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Sizing Circuit Breakers? |
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| Jun14-12, 09:51 AM | #18 |
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Sizing Circuit Breakers?Initially, [tex]P_{i} = S cos(\theta_{i})[/tex] Finally, [tex]P_{f} = S cos(\theta_{f})[/tex] If S is to remain the same, it must be such that, [tex]P_{i} \neq P_{f}[/tex] How can this be true if you claim that the motor is receiving the same KVA, voltage and current? Am I missing something? |
| Jun14-12, 10:01 AM | #19 |
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What was the voltage across the motor before the cap? What is the voltage across the motor after the cap? The voltage across the motor before and after is obviously 480 volts. The load of the motor has not changed.... How can the current and KVA change in regards to the motor? The current and reactive power coming from the source is now much different.....but...the motor doesn't have a clue this is happening....think about it. |
| Jun14-12, 10:06 AM | #20 |
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Where is the discrepancy? |
| Jun14-12, 10:29 AM | #21 |
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Your capacitor has a current vector of +90 degrees pointing straight up. The power company sees the combination of those two vectors turning the new single vector to lets say -5 degrees (not exact...but close enough) with a smaller magnitude. All that being said....your motor still has a lagging current vector at -20 degrees. The current and reactive power has now changed from the source and thru the capacitor.....but not thru the motor. I'm not sure how to fix your calcs....but if you understand what is happening....you should be able to adjust. Here's another way to think of it....change the voltage across a motor and it doesn't like that. Change the current thru a motor and it doesn't like that. Change the KVA for a motor and it doesn't like that! Motors hate change! |
| Jun14-12, 10:37 AM | #22 |
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| Jun14-12, 11:19 AM | #23 |
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I don't know how to help u with your calcs....sorry.
The power has definitely changed.....just not thru the motor. The power triangle of the motor has not changed. However, the power triangle of the entire system has changed. Sleep on it....you'll get it. Perhaps if you do actually do the calculations of the current thru the source before cap....and after cap. You will obviously have less current and lower power factor with the addition of the cap thru the source. Forget about the square root of 3 if it scares out.....just figure it at single phase. Although three phase power is simple P = I*V*√3 |
| Jun14-12, 11:39 AM | #24 |
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Nothing else in our circuit has any relevance to power. (REAL POWER) |
| Jun14-12, 11:45 AM | #25 |
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I don't think real power has changed. Only reactive power has changed. Picture the vector of the reactive power of the motor.....small vector pointing straight up. Picture the vector of the reactive power of the capacitor.....pointing straight down. Those two nearly cross eachother out.....All we are doing is making less reactive power....just not thru the motor. Lower reactive power in this case makes for lower current and less full load amps from the power company. |
| Jun14-12, 11:49 AM | #26 |
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| Jun14-12, 11:50 AM | #27 |
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Complex power and real power has stayed the same in motor........ See my post above......... |
| Jun14-12, 11:55 AM | #28 |
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Let's say the system is running.....source, cap and motor.
Let's say i remove the motor and leave the cap running. The current and reactive power thru the cap is the same........however...the power company now sees a much higher reactive power compared to the entire system running. |
| Jun14-12, 11:55 AM | #29 |
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These two are not the same. |
| Jun14-12, 11:59 AM | #30 |
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It's tricky to get....took me a while too. Like I said above....do all the calcs with current and voltage vectors....graph them out...... Graph out the current and voltage sin waves........ Calculate current and reactive power in motor before and after cap.... Calculate current and reacitve power thru source before and after cap..... Once you got it....you got it and will never forget. |
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