http://www.motorsanddrives.c=
om/cowern/motorterms12.html=0A=0A" =0ALOW VOLTAGE=0AWhen
electric motors ar=
e subjected to voltages, below the nameplate rating, some of
the characteri=
stics will change slightly and others will change more
dramatically. A basi=
c point is, to drive a fixed mechanical load connected to
the shaft, a moto=
r must draw a fixed amount of power from the power line. The
amount of powe=
r the motor draws is roughly related to the voltage times
current (amps). T=
hus, when voltage gets low, the current must get higher to
provide the same=
amount of power. The fact that current gets higher is not
alarming unless =
it exceeds the nameplate current rating of the motor. When
amps go above th=
e nameplate rating, it is safe to assume that the buildup of
heat within th=
e motor will become damaging if it is left unchecked. If a
motor is lightly=
loaded and the voltage drops, the current will increase in
roughly the sam=
e proportion that the voltage decreases.=0AFor example, a
10% voltage decre=
ase would cause a 10% amperage increase. This would not be
damaging if the =
motor current stays below the nameplate value. However, if a
motor is heavi=
ly loaded and a voltage reduction occurs, the current would
go up from a fa=
irly high value to a new value which might be in excess of
the full load ra=
ted amps. This could be damaging. It can be safely said that
low voltage in=
itself is not a problem unless the motor amperage is pushed
beyond the nam=
eplate rating.=0A=C2=A0=0AAside from the possibility of
over-temperature an=
d shortened life created by low voltage, some other
important items need to=
be understood. The first is that the starting torque,
pull-up torque, and =
pull-out torque of induction motors, all change based on the
applied voltag=
e squared . Thus, a 10% reduction from nameplate voltage
(100% to 90%, 230 =
volts to 207 volts) would reduce the starting torque,
pull-up torque, and p=
ull-out torque by a factor of .9 x .9. The resulting values
would be 81% of=
the full voltage values. At 80% voltage, the result would
be .8 x .8, or a=
value of 64% of the full voltage value.=0A=C2=A0=0AIn this
case, it is eas=
y to see why it would be difficult to start
=E2=80=9Chard-to-start=E2=80=9D=
loads if the voltage happens to be low. Similarly the
motor=E2=80=99s pull=
-out torque would be much lower than it would be under
normal voltage condi=
tions.=0A=C2=A0=0ATo summarize the situation, low voltage
can cause high cu=
rrents and overheating which will subsequently shorten motor
life. Low volt=
age can also reduce the motor=E2=80=99s ability to get
started and its valu=
es of pull-up and pull-out torque. On lightly loaded motors
with easy-to-st=
art loads, reducing the voltage will not have any
appreciable effect except=
that it might help reduce the light load losses and improve
the efficiency=
under this condition. This is the principle that is used in
the so-called =
Nola devices that are sold as efficiency improving add-on
equipment to moto=
rs."=C2=A0=0ALook at the digram in the article.=C2=A0 You
can see with a re=
duction in voltage, you get a corresponding increase in
current.=C2=A0 In t=
he given case, I suspect that the battery and/or the
alternator failed, wit=
h a resulting=C2=A0 voltage drop.=C2=A0 With several high
power devices, su=
ch as fans and lights, the current demands on the wiring
would have gone up=
.=C2=A0 Perhaps this caused overheating of the switch and
system failure.=
=C2=A0 This will only be answered when the original poster
tells what devic=
es failed.=C2=A0 State of the battery, alternator, switch
and other wirng.=
=0A=0AWhat do you think happened?=C2=A0 Your
thoughts......=0A=0ACheers!=0A=
Stuart=0A=0A________________________________=0AFrom: Larry
Alofs <lalofs@gm=
ail.com>=0ATo: Stuart <ve3smf@yahoo.com>
=0ACc: vanagonlist a <vanagon@gerr=
y.vanagon.com> =0ASent: Saturday, December 21, 2013
10:33:59 PM=0ASubject: =
Re: Van Dies while in motion=0A=0A=0A=0AStuart, =0A=C2=A0
I'm afraid you ar=
e too far out on your limb. =C2=A0Your basic starting point
that any device=
will use a fixed amount of power is incorrect.=0AThat makes
the rest of yo=
ur argument and conclusions false.=0A=C2=A0 I am wondering
where you came u=
p with that premise.=0A=0ALarry A.
Stuart, you are citing information for *A/C* motors. DC motors, and other resistive loads do not have power factors, and operate under simple Ohm's Laws. A DC motor has a fixed (abeit temperature variant to a degree) internal resistance, and when voltage drops, it will respond according to the relationship E=IR. R is fixed, thus when E drops, I drops as well. Power dissipation (Volt-Amps, i.e. E*I) decreases as voltage decreases. If DC power were invariant as you claim, then a headlamp bulb would glow with the same luminosity at 1 VDC as at 14VDC.
Keith Hughes
'86 Westy Tiico (Marvin)
