Voltage Drop Help

Which formula use depend on data provide in question
Total resistance-use ohm law-volt drop=IxR
Distance and conductor size -use circular mil formula -volt drop=2xKxIxD/CM (1.732 for2 in 3 phase)
Bother about resistance only. Ignore reactance for exam questions not ask with reactance
 
For 240v, you use one way resistance and 120v you use two way resistance, right?
No, the opposite, and you get the same %Vd number either way. 2DIR/240V = DIR/120V, just like DIR(sqrt(3))/480V = DIR/277V.
 
How does the OP apply the equations he has? Under what circumstances? We can only guess these are not on an EE exam.
For the balanced case with a neutral you only need Ohm's Law and the AC resistance of the conductors. The one way Vd in a single phase conductor compared to the phase to neutral voltage is the %Vd for the conductor set, whether it be single or three phase. Personally, when I can I would rather solve problems with first principles rather than by using formulas with derived constants in them.
 
For the balanced case with a neutral you only need Ohm's Law and the AC resistance of the conductors. The one way Vd in a single phase conductor compared to the phase to neutral voltage is the %Vd for the conductor set, whether it be single or three phase.
To be clear, by the "balanced case" you must mean a 2-wire single phase or a balanced (equal current) 3-wire 3 phase circuit, and the "neutral" you refer to would be a possibly fictitious neutral that may not be present in the circuit. E.g. for a 2-wire 120V L-N circuit, your fictitious neutral would be 60V from each conductor, and then you could either use 60V and the one-way resistance, or 120V and the two way resistance. Since multiplication is commutative.

Now you could have a 3-wire single phase circuit or a 4-wire 3-phase circuit, which have an actual neutral circuit conductor, and your statements still apply when the neutral current is zero. But if the neutral current were always going to be zero, most likely there wouldn't be a neutral conductor in the circuit. And if the neutral current is sometimes non-zero (obviously no longer "balanced") then you also need to consider the 2-wire L-N voltage drop.

Cheers, Wayne
 
No, the opposite, and you get the same %Vd number either way. 2DIR/240V = DIR/120V, just like DIR(sqrt(3))/480V = DIR/277V.

But if 240v and 120v supply same power., current 240v system lower, voltage drop lower, but in 120v system, current higher, voltage drop higher. Then how %vd same in 240v and 120v?
 
But if 240v and 120v supply same power., current 240v system lower, voltage drop lower, but in 120v system, current higher, voltage drop higher. Then how %vd same in 240v and 120v?
I believe all ggunn is saying is that if you have a 240V 2-wire load, and you want to compute the %Vd, you can use 240V along with the two-way resistance, or use 120V along with the one-way resistance. You'll get the same %Vd either way.

For absolute Vd, using 240V and two-way resistance gives you the absolute Vd between the 2 wires. And using 120V and the one-way resistance gives you the absolute Vd of one of the wires relative to a fixed neutral point. Which is just 1/2 of the absolute Vd between the 2 wires.

Cheers, Wayne
 
I believe all ggunn is saying is that if you have a 240V 2-wire load, and you want to compute the %Vd, you can use 240V along with the two-way resistance, or use 120V along with the one-way resistance. You'll get the same %Vd either way.

For absolute Vd, using 240V and two-way resistance gives you the absolute Vd between the 2 wires. And using 120V and the one-way resistance gives you the absolute Vd of one of the wires relative to a fixed neutral point. Which is just 1/2 of the absolute Vd between the 2 wires.

Cheers, Wayne
Voltage drop for different voltages, load power to consider to find current. So current not same in two cases. See #27
 
Voltage drop for different voltages, load power to consider to find current. So current not same in two cases. See #27
Yes, I don't believe that ggunn commented on that situation.

For a constant power 2-wire (or balanced 3-wire 3 phase) load, Vd absolute is proportional to current is proportional to 1/voltage, so %Vd is proportional to 1/voltage2. Halve the voltage, and %Vd goes up by a factor of 4.

Cheers, Wayne
 
But if 240v and 120v supply same power., current 240v system lower, voltage drop lower, but in 120v system, current higher, voltage drop higher. Then how %vd same in 240v and 120v?
Sorry, I misinterpreted your earlier question. For a 120V two wire run, yes, you use 2D.
 
Hi all,

Thank you so much for the help with understanding. Knowing that the 2kid is more of the equation that represents the round trip of the wire length was helpful. I am going to continue only with the this equation for my tests. The other seems to basic for what the tests are inquiring about. Here are two pics of the Q&A of these tests. They are both single phase questions but are using different methods of solving. Again i am just going to use 2KID as it seems more accurate.

 
See my questions in the link, hopefully this helps.
Two decimal places seems extreme.

The spreadsheet I use came close enough for me to pick an answer, but not specifically those listed. No doubt the way I set it up 30 years ago. I did not enter the exact resistance for the wire used.
Use the second formula that was in your 1st post for those questions.

One is a single phase question, the other is three phase.
 
Hi all,

Thank you so much for the help with understanding. Knowing that the 2kid is more of the equation that represents the round trip of the wire length was helpful. I am going to continue only with the this equation for my tests. The other seems to basic for what the tests are inquiring about. Here are two pics of the Q&A of these tests. They are both single phase questions but are using different methods of solving. Again i am just going to use 2KID as it seems more accurate.

It's accurate for sizes smaller than 1/0, and accurate for DC. For sizes 1/0 and larger when used with AC, other factors come into play, and the conductance per unit length isn't perfectly proportional to the kcmil area of the wire. As a second order effect, larger wires increase the effective AC resistance, along with the raceway type, whether it's plastic, non-ferrous metal (e.g. aluminum and stainless), or ferrous metal (e.g. steel). It has to do with inductance and the skin effect.

Chapter 9 Table 8 gives DC Resistance values, where it directly follows the formula you find in a Physics textbook (rewritten for the units we use in the NEC, as the 2*K*I*D/A formula we've been discussing). The K-value is its resistivity, typically called rho in a Physics textbook, translated from Ohm-meters to Ohm-kcmil/foot, and adjusted for 75 C temperature. Chapter 9 table 9, gives AC resistance as a function of conduit types and wire size, that accounts for the extra resistivity you get in larger sizes, also at 75C temperature.
 
Somewhat ironic this hasn't been posted after 35 posts but here is some background without getting too far into the weeds...


 
Somewhat ironic this hasn't been posted after 35 posts but here is some background without getting too far into the weeds...


...was... ?
 
Somewhat ironic this hasn't been posted after 35 posts but here is some background without getting too far into the weeds...


The more the merrier, thanks!
 
Correction:
Actually, the reactance between 2 parallel conductors it is:
X=2*pi*f*(0.1404*log10(2*S/dia)+0.0153)/10^3 [ ohm/1000ft] s=distance center-to-center
 
The key one based ohm law and other on circular mil formula
Ohm's law still applies. Precision may not be as high with one method vs another. Commonly used in the field calculations assume a certain temperature (30C I believe) to come up with a factor that is kind of universally used to come up with the resistance portion in Ohm's law which is considered good enough in most calculations. If a situation where more precision is desired or required then more goes into getting more precise on the actual resistance in said application.
 
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