# Watt Sec question

• posted
Just checking to see if I understand joules.
1 Joule = 1 Watt * 1 sec, or J = 1Ws
If 1.5M Ws is the energy dissapated by a resistor (with a value 2
Ohms), does this mean that there is 866 Amps every second thru the
resistor?
Thanks for any push in the right direction.
Bill
• posted
Amps is a measure of the current flow. So 'amps every second' is not appropriate. If a resistor is dissapating 1.5MW-seconds, we need to know over what time period it dissapates this energy. For example, it could be 1.5 MW of power for 1 second (one heck of a resistor), or it could be 1.5 watts of power over 1 million seconds. Both total 1.5MW-seconds of energy.
With a 2 ohm resistor, 1.5MW of power would require P=I^2*R or I = sqrt(P/R) = sqrt(1500000/ 2) = 866 amps. But if it was over a period of 1 million seconds, 1.5 watts would mean I = sqrt( 1.5/2) = 0.866 amps.
Hope this helps...
daestrom
• posted
Daestrom,
The ON time is a value that I wasn't told (well....sort of). The person that I talked to stated that the resistor had be rated for 1.5M Ws (at 2.3 Ohms) , which I interupted it as being a continous rating of 807 Amps (but that isn't the current rating required due to the fact of it being a dynamic braking application of a 250HP drive). After a few minutes of small talk, I asked how long the load would be applied, he said 6 seconds.
So I decided before I call him back and say, "Hey, do you mean 250kW second?", (250kW x 6s = 1.5M Ws)
• posted
The watt-second rating for a DB resistor is the maximum amount of energy that the device can absorb over a short time period without damage. The energy is stored as heat and transferred to the environment as the resistor cools off over a much longer period. Part of the heat energy is radiated, most is transferred by convection to nearby air.
If the motor has a fixed field, DB current jumps to approximately armature voltage divided by resistor ohms. It then decays toward zero along an exponential curve. If this is a voltage-and-field drive, DB current may be constant until speed drops below base speed ... depending on how the field regulator is set up. After that, current decays exponentially. In either case, DB current changes with time during the stop and the resistor's continuous-duty current rating can be much lower than peak DB current.
The resistance is chosen to limit peak current at the beginning of the stop and the watt-second rating is set by how much kinetic energy must be temporarily stored as tolerable temperature rise in the grid. The resistor must be mechanically strong enough to withstand the magnetic forces produced by the high current pulse as well.
Roby
• posted
| The ON time is a value that I wasn't told (well....sort of). The | person that I talked to stated that the resistor had be rated for 1.5M | Ws (at 2.3 Ohms) , which I interupted it as being a continous rating of | 807 Amps (but that isn't the current rating required due to the fact | of it being a dynamic braking application of a 250HP drive). After a | few minutes of small talk, I asked how long the load would be applied, | he said 6 seconds. | | So I decided before I call him back and say, "Hey, do you mean 250kW | second?", (250kW x 6s = 1.5M Ws)
What you need to find out is the actual current during the 6 seconds of braking. That could vary with the motor speed, too, so the worst case could be right at the start (when the resistor is cooler). Merely MWs doesn't give the whole picture. The thermal environment can matter, too, if the resistor will still have residule heat from previous application as now it will start at a higher temperature and decrease your margins. You need to know how often the brakes will be applied.

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