Great Catch! You are absolutely correct. For the max power to be 1.37MW the voltages would need to line values however the problem states they are not so this is actually equal to the single phase max power which needs to be multiplied by 3 to yield the three phase max power of 4.11MW. The units are also incorrectly labeled as "KW" instead of MW in the video. This has been flagged to be updated in the future.
My pleasure, glad it was helpful. This video was published 6 years ago before the change to the new CBT format of the power PE exam at the end of 2020, when you were expected to bring in outside references for formulas. This particular formula for mechanical torque is not included in the NCEES Reference Handbook. Since the NCEES Reference Handbook does not contain every formula, if a problem on the exam requires a formula that is not in the Reference Handbook, it is generally safe to assume that it will most likely be provided in the problem statement if it is required to solve the problem.
Xs is the synchronous reactance of the machine, typically determined from the results of the open and short circuit test by the manufacturer. The synchronous reactance of the machine cannot change, it's the inherent reactance of the machine at rated conditions. However, if you had a machine with equal ratings except for a different synchronous reactance value then yes, the max power would also be different as a result.
I have a question regarding the power calculation. Wouldn't you need to include angle of the induced starter voltage in the power calculation? (E < 0) and (Eo < 90)? Or is the equation using magnitudes, and i missed that part? Thanks again for working these out they really help! i am definitely signed up for your class!
Hi Spikes, by "angle of the induced starter voltage" are you referring to the applied voltage E, or the induced stator voltage Eo? For max power, you do not need to know the angles since max power is actually defined by an agle of 90º between the two. That's the trick to this question. For other questions, you would generally assume a reference angle of 0º for the applied line to neutral voltage (E), and solve from there. This results in the current angle being equal but opposite in polarity to the power angle theta which makes it easier to determine if the machine is leading or lagging.
Hi all, the speed given in the problem, 300 r.p.m , why it was considered to be the Ns, ( the speed of the stator magnetic field) normally the speed of the synch machine given in the problems is the speed of the rotor which is N, am I missing something?
Hi Razina, thanks for your comment. The conditions for maximum power of a synchronous machine (what is being asked for in this problem) is an electrical torque angle of 90 degrees. Sine of 90 degrees is equal to one: sin(90º) = 1
Clear, concise & complete. Excellent job!
Max power? More like max *learning* power! Thanks again for making and sharing such an excellent video series.
Clear explanation . Thank you Zach!!
Awesome video! Thank you!
Thanks AJ Glad you've been enjoying our videos
Thanks for working this example out.
Pmax=(1200V)(2400V)sin(90)/(2.1 ohm)= 1.37MW.
If this machine is three-phase (and they usually are) the maximum power would be 4.11MW.
Great Catch! You are absolutely correct. For the max power to be 1.37MW the voltages would need to line values however the problem states they are not so this is actually equal to the single phase max power which needs to be multiplied by 3 to yield the three phase max power of 4.11MW. The units are also incorrectly labeled as "KW" instead of MW in the video. This has been flagged to be updated in the future.
Thanks for the video. Was the equation of calculating mechanical torque angle from electrical torque angle given in the NCEES reference book?
My pleasure, glad it was helpful. This video was published 6 years ago before the change to the new CBT format of the power PE exam at the end of 2020, when you were expected to bring in outside references for formulas.
This particular formula for mechanical torque is not included in the NCEES Reference Handbook.
Since the NCEES Reference Handbook does not contain every formula, if a problem on the exam requires a formula that is not in the Reference Handbook, it is generally safe to assume that it will most likely be provided in the problem statement if it is required to solve the problem.
How do you decide on Xs? If it was different, say 5, wouldn't that change your max power?
Xs is the synchronous reactance of the machine, typically determined from the results of the open and short circuit test by the manufacturer. The synchronous reactance of the machine cannot change, it's the inherent reactance of the machine at rated conditions. However, if you had a machine with equal ratings except for a different synchronous reactance value then yes, the max power would also be different as a result.
if Xs was not given, can it be calculated from the givens?
it has to be provided
I have a question regarding the power calculation. Wouldn't you need to include angle of the induced starter voltage in the power calculation? (E < 0) and (Eo < 90)? Or is the equation using magnitudes, and i missed that part?
Thanks again for working these out they really help! i am definitely signed up for your class!
Hi Spikes, by "angle of the induced starter voltage" are you referring to the applied voltage E, or the induced stator voltage Eo? For max power, you do not need to know the angles since max power is actually defined by an agle of 90º between the two. That's the trick to this question. For other questions, you would generally assume a reference angle of 0º for the applied line to neutral voltage (E), and solve from there. This results in the current angle being equal but opposite in polarity to the power angle theta which makes it easier to determine if the machine is leading or lagging.
What makes you assume the problem is stating the line to neutral voltages?
I hope its not too late, but the question states that voltages are line to neutral values.
Hi all, the speed given in the problem, 300 r.p.m , why it was considered to be the Ns, ( the speed of the stator magnetic field) normally the speed of the synch machine given in the problems is the speed of the rotor which is N, am I missing something?
How did u get 1 for sin(x)? 1 was not given
Hi Razina, thanks for your comment. The conditions for maximum power of a synchronous machine (what is being asked for in this problem) is an electrical torque angle of 90 degrees. Sine of 90 degrees is equal to one:
sin(90º) = 1