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,
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- 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:
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)]
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)
From Rule#1 above, the Conductor resistance will increase with temperature.
Determine the resistance in Ω/kft of a 500kcmil, uncoated copper conductor at a temperature of 90°C.
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.
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.
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.
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
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
I is the magnitude of the line current flowing in the conductor, and
Ze is the effective impedance of the conductor in ohms.
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.
- Assume a 60°C conductor temperature.
- Ignore the disconnect, the fuse, and the overload relay when calculating voltage drop.
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.
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.
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
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?
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.