Breakdown Torque and HP

[nauticalx]

Is HP of a motor determined by breakdown torque or by (amps x volts)/745 = HP?
NEMA standards seem to state that it's determined by breakdown torque. How does this relate to the amps times volts equation above?
What is the actual maximum power generated by a single phase 120 V motor?
Thanks!

 

[GeorgeB]

Is HP of a motor determined by breakdown torque

no. The RATED output HP of a motor is determined by the manufacturer and is usually on its name tag/label. It may be measured at its RATED torque (and speed at that torque) which is (always IME) less than breakdown.

or by (amps x volts)/745 = HP?

Working INPUT power, expressed as HP, for a DC motor will be (amps x volts)/746. Working OUTPUT power, that which we usually call the motor's power rating, will be approximately INPUT power x efficiency.

With AC powered motors, these equations are for practical purposes worthless.

NEMA standards seem to state that it's determined by breakdown torque.

I've not ever noticed this; can you cite a reference?

How does this relate to the amps times volts equation above?

It doesn't.

What is the actual maximum power generated by a single phase 120 V motor?

In theory, there isn't any maximum. It's unusual to find one over 5HP which would require something over 4kW input power allowing for typical efficiency.
At 120V, that would consume 4000/(120 x PF) amps. ASSUMING a 0.7 PF for a single phase motor, that would be 45 to 50 amps.

Thanks!

You are welcome!

 

[nauticalx]

Thanks. Here it is (ANSI/NEMA MG 1-2016 (Revised 2018)...

10.34.1 Basis of Rating
The horsepower rating of a small or medium single-phase induction motor is based upon breakdown
torque (see 1.51). The value of breakdown torque to be expected by the user for any horsepower and
speed shall fall within the range given in Tables 10-5 and 10-6.

1.51 Pull-Out Torque
The pull-out torque of a synchronous motor is the maximum sustained torque which the motor will
develop at synchronous speed with rated voltage applied at rated frequency and with normal excitation.

 

[Jraef]

That said, consumer product marketing people will often make claims of higher “HP” ratings that are based on the Break Down Torque values. At first they would kind of allude to it by using terms like “Develops 2HP!” but then you look at the actual nameplate and see that draws 1000W and you question your understanding. Now they have dropped even that pretense and just state the bogus HP numbers that are incongruous to the electrical consumption ratings. Why can they get away with that? I guess because 99% of consumers don’t know.

 

[nauticalx]

no. The RATED output HP of a motor is determined by the manufacturer and is usually on its name tag/label. It may be measured at its RATED torque (and speed at that torque) which is (always IME) less than breakdown.

Working INPUT power, expressed as HP, for a DC motor will be (amps x volts)/746. Working OUTPUT power, that which we usually call the motor's power rating, will be approximately INPUT power x efficiency.

With AC powered motors, these equations are for practical purposes worthless.

I've not ever noticed this; can you cite a reference?

It doesn't.

In theory, there isn't any maximum. It's unusual to find one over 5HP which would require something over 4kW input power allowing for typical efficiency.
At 120V, that would consume 4000/(120 x PF) amps. ASSUMING a 0.7 PF for a single phase motor, that would be 45 to 50 amps.

You are welcome!

Thanks. Here it is (ANSI/NEMA MG 1-2016 (Revised 2018)...

10.34.1 Basis of Rating
The horsepower rating of a small or medium single-phase induction motor is based upon breakdown
torque (see 1.51). The value of breakdown torque to be expected by the user for any horsepower and
speed shall fall within the range given in Tables 10-5 and 10-6.

1.51 Pull-Out Torque
The pull-out torque of a synchronous motor is the maximum sustained torque which the motor will
develop at synchronous speed with rated voltage applied at rated frequency and with normal excitation.

 

[GeorgeB]

I think what they are saying is that I am allowed to rate a single-phase motor at what would be THERMALLY overloaded conditions. What I THINK is that the motor would not continuously provide that load. That would explain the ratings of "small" air compressor motors which cannot sustain the advertised ratings.

I'm in the industrial world with my motors almost always loaded by hydraulic pumps. I frequently designed systems that "overloaded" motors beyond their service factor, FOR SECONDS PER MINUTE. I NEVER would overload them to breakdown torque. They always had extended periods in an operating cycle at substantially less than 100% load, often less than 25% load. Motor failure from overheating was not a factor. I suspect that localized damage might have occurred given enough years of service. Some, but not many, used winding thermocouples, but those didn't show issues either.

Your pull-out torque is for synchronous motors which I've never used.

 

[winnie]

My understanding of the NEMA horsepower rating is that the motor must meet both continuous operating capacity _and_ meet minimum breakdown and pull up torque requirements.

This is where 'service factor' comes into play. If you have a 50hp motor with a 1.2 service factor, that means you can safely load it on a continuous basis up to 60hp, but the acceleration and breakdown torque are what you expect from a 50hp motor.

Motors can be intentionally overloaded by quite a bit if the time duration of the operation is limited. Sometimes it makes good sense to push a motor to 200% rated continuous torque for a few seconds at a time.

Jon

 

[retirede]

My understanding of the NEMA horsepower rating is that the motor must meet both continuous operating capacity _and_ meet minimum breakdown and pull up torque requirements.

This is where 'service factor' comes into play. If you have a 50hp motor with a 1.2 service factor, that means you can safely load it on a continuous basis up to 60hp, but the acceleration and breakdown torque are what you expect from a 50hp motor.

Motors can be intentionally overloaded by quite a bit if the time duration of the operation is limited. Sometimes it makes good sense to push a motor to 200% rated continuous torque for a few seconds at a time.

Jon

I’ll add that service factor may be one of the most misunderstood ratings on a motor.
When I designed OEM equipment, we worked closely with GE for our motors. Motor life depends on all sorts of things. GE engineers told us that their NEMA design motors had a 10year design life - X% failure rate in 10 years when operated at nameplate conditions. Operating at full service factor (usually 1.15) would cut the design life in half.

Since “most” motors will not be continuously operated at full nameplate HP AND maximum rated ambient temperature, they will last a very long time, even with periodic overloading.

 

[nauticalx]

That said, consumer product marketing people will often make claims of higher “HP” ratings that are based on the Break Down Torque values. At first they would kind of allude to it by using terms like “Develops 2HP!” but then you look at the actual nameplate and see that draws 1000W and you question your understanding. Now they have dropped even that pretense and just state the bogus HP numbers that are incongruous to the electrical consumption ratings. Why can they get away with that? I guess because 99% of consumers don’t know.

My understanding of the NEMA horsepower rating is that the motor must meet both continuous operating capacity _and_ meet minimum breakdown and pull up torque requirements.

This is where 'service factor' comes into play. If you have a 50hp motor with a 1.2 service factor, that means you can safely load it on a continuous basis up to 60hp, but the acceleration and breakdown torque are what you expect from a 50hp motor.

Motors can be intentionally overloaded by quite a bit if the time duration of the operation is limited. Sometimes it makes good sense to push a motor to 200% rated continuous torque for a few seconds at a time.

Jon

I’ll add that service factor may be one of the most misunderstood ratings on a motor.
When I designed OEM equipment, we worked closely with GE for our motors. Motor life depends on all sorts of things. GE engineers told us that their NEMA design motors had a 10year design life - X% failure rate in 10 years when operated at nameplate conditions. Operating at full service factor (usually 1.15) would cut the design life in half.

Since “most” motors will not be continuously operated at full nameplate HP AND maximum rated ambient temperature, they will last a very long time, even with periodic overloading.

Great points.

That said, Jraef makes an interesting point. What actual standard accounts for these "maximum horsepower" claims in consumer products? Service factor is for short durations of time. But does the device ever operate at that maximum horsepower?

 

[winnie]

I don't know the details of how consumer devices derive their inflated numbers. Probable something like expressing the locked rotor power consumption in horsepower.

Short term overload power (and torque) ratings make good sense for things like electric vehicles, where the power to cruise down the highway is significantly lower than the power needed to accelerate to highway speeds. But it is really necessary to be honest about such short term ratings.

-Jon

 

[CrazyWabbit]

my 3hp well pump draws 3800 watts, some of that has to be due to cable length.
746w/hp x 3 = 2238 watts
2238/3800=59% efficient

 

[Jraef]

No. Something is wrong. How are you determining that it is pulling 3800W?

 

[CrazyWabbit]

real power meter (amp clamp and voltage probes) and it gives power factor. i've also measured other wells in my area and they pull the same.

 

[kwired]

my 3hp well pump draws 3800 watts, some of that has to be due to cable length.
746w/hp x 3 = 2238 watts
2238/3800=59% efficient

746W = 1 HP. A three HP motor would be rated 3 HP at the output shaft, no motor is 100% efficient so any excess watts consumed is factored toward the efficiency of the motor.

Now what I may or may not be correct with is 3 HP submersible pump - is that the motor rating or the rating of the entire assembly? There is inefficiencies in the pump also which could possibly explain a 59% efficiency vs the motor itself that maybe is 85-95% efficient.

 

[Carultch]

Is HP of a motor determined by breakdown torque or by (amps x volts)/745 = HP?
NEMA standards seem to state that it's determined by breakdown torque. How does this relate to the amps times volts equation above?
What is the actual maximum power generated by a single phase 120 V motor?
Thanks!

Usually, the horsepower refers to the mechanical power output of the motor, as in how much mechanical work it can do per unit time. While if there is an electrical input power given, it would be in Watts or kW. I'd prefer to just have them both in kW, and not even need to think about horsepower, since I have a more instinctive idea of what a kilowatt is, than what a horsepower is, and the Watt makes for a lot more elegant calculations. It would also make the efficiency more obvious if the input and output had the same units.

When rotation rate is in radians per second, and torque is in Newton-meters, multiplying the two of them will yield the mechanical output power in Watts. The input Watts would be Volts * Amps * power factor (and *sqrt(3) when applicable for 3-phase). The efficiency is output power divided by input power.

You need conversion factors if you work with other units. For torque in ft-lbs and rotation rate in RPM, you divide by (5252 ft-lbf-rpm/hp) to get horsepower.

 

[kwired]

Usually, the horsepower refers to the mechanical power output of the motor, as in how much mechanical work it can do per unit time. While if there is an electrical input power given, it would be in Watts or kW. I'd prefer to just have them both in kW, and not even need to think about horsepower, since I have a more instinctive idea of what a kilowatt is, than what a horsepower is, and the Watt makes for a lot more elegant calculations. It would also make the efficiency more obvious if the input and output had the same units.

When rotation rate is in radians per second, and torque is in Newton-meters, multiplying the two of them will yield the mechanical output power in Watts. The input Watts would be Volts * Amps * power factor (and *sqrt(3) when applicable for 3-phase). The efficiency is output power divided by input power.

You need conversion factors if you work with other units. For torque in ft-lbs and rotation rate in RPM, you divide by (5252 ft-lbf-rpm/hp) to get horsepower.

My experience is they usually rated in according to output power, NEMA motors usually in HP, others usually in watts/kW.

input power is seldom directly marked though you can normally calculate it using V, A, PF and efficiency, as those are all input values.

you must assume you are operating at nameplate volts, frequency, and have rated output is being delivered for all the nameplate data to measure as marked.