Motor locked rotor current

[Jpflex]

I’m not sure what I’m doing wrong

I have a motor with specs 5 HP, 460 volts, 3 phase, 3.7 watts, 6.33 FLA, power factor 0.82, DES B, ambient temp rise 40 deg cel, service factor 1.15, efficiency 88.5 percent


I use the formula in ugly book to find locked rotor current

KVA not given on label so I figured 3700 watts / KVA (unknown variable x) = 82 percent power factor

so solving for x or unknown gives a KVA of 4,512.19


Formula to find locked rotor current:



1000 x hp x KVA / HP

Top results over or Divided by

Volts (460) x power factor (0.82 percent) x efficiency 88.5 percent x 3 phase (1.732)

= locked rotor current of 8,158.89 amperes

However NEC code book says 46 amperes under TABLE 430.245?

 

[jim dungar]

Why do you need the locked rotor current?
For motors this size it will rarely be more than 10x the full loadcurrent.

The formula is kVA = kW/PF.
Your values are 3.7W with PF = 88.5 (power factor has no dimension, it is not %)

Motors are rated in output power, but you need into power.
1HP output = 746W output, kW input = kW output/efficiency.

Watch your multipliers 3.7W = .037kW

 

[augie47]

Look at the example in your Ugly's book. Where you using KVA the formula uses the KVA/HP for the given Code letter.
I don't know where LRA would be in question for 5HP
In your case I think it's 21 amps.

 

[don_resqcapt19]

You use the code letter from the nameplate and Table 430.7(B) to get a locked rotor current. 430.7(A)(8) requires the code letter or the locked rotor amps to be on the nameplate.

 

[Jpflex]

Why do you need the locked rotor current?
For motors this size it will rarely be more than 10x the full loadcurrent.

The formula is kVA = kW/PF.
Your values are 3.7W with PF = 88.5 (power factor has no dimension, it is not %)

Motors are rated in output power, but you need into power.
1HP output = 746W output, kW input = kW output/efficiency.

Watch your multipliers 3.7W = .037kW

An engineer was worried that if the motor used to crush rocks seized up that it would burn the motor as a disconnect within sight was recently added but the fuses are 5 amps larger than specs. This is because the disconnect was taken from company junk pile. Also overload is not likely set correct by original cabinet installer

Power factor is a percent. It is a ratio of true power watts over apparent power VA

That’s why when the motor label did not provide the KVA rating I figured I would solve it algebraicly by taking true power watts over x reactive power = 88.5 percent and solve

 

[jim dungar]

Power factor is a percent. It is a ratio of true power watts over apparent power VA

That’s why when the motor label did not provide the KVA rating I figured I would solve it algebraicly by taking true power watts over x reactive power = 88.5 percent and solve

You are correct PF is a ratio. However it is not a percentage, even though industry slang uses that terminology.
PF can be lagging or leading which are often expressed as being positive or negative.

If you start with 100kVA and multiply it by 85%, you end up with 85kVA
However if you start with 100kVA and multiply it by .85 PF you end up with 85kW.

I would almost never rely on my branch protection device to protect my motor. For a crusher application I would use an electronic overload relay, maybe even one with jam protection.

 

[Jpflex]

You are correct PF is a ratio. However it is not a percentage, even though industry slang uses that terminology.
PF can be lagging or leading which are often expressed as being positive or negative.

If you start with 100kVA and multiply it by 85%, you end up with 85kVA
However if you start with 100kVA and multiply it by .85 PF you end up with 85kW.

I would almost never rely on my branch protection device to protect my motor. For a crusher application I would use an electronic overload relay, maybe even one with jam protection.

However I still disagree because power factor is still a percentage whether voltage is leading or lagging current. Voltage is either leading or lagging by a percent or degree of current, but perhaps you’re looking at it in another way?

Here’s a pic of formula for rotor locker current I used in ugly book

 

[Jpflex]

You are correct PF is a ratio. However it is not a percentage, even though industry slang uses that terminology.
PF can be lagging or leading which are often expressed as being positive or negative.

If you start with 100kVA and multiply it by 85%, you end up with 85kVA
However if you start with 100kVA and multiply it by .85 PF you end up with 85kW.

I would almost never rely on my branch protection device to protect my motor. For a crusher application I would use an electronic overload relay, maybe even one with jam protection.

You said “If you start with 100kVA and multiply it by 85%, you end up with 85kVA”

However the answer would not be 85KVA but it would be 85K WATTS (true power).

And 1.732 would be factored for 3 phase

The branch circuit OCPD (over current protective device) would be set to 300 percent of full load current if using inline non time delay fuses

 

[jim dungar]

However I still disagree because power factor is still a percentage whether voltage is leading or lagging current. Voltage is either leading or lagging by a percent or degree of current, but perhaps you’re looking at it in another way?

Here’s a pic of formula for rotor locker current I used in ugly book

Thanks, but I will stick with what I learned in my engineering class for Power Systems and Rotating Machines.

BTW, Wikipedia also agrees with me.

Since the units are consistent, the power factor is by definition a dimensionless number between -1 and 1.

 

[jim dungar]

You said “If you start with 100kVA and multiply it by 85%, you end up with 85kVA”

However the answer would not be 85KVA but it would be 85K WATTS (true power).

No.

It would be watts only if I used PF instead of %.

Power Factor is the Cosine of the angle between Real and Apparent power. Trigonometric function results are not expressed as percentages.

 

[Jpflex]

You are correct PF is a ratio. However it is not a percentage, even though industry slang uses that terminology.
PF can be lagging or leading which are often expressed as being positive or negative.

If you start with 100kVA and multiply it by 85%, you end up with 85kVA
However if you start with 100kVA and multiply it by .85 PF you end up with 85kW.

I would almost never rely on my branch protection device to protect my motor. For a crusher application I would use an electronic overload relay, maybe even one with jam protection.

You said “If you start with 100kVA and multiply it by 85%, you end up with 85kVA”

However the answer would not be 85KVA but it would be 85K WATTS (true power).

And 1.732 would be factored for 3 phase

The branch circuit OCPD (over current protective devicewould be set to 300 percent of full load current if using inline non time delay fuses

No.

It would be watts only if I used PF instead of %.

Power Factor is the Cosine of the angle between Real and Apparent power. Trigonometric function results are not expressed as percentages.

yes you are correct power factor is the cosine angle between a triangles adjacent leg over its hypotenuse leg using “SOHCAHTOA”

This again represents a fraction which when calculated results in a decimal figure such as 0.64 or 0.64% or another decimal

So using the power factor in percent or the decimal without the percent figure to the right of the decimal still results in true power

KVA apparent power excludes using apparent power or a percent because this is the value with the circuit at unity power factor or a power factor of 1, 100%

 

[Jpflex]

However, why can’t the formula for motor locked rotor current be used and get the same result of using the motor name plate code letter along with the NEC table for KVA per HP to figure locked rotor current values? Shouldn’t results be the same?

Ugly manual says formula is


1000 x HP x KVA/HP
—————————-
E(volts) x 1.732(3 phase) x efficiency% x power factor%

 

[Jpflex]

Look at the example in your Ugly's book. Where you using KVA the formula uses the KVA/HP for the given Code letter.
I don't know where LRA would be in question for 5HP
In your case I think it's 21 amps.

Does KVA/HP in formula mean KVA Divided by Horsepower or KVA per Horsepower as determined from NEC book?

The first would not make as much sense because HP would cancel out if first multiplying HP by KVA and then dividing by HP. The result would be just KVA

However as it’s written the first looks as it’s to be applied

RLC = HP X KVA/HP
———————
E(volts) x 1.732( 3 phase) x power factor % x efficiency %

 

[Jpflex]

TOP formula correction

1000 x HP X KVA/HP

 

[wwhitney]

BTW, Wikipedia also agrees with me.

But % is a dimensionless modifier. It just means "/100", nothing else. 85% = 85/100.

Cheers, Wayne

 

[augie47]

However, why can’t the formula for motor locked rotor current be used and get the same result of using the motor name plate code letter along with the NEC table for KVA per HP to figure locked rotor current values? Shouldn’t results be the same?

Ugly manual says formula is


1000 x HP x KVA/HP
—————————-
E(volts) x 1.732(3 phase) x efficiency% x power factor%

If you look closely at the formula in Uglys it says "KVAhp" which references the Table with Code Letters (above the formula). You might look at the formula as LRA= HP x Code Letter X 1000
E

 

[Jraef]

However I still disagree because power factor is still a percentage whether voltage is leading or lagging current. Voltage is either leading or lagging by a percent or degree of current, but perhaps you’re looking at it in another way?

Here’s a pic of formula for rotor locker current I used in ugly book

When it says the .85PF on the nameplate, that is the PF at full load on the motor. The PF when the rotor is locked is much much lower, often around a .2PF.

 

[Jpflex]

But % is a dimensionless modifier. It just means "/100", nothing else. 85% = 85/100.

Cheers, Wayne

When it says the .85PF on the nameplate, that is the PF at full load on the motor. The PF when the rotor is locked is much much lower, often around a .2PF.

Yea there should not be much reactive power because there is little counter EMF without windings spinning within magnetic fields

However, I wasn’t trying to find power factor during motor locked current. I was just trying to learn how to calculate LRC because engineer was worried motor would lock up while female helper operated motor to crush rocks and burn motor. Just wanted to assure him fuses would blow before this could happen

 

[Jpflex]

If you look closely at the formula in Uglys it says "KVAhp" which references the Table with Code Letters (above the formula). You might look at the formula as LRA= HP x Code Letter X 1000
E

Yes I think this is what was throwing me off. I think uglies was saying KVA per horsepower but not KVA divided by horsepower as they write it

So do I 1000 x KVA (of motor) x hp? As indicated in NEC chart?

 

[Jpflex]

When it says the .85PF on the nameplate, that is the PF at full load on the motor. The PF when the rotor is locked is much much lower, often around a .2PF.

Also when you say this is the power factor at full load of the motor, how do you know how much load is placed on conveyer, pullies, motor? There is a point of torque pullout of a motor or point a motor stalls based on weight or resistance of load