NCEES 531
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Thanks for your reply. I agree with you 100%. However if thats the case then the Voltages (Phase to neutral) should be: Va=500/sqrt(3) angle 60 and Vb=500/sqrt(3) angle -30. However the soultion has Va=500/sqrt(3) angle 90 and Vb=500/sqrt(3) angle 0
Am I missing something here?
What you are missing is that you are mixing up phase voltages with voltages at certain locations. The voltage at location A is V_A, not Va. The problem states that at point A, the line to line voltage (V_A) is 500kV with angle "sigma". It goes on to say that at location B (some distance away from location A), the line to line voltage (V_ B) is 500kV with angle "0". The other given information is a bunch of explanation about the maximum power transfer, etc., but the bottom line is it telling you to set "sigma" equal to 90 degrees. This is basically a power flow problem involving a transmission line and reactive power losses.
The way I solved this problem was like this:
1st.
Calculate the per phase current (V_A/sqrt3 - V_B/sqrt3)/X. Why per phase? because don't forget when we talk about reactances of lines, we are in general talking about per phase reactances, and you can't very well divide line to line voltage by per phase reactance, that would be "apples divided by oranges" and we want to divide "apples by apples". Make sense? Also, when you do the math don't forget that inductive reactance always has 90 degree impedance angle, otherwise you end up with a negative real current, which is well... just wrong. Ater you do the math you should get I=8.17 kAmps (per phase) with angle -45 degrees. In this problem, the angle of the current is irrelevant bcs the impedance of the line is all reactive, so we need only the magnitudes.
2nd.
Calculate the per phase reactive power I^2 * X (per phase) = 3,333
3rd.
Calculate the 3-phase reactive power 3* per phase reactive power = ~ 10,000
Notes and Comments:
1. The fact that NCEES gave you information regarding 5,000 MW is irrelavent. This is quite common with NCESS to give you more information than you actually need.
2. I've noticed that NCEES likes to mix up "apples and oranges" in three phase problems, that is, they'll give part of the information you need in per phase and other parts in 3-phase... it's up to you to do all the conversions into one or the other prior to starting the math.
3. NCEES will almost always put in the answer both the single phase and three phase result, just to check whether or not you are paying attention..
I get what your saying however you are still converting from Line to Line Voltage (500kV) to Line to Neutral Voltage (500kv/sqrt(3)). As a result when you convert there should be a 30 degree phase shift, if you recall Line to Line voltage lead Line to Neutral voltage by 30 degrees. I understand how the problem was solved, I just dont understand why the 30 degree phase shift was not taken into accound when converting the voltages.
What you are missing is that you are mixing up phase voltages with voltages at certain locations. The voltage at location A is V_A, not Va. The problem states that at point A, the line to line voltage (V_A) is 500kV with angle "sigma". It goes on to say that at location B (some distance away from location A), the line to line voltage (V_ B) is 500kV with angle "0". The other given information is a bunch of explanation about the maximum power transfer, etc., but the bottom line is it telling you to set "sigma" equal to 90 degrees. This is basically a power flow problem involving a transmission line and reactive power losses.
The way I solved this problem was like this:
1st.
Calculate the per phase current (V_A/sqrt3 - V_B/sqrt3)/X. Why per phase? because don't forget when we talk about reactances of lines, we are in general talking about per phase reactances, and you can't very well divide line to line voltage by per phase reactance, that would be "apples divided by oranges" and we want to divide "apples by apples". Make sense? Also, when you do the math don't forget that inductive reactance always has 90 degree impedance angle, otherwise you end up with a negative real current, which is well... just wrong. Ater you do the math you should get I=8.17 kAmps (per phase) with angle -45 degrees. In this problem, the angle of the current is irrelevant bcs the impedance of the line is all reactive, so we need only the magnitudes.
2nd.
Calculate the per phase reactive power I^2 * X (per phase) = 3,333
3rd.
Calculate the 3-phase reactive power 3* per phase reactive power = ~ 10,000
Notes and Comments:
1. The fact that NCEES gave you information regarding 5,000 MW is irrelavent. This is quite common with NCESS to give you more information than you actually need.
2. I've noticed that NCEES likes to mix up "apples and oranges" in three phase problems, that is, they'll give part of the information you need in per phase and other parts in 3-phase... it's up to you to do all the conversions into one or the other prior to starting the math.
3. NCEES will almost always put in the answer both the single phase and three phase result, just to check whether or not you are paying attention..
You can do it that way, and you will still get the same answer. It's just another step you don't need which can potentially be an error while you are performing the calculations. There are certain instances for which the -30 phase shift from line-line to line-phase is really important, for example when the question asks you for the phase angle, and the phase angle is included in the answer choices. But in general, so long as you are consistent in keeping 120 degrees between the phases, the magnitude will be the same. Plug in the -30 on both V_A and V_B voltages and you'll see you get exactly the same answer after you do the math.I get what your saying however you are still converting from Line to Line Voltage (500kV) to Line to Neutral Voltage (500kv/sqrt(3)). As a result when you convert there should be a 30 degree phase shift, if you recall Line to Line voltage lead Line to Neutral voltage by 30 degrees. I understand how the problem was solved, I just dont understand why the 30 degree phase shift was not taken into accound when converting the voltages.
What you are missing is that you are mixing up phase voltages with voltages at certain locations. The voltage at location A is V_A, not Va. The problem states that at point A, the line to line voltage (V_A) is 500kV with angle "sigma". It goes on to say that at location B (some distance away from location A), the line to line voltage (V_ B) is 500kV with angle "0". The other given information is a bunch of explanation about the maximum power transfer, etc., but the bottom line is it telling you to set "sigma" equal to 90 degrees. This is basically a power flow problem involving a transmission line and reactive power losses.
The way I solved this problem was like this:
1st.
Calculate the per phase current (V_A/sqrt3 - V_B/sqrt3)/X. Why per phase? because don't forget when we talk about reactances of lines, we are in general talking about per phase reactances, and you can't very well divide line to line voltage by per phase reactance, that would be "apples divided by oranges" and we want to divide "apples by apples". Make sense? Also, when you do the math don't forget that inductive reactance always has 90 degree impedance angle, otherwise you end up with a negative real current, which is well... just wrong. Ater you do the math you should get I=8.17 kAmps (per phase) with angle -45 degrees. In this problem, the angle of the current is irrelevant bcs the impedance of the line is all reactive, so we need only the magnitudes.
2nd.
Calculate the per phase reactive power I^2 * X (per phase) = 3,333
3rd.
Calculate the 3-phase reactive power 3* per phase reactive power = ~ 10,000
Notes and Comments:
1. The fact that NCEES gave you information regarding 5,000 MW is irrelavent. This is quite common with NCESS to give you more information than you actually need.
2. I've noticed that NCEES likes to mix up "apples and oranges" in three phase problems, that is, they'll give part of the information you need in per phase and other parts in 3-phase... it's up to you to do all the conversions into one or the other prior to starting the math.
3. NCEES will almost always put in the answer both the single phase and three phase result, just to check whether or not you are paying attention..
Here is another instance where there was the same confusion and I answered it similarly:
/>http://engineerboards.com/index.php?showtopic=19446
You’re correct. I get the same magnitude when I take into account the 30 degree shift; however the overall angle is different. I guess in this case its best to leave the shift out.You can do it that way, and you will still get the same answer. It's just another step you don't need which can potentially be an error while you are performing the calculations. There are certain instances for which the -30 phase shift from line-line to line-phase is really important, for example when the question asks you for the phase angle, and the phase angle is included in the answer choices. But in general, so long as you are consistent in keeping 120 degrees between the phases, the magnitude will be the same. Plug in the -30 on both V_A and V_B voltages and you'll see you get exactly the same answer after you do the math.I get what your saying however you are still converting from Line to Line Voltage (500kV) to Line to Neutral Voltage (500kv/sqrt(3)). As a result when you convert there should be a 30 degree phase shift, if you recall Line to Line voltage lead Line to Neutral voltage by 30 degrees. I understand how the problem was solved, I just dont understand why the 30 degree phase shift was not taken into accound when converting the voltages.
What you are missing is that you are mixing up phase voltages with voltages at certain locations. The voltage at location A is V_A, not Va. The problem states that at point A, the line to line voltage (V_A) is 500kV with angle "sigma". It goes on to say that at location B (some distance away from location A), the line to line voltage (V_ B) is 500kV with angle "0". The other given information is a bunch of explanation about the maximum power transfer, etc., but the bottom line is it telling you to set "sigma" equal to 90 degrees. This is basically a power flow problem involving a transmission line and reactive power losses.
The way I solved this problem was like this:
1st.
Calculate the per phase current (V_A/sqrt3 - V_B/sqrt3)/X. Why per phase? because don't forget when we talk about reactances of lines, we are in general talking about per phase reactances, and you can't very well divide line to line voltage by per phase reactance, that would be "apples divided by oranges" and we want to divide "apples by apples". Make sense? Also, when you do the math don't forget that inductive reactance always has 90 degree impedance angle, otherwise you end up with a negative real current, which is well... just wrong. Ater you do the math you should get I=8.17 kAmps (per phase) with angle -45 degrees. In this problem, the angle of the current is irrelevant bcs the impedance of the line is all reactive, so we need only the magnitudes.
2nd.
Calculate the per phase reactive power I^2 * X (per phase) = 3,333
3rd.
Calculate the 3-phase reactive power 3* per phase reactive power = ~ 10,000
Notes and Comments:
1. The fact that NCEES gave you information regarding 5,000 MW is irrelavent. This is quite common with NCESS to give you more information than you actually need.
2. I've noticed that NCEES likes to mix up "apples and oranges" in three phase problems, that is, they'll give part of the information you need in per phase and other parts in 3-phase... it's up to you to do all the conversions into one or the other prior to starting the math.
3. NCEES will almost always put in the answer both the single phase and three phase result, just to check whether or not you are paying attention..
Here is another instance where there was the same confusion and I answered it similarly:
http://engineerboard...showtopic=19446
Thanks for all your help!
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