braking torque with VFD 1

[fab1961]

Dear all,

in our applications we use braking resistors connected to VFDs.
For a new plant I have to check more precisely how much braking torque is available.
I consider the braking torque as [ ( V DCbus / resistor ) / rated motor current ] * motor torque at rated current * K
where K is the torque derating when frequency is over 50-60 Hz (torque being inversely proportional to frequency)
Am I missing something ?

Thank you for your help

 

[waross]

Torque doesn't care much about frequency.
It's all about current.
The torque is related to the square of the current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

I don't understand your K factor.

Basically, the resistor can dissipate a certain wattage (or HP). You would have to calculate the torque this equates to at each motor speed. Torque = HP x 5250 / rpm

As the speed slows down, you'll hit a point where the possible torque will be drive or motor limited.

 

[fab1961]

Consider running the motor instead of braking it. When above base frequency (50 Hz Europe 60 Hz USA) the torque
is reduced, we enter the constant-power area. If torque were not reduced, then, since power = torquexspeed, we would
have infinite power since both speed and torque increase. Also you have the magnetic saturation etc. So I thought that
braking is just the opposite of accelerating and therefore the torque available is the rated one until base frequency
and above that it is reduced. I hope I made myself understood, English is not my mother language...

 

[jraef]

Are you running the motor over its base frequency? You imply this with your "K" factor description, but is that really the case? If so, then yes your initial braking torque capability will be lower than peak possible because the motor negative torque capacity is reduced. But once typoy get to the normal motor base speed or below, you return to full torque capability.

But the braking torque capability in the SYSTEM is limited by the motor peak torque capability (which is what the motor power rating is based on). What you are doing with dynamic braking is to transmute the kinetic energy in the rotating mass into electrical energy in the motor, then removing it from the motor and transmuting it again into heat energy in the resistor. That energy comes back at the rate of current flow possible in the system, and current equates almost directly to torque, in this case negative torque. So whatever the peak running torque capability is in the motor dictates the peak braking torque POSSIBLE. But that is constrained by the ability of the chopper transistor in the VFD to handle the current, which sets the ohm value of the resistor (most VFDs will give you that ohm value; NEVER use a lower value). Then the total braking capacity of the system is also limited by the ability of the resistor to dissipate that energy, which is constrained by the watt rating of the resistor, the duty cycle of the process and cooling time between braking action. It's actually a very complex and involved process, but most of the time, the VFD mfr will give you "shortcuts" of ratings and duty cycles that will help you determine this, based on the components in their particular VFD. That's really where you should start.

Just to give you an example, this is BEGINNING of a process for sizing a dynamic braking module for a VFD that does not have one built-in. I was going to try to post the entire section, but it just became too tedious. You can read the whole thing starting on page 143 of this document.

Sizing the Dynamic Brake Module. Gather the following information.
1. The nameplate power rating of the motor in watts, kilowatts, or horsepower.
2. The nameplate speed rating of the motor in rpm or rps.
3. The motor inertia and load inertia in kilogram-meters², or lb•ft².
4. The gear ratio, if a gear is present between the motor and load, GR.
5. Review the Speed, Torque Power profile of the application.

Equations used for calculating Dynamic Braking values will use the following variables.
ω_(t) = The motor shaft speed in Radians/second, or
N_(t) = The motor shaft speed in Revolutions Per Minute, or RPM
T_(t) = The motor shaft torque in Newton-meters, 1.01 lb•ft - 1.355818N•m
-P_(t) = The motor shaft power in Watts, 1.0HP = 746 Watts -Pb = The motor shaft peak regenerative power in Watts

Step 1 – Determine the Total Inertia

J_T = J_m + GR² x J_L

J_T = Total inertia reflected to the motor shaft, kilogram-meters2, kg•m², or pound-feet2, lb•ft²
J_m = Motor inertia, kilogram-meters², kg•m², or pound-feet², lb•ft²
GR = The gear ratio for any gear between motor and load, dimentionless
J_L = Load inertia, kilogram-meters2, kg•m2, or pound-feet², lb•ft² – 1 lb•ft² = 0.04214011 kg•m²

Step 2 – Calculate the Peak Braking Power

P_b = J_T × ω² / t₃-t₂

J_T = Total inertia reflected to the motor shaft, kg•m2
ω = rated angular rotational speed, Rad⁄s = 2πN /60
N = Rated motor speed, RPM
t₃ - t₂ = total time of deceleration from rated speed to 0 speed, in seconds
P_b = peak braking power, watts (1.0 HP = 746 Watts)

Compare the peak braking power to that of the rated motor power, if the peak braking power is greater that 1.5 times that of the motor, the deceleration time, (t₃ - t₂), needs to be increased so that the drive does not go into current limit. Use 1.5 times because the drive can handle 150% current maximum for 3 seconds.

Step 3 – Calculating the Maximum Dynamic Brake Resistance Value

R_db1 = V_d² / P_b

V_d = The value of DC bus voltage that the chopper module regulates at
P_b = The peak braking power calculated in Step 2
R_db1 = The maximum allowable value for the dynamic brake resistor in Step 2

The choice of the Dynamic Brake resistance value should be less than the value calculated in Step 3. If the value is greater than the calculated value, the drive can trip on DC bus overvoltage. Remember to account for resistor tolerances.

There are 5 more steps...

Note: Step 3 is what the VFD mfr will give you if the brake chopper is built into the VFD.



" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[LionelHutz]

Sure, I guess the confusion is because your formula is fundamentally flawed.

You can't calculate torque from wattage without considering rpm.

The motor reactive current doesn't make it to the resistor, so using motor current to calculate motor power dissipated in the resistor isn't accurate.

The possible braking torque that a fixed resistor will give you goes down as the motor speed increases. A resistor capable of 100% braking torque at rated speed will be capable of 50% braking torque at 200% of rated speed or capable of 200% braking torque at 50% of rated speed. Doesn't mean you can get those torques, just what the resistor is capable of.