PA28-140 Starting Issues

[AV8R_87]

You don't seem to understand the series-wound electric motor as used in a starter.

Trust me, I do.
Switching current through an armature, with actual gaps between the reversal, as in the case of a commutated DC motor, is not the same thing as a sinusoidal wave (* I'll give you another example at the end on how pulsed currents completely modifiy the reactive behavior of capacitors and inductors). Allowing the magnetic field to collapse between polarity reversals changes things a lot.

And, as I mentioned before, which you seemed to ignore, if your theory was correct then field weakening wouldn't work, because the reactance of the armature wouldn't change just because you reduce the current in the field winding.

Yet I task you with another experiment, which you will probably ignore: take a starter motor and measure its field winding resistance. Then apply power to it and measure (subjective measurement is ok if you don't have the proper tools) its no load RPM. Then install a resistor in parallel with the field winding, of a similar value to the field winding resistance. Power the motor again and measure the RPM. Feel free to repeat that under a load, if you don't think the no load results are relevant.

Explain how the RPM significantly increased when the only change was to the field current, when according to your inductive reactance, with an increase in RPM there should've been an increase in reactance which should've limited the current even more.

Also, please use your own words. Copy/pasting three pages of stuff that mostly doesn't apply does't prove anything. If you still think I'm a complete idiot that does't have a clue about how electrical machinery works, explain it to me like I'm 5, in four paragraphs or less. See above for example.

And as far as 77/80 compression being the same as a small resistance in a circuit? That's a false equivalency.

It's a simple comparison of what 3% degradation would look like in typycal maintenance terms that are non-electrical. Think of it like a methaphor. Think of a rubber fuel hose that swelled a bit as it aged, reducing its internal cross-section by 3%, causing fuel starvation on take-off power. That would be a pretty marginal fuel system design, wouldn't it?

*Here's your bonus reactance example, completely off-topic otherwise, which is also a good PSA for anyone using small electronics with inverters:
Most small electronics use what's called a capacitive droper in their power supply, where its reactance at 60Hz causes it to behave like a resistor, but without the heat dissipation. All fine with a nice sine wave from the outlet, and even with a typical generator that might output a sine wave with 10% THD (Total Harmonic Distorsions).
When used with a "Modified Sine Wave" inverter, which is fancy name for a square wave with a reduced duty cycle (quite similar to how the current looks in your starter motor armature, by the way), the inductive reactance is much lower than when used on a sine wave which causes them to overload and burn out the rest of the circuit.
Even worse, most "Pure Sine Wave" inverters will use some form of rapid waveform switching to control overloads. So while your capacitive dropper power supply circuit functions fine, the moment you subject the inverter to a momentary overload (think power tools or other motor loads with a large inrush current), it will start to rapidly pulse the output, causing that capacitor circuit to not function as designed (because reactance drops significantly with pulsed, non-sinusoidal power), overloading and destroying your device. Think of a Kill-A-Watt power monitor plugged into your inverter. Or some other low draw device. Phone and laptop chargers should handle this a lot better.

Here's how your typical generator power output looks like (about 7% THD):

 

[Dan Thomas]

It's a simple comparison of what 3% degradation would look like in typycal maintenance terms that are non-electrical. Think of it like a methaphor. Think of a rubber fuel hose that swelled a bit as it aged, reducing its internal cross-section by 3%, causing fuel starvation on take-off power. That would be a pretty marginal fuel system design, wouldn't it?

An increase in resistance across a connection or set of contactor contacts from zero to .02 ohms is not a 3% increase. Or, if you want to get fussy, an increase from .0001 ohms to .02 ohms. That's what hurts. It's a 200-fold increase, or a 19,900% increase. It is not tolerable, and installing a bigger battery will not fix it. It's not the battery that is marginal; it's the circuitry itself because it was not maintained. I don't understand why you cannot see it.

Switching current through an armature, with actual gaps between the reversal, as in the case of a commutated DC motor, is not the same thing as a sinusoidal wave

An inductor has the effect of smoothing choppy stuff like that, turning it into more like a sine wave. The inductor resists an increase in current, creating an upward slope, and it also resists a decreasing current, with the collapsing field boosting the flow, creating a downward curve.

In any case, it IS AC and it does create inductive reactance, which is why the increasing RPM reduces the draw. And in the series-wound motor, the field current equals the armature current, as with any other series circuit, so why all the stuff about sticking a resistor across the field to prove something? If load increases, RPM falls, current in BOTH the field and armature increase, and torque increases. It's not magic.

 

[bentmettle]

Which one of these posts explains how/why the internal resistance of the battery of the battery can be ignored when looking at the current supply loop?

There was a lot of back and forth over the word "marginal" and I think had someone used the term "robust" instead, the conversation might have converged a bit.

 

[Dan Thomas]

Which one of these posts explains how/why the internal resistance of the battery of the battery can be ignored when looking at the current supply loop?

There was a lot of back and forth over the word "marginal" and I think had someone used the term "robust" instead, the conversation might have converged a bit.

From Quora:


Distinguishing between good and bad 12V lead-acid batteries using internal resistance involves measuring the internal resistance and comparing it to acceptable values. Here’s a step-by-step guide:

1. Understanding Internal Resistance

• Internal Resistance (IR): This is the opposition to current flow within the battery. A lower internal resistance indicates a healthier battery, while a higher value suggests aging or damage.

2. Measuring Internal Resistance

• Equipment Needed: Use a battery tester or a multimeter with a specific function to measure internal resistance.
• Procedure:

1. Fully charge the battery before testing.
2. Disconnect the battery from any load.
3. Use the tester to measure the internal resistance, following the manufacturer’s instructions.

3. Interpreting the Results

• Typical Values:
  
  • Good Battery: Generally, a healthy 12V lead-acid battery will have an internal resistance of around 5 to 10 milliohms (mΩ) when fully charged.
  • Moderate Resistance: Values between 10 to 20 mΩ may indicate some deterioration but could still be usable.
  • Poor Battery: Values above 20 mΩ usually indicate a failing battery that may not hold a charge effectively.

So, 5 to 10 milliohms in a good battery. That is .005 to .010 ohms. That causes some of the voltage drop under high current draws. But the hypothetical .02 ohms I postulated in a bad contactor or cable is double the worst-case reading for a good battery, four times worse than the best-case battery, and it is something we can fix by cleaning/repairing connections or replacing aged contactors or cables, once we know where the fault lies. Like any other resistance in a series circuit, the battery's internal resistance limits its performance and we have to accept it, but we sure don't have to accept rotten circuitry, and we can't fix the rotten circuity by replacing a good battery with another good battery or even a bigger battery.

There are fairly young batteries out there that have been ruined by poor maintenance practices. Every new battery comes with a brochure that tells the maintainer how to fill it with acid and how much acid to put in, how to leave it sit for awhile and then adjust the level again, then put it on charge before adding any more acid up to the split ring. But so many guys just fill it to the split ring, whereupon the heating during charge, and the hydrogen bubbles forming on the plates, expand that electrolyte and it overflows once it reaches the top ends of the splits in the ring, as the gas can no longer escape. I used to see it all the time: battery boxes eaten out by overflowed acid from that, and from guys topping off that battery to the split ring while it was not being charged as per the instructions. I used to top them off to halfway between the tops of the plates and the bottom of the split rings, and they never overflowed. With customer's airplanes at annual I took the battery out and charged it and adjusted the electrolyte as per the manuals, tested the battery for capacity, then recharged it.

See now what happens to the pH of the battery acid when it overflows and is replaced topping off with distilled water? It gets weaker. Do that a few times and the battery's capacity is shot because the electrolyte is more water than acid, and the hydrometer shows it as a lower specific gravity.

Read the manuals. The makers of the stuff know way more about their products than we ever will. Batteries aren't cheap.

 

[Albany Tom]

It's great to understand the theory behind how a DC motor works, but it doesn't take away from the fact that bad wiring in a starter circuit is a DC, not AC, problem. As Dan has pointed out, the problem is at the sub ohm level, and you're not going to measure that with any normal DMM by itself, or even accurately with a 4 wire ohms DMM. IMHO, you need to put about 10 amps over the wiring in question, and measure the DC voltage drop over it. That, and only that or an equivalent or larger load, is going to tell you if the contacts, terminals, and wire are OK and if not where the problem is.

As at least one person pointed out, some old PA-28's used to have aluminum wiring. Now aluminum isn't inherently bad....my service entrance is Al, as is pretty much all the distribution wiring in the country. But I've seen what 40 year old aluminum wiring in an industrial setting can look like, and it's not pretty. Might be fine, might not be. It has been known to cause problems.

As far as a bigger battery goes? I'm thinking of the recent Pink Floyd quotes. If the lead singer has the flu, could you give him some speed and continue with the show? Sure. Might not be what I'd do if he were my kid, though.

Just my 2 cents. Not an A&P. Have fixed a few industrial electrical things over the years.

 

[Bell206]

IMHO, you need to put about 10 amps over the wiring in question, and measure the DC voltage drop over it. That, and only that or an equivalent or larger load, is going to tell you if the contacts, terminals, and wire are OK and if not where the problem is.

And that same for checking voltage drops on the ground side as well. There are even times when checking the ground side 1st can get you to the problem quicker.

 

[AV8R_87]

And in the series-wound motor, the field current equals the armature current, as with any other series circuit, so why all the stuff about sticking a resistor across the field to prove something? If load increases, RPM falls, current in BOTH the field and armature increase, and torque increases. It's not magic.

It's also something you fail to understand. I've given you a practical example that is easy to replicate which will prove your inductive reactance assumption wrong.
In any real (over 5kW or so) series wound motor application you will see resistors that bypass a portion of the field current to counteract the back-EMF and attain a higher motor speed while still providing excellent starting torque. If your therory was correct this wouldn't be possible.

In any case, it IS AC and it does create inductive reactance, which is why the increasing RPM reduces the draw.

You're literally claiming the earth is flat in electrical terms.

An inductor has the effect of smoothing choppy stuff like that, turning it into more like a sine wave. The inductor resists an increase in current, creating an upward slope, and it also resists a decreasing current, with the collapsing field boosting the flow, creating a downward curve.

Do you even have any idea what inductance level we're talking about?

About 0.3 to 0.5 mH for each of the 12 or so windings in the rotor. The brushes usually contact more than one segment on the commutator, so you end up with the total inductance being half of that, about 0.2mH
At your typical cranking speeds (let's be generous and say 120 RPM), the starter, turning at roughly 10 times that speed, is at 1200RPM.That's 20 revolutions per second. You have 12 or so pole pairs in the rotor, so the pole switching frequency is 1.66Hz. Let's be generous again and round it up to 2Hz. Even if that happened in clear air, and not in a rotating magnetic field, and it was a pure sine wave and not a weird chopped up reduced duty cycle, your inductive reactance for 0.2mH would add up to a whopping 2.5 milliohms.
Your example starter (560A stall current at 4 volts) has an internal resistance of 7.1 milliohms. Assuming 6V at the starter, at 0 RPM it would draw 845A, and if your theory was correct, it would draw 625A at nominal cranking RPM.
Yet we all know the starter isn't pulling that much. It would completely deplete the battery after about two minutes of cumulative cranking (more like a minute, minute and a half in practice), and we've all seen people with flooded engines cranking them (sometimes with brief pauses, although a lot of people seem to ignore the duty cycle requirements) for more than that.

Now, going by your no-load numbers, (75A at 10V, 5500RPM), and knowing the resistance of the windings (and I will humor you by including the very optimistic inductive reactance numbers, multiplied by 4) will give us an effective voltage across the windings of only 0.95V. Where are the remaining 9 volts?
That is your back-EMF, at a field current of 75A. You decrease the RPM and your back-EMF decreases proportionally, but because your field current increases that will restore a good proportion of the back-EMF. At about 220A and 1200RPM, you end up with about 6V of back-EMF. The battery is around 8V at that current, the copper wiring (including contactors) is 2-3 milliohms, plus the 7.1 milliohms in the starter. So you apply an effective 2 volts across 10 milliohms, which gives you a steady state cranking current of 200A. Now isn't that more in line with reality?
Especially since the master/start solenoids are usually rated for 150A continuous and 300A inrush.

An increase in resistance across a connection or set of contactor contacts from zero to .02 ohms is not a 3% increase. Or, if you want to get fussy, an increase from .0001 ohms to .02 ohms. That's what hurts. It's a 200-fold increase, or a 19,900% increase.

You refuse to see the forrest because of the trees, confusing component-level degradation margin with system-level degradation margin, which is what we were talking about.

It is not tolerable, and installing a bigger battery will not fix it. It's not the battery that is marginal; it's the circuitry itself because it was not maintained. I don't understand why you cannot see it.

You have a 12V system where a 3% voltage drop in the wiring and contactors is causing starting problems, because the battery drops to 6V during cranking.

I understand why the system is like that, from a design perspective.
I understand that your hands, as a maintainer, are tied and you can't do anything about the battery, except replace it when it's degraded enough. Your choices are limited to doing what you're doing, ensuring that the solenoids and wiring are in good shape, and that all connections are clean and tight.

But just because you do that and that restores the system to (a sort of) operational state doesn't negate my original statement that got you on this thread drift crusade to prove me wrong.
The battery is marginal, performance-wise. It is less of an issue in installations where it is mounted on the firewall, because you get a couple milliohms less of resistance due to the shorter wiring. At that point you can afford an extra milliohm or two of resistance due to normal aging and wear of the wiring and contactors. But in a PA28, the extra wire length eats into that narrow margin even more. You don't really have any margin when that battery sags to 6-7 volts when you hit the starter switch. Now if the battery was larger and able to hold voltage better under load, then an extra milliohm or two in the wiring and contactors wouldn't be a problem.
It might even be marginal when it comes to capacity, once you apply the derating factors that are normally used for IFR requirements. The total load can't be more than 50A to ensure you get 30 minutes of battery operating time if you lost the alternator in IMC. By the time you add pitot heat and other high loads that are typical in modern instrument panels (LCD screens have heaters for low temperature operation, and you have to account for that in the calculations) things might get pretty close.

Anyway, enough of this. I do appreciate the fact that we've kept this disagreement civilized, and I hope others have learned a thing or two from it, even if it wasn't the original intent of this thread. You do have a lot of knowledge (and I appreciate the fact that you share it with us) and a good grasp on basic electrical systems knowledge, but there are some misconceptions in the electric machines theory part. Thank you for the challenge, though - it was nice to dig through stuff that I haven't touched in a while.