Voltage Drop

[Unbridled]

Using Vd= 1.73*K*I*L/CM

Load= 3 phase 3 Wire.
Feeder Length 150 FT

Question: In regards to total length to be used in the formula, do you consider the sum of all 3 ungrounded conductors length. IE. 150 FT X 3 (250).

Just trying to brush up.:thumbsup::thumbsup:

 

[Ingenieur]

No
one way length
the 1.73 accounts for the supply/rtn

 

[Unbridled]

No
one way length
the 1.73 accounts for the supply/rtn

Thanks Ingenieur , I assume that single phase length then, is the total of both of the current carrying conductors?

Why is that, do you know?
Do 3 PH systems harmonically cancel in some way as to only consider one conductor length for Vd calculations?

Please Advise,
:thumbsup:

 

[oldsparky52]

Thanks Ingenieur , I assume that single phase length then, is the total of both of the current carrying conductors?

Why is that, do you know?
Do 3 PH systems harmonically cancel in some way as to only consider one conductor length for Vd calculations?

Please Advise,
:thumbsup:

I think it has to do with the phases being 120 degrees out from each other.

 

[kwired]

With single phase you have to double length because same current is traveling in one conductor and out the other. Single phase is simple - what comes in on one has to go out on the other.

With balanced three phase load the current returning on A is partially coming from both B and C, current returning on B is partially coming from A and C, current returning on C is partially returning on A and B. All that happening so fast it is nearly the same time from a human perspective. Square root of 3 is the key factor that results in the amount of RMS current that flows and is associated with the fact the phase angle is 120 degrees.

 

[Unbridled]

With single phase you have to double length because same current is traveling in one conductor and out the other. Single phase is simple - what comes in on one has to go out on the other.

With balanced three phase load the current returning on A is partially coming from both B and C, current returning on B is partially coming from A and C, current returning on C is partially returning on A and B. All that happening so fast it is nearly the same time from a human perspective. Square root of 3 is the key factor that results in the amount of RMS current that flows and is associated with the fact the phase angle is 120 degrees.

Yes, I understand phase 3 PH currents being 120 Degrees apart. So you are saying that this also true with voltage as used in Vd calc's?

 

[Ingenieur]

As others have said
it is the vector or phasor sum of 2 phase lines (if you sum 3 = 0 in a balanced system)
ph ia = 1/0 deg = 1 + 0j
ph ib = 1/120 deg = -0.50+ 0.866j
sign convention is opposite, one is entering a node the other leaving
1 + 0j -(-0.50 + 0.866j) = 1.5 - 0.866j = 1.732/-30 deg
1.732 is sqrt3
so magnitude of the supply/'rtn' currents is sqrt3 x 1 current

 

[Ingenieur]

Yes, I understand phase 3 PH currents being 120 Degrees apart. So you are saying that this also true with voltage as used in Vd calc's?

Yes, basically (but offset from i based on load type)
but in vdrop calcs you are actually using current
vdrop = 1.732 x i x conductor R
the equation you used substitues Area (circ mils) and resistivity (K) for R
R = resistivity x L/A

 

[Smart $]

Yes, I understand phase 3 PH currents being 120 Degrees apart. So you are saying that this also true with voltage as used in Vd calc's?

Yes, because the voltage drop always obeys ohm's law: E=I×R. Outgoing current on one line returns on the other two lines, varying proportionally to the respective voltage across the lines. Incoming current is just the reverse. This is true of all three lines... but 120° phasing shifts the timing of how it occurs.

 

[Ingenieur]

We assume a balanced system
so we only calculate drop on one line-line v, 2 lines
and we assume the other 2 v's will be the similar

 

[Russs57]

In the interest of being complete.....should we mention that power factor of the load (temperature too) has a bearing?

 

[Ingenieur]

In the interest of being complete.....should we mention that power factor of the load (temperature too) has a bearing?

good points
in this case the i is given so it is safe to assume it includes real and reactive components
that is why I like using values for Z (pf 0.85) from the tables
imo a more conservative approach
either way though the difference is moot in most cases
one reason v is usually higher than rated
208 is 212 range
480 is 490 or so
a bit higher often
and motors are rated 460 for example

temperature is another matter
if not given I guess safe to assume std conditions of amb 40 C and a rise of 30-35???
and more to do with insulation rating than change in resistivity

 

[Smart $]

...
temperature is another matter
if not given I guess safe to assume std conditions of amb 40 C and a rise of 30-35???
and more to do with insulation rating than change in resistivity

The values given in the Chapter 9 Table 9 are at 75°C. We seldom 'drive' a conductor to 75°C. Using the case provided in this thread, 27A through a #8 copper conductor in 30°C ambient will not raise the temperature to anywhere near 75°C in typical operating conditions.

 

[Ingenieur]

Power lost in conductor = 27 x 6.2 = 168 W over 300 ft ~ 0.56 W/LF
the surface area of 1 LF #8 ~ 0.0107 sq ft
or 0.56/0.0107 = 52 w/sq ft
that will get warm, no?
think of a large 25 W bulb with ~72 sq in (1/2 sq ft) of area
maybe not hot but warm

if amb is say 70F wouldn't the cond get at least that warm and equalize?
or warmer to radiate to the amb air since heat is continously generated in the conductor as long as current flows?

but I agree on his scale the effects are moot

 

[Smart $]

Power lost in conductor = 27 x 6.2 = 168 W over 300 ft ~ 0.56 W/LF
the surface area of 1 LF #8 ~ 0.0107 sq ft
or 0.56/0.0107 = 52 w/sq ft
that will get warm, no?
think of a large 25 W bulb with ~72 sq in (1/2 sq ft) of area
maybe not hot but warm

if amb is say 70F wouldn't the cond get at least that warm and equalize?
or warmer to radiate to the amb air since heat is continously generated in the conductor as long as current flows?

but I agree on his scale the effects are moot

52W sounds like it would get very warm... but it's just a play on numbers.
52W*300ft/168W = 93ft of conductor dissipating that 52W.

 

[Ingenieur]

52W sounds like it would get very warm... but it's just a play on numbers.
52W*300ft/168W = 93ft of conductor dissipating that 52W.

300 LF = 3.2 sq ft of surface area
168/3.2 = 52 W/sq ft
same thing
I would think as i approaches rated ampacity temp approaches insul rating?
for a given amb temp and install method, ie, conduit, buried, free air, etc

I know there is a calculation for temp rise, too lazy to find it lol

 

[Smart $]

300 LF = 3.2 sq ft of surface area
168/3.2 = 52 W/sq ft
same thing
I would think as i approaches rated ampacity temp approaches insul rating?
for a given amb temp and install method, ie, conduit, buried, free air, etc

I know there is a calculation for temp rise, too lazy to find it lol

Well, as it approaches the ampacity given in the 75°C column of Table 310.15(B)(16) it will approach 75°C in an enclosure. At least that's the theory behind terminal temperature limitation coordination. Somewhat the same as going by the ampacity adjusted for the conditions of use (installation method). But going by the 90°C ampacity won't do us any good if we can't go over 75°C in general.

Been down that road trying to compensate for temp rise before. It amounts to being too complex for something that isn't required. Perhaps it would be if all you installed were fire pumps. :huh:

 

[Ingenieur]

Here's another one
there was an error in my calcs (surface area = dia x Pi x L)
I forgot the Pi
so W/sq ft ~ 1/Pi x 52 calculated ~ 15 or so

30 A at 208/1
10 awg
100 LF one way
3%, 6.2 drop, 186 W
surface area for 200 LF = 5.3
W/sq ft = 35 W/sq ft

curious what temp rise would be in amb of 30 C

 

[Smart $]

Here's another one
there was an error in my calcs (surface area = dia x Pi x L)
I forgot the Pi
so W/sq ft ~ 1/Pi x 52 calculated ~ 15 or so

30 A at 208/1
10 awg
100 LF one way
3%, 6.2 drop, 186 W
surface area for 200 LF = 5.3
W/sq ft = 35 W/sq ft

curious what temp rise would be in amb of 30 C

I'm not sure how to calculate temp rise accurately. Neher-McGrath?
As to estimating, can we assume delta t is proportional to delta I².
Table 310.15(B)(16) gives 35A for #10 at 75°C.
30²/35² = 73.5%
... × (75°C - 30°C) = 33°C