In article " Voltage drop calculations Part One ", I indicated that there are eight methods for Voltage Drop Calculations as follows:
1 Ohm’s Law Method,
2 European method,
3 U.S method which divided to:
 Circularmils method,
 Chapter (9) tables method.
4 Approximate Mathematical method,
5 Quick OnLine method,
6 AmpereFeet method,
7 Lookup tables,
8 Excel spreadsheets.
And I explained the first two methods in this Article.
Also, In Article " Voltage drop calculations Part Two ", I explained the third method: U.S Method and today I will continue explaining how to use Chapter (9) tables method for Voltage Drop Calculations as follows.
3.2 NEC Chapter 9 – “Tables 8 and 9”
NEC Chapter 9 – “Tables 8 and 9”
Conductor
Characteristics, Conductor resistance (DC) or impedance (AC) information is
provided by the NEC in Chapter 9 – “Tables” specifically within Tables 8 and
9 as follows:

Table 8: Conductor Properties
Table
8 (in below image) provides the following information:
1
Conductor Characteristics like:
2
DirectCurrent Resistance based on conductor size for coated and uncoated
copper conductors and for Aluminum conductors.
The
Resistance is defined in terms of Ω/km and Ω/kft, and is based on a 75°C
conductor temperature.
3
(5) numbers of Foot Print Notes (FPNs) which are:

Rule#1: Temperature Effect on Resistance
A
Foot Print Note (FPN) number (2) below Table 8 provides a method for
determining resistance at temperatures other than 75°C.
As
per this FPN#2, resistance at temperatures other than 75°C may be determined
by:
R_{2 }= R_{1 }[1+α (T_{2} –
75)]
Where:
R1
is the resistance at 75°C,
R2
is the resistance at temp T2,
T2
is the new conductor temp, and
α
is the conductor material coefficient (αcu = 0.00323 and αal = 0.00330)

Important!!!
From
Rule#1 above, the Conductor resistance will increase with temperature.

Example#1:
Determine the resistance in Ω/kft of a 500kcmil, uncoated copper conductor at a temperature of 90°C.
Solution:
As per Table 8, the DC resistance at 75°C is: R1 = 0.0258 Ω/kft
Using αcu = 0.00323, the resistance at 90°C is: R2 = R1 [1+α (T2 – 75)] = 0.0258 [1+0.00323 (90 – 75)] = 0.0258[1+0.04845] = 0.02705 Ω/kft
Table 9: AC Resistance, Reactance and Effective
Impedance
Table
9 (in below image) provides the following information:
1
Conditions for application:
The
resistance and reactance values provided by Table 9 relate to 600V cables
used in 3Φ, 60Hz, 75°C (167°F) circuits that have three single conductors in
a conduit.
The
values are representative for of other 600V cables operating at 60Hz.
2
AC Impedance Values:
Similar
to Table 8, the (AC) resistance and reactance is defined in terms of Ω/km and
Ω/kft, and is based on a 75°C conductor temperature.
3
Effective Impedance:
Table
9 also provides an Effective Impedance (Ze) for uncoated copper and aluminum
conductors installed in PVC conduits, Aluminum Conduits or Steel Conduits,
both in Ω/km and Ω/kft, based on a 0.85 lagging operational power factor.
4
(2) Numbers of Foot Print Notes (FPNs) which are:
5
The basic assumptions and the limitations of Table 9 are as follows:

Temperature Effect on AC Resistance
Note
that the resistance values provided within Table 9 are based on a 75°C
temperature. For temperatures other than 75°C, the resistance values can be
adjusted in the same manner as the DC resistances provided by Table 8.

Important!!!
The
reactance values provided within Table 9 are independent of temperature, and
thus do not need to be adjusted for temperatures other than 75°C.

Example#2:
Determine both the resistance and the reactance in Ω/kft of the 600V, 500kcmil, copper cables used within a 3Φ circuit at a temperature of 75°C if the cables are enclosed in an aluminum conduit.
Solution:
As per Table 9, the resistance/reactance at 75°C for 500kcmil cable is: R = 0.032 Ω/kft
X = 0.039 Ω/kft
Voltage Drop Calculations Using Table (9)'s Effective Impedance values
Z = R cosθ + X sinθ
Where: R is the conductor resistance (Ω/length), X is the conductor reactance (Ω/length), and cosθ is the circuit’s power factor. Note that sinθ may be determined from power factor. Voltage Drop from Effective Impedance may be calculated by:
V drop = I Ze
Where: I is the magnitude of the line current flowing in the conductor, and Ze is the effective impedance of the conductor in ohms. 
Example#3:
In the above figure, Determine the operational linevoltage seen at the terminals of the motor when the motor is drawing rated current at a power factor of 0.92 lagging, assuming rated linevoltage at the secondary terminals of the transformer.
Notes:
 Assume a 60°C conductor temperature.
 Ignore the disconnect, the fuse, and the overload relay when calculating voltage drop.
Solution:
Step#1: Determine the Impedance Characteristics of the feeder conductors
From NEC Chapter 9  Table 9:
R = 0.045 (Ω/kft) (at 75°C)
X = 0.051 (Ω/kft)
Adjust the resistance for 60°C (R75°C = 0.045) R2 = R1 [1+α (T2 – 75)] = 0.045[1+0.00323 (60 – 75)] = 0.045[1 0.04845] = 0.04282 Ω/kft
Conductor Impedance Values at 60°C:
R = 0.04282 (Ω/kft)
X = 0.051 (Ω/kft)
Step#2: Determine the Effective Impedance (Ze) of the feeder conductors which is a function of the operating power factor of the circuit. Since the p.f. of the load ≠ 0.85, the effective impedance must be calculated using:
Solving for sinθ:
θ = cos1 (pf) = cos1 (0.92) = +23.074°
sin θ = sin (23.074°) = 0.392
Note that θ is “positive” since it is a “lagging” p.f. Thus, given R = 0.04282 and X = 0.051, the effective impedance in Ω/kft is:
Z = R cosθ + X sinθ = (0.04282) (0.92) + (0.051) (0.392) = 0.0394 + 0.02 = 0.0594 Ω/kft Taking into account the length of the feeder, the effective impedance in Ω is:
Ze = 0.0394+ 0.02 = 0.0594 Ω/kft x (140ft/1000) = 0.008316Ω
Step#3: Determine the voltage drop across each conductor due to the line current V drop = I Ze = (227 A) (0.008316Ω) = 1.888 V
Note that this is the perconductor voltage drop, which relates to a lineneutral drop perphase.
Step#4: Determine the linevoltage at the terminals of the motor Since the calculated voltage drop relates to a decrease in the lineneutral (phase) voltage magnitude, the line voltage will experience a √3 x greater decrease in magnitude.
V drop (Line) =√3 V drop = √3 (1.888 ) = 3.27 V Resulting in a motorterminal linevoltage:
V line( loadend) = V line (service end)  Vdrop (Line) = 480  3.27 = 476.7 V
Note that a 3.27V drop in the linevoltage magnitude relates to a 0.68% decrease.
Example#4:
A feeder has a 100ampere continuous load. The system source is 240 volts, 3 phase, and the supplying circuit breaker is 125 amperes. The feeder is in a trade size 1.25 aluminum conduit with three 1 AWG THHN copper conductors operating at their maximum temperature rating of 75°C. The circuit length is 150 ft, and the power factor is 85 percent.
Using Table 9, determine the approximate voltage drop of this circuit.
Solution:
Step#1: Find the approximate linetoneutral voltage drop.
Using the Table 9 column “Effective Z at 0.85 PF for Uncoated Copper Wires,” select aluminum conduit and size 1 AWG copper wire. Use the given value of 0.16 ohm per 1000 ft in the following formula:
Voltage drop (linetoneutral) = table value x (circuit length /1000 ft) x circuit load
= 0.16 ohm x (150 ft /1000 ft) x 100 A = 2.40 V
Step# 2: Find the linetoline voltage drop:
Voltage drop (linetoline) = voltage drop (linetoneutral) x √3 = 2.40 V x 1.732 = 4.157 V
Step# 3: Find the voltage present at the load end of the circuit = 240 V – 4.157 V = 235.84 V
Example#5:
A 270ampere continuous load is present on a feeder. The circuit consists of a single 4in. PVC conduit with three 600 kcmil XHHW/USE aluminum conductors fed from a 480 volt, 3phase, 3wire source. The conductors are operating at their maximum rated temperature of 75°C. If the power factor is 0.7 and the circuit length is 250 ft, is the voltage drop excessive?
Solution:
Step#1: Using the Table 9 column “XL (Reactance) for All Wires,” select PVC conduit and the row for size 600 kcmil. A value of 0.039 ohm per 1000 ft is given as this XL. Next, using the column “AlternatingCurrent Resistance for Aluminum Wires,” select PVC conduit and the row for size 600 kcmil. A value of 0.036 ohm per 1000 ft is given as this R.
Step# 2: Find the angle representing a power factor of 0.7.
Using a calculator with trigonometric functions or a trigonometric function table, find the arccosine (cos1 θ) of 0.7, which is 45.57 degrees.
Step# 3: Find the impedance (Z) corrected to 0.7 power factor
Zc = (R x cos θ) + (XL x sin θ)
= (0.036 x 0.7) + (0.039 x 0.7141) = 0.0252 + 0.0279 = 0.0531 ohm to neutral
Step# 4: find the approximate linetoneutral voltage drop:
Voltage drop (linetoneutral) = Zc x (circuit length / 1000 ft) x circuit load
= 0.0531 x (250 ft / 1000 ft) x 270 A = 3.584 V
Step# 5: Find the approximate linetoline voltage drop:
Voltage drop (linetoline) = voltage drop(linetoneutral) x √3 = 3.584 V x 1.732 = 6.208 V
Step# 6: Find the approximate voltage drop expressed as a percentage of the circuit voltage:
Percentage voltage drop (linetoline) = 6.208 V x 100 / 480 V = 1.29% VD
Step#7: Find the voltage present at the load end of the circuit:
480 V  6.208 V = 473.8 V
Conclusion: According to 210.19(A)(1), Informational Note No. 4, this voltage drop does not appear to be excessive.
In the next Article, I will continue explaining Other Methods for Voltage Drop Calculations. Please, keep following.
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