a one ohm resister to it. You measure 14 volts across the resister (and the battery). Based on ohm's law, you know that 14 A is flowing through the resister. Then, use any of the equations (P=I*V, P=E^2/R, or P=I^2R) to calculate that 196 watts are being dissapated by the resister. The resister is obviously getting hot. How hot is the battery getting? It has the same 14 amps of current flowing through it. It has the same 14 volts of voltage across it. Why isn't it dissapating 196 watts? Because it is a source. If you had an ideal source, there would be NO power dissapated by the battery. Anyone who has tried to start a stubborn engine will tell you, however that the battery will get warm. It won't get near as warm as the starter, however. A real life battery is represented by a perfect voltage source in series with a resistance. This resistance is refered to as the "internal resistance" of a battery. The internal resistance of a car battery is some small fraction of an ohm. Now, let's get back to our example. The power dissapated by the resister is 196 watts. The power dissapated (not supplied) by the battery is I^2R, where R is the internal resistance of the battery. If the internal resistance of the battery is 0.01 ohm, the battery will dissapate 1.96 watts while it supplies 196 watts to the external load (the resister). Let's expand our model a little more. If we have a battery with an internal resistance of 0.01 that is supplying 14 V at 14 A into an external load, the source voltage is 14.14V. If you increase the load (decrease the resistance), the output voltage will drop because the increased current will increase the voltage drop across the internal resistance of the battery. Did you ever notice that putting a load on a battery causes its output voltage to drop? If not, take a look at the voltmeter next time you start your car. Now, back to this battery that is supplying 14V at 14A. If you could actually directly measure the voltages inside the battery, you would find a 14.14V source and 0.14V dropped by the internal resistance. Subtract them and you get the 14V. THE POWER DISSAPATED BY THE INTERNAL RESISTANCE OF THE BATTERY IS NOT THE PRODUCT OF THE OUTPUT VOLTAGE AND CURRENT OF THE BATTERY. IT IS THE PRODUCT OF THE OUTPUT CURRENT AND THE VOLTAGE DROP OF THE INTERNAL RESISTANCE. Now, let's do something different. Increase the resistance to 4 ohms and reduce the current to 7 amps, and increase the voltage supplied by the battery to 28 V. You still have 196 watts being dissapated by the load. The current has been cut in half, however. Doing some calculations, you'll find (with the same 0.01 ohm internal resistance) that the voltage of the ideal source is 28.07 volts, so the voltage across the internal resistance is 0.07V. Any way you slice it, using any of the three power equations discussed, the power dissapated by the internal resistance is 0.49 Watts. Compare this to the 1.96 watts in the earlier example. THE POWER DISSAPATED BY THE INTERNAL RESISTANCE REPRESENTS THE "LOSS" OF THE BATTERY. OK... so we see that doubling the voltage while supplying the same power will cut the losses by a factor of 4. So, the alternater will run cooler while providing the same power (at a fraction of the current). Now, for a final example, we'll keep the origional rated current of 14 amps and double the voltage to 28 volts. The output power is now 392 watts, or twice the origional 196 watts. The power dissapated by the 0.01 ohm internal resistance is 14*14*.01=1.96 watts. We have twice the power supplied to the load, but the same internal losses, which means the same heat generated. Our efficiency has gone up. Try that with a tenfold increase in voltage and you'll find an even bigger increase in efficiency and capacity. To apply this to an alternater, we need to either look at the average output of the three stator windings, or an instantaneous output. It doesn't matter - it applies equally well. The ideal source voltage is the voltage supplied by the moving magnetic field. The internal resistance is the resistance of the windings and the diodes. There is some heat generated by the field coil and the hysteresis losses in the iron, so we won't be able to get quite as much increase as calculated. We engineers are used to things being less than ideal in practice, however. So... I have spent too much time writing this treatise. If anyone wants to argue, I would recommend that they have the courtasy of reading the entire thing first. If a point is brought up that has been handled in this message, I'll simply point back to the message. I have better things to do with my time than to repeat myself. Ray Drouillard, BSEE On Sat, 06 Feb 1999 16:05:51 -0600 John Hess writes: >Actually, you haven't really considered that: > >while P=IE (the basic formula) >E=IR, therefore substituting IR for E: >P=I*IR or I^2R >I=E/R, therefore substituting E/R for I >P=E*E/R or E^2/R, etc. > >To say that one is not related to the other is to refute Omm's Law. >Note >that this is not electronics theory, it is _LAW_! > > >Raymond C Drouillard wrote: > >> >requires closer to 2.5 HP. Pushing the same current at higher voltage >> >quickly oveheats the alternator due to the increase in these I^2 R >> >losses, which increase at the rate of a power of two with voltage >> >increase - double the voltage, 4 times the heat with the same >current. >> >> I beg to differ, but the I^2 R losses are independant of the output >> voltage of the device >> >> P = I^2 R (hense the name) >> >> That is why power is transmitted using the "high tension" (High Voltage) >> lines. >> >> If you are stuffing current through a resister, the power dissapated by >> that resister is dependant only on the current through the resister and >> the resistance. >> >> P = I^2 R >> >> Let me explain... >> >> Power is defined as voltage * current >> >> P = IV >> >> That is, the current through the resister * the voltage across the >> resister. The voltage across the resister can be calculated using Ohm's >> law. >> >> V = IR >> >> Taking the origional P = IV equation and substituting IR for the V yields >> >> P = I * (IV) = I^2 R >> >> What I am proposing uses exactly the same princable that has always been >> used by the electric suppliers to get power from the source to the >> substations, and finally to your home. >> >> One thing that neither of us mentioned is that some of the heat is >> generated by hysteresis losses in the iron core. That will somewhat >> reduce the ultimate power that can be supplied by the system, but it'll >> still be more than the unmodified system will provide. >> >> Ray Drouillard, BSEE ___________________________________________________________________ You don't need to buy Internet access to use free Internet e-mail. Get completely free e-mail from Juno at http://www.juno.com/getjuno.html or call Juno at (800) 654-JUNO [654-5866] ------------------------------ From: KD6JDJ@xxx.com Date: Sun, 7 Feb 1999 00:52:29 EST Subject: Re: fusible link questions You can buy a box that hooks up to the alternator on most cars that > provides 110VAC for running power tools remotely, while still charging the > battery. I really dont think that this is ever true. It Would be VERY difficult to simultaneously produce 110 VAC and 14 VDC from one alternator. I think that it has never been done with a box hooked up to an alternator. It would be simple to generate 110 vac from an automotive alternator, but it would be 3 phase.(unless you found one of the few single phase alternators ( French for instance) . It is very easy to make 110VDC with a conventional automotive alternator. (the BOX for instance) { The BOX disconnects the alternator output from the 12 VDC battery and connects the field (rotor) to the batterys 12 VDC} . With the BOX in the 110 VDC position , the output terminal (which is now disconnecred from the battery) will go to a higher voltage as the RPM is increased. If the RPM exceeds some limit , the alternator's VDC will destroy the alternator's diodes , so be careful not to overrev the engine. Some companies have wound special stators that fit into Delco alternators so as to make both welding voltage and 110 VDC. I am sure that you could , at home, mahe a very interesting electrical unit from an automotive alternator by winding a special stator. But , it may likely be impractical when you consider the vast number of other sources of project materials are available these days. This information is presented here to (hopefully) put some reality into the thoughts about the use of common automotive alternators. I sure dont want to get into any discussions about when the above information may not necessarily be true. Jerry ------------------------------ From: Raymond C Drouillard Date: Sat, 6 Feb 1999 02:05:20 -0500 Subject: Re: fusible link questions >What is the current in and voltage across the field windings necessary to >generate the power proposed? I'm willing to bet that it isn't that huge. >Certainly the full power output proposed by the original poster will not be >dissapated in the field windings. > >I've no idea what current is normally presant in the field windings, but I >wouldn't expect it to be large, and I would expect that it could be increased >significantly before you have problems with overheating. I hadn't even considered the power dissapated by the field coil. It should be small compared to the power dissapated by the rest of the system. If not, it can be replaced with a ceramic or rare earth permenent magnet. > >The other issue noone has mentioned is the voltage output, can the diodes and >other components survive the high voltage? I didn't mention it, but my plan is actually to remove the diodes when I remove the regulater, and bring three or four wires (depending on if I go with Y or Delta) out and mounting the rectifiers and switching regulater in a friendlier location. I would use rectifiers with the appropriate ratings. Ray Drouillard > >-- >Tom Parker - tparker@xxx.nz > - http://www.geocities.com/MotorCity/Track/8381/ > > ___________________________________________________________________ You don't need to buy Internet access to use free Internet e-mail. Get completely free e-mail from Juno at http://www.juno.com/getjuno.html or call Juno at (800) 654-JUNO [654-5866] ------------------------------ From: Chris Conlon Date: Sun, 07 Feb 1999 02:48:07 -0500 Subject: Re: Digital: fuel pressure gage; boost gage At 05:00 PM 2/6/99 -0600, Clarence Wood wrote: > Is it possible to use the pressure sensing device in a digital scale to > build such things as an electronic fuel pressure gage, or an electronic > boost gage, or a boost control? My guess is that a scale uses a strain gauge of some sort. To use it as a pressure sensor you'd have to have some kind of A/B diaphragm arrangement. Can be done, but ick. If you want to measure *air* pressure you have lots of choices. Grab yer Digikey catalog and look up "pressure sensors". Or my own personal faves, Motorola's MPX line, which are oodles cheaper. If you need to measure water/fuel/oil pressure, Digikey is still yer buddy, they have stainless steel units in pipe thread packages. Think oil pressure sender, I guess. I could use one of these as my fuel pressure sensor, but I'm an effin' cheap bastard and they're like $80-$200 depending. The Motorola air-only units are like ~$15, the new automotive specialty ones that Tom mentioned are, uhh, did anyone see a part number or a price? I'm desperate tho, and Moto likes me so I'll try to call Monday and bum a few offa them. Hope this helps, Chris C. ------------------------------ End of DIY_EFI Digest V4 #95 **************************** To subscribe to DIY_EFI-Digest, send the command: subscribe diy_efi-digest in the body of a message to "Majordomo@xxx. A non-digest (direct mail) version of this list is also available; to subscribe to that instead, replace "diy_efi-digest" in the command above with "diy_efi".