Voltage Drop Calculator

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Ingenieur said:
am I missing something?
the pf ie 0.9, 0.85, etc is for the load, not conductor
Click to expand...


The significance being that because of vector addition the voltage drop across the wire impedance will cause a smaller change in the voltage seen by the load.

You can justify calling either of those "voltage drop", but AFAIK the second one has more importance for equipment operation.

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Ingenieur said:
no
Click to expand...

???

I was hoping to work on the MV VD equations from a weeks back today, but instead I have been trying to convince myself the earlier spreadsheet is correct.

If we break this down into a straight forward V=IZ equation....from your earlier notes.
Define the LOAD current as Imag <ph ang.
If I calculate Zline = Z mag<ph ang
Vdrop or Vline = Imag x Z mag ....ph ang irrelevant to MAG of Vdrop
No multiplying by SQRT-3
Multiply by 2 if it is a single phase circuit.
Agree or disagree ?
 
MyCleveland said:
???

I was hoping to work on the MV VD equations from a weeks back today, but instead I have been trying to convince myself the earlier spreadsheet is correct.

If we break this down into a straight forward V=IZ equation....from your earlier notes.
Define the LOAD current as Imag <ph ang.
If I calculate Zline = Z mag<ph ang
Vdrop or Vline = Imag x Z mag ....ph ang irrelevant to MAG of Vdrop
No multiplying by SQRT-3
Multiply by 2 if it is a single phase circuit.
Agree or disagree ?
Click to expand...
3 ph one way L x sqrt3 for total drop


the NEC values are based on a CKT pf = 0.85 at rated ampacity (not cond Z ph ang)
assume 500 kcmil steel, 360 A
X = 0.048
R = 0.029
X/R = 1.6552 so ang = 58.86 deg, ph ang = 0.517 NOT 0.85

Z = R x 0.85 + X x sin(arccos 0.85) = 0.0499 (table 0.050)
if the ckt pf is different it can corrected
Z new pf = R x pf + X x sin(arccos pf)

v = 600/0 vac, i = 360/-31.8 A (ckt pf = 0.85) and L = 500'
Z cond = sqrt3 x 500/1000 x (0.029 + j 0.048) = 0.0251 + j 0.0416
= 0.0486/58.9 deg
Z load = 1.6236/31.01 deg, load pf 0.8571
Z ckt = 1.6667/31.8 deg, pf 0.85
the sum of Z cond + Z load yields the CKT pf 0.85
 
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What is the actual distance you have to go? It seems to me that with that much amperage, copper wire, and large conduit, stepping up to 480 volts and back down maybe more cost-effective, especially if one side is off a not yet ordered utility Transformer, where there would be no difference in cost between XXXX volts to 480V vs 208V.
 
Ingenieur said:
the NEC values are based on a CKT pf = 0.85 at rated ampacity
Z = R x pf + X x sin(arccos pf) ..etc = 0.0486
Click to expand...

NEC table values require adjustment for Derating, Ambient, Motors, Termination Temps, and (Pf at ampacity). All are critical variables.

To actually see conductor Temp.Rise with changes in Ambient and Load:
Termination Temperature (RT) can replace (R) without Phase-Angle Vectors

Z = (RT 0.02872 * pf 0.85 + X 0.048 * sin(arccos pf)) = 0.0497 / 1000 * L 232 = 0.01153

RT = R 0.029 * (1 + IF(EXACT(Wire,"Copper"),0.00323,0.0033) * (Temp.Rise - 75))

Temp.Rise = Ambient 30 + (TR 90 - Ambient 30) * (Load 360)^2 / (Imax 430)^2

TR = Conductor Max Temperature Rating
Imax = Max Amps at TR * any derating for > 3 ccc's

1) In conduit with > 3 ccc's. NEC table values must be adjusted by a Derating Factor
2) I-Rise = Motor load adjustment: Load * 1-(1-PF)*(% Motors)
 
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ramsy said:
NEC table values require adjustment for Derating, Ambient, Motors, Termination Temps, and (Pf at ampacity). All are critical variables.

To actually see conductor Temp.Rise with changes in Ambient and Load:
Termination Temperature (RT) can replace (R) without Phase-Angle Vectors

Z = (RT 0.02872 * pf 0.85 + X 0.048 * sin(arccos pf)) = 0.0497 / 1000 * L 232 = 0.01153

RT = R 0.029 * (1 + IF(EXACT(Wire,"Copper"),0.00323,0.0033) * (Temp.Rise - 75))

Temp.Rise = Ambient 30 + (TR 90 - Ambient 30) * (Load 360)^2 / (Imax 430)^2

TR = Conductor Max Temperature Rating
Imax = Max Amps at TR * any derating for > 3 ccc's

1) In conduit with > 3 ccc's. NEC table values must be adjusted by a Derating Factor
2) I-Rise = Motor load adjustment: Load * 1-(1-PF)*(% Motors)
Click to expand...

The Chap 9 table 9 values are for MAX temp of conductors...75C.
Who would adjust down ?

Table 9 method, note 2 states "a good approximation"

I will try and rephrase my earlier question.
For a three phase application.
Given load current and PF. Use this to calculate load Z(load) of a single phase model of the circuit.
Calculate Z(line) from BASE R & X values from table 9....NO Zeffective.
Calculate Z(total) = Z(load) + Z(line)
Calculate I cct = V(base) x Z(total)
Calculate Vdrop = I cct x Z(line)

Is this NOT an exact value ? No multiplying by sqrt(3) or 2.
 
MyCleveland said:
The Chap 9 table 9 values are for MAX temp of conductors...75C.
Who would adjust down ?

Table 9 method, note 2 states "a good approximation"

I will try and rephrase my earlier question.
For a three phase application.
Given load current and PF. Use this to calculate load Z(load) of a single phase model of the circuit.
Calculate Z(line) from BASE R & X values from table 9....NO Zeffective.
Calculate Z(total) = Z(load) + Z(line)
Calculate I cct = V(base) x Z(total)
Calculate Vdrop = I cct x Z(line)

Is this NOT an exact value ? No multiplying by sqrt(3) or 2.
Click to expand...

i = v/z, a typo
you must factor in length since the x and r values are per unit length values
z line = L/1000 x (r + j x)
to convert to 3 ph x sqrt
to 1 ph x 2
 
Ingenieur said:
i = v/z, a typo Yes...sorry
you must factor in length since the x and r values are per unit length values Understood and done
z line = L/1000 x (r + j x)
to convert to 3 ph x sqrt I am converting to a single phase model...no return Z if assumption is balanced 3p load
to 1 ph x 2
Click to expand...
Agreed for a single phase calc I would need to x 2...I stated 3-phase above

So you disagree with the logic ?...trying to make sure I am explaining correctly.

If the Zeff method in chap 9 is an APPROXIMATION...is this not an EXACT method ?
 
the i has to rtn somehow?
if balance it rtns on a phase line

there is no 'exact'
all are approximations of sorts
yours will be close enough

if one uses the nec values and ampacity ratings (>3 cond, amb temp, etc) it will be very close

consider
calc is done at 100% load, which is seldom the case
the util v is higher than nominal, ie, 208 is 210-212, 480 is ~490
and end devices are rated lower than nominal, 120 is 110/115, 480 is 460, etc
 
don_resqcapt19 said:
The online calculator that I typically use shows 170' at 30°C will have a 3% voltage drop with those conductors and 1600 amp load.
Click to expand...

Don
Please help me with ambient T vs Conductor max operating T.
Examples in Handbook "Conductors are operating at maximum rated temp of 75C".
I understand how ambient can add to operating, but is not operating subjective to what the feeder/branch circuit is operating at relative to percentage of circuit capacity.

How would you come up with an operating temp starting point ? or because wire is 75C rated, this is your base ?
 
the base temp is the one the cable is rated at, iirc 30 C (40?)
it will rise to 75 at this ambient at rated ampacity
if actual amb > rated amb the ampacity is derated (or if act < rated uprated)
 
MyCleveland said:
Please help me with ambient T vs Conductor max operating T.
How would you come up with an operating temp starting point ? or because wire is 75C rated, this is your base ?
Click to expand...

Conductor Insulation Temperature must be > Temp.Rise + Ambient

T2 = Estimated Conductor Temperature = (T1 Ambient + Temp.Rise)

Where:
T2 = T1 + (TR-T1) * Load^2 / Imax^2
TR = Conductor Max Temperature Rating
Load = Circuit Amps
Imax = Max Amps at TR, derated for ccc's
In conduit with > 3 ccc's. NEC table values must be adjusted
 
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For Ampacity based on NEC tables, adjustments are required for motors, and derating conditions

Temperature calc is worthless if Motor Load (I-Rise) adjustments are ignored.

Motor load adjustment: Load + I-Rise
Load = Load * 1-(1-PF)*(% Motors)
 
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