 |
Matronics Email Lists
Web Forum Interface to the Matronics Email Lists
|
| View previous topic :: View next topic |
| Author |
Message |
user9253
Joined: 28 Mar 2008
Posts: 1935
Location: Riley TWP Michigan
|
 Posted: Tue Aug 19, 2008 8:07 am Post subject: High resistance pops breaker? |
 |
|
There was a "TECHNICAL COUNSELOR" article written in the April 2008 issue of Sport Aviation about the landing light tripping a circuit breaker due to high resistance connections. You can read the complete article here:
http://www.eaa.org/sportaviation/2008/april/0804_tc.pdf
In the August 2008 issue, a letter to the editor questions why a higher resistance would result in higher current. Following is the "TECHNICAL COUNSELOR" author's reply in the Aug 2008 Sport Aviation:
George's Reply:
Using Ohm's and Watt's laws, if a connection resistance increases, the voltage drop across that resistance increases, which increases the current draw and power demands on the circuit. The power draw of the engine and the lamp increase as voltage drops. Adding to this issue is the fact that an engine-operated landing light is marginal at best in terms of the circuit breaker rating; that is, with the motor drive and the lamp, you're already close to the trip set point. As the resistance increases in the contact stacks—in this case because of corrosion ... the voltage drops across each of these series connections and in turn causes the motor to see less voltage and draw more current to reach its power, which in turn would trip the breaker in concert with the light turning on, using voltage from the same source. Since the watts for the motor and lamp remain relatively constant, the current rises.
As such, when you add in the voltage drop across even minor resistances in a stack of contacts, due to the voltage drop, the current draw over the entire circuit increases to the point where the circuit breaker trips.—George Wilhelmsen
George Wilhelmsen holds a commercial certificate with an instrument rating and has more than 1,000 hours of flight experience. He has a bachelor’s degree in engineering technology with a background in DC, analog, and digital controls. He flies a Beech Debonair.
End of quotes. Now this is Joe writing. The landing light in question had an electric motor to retract the light into the wing when not in use. Motors do draw more current when operated at less than design voltage. Perhaps the author understands ohm's law but just did not explain it very well. I got the impression that he was saying that a lamp uses more current if the electrical connections have higher resistance. That is not true. I wonder, could this possibly be the same George?
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
http://www.matronics.com/Navigator?AeroElectric-List |
|
_________________
Joe Gores |
|
| Back to top |
|
 |
dale.r(at)cox.net
Guest
|
| Posted: Tue Aug 19, 2008 9:00 am Post subject: High resistance pops breaker? |
|
|
user9253 wrote:
| Quote: |
There was a "TECHNICAL COUNSELOR" article written in the April 2008 issue of Sport Aviation about the landing light tripping a circuit breaker due to high resistance connections.
..
|
| Quote: | George's Reply:
Using Ohm's and Watt's laws, if a connection resistance increases, the voltage drop across that resistance increases, which increases the current draw and power demands on the circuit. The power draw of the engine and the lamp increase as voltage drops.
..
|
| Quote: | George Wilhelmsen holds a commercial certificate with an instrument rating and has more than 1,000 hours of flight experience. He has a bachelor’s degree in engineering technology with a background in DC, analog, and digital controls. He flies a Beech Debonair.
End of quotes. Now this is Joe writing. The landing light in question had an electric motor to retract the light into the wing when not in use. Motors do draw more current when operated at less than design voltage. Perhaps the author understands ohm's law but just did not explain it very well. I got the impression that he was saying that a lamp uses more current if the electrical connections have higher resistance. That is not true. I wonder, could this possibly be the same George?
--------
Joe Gores
|
This is what happens when one writes an answer before s/he's had hir
morning cocoa/coffee (I've done that a few times myself.)
True enough, the high resistance connection does increase the voltage
~drop~ across the connection, but it is - in effect - subtracted from
the voltage across the total circuit; it becomes a circuit of series
loads. It's a shame that got past the proof-readers into a magazine with
world-wide distribution.
Dale R.
COZY MkIV #0497
Ch.12; Ch's 13, 16, 22 & 23 in-progress
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
http://www.matronics.com/Navigator?AeroElectric-List |
|
|
|
| Back to top |
|
|
nuckolls.bob(at)cox.net
Guest
|
| Posted: Tue Aug 19, 2008 5:48 pm Post subject: High resistance pops breaker? |
|
|
At 09:55 AM 8/19/2008 -0700, you wrote:
| Quote: |
user9253 wrote:
>
>
>There was a "TECHNICAL COUNSELOR" article written in the April 2008 issue
>of Sport Aviation about the landing light tripping a circuit breaker due
>to high resistance connections.
...
>George's Reply:
>Using Ohm's and Watt's laws, if a connection resistance increases, the
>voltage drop across that resistance increases, which increases the
>current draw and power demands on the circuit. The power draw of the
>engine and the lamp increase as voltage drops.
...
>George Wilhelmsen holds a commercial certificate with an instrument
>rating and has more than 1,000 hours of flight experience. He has a
>bachelor’s degree in engineering technology with a background in DC,
>analog, and digital controls. He flies a Beech Debonair.
>
>End of quotes. Now this is Joe writing. The landing light in question
>had an electric motor to retract the light into the wing when not in
>use. Motors do draw more current when operated at less than design
>voltage. Perhaps the author understands ohm's law but just did not
>explain it very well. I got the impression that he was saying that a
>lamp uses more current if the electrical connections have higher
>resistance. That is not true. I wonder, could this possibly be the same
>George.
>
>--------
>Joe Gores
>
This is what happens when one writes an answer before s/he's had hir
morning cocoa/coffee (I've done that a few times myself.)
True enough, the high resistance connection does increase the voltage
~drop~ across the connection, but it is - in effect - subtracted from the
voltage across the total circuit; it becomes a circuit of series loads.
It's a shame that got past the proof-readers into a magazine with
world-wide distribution.
|
It is not an automatic thing that motors draw more
current as the voltage goes down. There is a motor
torque constant (Kt) which is the effort delivered
at the shaft stated as a ratio of torqe/amps.
See:
http://www.portescap.com/catalog/35GLT%20spec%20page%207.08.pdf
Note the seven items under "Intrinsic Parameters". These
numbers enable the designer to predict any motor's performance
in a design. The item were interested in is the line
labeled mNm/A or millinewton-meters per amp.
Now, if the motor is driving a load where torque demand
is a function of speed, then the current draw of the motor
will be right in step with that function. So, let's visualize
a motorized landing light. If we crank it slower, would we
expect it to take more torque? Similarly, hydraulic pumps
resist rotation pretty much in proportion to the pressure
being exerted by the pump . . . if the flow in the system
goes down due to reduced speed, would we expect the pressure
to automatically get bigger?
I won't say that ALL motor driven systems demand the
same or more torque as speed goes down . . . but it's
an unusual system. Further, it's definitely not a system that
is raising or lowering a landing light against friction,
gravity and air-loads.
Motors slow down for two reasons, reduction in supply voltage
which is linked to the Back EMF constant. If torque demands
are the same, current will be the same.
The motor slows down because it is loaded heavier. Torque
goes up, losses in the motor terminal resistance goes up
and what's left over is treated just like a reduction
in supply voltage. Kt is still in play here and if the
torque is high enough, it may very well cause a breaker
to open.
But increases in series resistance alone are almost
never a recipe for tripping breakers.
Now, it occurs to me that the "high resistance joint"
was in fact a loose connection . . . i.e. an "intermittent
low resistance joint". Here's a case where rattling a
motor + lamp inrush currents MIGHT account for a
continuous stressing of a breaker with a series
of high current loads with a duty-cycle too low to
heat up the lamp+spin the motor but sufficiently
irritating to a breaker to trip it.
In this case, calling it a "high-resistance joint"
causes the wrong physics to be applied to the hypothetical
analysis. Low average duty cycle voltage to a motor-
lamp combination not only piles the admittedly
high lamp inrush on top of the low back-emf of a
non or very slowly spinning motor.
Bob . . .
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
http://www.matronics.com/Navigator?AeroElectric-List |
|
|
|
| Back to top |
|
|
user9253
Joined: 28 Mar 2008
Posts: 1935
Location: Riley TWP Michigan
|
| Posted: Wed Aug 20, 2008 11:04 am Post subject: High resistance pops breaker? |
|
|
Bob,
If I understand correctly, as the landing-light-retract motor slows due to increased circuit resistance (and thus lower voltage at the motor), the motor current will not increase because the torque required to operate the retract mechanism is the same as before. The power (watts) of the motor is less because power equals speed times torque, and the speed has decreased. The total energy used to operate the mechanism will be about the same because it is a function of power and time. Even though the motor is less powerful operating at a lower voltage, it runs for a longer time and thus uses about the same total power as it would operating at a higher voltage for a shorter time.
It seems to me that a slower motor develops less back-EMF and therefore would allow more current to flow. But on the other hand, higher resistance in the circuit counteracts that. I do not know how to calculate that, so I will take your word for it that the current stays the same. Suppose there are no high resistance connections and one compares the motor current at two different voltages, say 14 volts compared to 11 volts with a failed alternator. What happens to the motor current as the supply voltage drops? From what you wrote, I assume that the current will not increase.
I am not trying to contradict anything that you said. I just want to understand. My experience has been with AC motors that try to maintain their synchronous speed by increasing current draw when heavily loaded or when operated at a lower voltage. DC motors do not have any certain speed that they strive for.
Joe Gores
[quote][b]
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
http://www.matronics.com/Navigator?AeroElectric-List |
|
_________________
Joe Gores |
|
| Back to top |
|
|
Eric M. Jones
Joined: 10 Jan 2006
Posts: 565
Location: Massachusetts
|
| Posted: Wed Aug 20, 2008 4:24 pm Post subject: Re: High resistance pops breaker? |
|
|
I was going to test this but it soon became apparent that the problem was beyond mere testing.
One consideration that I have not seen is that breakers wear out. "I squared T" or current squared X time" is how breakers blow, but the problem has many back alleys of complication. Once a beaker has tripped at above a specified current, the breaker will be degraded and trip thereafter usually at a lower current. All breakers have a one-time maximum trip current. (This is what is wrong with crowbar OVPs).
So--can a landing light blow a circuit breaker below the specified lamp current if there are some high-resistance connections? My guess is "yes" but it depends on the breaker, the lamp, the nature of the loose connections, the wire, and myriad other variables. The likely culprit is the breaker.
"When trouble arises and things look bad,
there is always one individual who perceives
a solution and is willing to take command.
Very often, that individual is crazy.
--Dave Barry"
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
http://www.matronics.com/Navigator?AeroElectric-List |
|
_________________
Eric M. Jones
www.PerihelionDesign.com
113 Brentwood Drive
Southbridge, MA 01550
(508) 764-2072
emjones(at)charter.net |
|
| Back to top |
|
|
nuckolls.bob(at)cox.net
Guest
|
| Posted: Sat Aug 23, 2008 9:13 am Post subject: High resistance pops breaker? |
|
|
At 05:24 PM 8/20/2008 -0700, you wrote:
| Quote: |
I was going to test this but it soon became apparent that the problem was
beyond mere testing.
One consideration that I have not seen is that breakers wear out. "I
squared T" or current squared X time" is how breakers blow, but the
problem has many back alleys of complication. Once a beaker has tripped at
above a specified current, the breaker will be degraded and trip
thereafter usually at a lower current. All breakers have a one-time
maximum trip current. (This is what is wrong with crowbar OVPs).
|
I've heard this idea handed down from sage-to-
acolyte over campfire and beers for decades. My
brother-in-law was told this by a journeyman electrician
while he was getting his ticket to string wire . . .
but my friends at Cutler-Hammer, Klixon, Mechanical
Products and others cannot embrace this legend.
I'll refer the readers to Mil-C-5809G available at:
http://www.aeroelectric.com/Reference_Docs/Mil-Specs/5809G.pdf
and specifically to paragraph 4.7.14 where products
qualified to military applications (and most commercial
aircraft applications) are subjected to huge interruption
current tests from both a closed condition (breaker opens
due to operational fault) and open condition (breaker
is closed into a pre-existing fault) trip test.
Referring to one of many specs for the breakers
recommended for aircraft at:
http://www.aeroelectric.com/Reference_Docs/Circuit_Breaker/MS3320.pdf
In Table III we find the range of interrupt current tests
that each size breaker is subjected to. Note that
the numbers (particularly categories C & D for 28v
systems) are in the thousands of amps. Many times higher
than stress to a crowbar-tripped breaker. Yet after being
subject to these tests, they are required to past the
200% overload trip calibration.
An interesting feature of the interrupt current test,
only one or two devices out of a larger QPL sample
are subjected to this test while greater numbers of
parts are tested more rigorously for other qualities.
Could it be that risk for failure due to interruption
current stress is so low that folks never seem to fail
it . . . while other qualities of a breaker demand
closer monitoring?
For the benefit of our readers, know that for a
manufacturer to sustain membership in the club of qualified
suppliers, a sampling of their continuous production output
receives the same battery of tests every so often in their
production cycle. For devices strongly affected variables
in materials and process, qualified products lists (QPL)
are maintained wherein the various manufacturer's pay their
procedural dues to stay listed. Fall off that list and the
military cannot purchase your product no matter how good
you were when initiated into the club.
I have offered this documentation before and not
one of my detractors had commented on it. For folks
who are so invested in manufacturer's data sheets
as infallible, I'm mystified as to the ease with which
repeatable experiments levied by qualification test
plans and membership on qualified products list are
so easily ignored . . .
Finally, 5A Klixon breakers incorporated into production
test stands for alternators/regulators fitted with
crowbar OV protection were subjected to hundreds
of crowbar events over a period of years . . . with
no observed degradation of breaker performance. It would
be interesting to retrieve one of those breakers and
run the 200% trip time test on them.
| Quote: | So--can a landing light blow a circuit breaker below the specified lamp
current if there are some high-resistance connections? My guess is "yes"
but it depends on the breaker, the lamp, the nature of the loose
connections, the wire, and myriad other variables. The likely culprit is
the breaker.
|
Hmmmm . . . . connections, lamp, breaker, wire, and
"myriad of other variables". This shopping list of
'maybes' implies a sage understanding but without
transfer of understanding to your readers.
How can the condition of wire exacerbate breaker
tripping? If the breaker is not undersized by error of
system design, how does the breaker itself participate
in a scenario of nuisance tripping? 99.999% of all
breakers run the lifetime of an airplane never being
called upon to do their job. Qualification demands
thousands of demonstrated operations before they are
allowed on the airplane. Share with us a rationale
for your assertion: "likely culprit is the breaker".
If we are deprived of some knowledge/understanding
of physics, repeatable experiments or artfully analyzed
failures, please help us out.
Bob . . .
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
http://www.matronics.com/Navigator?AeroElectric-List |
|
|
|
| Back to top |
|
|
nuckolls.bob(at)cox.net
Guest
|
| Posted: Sat Aug 23, 2008 10:35 am Post subject: High resistance pops breaker? |
|
|
At 03:03 PM 8/20/2008 -0400, you wrote:
| Quote: | Bob,
If I understand correctly, as the landing-light-retract motor slows due to
increased circuit resistance (and thus lower voltage at the motor), the
motor current will not increase because the torque required to operate the
retract mechanism is the same as before.
|
Correct.
| Quote: | The power (watts) of the motor is less because power equals speed times
torque, and the speed has decreased. The total energy used to operate
the mechanism will be about the same because it is a function of power
and time.
|
Correct.
| Quote: | Even though the motor is less powerful operating at a lower voltage, it
runs for a longer time and thus uses about the same total power as it
would operating at a higher voltage for a shorter time.
|
Correct.
| Quote: | It seems to me that a slower motor develops less back-EMF and therefore
would allow more current to flow.
|
If all other variables are the same, then the only way the motor
slows down is by increasing the load on the motor. Increased
torque translates to increased current. Now, the only reason
the motor slows down is because of resistance in the loop drops
the effective motor-supply voltage due to losses in the system.
| Quote: | But on the other hand, higher resistance in the circuit counteracts
that. I do not know how to calculate that, so I will take your word for
it that the current stays the same.
|
I appreciate your profession of faith but it is
not necessary to make your own understanding
subservient to mine. Take a peek at:
http://www.aeroelectric.com/Reference_Docs/Misc_PDF/Motors_as_Energy_Converters.pdf
This is but one of thousands of down-loadable
treatments of this topic available on the 'net.
I just got home from a three day trip to a client's
facility to bring this same information to their
non-electrical engineering staff. Everybody walked
out of the discussions with a good working knowledge
of motor performance.
Note that the speed torque curve for this motor
has SLOPE. I.e, more torque results in increased
current flow which increases the drop in effective
applied voltage that is manifested by a lower RPM.
The speed torque curves for a motor ASSUME constant
terminal voltage at the motor. This is a necessary
constraint on scope of data because the motor
manufacturer has no control over EXTERNAL resistance
in the system. However the system designer is
obligated to include both internal and external
resistance to anticipate as-installed motor
performance.
Stall torque (and current) is a function of applied
voltage divided by total resistance. This is plain
ol' Ohm's law. One may readily deduce that the slope
of the curve is made steeper by increasing total
resistance; flatter by reducing total resistance. But
the current demanded by the motor for any condition
is firmly locked to effort in the motor shaft.
Note that for any given torque, there is a predictable
level of current demanded by the motor. Whether or not
the motor can sustain speed necessary for performance
at that load is directly affected by TOTAL circuit
resistance. The astute system designer draws a NEW
speed torque curve that has the same no-load speed
but a steeper slope and therefore lower stall torque
combined with a higher sensitivity of speed to torque.
If we could build a super-conducting motor, the
speed-torque plot would be a horizontal line. But
if the motor were super-conducting and external
wiring was not, the "flat line" motor would perform
to a new curve that has some small, but predictable
and perhaps significant slope.
| Quote: | Suppose there are no high resistance connections and one compares the
motor current at two different voltages, say 14 volts compared to 11 volts
with a failed alternator. What happens to the motor current as the supply
voltage drops? From what you wrote, I assume that the current will not
increase.
|
Correct. The speed of the motor is given by:
RPM = (Eapplied - I*Rtotal)/(Volts/RPM)
Where
I*Rtotal describes depression of effective voltage
due to losses.
Volts/RPM describes the CEMF constant Ke
and I is a function of torque at the motor shaft.
One may easily deduce that with all other things
constant, an increase in torque reflects as a
proportionate increase in I resulting in more
losses through R. Speed MUST therefore go down
as the effect of a decrease in effective operating
voltage.
Here's a more detailed discussion of motor
characteristics for those who are interested.
http://www.aeroelectric.com/Reference_Docs/Misc_PDF/Motors_as_Energy_Converters.pdf
| Quote: | I am not trying to contradict anything that you said. I just want to
understand. My experience has been with AC motors that try to maintain
their synchronous speed by increasing current draw when heavily loaded or
when operated at a lower voltage.
|
AC motors are more like a transformer-in-motion. Shaft
effort is a function of voltage induced in rotor conductors
by the low frequency AC difference between rotating
magnetic field in stator windings as "primaries" and
conductors in the squirrel cage as "secondary" windings.
The coupling coefficient for energy rises as the DIFFERENCE
frequency (hence increase in delta-velocity between ac
mains stator and induced currents in rotor) get larger
when rotor speed goes down due to loading. The physics which
control behavior of the induction motor are quite apart from
that of DC motors. While the two technologies operate on the
same rudimentary rules of magnetics and current flow, one
cannot accurately predict all behaviors of one from even the most
complete understanding of the other.
| Quote: | DC motors do not have any certain speed that they strive for.
|
Yes, they do. In AC motors, no-load (synchronous) speed
is driven by line frequency and numbers of poles in the
stator windings. In a DC motor, no-load speed is a
function of applied voltage divided by Ke when losses
are quite low. As I mentioned earlier, if losses could be
driven to zero, then speed torque would be a flat line
defined by Eapplied/Ke. I think there's a TON of confusion out
there about DC motors based on folks real life experiences
and observations of AC motors. They're entirely different
technologies.
Bob . . .
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
http://www.matronics.com/Navigator?AeroElectric-List |
|
|
|
| Back to top |
|
|
user9253
Joined: 28 Mar 2008
Posts: 1935
Location: Riley TWP Michigan
|
| Posted: Mon Aug 25, 2008 8:08 am Post subject: High resistance pops breaker? |
|
|
DC motors have the greatest torque and current when they are standing still and voltage is first applied. According to my old college book on basic electricity, a DC motor (with shunt winding) starting current can be up to 100 times greater than the full load running current. For larger motors, a resistance type motor-starter is used to limit the starting current to prevent damage to the motor and also to prevent excessive acceleration that could damage the load. Adding resistance to the circuit during starting slows the acceleration and prolongs the time that it takes for a motor to attain normal running speed.
Getting back to that Sport Aviation article that claimed that "High Resistance Pops Breaker". The circuit breaker in that landing light circuit might have been sized to trip if the motor current was high due to a jammed landing-light retract mechanism. But due to the time-delay characteristics of circuit breakers, the breaker would not trip during normal motor starting even though the starting current is above the breaker trip point for a short time. Is it possible that there is some value of resistance (due to bad connections) that, when added to the circuit, could allow motor-starting current to flow that is still above the circuit breaker trip point, but for a prolonged period of time? Could the author of that Sport Aviation article be correct that bad connections did cause the circuit breaker to trip, but for entirely different reasons than he stated. He said that bad connections made the current go up. Perhaps the resistance made the current go down (but not below the circuit breaker trip point) and prolonged the motor-starting time long enough for breaker to trip.
Joe Gores
[quote][b]
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
http://www.matronics.com/Navigator?AeroElectric-List |
|
_________________
Joe Gores |
|
| Back to top |
|
|
nuckolls.bob(at)cox.net
Guest
|
| Posted: Mon Aug 25, 2008 3:16 pm Post subject: High resistance pops breaker? |
|
|
At 12:06 PM 8/25/2008 -0400, you wrote:
| Quote: | DC motors have the greatest torque and current when they are standing
still and voltage is first applied.
|
Yes, zero velocity = zero counter emf. Therefore in the instant of time
were the armature is just spinning up, current draw is applied voltage
divided by internal resistance. That's the lower most end of the speed-
torque curve and is usually called "stall current".
| Quote: | According to my old college book on basic electricity, a DC motor
(with shunt winding) starting current can be up to 100 times greater than
the full load running current.
|
Yes, a shunt winding has a great deal in common with
modern PM motors for starters. The magnets are so strong
that fewer turns are needed in the armature to achieve
desired performance . . . the speed torque curve is
flatter and the stall current VERY significant.
| Quote: | For larger motors, a resistance type motor-starter is used to limit the
starting current to prevent damage to the motor and also to prevent
excessive acceleration that could damage the load. Adding resistance to
the circuit during starting slows the acceleration and prolongs the time
that it takes for a motor to attain normal running speed.
|
Correct . . . artificial increase in slope of speed-torque
curve. Torque AND inrush current are reduced.
| Quote: | Getting back to that Sport Aviation article that claimed that "High
Resistance Pops Breaker". The circuit breaker in that landing light
circuit might have been sized to trip if the motor current was high due to
a jammed landing-light retract mechanism. But due to the time-delay
characteristics of circuit breakers, the breaker would not trip during
normal motor starting even though the starting current is above the
breaker trip point for a short time. Is it possible that there is some
value of resistance (due to bad connections) that, when added to the
circuit, could allow motor-starting current to flow that is still above
the circuit breaker trip point, but for a prolonged period of time? Could
the author of that Sport Aviation article be correct that bad connections
did cause the circuit breaker to trip, but for entirely different reasons
than he stated. He said that bad connections made the current go
up. Perhaps the resistance made the current go down (but not below the
circuit breaker trip point) and prolonged the motor-starting time long
enough for breaker to trip.
|
I pondered that for a bit but I don't think so. Acceleration
of a motor to full speed takes but tens of milliseconds. The
only landing light motors I've seen were split series field
motors . . . apply (+) to one field to retract, the other
field to extend. The effect of series windings is to limit
stall current (although torque tends to get better due to
increased field flux - assuming the magnetics are not saturated).
Starting current on these relatively small mechanisms is
short in duration and not spectacular.
If the motor were jammed . . . yeah, I can see this generating
a trip but a simple cleaning/tightening of joints in the wiring
can't account for an increase of current to an I(squared)*T
value necessary to trip a breaker.
Bob . . .
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
http://www.matronics.com/Navigator?AeroElectric-List |
|
|
|
| Back to top |
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You can download files in this forum
|
Powered by phpBB © 2001, 2005 phpBB Group
|
Get Email Distribution Too!
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
