# Voltage Drop Calculations- Part Three

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:

• Circular-mils method,
• Chapter (9) tables method.

4- Approximate Mathematical method,

5- Quick On-Line method,

6- Ampere-Feet method,

7- Lookup tables,

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” (DC Resistance) with (5) numbers notes, Table 9: “Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75°C (167°F) — Three Single Conductors in Conduit”

 Table 8: Conductor Properties Table 8 (in below image) provides the following information: 1- Conductor Characteristics like: Conductor size in AWG or Kcmil, Conductor Area in mm2 and Circular Mils, No of strands per conductor and diameter in mm and inch for each strand Diameter in mm and inch for overall conductor, Area in mm2 and inch2 for overall conductor, 2- Direct-Current 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: These resistance values are valid only for the parameters as given. Using conductors having coated strands, different stranding type, and, especially, other temperatures changes the resistance. Equation for temperature change: R2 = R1 [1+ α (T2 -75)] where αcu = 0.00323, αAL = 0.00330 at 75°C. Conductors with compact and compressed stranding have about 9 percent and 3 percent, respectively, smaller bare conductor diameters than those shown. See Table 5A for actual compact cable dimensions. The IACS conductivities used: bare copper = 100%, aluminum = 61%. Class B stranding is listed as well as solid for some sizes. Its overall diameter and area is that of its circumscribing circle.

 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: R2 = R1 [1+α (T2 – 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: (AC) reactance XL for all wires installed in PVC/ Aluminum Conduits or Steel Conduits. (AC) resistance for uncoated copper conductors installed in PVC conduits, Aluminum Conduits or Steel Conduits. (AC) resistance for Aluminum conductors installed in PVC conduits, Aluminum Conduits or Steel Conduits. 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: These values are based on the following constants: UL-Type RHH wires with Class B stranding, in cradled configuration. Wire conductivities are 100 percent IACS copper and 61 percent IACS aluminum, and aluminum conduit is 45 percent IACS. Capacitive reactance is ignored, since it is negligible at these voltages. These resistance values are valid only at 75°C (167°F) and for the parameters as given, but are representative for 600-volt wire types operating at 60 Hz. Effective Z is defined as R cos(θ) + X sin(θ), where θ is the power factor angle of the circuit. Multiplying current by effective impedance gives a good approximation for line-to-neutral voltage drop. Effective impedance values shown in this table are valid only at 0.85 power factor. For another circuit power factor (PF), effective impedance (Ze) can be calculated from R and XL values given in this table as follows: Ze = R x PF +XL sin[arccos(PF)]. 5- The basic assumptions and the limitations of Table 9 are as follows: Capacitive reactance is ignored. The calculated voltage-drop values are approximate. For circuits with other parameters, the Neher–McGrath ac-resistance calculation method is used (will be explained in next Article).

 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 The Effective Impedance value provides a simple method for calculating the voltage-drop that will occur across the conductors due to the load current flowing in the circuit.  The magnitude of the voltage drop across in a circuit is determined by multiplying the effective impedance of the circuit conductors in Ω by the magnitude of the line current flowing in the circuit.  The Effective Impedance of the conductors in Ω is found by multiplying the Ω/length value provided in Table 9 by the length of the conductors (in the appropriate units).  Note that for power factors other than 0.85 lagging, the effective impedance may be calculated using a formula provided in the FPNs for Table 9:  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 line-voltage seen at the terminals of the motor when the motor is drawing rated current at a power factor of 0.92 lagging, assuming rated line-voltage 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 = cos-1 (pf) = cos-1 (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 per-conductor voltage drop, which relates to a line-neutral drop per-phase.

Step#4: Determine the line-voltage at the terminals of the motor
Since the calculated voltage drop relates to a decrease in the line-neutral (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 motor-terminal line-voltage:
V line( load-end) = V line (service- end) - Vdrop (Line) = 480 - 3.27 = 476.7 V

Note that a 3.27V drop in the line-voltage magnitude relates to a 0.68% decrease.

Example#4:

A feeder has a 100-ampere 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 line-to-neutral 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 (line-to-neutral) = 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 line-to-line voltage drop:
Voltage drop (line-to-line) = voltage drop (line-to-neutral) 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 270-ampere continuous load is present on a feeder. The circuit consists of a single 4-in. PVC conduit with three 600 kcmil XHHW/USE aluminum conductors fed from a 480- volt, 3-phase, 3-wire 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 “Alternating-Current 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 (cos-1 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 line-to-neutral voltage drop:
Voltage drop (line-to-neutral) = 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 line-to-line voltage drop:
Voltage drop (line-to-line) = voltage drop(line-to-neutral) 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 (line-to-line) = 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.