0:33 - Problem statement 2:10 - Equivalent Circuit 3:27 - Solve for the resistance per phase (r1) 4:36 - Solve for the total no load apparent power |SNL| 6:32 - Solve for the Magnetizing reactance (Xm) 11:12 - Solve for the core, winding, and friction resistance (Rm) 6:16 - Solve for the total locked rotor apparent power |SLR| 18:20 - Solve for the total rotor reactive power (QLR) 20:03 - Solve for the leakage reactance (x) 24:48 - Solve for the rotor resistance at locked rotor (R2/s slip resistance)
I have watched this video over and over. I'm happy to say that I no longer have issues setting up the equivalent circuits and ready to solve this type of problem on exam day. I can't thank you enough Zach, keep up the good work!
Can you please clarify why during when you were solving for core, winding, and friction resistance (Rm) you did not use resistance R2 (rotor resistance). I see that it is in series with R1 (stator resistance). However, when you were plugin in the varies power loss ie Pnl=Prm + 3I^2R1 but did not add 3I^2R2. What is the problem had said core, winding, and friction resistance (Rm) + rotor resistance. Would it be fair to say that at the juncture then Pr2 would be added. I though frictional losses occur at the rotor.
One of my favorite Zach's video is this one. Induction motor no load and locker rotor test. I don't think there is anyone could explain this type of problem thoroughly aside from him. Kudos!
Amazing video. Zach lessons are the best out there. Locked rotor test is a topic that you may find on the exam and personally i dint have enough study materials about it until I found Zach materials.
This is another one of Zach's superb videos. This is probably in my top 5 of all his explanations, because he does such a great job breaking down the concepts to be truly understandable. I've always struggled with locked rotor test, and I feel much more confident after watching this video.
Hi Zach. I noticed that you subtracted the stator copper losses from the NL real power measurement (5300 W) in order to calculate the iron losses. Why is this the case? For the NL test, don't we ignore everything besides the magnetizing branch (magnetizing reactance and iron resistance)? Therefore, I thought that the entire 5300 W represents the power lost in RM.
This was bugging me as well. I know you're comment was a year ago, but I'll post this in case it helps someone else. The stator copper losses are actually in series before the parallel magnetizing branch in the full equivalent circuit model and are moved to the right to make other calculations easier as it is normally irrelevant to the accuracy of the model. The stator will certainly have current through it during the no load test and will produce losses, so for the no load test only, imagine r1 to the left of the magnetizing branch, which is why it needs to be accounted for.
@@yuechen2421 I can't give you an answer on that one. If you do calculate it that way the result is fairly close at 6.14, but I don't have a text book that actually had a good explanation for why you use the simplified model for NL reactance and the normal equivalent for the NL resistance. The equivalent circuits are approximations though and if you solve for jx first you neglect the magnitizing branch anyways, so attempting to solve it that way isn't 100 percent accurate and the normal procedure that is followed to solve for all of the unknowns is likely the best model we can get from these two tests if I were to take a guess at it.
Sir, I do have a question. For the induction motor equivalent circuit, normally the stator resistance and reactance come before the magnetizing resistance and reactance. For question 3, I see you assumed that all reactive power is consumed by the magnetizing reactance, should that reactive power includes the stator reactance also?
Excellent group of questions. Just wondering though for the final question (VIII) are you solving for R2 or R2/s? If R2/s, then if the slip was not =1 then the value at the end would be multiplied by s (s*0.061 ohms) vs being divided by s (0.061 ohms/s) as you stated at the end. Thanks again for the content.
For (IV) core, winding, and friction resistance, if the no-load test is equivalent to the open circuit test, why do you consider R1. is that supposed to be Pnl=PRm?
Hi Zach, thanks for this information is very valuable. Question. I have a voltage and amperes from leads A-B, B-C, A-C, and their corresponding amperes, I think I just need to calculate the VA that I get from those measurements and use the power factor estimated to get a calculated watts, will that work as simple as that?
Because apparent power (S) is the hypotenuse (largest leg) of the triangle. There is only a positive sign under the square root if you are solving for apparent power (S). a^2 + b^2 = c^2 Is the same thing as: P^2 + Q^2 = S^2. As long as you know two legs of the triangle you can solve for the third unknown: S = √(P^2 + Q^2) P = √( S^2 - Q^2 Q = √(S^2 - P^2) Try deriving each of these by hand from the original formula and it will be nice and clear. No such thing as a dumb question.
I am currently preparing for Apr PE exam and your examples really help in understanding the concepts without any difficulties. For this example should the stator resistance per phase = 1.3/3 instead of 1.3/2? as it appears to be a delta to wye conversion? Also the Reative NL power would be = magnetizing reactance power + stator reactance power? Also, there would be no Slip resistance at NL since you didnt add it in Total NL real power eq. And the current through the magnetizing resistance would almost close to zero, how do i find it if i were to just use the eq. power = I^2 * Z?
Hi Christian. It is a wye connected motor. If you measure the resistance using a meter across two terminals, you are effectively measuring two phases of the wye motor, so you need to divide this value by 2 to end up with the per phase resistance. Yes. The reactive no-load power drawn by the motor is the sum of every inductive aspect of the motor. Correct, there is no rotor resistance at no load. There is no mechanical force that the rotor is driving. It because negligible.
I'm confused about when to use P=V^2/R and Q=V^2/X. I notice you use P and Q 3 phase and the voltage line to line, and that gives you the R and X values from line to neutral. Why is that?
Hi Sailus, impedance is always a per phase quantity even in three phase systems. This is unlike line and voltage quantities that may be either per phase or a line quantity.
En is the phase voltage, also known as the line-to-neutral voltage or the per phase voltage. The circuit above is a single phase equivalent circuit to help us analyze a three phase system. Single phase equivalent circuits *always* use the per phase quantities of a wye source wye load. This is similar to standard three phase circuit analysis when you must convert any delta connections (source or load) to their wye equivalent, before being able to use the per phase line to neutral quantities in a single phase equivalent circuit.
If you could include time stamps of each step you are referring to, it would make it much easier for me to answer your question. The line voltage is used to calculate Rm because the power valuing being used is three phase.
@@electricalpereview Sorry I thought I captured this. You state stator resistance is 0.65ohm at 4:42 that seems to be a value at L-N. The around 15:58 it seems that Rm=778. You use 575V, should this not be 331V, as the diagram is voltage to neutral? I wish I would of found your course earlier, your material is amazing. But given I take my test at the end of the month. I dont think I can cram all your course work in that time frame.
@@cojeff5688 Thanks Jeff, now I understand your question. Earlier, when we calculated PRM (425 W), we calculated the total three-phase power value, and not just the "per-phase" power value absorbed across one phase of Rm by the line to neutral voltage En. Because PRm is already a three-phase value, I used the line voltage of the motor (575V) and not the line to neutral voltage (En) to calculate Rm. The neat thing is you can do either and the math works out just the same. If you want to use the line to neutral voltage in this calculation (En = 575V/√3), then just be sure to use the per-phase value of PRm along with it instead of the three-phase value of PRm that we used along with the full line voltage. We can find the per-phase value of PRm by dividing the three-phase value by three: PRm-1ø = PRM -3ø/3 PRm-1ø = 425 W/3 PRm-1ø = 141.67 W Next, use this value along with the line to neutral voltage (EN = 575V/√3) to calculate the same per-phase value of Rm: Rm = En^2/PRm-1ø Rm = (575V/√3)^2 / 141.67 W Rm = 778 Ω Same value that we get at 15:58 using line voltage and three-phase power! Don't forget to square the line to neutral value *after* you divide by the square root of three.
0:33 - Problem statement
2:10 - Equivalent Circuit
3:27 - Solve for the resistance per phase (r1)
4:36 - Solve for the total no load apparent power |SNL|
6:32 - Solve for the Magnetizing reactance (Xm)
11:12 - Solve for the core, winding, and friction resistance (Rm)
6:16 - Solve for the total locked rotor apparent power |SLR|
18:20 - Solve for the total rotor reactive power (QLR)
20:03 - Solve for the leakage reactance (x)
24:48 - Solve for the rotor resistance at locked rotor (R2/s slip resistance)
I have watched this video over and over. I'm happy to say that I no longer have issues setting up the equivalent circuits and ready to solve this type of problem on exam day. I can't thank you enough Zach, keep up the good work!
Can you please clarify why during when you were solving for core, winding, and friction resistance (Rm) you did not use resistance R2 (rotor resistance). I see that it is in series with R1 (stator resistance). However, when you were plugin in the varies power loss ie Pnl=Prm + 3I^2R1 but did not add 3I^2R2.
What is the problem had said core, winding, and friction resistance (Rm) + rotor resistance. Would it be fair to say that at the juncture then Pr2 would be added.
I though frictional losses occur at the rotor.
One of my favorite Zach's video is this one. Induction motor no load and locker rotor test. I don't think there is anyone could explain this type of problem thoroughly aside from him. Kudos!
Very useful. I watched several times taking all the notes. The level how such complicated study is broken down step-by-step is superb!
Glad it was helpful!
Another extremely helpful video. I've gone back to this one several times during my last few months of studying.
Amazing video. Zach lessons are the best out there. Locked rotor test is a topic that you may find on the exam and personally i dint have enough study materials about it until I found Zach materials.
Thank you Zach , your course made it very comfortable preparing for PE exam , you covered all the hard and confusing material
Thanks Rania. Best of luck on the PE exam, we greatly enjoyed having you as a student in our online program this semester.
Great video that really simplifies the different types questions you may be asked regarding motor no load and locked rotor tests. Thanks!
Great video, Zach explained in a simpler way this important but difficult topic.
After struggling through the few practice test I can find, the online class and videos have prepared me for the April PE EXAM .
this is a great Motor no load and locked rotor test. a great video if you are struggling on this subject. thanks zach
I am still trying to fully understand how motor and NEC code questions can be worded, but this definitely helped me a lot!
It is a great video. Explains all the details on how locked rotor state of the motor occurs and locked rotor test.
Thank you! This is great. I like how you go step by step in such a way to make this very clear.
This is another one of Zach's superb videos. This is probably in my top 5 of all his explanations, because he does such a great job breaking down the concepts to be truly understandable. I've always struggled with locked rotor test, and I feel much more confident after watching this video.
Awesome video! Thank you!
Thanks for commenting, Glad you enjoyed it!
Great instruction video. Easy to understand and easy to follow.
Hi Zach. I noticed that you subtracted the stator copper losses from the NL real power measurement (5300 W) in order to calculate the iron losses. Why is this the case? For the NL test, don't we ignore everything besides the magnetizing branch (magnetizing reactance and iron resistance)? Therefore, I thought that the entire 5300 W represents the power lost in RM.
This was bugging me as well. I know you're comment was a year ago, but I'll post this in case it helps someone else. The stator copper losses are actually in series before the parallel magnetizing branch in the full equivalent circuit model and are moved to the right to make other calculations easier as it is normally irrelevant to the accuracy of the model. The stator will certainly have current through it during the no load test and will produce losses, so for the no load test only, imagine r1 to the left of the magnetizing branch, which is why it needs to be accounted for.
@@jaymoseley6216 How about for the magnetizing reactance? the reactive power he calculated was only for Xm, did not include jX. I am kind of confused.
same question here.
@@yuechen2421 I can't give you an answer on that one. If you do calculate it that way the result is fairly close at 6.14, but I don't have a text book that actually had a good explanation for why you use the simplified model for NL reactance and the normal equivalent for the NL resistance. The equivalent circuits are approximations though and if you solve for jx first you neglect the magnitizing branch anyways, so attempting to solve it that way isn't 100 percent accurate and the normal procedure that is followed to solve for all of the unknowns is likely the best model we can get from these two tests if I were to take a guess at it.
@@jaymoseley6216 thanks jay!
Sir, I do have a question. For the induction motor equivalent circuit, normally the stator resistance and reactance come before the magnetizing resistance and reactance. For question 3, I see you assumed that all reactive power is consumed by the magnetizing reactance, should that reactive power includes the stator reactance also?
Excellent group of questions. Just wondering though for the final question (VIII) are you solving for R2 or R2/s? If R2/s, then if the slip was not =1 then the value at the end would be multiplied by s (s*0.061 ohms) vs being divided by s (0.061 ohms/s) as you stated at the end. Thanks again for the content.
Solving for just the rotor resistance R2, and not the rotor slip resistance R2/s
At 13:13 how do you get that the current in R1 is 50A? Isn't that the line current? How about the current in the Xm || Rm branch?
Thanks Zach, this is a great explanation.
For (IV) core, winding, and friction resistance, if the no-load test is equivalent to the open circuit test, why do you consider R1. is that supposed to be Pnl=PRm?
Great video dude! Thank you!🙇♂️
Glad you found it helpful
Hi Zach, thanks for this information is very valuable. Question. I have a voltage and amperes from leads A-B, B-C, A-C, and their corresponding amperes, I think I just need to calculate the VA that I get from those measurements and use the power factor estimated to get a calculated watts, will that work as simple as that?
That should work.
Very helpful tutoring
might be a dumb question, but why did you use "-" in the pythagorean equation for the locked rotor reactive power?
Because apparent power (S) is the hypotenuse (largest leg) of the triangle. There is only a positive sign under the square root if you are solving for apparent power (S).
a^2 + b^2 = c^2
Is the same thing as:
P^2 + Q^2 = S^2.
As long as you know two legs of the triangle you can solve for the third unknown:
S = √(P^2 + Q^2)
P = √( S^2 - Q^2
Q = √(S^2 - P^2)
Try deriving each of these by hand from the original formula and it will be nice and clear.
No such thing as a dumb question.
I am currently preparing for Apr PE exam and your examples really help in understanding the concepts without any difficulties. For this example should the stator resistance per phase = 1.3/3 instead of 1.3/2? as it appears to be a delta to wye conversion? Also the Reative NL power would be = magnetizing reactance power + stator reactance power? Also, there would be no Slip resistance at NL since you didnt add it in Total NL real power eq. And the current through the magnetizing resistance would almost close to zero, how do i find it if i were to just use the eq. power = I^2 * Z?
Hi Christian.
It is a wye connected motor. If you measure the resistance using a meter across two terminals, you are effectively measuring two phases of the wye motor, so you need to divide this value by 2 to end up with the per phase resistance.
Yes. The reactive no-load power drawn by the motor is the sum of every inductive aspect of the motor.
Correct, there is no rotor resistance at no load. There is no mechanical force that the rotor is driving. It because negligible.
I'm confused about when to use P=V^2/R and Q=V^2/X. I notice you use P and Q 3 phase and the voltage line to line, and that gives you the R and X values from line to neutral. Why is that?
Ok, I think I see now, it is the voltage across the resistor. My other question then is why is the source En and not Ephase?
Hi Sailus, impedance is always a per phase quantity even in three phase systems. This is unlike line and voltage quantities that may be either per phase or a line quantity.
En is the phase voltage, also known as the line-to-neutral voltage or the per phase voltage. The circuit above is a single phase equivalent circuit to help us analyze a three phase system. Single phase equivalent circuits *always* use the per phase quantities of a wye source wye load. This is similar to standard three phase circuit analysis when you must convert any delta connections (source or load) to their wye equivalent, before being able to use the per phase line to neutral quantities in a single phase equivalent circuit.
I thought the equation would have been sqrt(3) * V^2/X or R. Was the equation derived from the 3 Phase apparent power equation S=sqrt(3)*V*I?
In step IV should you of not divide by root 3 as the voltage is L-N?
If you could include time stamps of each step you are referring to, it would make it much easier for me to answer your question. The line voltage is used to calculate Rm because the power valuing being used is three phase.
@@electricalpereview Sorry I thought I captured this.
You state stator resistance is 0.65ohm at 4:42 that seems to be a value at L-N. The around 15:58 it seems that Rm=778. You use 575V, should this not be 331V, as the diagram is voltage to neutral?
I wish I would of found your course earlier, your material is amazing. But given I take my test at the end of the month. I dont think I can cram all your course work in that time frame.
@@cojeff5688 Thanks Jeff, now I understand your question.
Earlier, when we calculated PRM (425 W), we calculated the total three-phase power value, and not just the "per-phase" power value absorbed across one phase of Rm by the line to neutral voltage En.
Because PRm is already a three-phase value, I used the line voltage of the motor (575V) and not the line to neutral voltage (En) to calculate Rm. The neat thing is you can do either and the math works out just the same.
If you want to use the line to neutral voltage in this calculation (En = 575V/√3), then just be sure to use the per-phase value of PRm along with it instead of the three-phase value of PRm that we used along with the full line voltage.
We can find the per-phase value of PRm by dividing the three-phase value by three:
PRm-1ø = PRM -3ø/3
PRm-1ø = 425 W/3
PRm-1ø = 141.67 W
Next, use this value along with the line to neutral voltage (EN = 575V/√3) to calculate the same per-phase value of Rm:
Rm = En^2/PRm-1ø
Rm = (575V/√3)^2 / 141.67 W
Rm = 778 Ω
Same value that we get at 15:58 using line voltage and three-phase power!
Don't forget to square the line to neutral value *after* you divide by the square root of three.
Insano