DC MOTORS OVERLOAD PROTECTION

[revdude]

IN GENERAL,DO DC MOTORS WITH FIXED MAGNET FIELDS REQUIRE
OVERLOAD SIZING TO BE 125% OF THE FLA OR DO THEY GET FUSED
@ 100% BECAUSE THERE IS NO INDUCTIVE SURGE AS IN A AC APPLICATION.

 

[cbarn24050]

Hi, there can be a surge with a dc motor if you apply the arm voltage directly while the motor is stationary. In some cases the current flow is sufficient to demagnetise the magnets.

 

[electricuwe]

cbarn24050 statement on demagnetising is correct, but the current necessary is usually much higher than 125% FLA. But thermal issues also have to be consideed.

 

[jbartos]

Suggestion: The DC current is 0.9 of AC rms current. If you have a thermal overload rated at 1.25 x FLA rms current, then the same DC FLA has to have its overload rating 1.25 x FLA x (1/0.9) x (1/0.9) since the thermal overload heater is actuated over R x I**2 unless you have thermal overload calibrated for DC current. In that case 1.25 x FLA (DC) would approximately correct, depending on the motor design and duty (check the manufacturer's recommendations). The thermal overload is rated 125% of FLA not only because of inrush or starting motor current but also to protect motor against the motor shaft overloads while motor is running. The reason for 125% figure is that the motor has some manufacturing tolerance as well as the overload relay heaters plus some design margin is imbedded.

 

[electrageek]

Most DC motor manufacturers allow their motors to carry 150% armature current for 1 minute before tripping. The thermal protection needs to take this into account.

 

[jbartos]

Suggestion: I tend to agree with the previous posting since 1.25 x (1/0.9) x (1/0.9) = 1.543

 

[buzzp]

Note: OL setting is 125% of HP rating not FLA for AC motors. See previous posts.

 

[jbartos]

Suggestions:
1) Reference:
1. NFPA 70-2002 National Electrical Code indicates:
430.32 Continuous-Duty Motors.
(A) More Than 1 Horsepower. Each continuous-duty motor
rated more than 1 hp shall be protected against overload
by one of the means in 430.32(A)(1) through (A)(4).
(1) Separate Overload Device. A separate overload device
that is responsive to motor current. This device shall
be selected to trip or shall be rated at no more than the
following percent of the motor nameplate full-load current
rating:
Motors with a marked service factor 1.15 or greater 125%
Motors with a marked temperature rise 40°C or less 125%
All other motors 115%

2. Robert W. Smeaton "Switchgear and Control Handbook," McGraw-Hill Book Co., 1987
page 26-2 The DC Motors
Standard Nema industrial motors are rated to carry 100% load continuously, 150% for 1 minute. Heavy-duty mill motors may carry up to 300% for 1 minute. Other applications require motors to carry overload currents within these extremes. A current-limiting function is provided within the converter controller to limit Ia within the motor-overload rating. For example, with a general purpose motor, Ia=(V-E)/Ra would be limited to 150%, where
Ia is armature current
V is terminal voltage
E is counter EMF
Ra is armature resistance
(all for DC shunt motors).

 

[buzzp]

This post addresses the post by jbartos on May 5.

The heating affect of DC current vs AC current is compared as follows: The true rms AC current has the same heating affect as the DC current at the same level as the TRMS AC current. I would think this would be the way to compare the two. Also, you do not want to use an AC overload in a DC application for the following reasons:
1. MFG does not specify this
2. Device might not monitor DC
3. Make and Break ratings of starters (contactors with OL protection built in) has to be higher in DC applications
4. Arc suppression (no zero cross to quench the arc so other methods are used in switching DC loads)circuitry has to be used (built into DC relays, contactors, starters).

 

[jbartos]

Suggestions to buzzp (Electrical) May 29, 2002 marked by ///\\The heating affect of DC current vs AC current is compared as follows: The true rms AC current has the same heating affect as the DC current at the same level as the TRMS AC current.
///Please, can you prove it mathematically such that it corresponds to the physical principle/reality of thermal dissipation of heat in the overload or resistor or refer to suitable literature?\\ I would think this would be the way to compare the two. Also, you do not want to use an AC overload in a DC application for the following reasons:
1. MFG does not specify this
///At some point in the past, there were no dc overloads however, the dc machines were approximately protected by ac overload. Similar holds true for ac fuses that were used to protect dc loads. Obviously, nowadays things are different.\\2. Device might not monitor DC
///Yes, agree. This would have to be adjusted or customized, which would be approximate as it was done in the past.\\3. Make and Break ratings of starters (contactors with OL protection built in) has to be higher in DC applications
///This is known fact for all contacts, e.g. relays, contactors, switches, etc.\\4. Arc suppression (no zero cross to quench the arc so other methods are used in switching DC loads)circuitry has to be used (built into DC relays, contactors, starters).
///The dc arc suppression has been around longer than the ac arc suppression at coils, and contacts due to more severe nature of dc arc.\\

 

[buzzp]

See comments below in **
Suggestions to buzzp (Electrical) May 29, 2002 marked by ///\\The heating affect of DC current vs AC current is compared as follows: The true rms AC current has the same heating affect as the DC current at the same level as the TRMS AC current.
///Please, can you prove it mathematically such that it corresponds to the physical principle/reality of thermal dissipation of heat in the overload or resistor or refer to suitable literature?\\**This has been proven many times. Where do you think TRMS came from? There is no need for me to prove this. If you want, set up an experiment with a resistor and a thermocouple. Measure the heat on the resistor with, say 5 amps DC going through it and then the same thing with 5amp TRMS goiing through it. You will measure the same.

I would think this would be the way to compare the two. Also, you do not want to use an AC overload in a DC application for the following reasons:
1. MFG does not specify this
///At some point in the past, there were no dc overloads however, the dc machines were approximately protected by ac overload. Similar holds true for ac fuses that were used to protect dc loads. Obviously, nowadays things are different.\\2. Device might not monitor DC
///Yes, agree. This would have to be adjusted or customized, which would be approximate as it was done in the past.\\** There is no adjusting an AC device to monitor DC if it was not designed that way. A CT for example will saturate if DC current runs through it and the readings will be meaningless. The only way to interchange an AC monitoring device with DC is to have a shunt, hall-effect sensor, or similar technology.

3. Make and Break ratings of starters (contactors with OL protection built in) has to be higher in DC applications
///This is known fact for all contacts, e.g. relays, contactors, switches, etc.\\** This is why you can NOT use an AC starters in DC applications

4. Arc suppression (no zero cross to quench the arc so other methods are used in switching DC loads)circuitry has to be used (built into DC relays, contactors, starters).
///The dc arc suppression has been around longer than the ac arc suppression at coils, and contacts due to more severe nature of dc arc.\\

 

[jbartos]

Suggestion to buzzp (Electrical)May 30, 2002 marked ///\\**This has been proven many times.
///This answer is missing reference or mathematical proof.
E.g. mathematical proof of the difference between the AC RMS Power Prms and DC Power Pdc from pure sinusoidal voltage v=V sinwt and current i=Isinwt:
Prms={(1/T) integral from 0 to T of [(v x i)**2 dt]}**0.5= (V x I)/2 = Vrms x Irms
since Vrms=V/sqrt2 and Irms=V/sqrt2
Pdc = (1/T) x Integral from 0 to T of [(|v| x |i|) dt]= (2 V/ pi) x (2 I/pi) = (0.64 x V) x (0.64 x I) = (0.9 Vrms) x (0.9 x Irms) = 0.81 Vrms x Irms =0.81 Prms.
The resistor can generate Pdc watts only which are different watts from Prms Watts. IEEE and Standard Handbook for Electrical Engineers downplayed this fact significantly, however, it stays here as you can see.\\ Where do you think TRMS came from?
///TRMS came from RMS system the power producers and distributors established. Namely, from the definition of Vrms and Irms, consequently power Prms=Vrms x Irms. When there were DC distribution systems in the beginning, the customer got 19% more heat which nowadays is not available. If the Utility provided dc power supply and charged the same rate for DC kWhr at Vdc (=Vrms) and AC kWhr at Vrms I would definitely preferred DC kWhr. If you preferred AC kWhr, that would be your prerogative based on your statements.\\

 

[buzzp]

Okay, now take the mathematically theroy and perform the experiment I told you too. I am not sure your calculations proved anything since DC is constant(not a periodic function). However, when you take an integral of something you are sort of dividing it into its DC components with a given harmonic. It sounds as though you are talking about a DC component on an AC system or something else.

The RMS value of sinusoidal source is also know as an 'effective' value. This is because of the need to compare the amount of energy delivered from a sinusoidal source vs. a DC source. The DC value delivers the same amount of energy every T seconds as an AC RMS source of the same magnitude. Where T is the cycle time. The above is fact and I hope your not in dispute with this. If you are then you are wrong. Perhaps, we misunderstood one another with this electronic communication.

 

[wired1]

I agree with buzzp. The experiment will show that the rms (effective) current has the same heating effect as dc. If you integrate a full-wave-rectified sinwave over time, you get an average of .636xVpeak. This is the dc level obtainable from the line for power supply design. If you integrate the square of a sinwave then unsquare it, you have the rms level which is .707Vpeak. This is the level that has the same heating effect as dc. This is a statement that I have accepted on faith and can't give you the theory of why.

 

[cbarn24050]

Hi, the reason is because heating is proportional to power, power is proportional to the square of the voltage for a resistive load.

 

[redtrumpet]

jbartos - reference "Electric Circuit Analysis", Robert A. Bartkowiak, Harper and Row Publishers:

Section 10.4, "The Effective or Root-Mean-Square Value". states

"An alternating current is said to have an effective value of 1 A when it develops the same amount of heat in a given resistance as would be produced by a direct current of 1 A in the same resistance over the same time."

The effective value of a sine wave is 0.707 of the peak value. See the above reference for the proof.

I agree with buzzp and wired1.

Now, if you would be so kind as to state the reference for your assertion that the rms value does not have the same heating effect as the dc value. In your proof you appear to have derived the average value of a sinusoid, not the effective value.