winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
Smart$,
I believe that you are not expressing the total voltage drop correctly. My reasoning is that the voltage drop at the 'hot' end of the load and the voltage drop at the 'neutral' end of the load have the vector for the load voltage between them. You get a vector that is the correct total voltage across the conductors feeding the load, but this is not necessarily the percentage change in voltage seen by the load. IMHO you have to add the voltage dropped across the conductors to the supply voltage to get the voltage seen by the load, and the % voltage drop is then (supply voltage - load voltage)/supply voltage.
Lou,
In a three phase system, as current flow is increased on the B leg, the voltage seen by loads on the A leg will go up. But you need to look at the details of the vectors.
Here is my first attempt at posting diagrams, previously I've simply tried to describe the diagrams.
1) The circuit. I am assuming a perfect 120/208V supply, and I am making the simplifying approximation that the currents in the loads remain constant and in phase with the supply. This is strictly not correct, since the loads will actually draw current that depends upon the supply voltage and phase angle, however for situations where the voltage drop is small with respect to the load voltage, this simplifying approximation does not introduce a significant error.
http://www.mikeholt.com/code_forum/attachment.php?attachmentid=261&d=1165082617
2) Now, the first thing that I will do is calculate the current through the neutral resistance. First we use vector math to add 50A at 0 degrees to 25A at 120 degrees, to get a net current through the neutral of 43.3A at 30 degrees. Since phase A is carrying more current than phase B, the current is more nearly in phase with phase A.
http://www.mikeholt.com/code_forum/attachment.php?attachmentid=262&d=1165082673
(You will note that this is essentially the same diagram that Smart$ posted, except that I am using a different sign convention. I am probably using the 'wrong' sign convention; it shouldn't really matter as long as you remain internally consistent.)
3) Next I will calculate the voltage at 'Node N'. This is the point where the two circuits combine, and the net current is flowing through the neutral resistance. The voltage across an impedance is given by Ohm's law, E=IZ. However you have to remember that E (voltage), I (current) and Z (impedance) are all vector values. For now Z is a pure resistance, which means that E will be in phase with I, so we don't have to worry about how to _multiply_ vector quantities; instead we will simplify and say that the magnitude of E is simply the magnitude of I times Z, and that the direction of E is simply the direction of I.
The voltage at Node N is thus 43.3A at 30 degrees * 0.03 Ohms = 1.3V at 30 degrees.
http://www.mikeholt.com/code_forum/attachment.php?attachmentid=263&d=1165082682
4) We also have to calculate the voltages at 'Node A load' and 'Node B load'. These are the voltages where the circuit conductors feed the load, after the current has gone through the conductor resistance. One point that is very important to note: you have to get the sign correct. I don't have a very good methodology for this (probably the result of my bad sign convention), and instead I intuit this out every time. But we know that the voltages at the load points are lower than the supply points, so clearly the E=IR voltage at the load points has to 'fight' the supply voltage. Since we made the simplifying assumption that the load currents were in phase with the _supply_, then the supply voltages simply get reduced by the voltage drop in the supply conductors.
At the source, we had three voltage nodes, 'Node 0' at 0V, 'Node A sup' at 120V 0 degrees, and 'Node B sup' at 120V 120 degrees. At the load we have three voltage nodes, 'Node N' at 1.3V 30 degrees, 'Node A load' at 118.5V 120 degrees, and 'Node B load' at 119.25V 120 degrees. We do the vector subtraction (simply measure the appropriate distances on the 2D voltage/phase plane) to get the voltage delivered to the load. I've distorted the below drawing to make it fit.
http://www.mikeholt.com/code_forum/attachment.php?attachmentid=264&d=1165082691
5) The net result is that the voltage delivered to the 25A load is 119.26V, to the 50A load is 117.37V.
-Jon
I believe that you are not expressing the total voltage drop correctly. My reasoning is that the voltage drop at the 'hot' end of the load and the voltage drop at the 'neutral' end of the load have the vector for the load voltage between them. You get a vector that is the correct total voltage across the conductors feeding the load, but this is not necessarily the percentage change in voltage seen by the load. IMHO you have to add the voltage dropped across the conductors to the supply voltage to get the voltage seen by the load, and the % voltage drop is then (supply voltage - load voltage)/supply voltage.
Lou,
In a three phase system, as current flow is increased on the B leg, the voltage seen by loads on the A leg will go up. But you need to look at the details of the vectors.
Here is my first attempt at posting diagrams, previously I've simply tried to describe the diagrams.
1) The circuit. I am assuming a perfect 120/208V supply, and I am making the simplifying approximation that the currents in the loads remain constant and in phase with the supply. This is strictly not correct, since the loads will actually draw current that depends upon the supply voltage and phase angle, however for situations where the voltage drop is small with respect to the load voltage, this simplifying approximation does not introduce a significant error.
http://www.mikeholt.com/code_forum/attachment.php?attachmentid=261&d=1165082617
2) Now, the first thing that I will do is calculate the current through the neutral resistance. First we use vector math to add 50A at 0 degrees to 25A at 120 degrees, to get a net current through the neutral of 43.3A at 30 degrees. Since phase A is carrying more current than phase B, the current is more nearly in phase with phase A.
http://www.mikeholt.com/code_forum/attachment.php?attachmentid=262&d=1165082673
(You will note that this is essentially the same diagram that Smart$ posted, except that I am using a different sign convention. I am probably using the 'wrong' sign convention; it shouldn't really matter as long as you remain internally consistent.)
3) Next I will calculate the voltage at 'Node N'. This is the point where the two circuits combine, and the net current is flowing through the neutral resistance. The voltage across an impedance is given by Ohm's law, E=IZ. However you have to remember that E (voltage), I (current) and Z (impedance) are all vector values. For now Z is a pure resistance, which means that E will be in phase with I, so we don't have to worry about how to _multiply_ vector quantities; instead we will simplify and say that the magnitude of E is simply the magnitude of I times Z, and that the direction of E is simply the direction of I.
The voltage at Node N is thus 43.3A at 30 degrees * 0.03 Ohms = 1.3V at 30 degrees.
http://www.mikeholt.com/code_forum/attachment.php?attachmentid=263&d=1165082682
4) We also have to calculate the voltages at 'Node A load' and 'Node B load'. These are the voltages where the circuit conductors feed the load, after the current has gone through the conductor resistance. One point that is very important to note: you have to get the sign correct. I don't have a very good methodology for this (probably the result of my bad sign convention), and instead I intuit this out every time. But we know that the voltages at the load points are lower than the supply points, so clearly the E=IR voltage at the load points has to 'fight' the supply voltage. Since we made the simplifying assumption that the load currents were in phase with the _supply_, then the supply voltages simply get reduced by the voltage drop in the supply conductors.
At the source, we had three voltage nodes, 'Node 0' at 0V, 'Node A sup' at 120V 0 degrees, and 'Node B sup' at 120V 120 degrees. At the load we have three voltage nodes, 'Node N' at 1.3V 30 degrees, 'Node A load' at 118.5V 120 degrees, and 'Node B load' at 119.25V 120 degrees. We do the vector subtraction (simply measure the appropriate distances on the 2D voltage/phase plane) to get the voltage delivered to the load. I've distorted the below drawing to make it fit.
http://www.mikeholt.com/code_forum/attachment.php?attachmentid=264&d=1165082691
5) The net result is that the voltage delivered to the 25A load is 119.26V, to the 50A load is 117.37V.
-Jon
Last edited:
