Voltage Drop Calculations- Part Two



In Article " Voltage drop calculations- Part One ", I explained the following points:

  • What is the Voltage Drop? 
  • Why to calculate the Voltage Drop? 
  • What are the factors affecting the voltage drop? 
  • NEC voltage drop Recommendations 


Also, in this Article, 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.



I explained the first two methods and today I will continue explaining other Voltage Drop Calculations Methods as follows.




3- U.S Method



This method can be divided to two sub-methods as follows:

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



3.1 Circular-Mils Method








Difference between Circular-Mils Method and Ohms Law Method

This method is a bit more involved than the Ohms Law method, but the big advantage is you can use it for three-phase or single-phase.






Rule#1: Using Circular-Mils Method for Voltage Drop Calculations


In this method, the voltage drop is calculated by using the following formulas:


For Single-phase
VD = 2 x K x I x D/CM
For Three-phase
VD = 1.732 x K x I x D/CM


Where:

K = Direct-Current Constant.
K represents the dc resistance for a 1,000-circular mils conductor that is 1,000 ft long, at an operating temperature of 75؛C. K is 12.9 ohms for copper and 21.2 ohms for aluminum.

I = Load Amperes
The load in amperes at 100% (not at 125% for motors or continuous loads).

D = Distance
The distance the load is from the power supply. When calculating conductor distance, use the length of the conductor—not the distance between the equipment connected by the conductor. To arrive at this length, add distance along the raceway route to the amount of wire sticking out at each end. An approximation is good enough. Where we specify distances here, we are referring to the conductor length.

CM = Circular-Mils
The circular mils of the circuit conductor as listed in NEC Chapter 9, Table 8 (see below image).










Important!!!

For ac circuits with conductors 2/0 AWG and larger, you must adjust the dc resistance constant K for the effects of self-induction (eddy currents). This is done by using the Adjustment Factor ( Q ).

Q = Alternating-Current Adjustment Factor

Calculate the “Q” Adjustment Factor by dividing the ac ohms-to-neutral impedance listed in Chapter 9, Table 9 by the dc resistance listed in Chapter 9, Table 8.






Example#1:


A 3 phase 36kVA load rated 208 V is wired to the panelboard with 80 ft lengths of 1 AWG THHN aluminum. What is the approximate voltage drop of the feeder circuit conductors?



Solution:




Applying the three-phase formula: 

VD = 1.732 x K x I x D/CM 

Where: 

K = 21.2 ohms, aluminum

I = (36,000/(208 x 1.732)) = 100A

D = 80 ft

CM = 83,690 (Chapter 9, Table 8)

VD = 1.732 x 21.2 x 100A x 80/83,690 = 3.51 V


VD Percent = 3.51 / 208 = 1.69 %







Rule#1 Algebraic variations

Using basic algebra, you can apply the same basic formula to find one of the other variables (I, D or CM) if you already know the voltage drop.

Suppose you want to know what size conductor you need to reduce the voltage drop to the desired level. Simply rearrange the formula.

For three-phase, it would look like this:

CM (3 phase) = 1.732 x K x I x D/VD

For single-phase calculations, you would use 2 instead of 1.732.






Example#2:

Suppose you have a 3 phase, 15 kVA load rated 480V and 390 ft of conductor. What size conductor will prevent the voltage drop from exceeding 3% percent?



Solution: 








Applying the three-phase formula:

CM (3 phase) = 1.732 x K x I x D/VD

Where:

K = 12.9 ohms, copper

I = (15,000/(480 x 1.732)) = 18 A

D = 390 ft

VD = 480V x 0.03 = 14.4V

CM = 1.732 x 12.9 x 18 x 390/14.4V = 10,892

From Chapter 9, Table 8, you will find that the conductor that has CM equal to or next larger than 10,892 is 8 AWG.





Example#3:


What is the maximum length of 6 AWG THHN you can use to wire a 480V, 3 phase, 37.5 kVA transformer to a panelboard so voltage drop does not exceed 3% percent?



Solution:

Applying the three-phase formula: 


D (3 phase) = CM x VD/(1.732 x K x I) 

Where:

CM = 26,240, for 6 AWG Chapter 9, Table 8

VD = 480V x 0.03 = 14.4 V

K = 12.9 ohms for copper

I = (37,500/480 x 1.732)) = 45A 


D = 26,240 x 14.4/1.732 x 12.9 x 45 = 376 ft 

Sometimes, the only method of limiting voltage drop is to limit the load. Again, we can rearrange the basic formula algebraically: I = CM x VD/1.732 x K x D. 




Example#4:

An installation contains 1 AWG THHN conductors, 300 ft long in an aluminum raceway, to a 3 phase 460/230V power source. What is the maximum load the conductors can carry, without exceeding the NEC recommendation for voltage drop?



Solution:

Applying the three-phase formula:

I = CM x VD/(1.732 x K x D)

Where:

CM = 83,690 for (1 AWG), Chapter 9, Table 8

VD = 460V x 0.03 = 13.8V

K = 12.9 ohms, for copper

D = 300 ft 


I = 83,690 x 13.8/1.732 x 12.9 x 300 = 172A 


But as per NEC 110.14(C) and Table 310.15(B)(16), the maximum load permitted on 1 AWG THHN at 75C is 130A.


So, as shown in the above example, 
You maintain the voltage drop within 3% to meet NEC Recommendation But you violate NEC Requirements for conductor Ampacity. 







Important!!!

In any conductors/cables selection or design process, verify the NEC requirements first for Conductor amapcity and others then perform the recommended check for voltage drop to ensure best performance and low cost.






3.2 Chapter (9) Tables Method







Rule#2: Using Chapter (9) Tables Method for Voltage Drop Calculations


In this method, the voltage drop is calculated by using the following formulas:


For Single-phase
VD = 2 x L x R x I / 1000
For Three-phase
VD = 1.732 x L x R x I / 1000


Where:

VD: The voltage drop (V),

L : The length of conductor (m),

R: The resistance value from Chapter 9, Table 8 (ohm/km) or (ohm/kFT)
The value R is determined from the National Electrical Code (NEC), Chapter 9, Table 8 column Direct Current Resistance at 75 degrees C/ Copper/ Uncoated.

I: The load current (A).






Example#5:


Determine the voltage drop of a 380V, 3 phase circuit with a current of 100A and a length of 150 m and a conductor size of 3 AWG.



Solution: 






This is a secondary service feed so, recommended voltage drop = 3%

Applying the three-phase formula: 


VD = 1.732 x L x R x I / 1000 

Where:

L = 150 m

R = 0.802 for (3 AWG), Chapter 9, Table 8

I = 100 A 


VD = 1.732 x 150 x 0.802 x 100 /1000 = 20.834 V 

But, the maximum voltage drop allowed in this case = 0.03 x 380 V = 11.4 V. 

The voltage drop value must be minimized. 





Example#6:


For example#5, determine the size of cable that will be required to meet the voltage drop requirement (3%).



Solution:


Since the maximum voltage drop allowed in this case = 0.03 x 380 V = 11.4 V.

So, you need to determine the value of R that will meet the following formula: 


11.4 = 1.732 x 150 x R x 100 / 1000 

R = 11.4 x 1000/(1.732 x 150 x 100) = 0.438 ohm/km


Referencing the NEC Chapter 9, Table 8 indicates that the cable size with a voltage drop of 0.438 ohm/km or less is 1/0 AWG (70 mm) cable with a resistance of 0.399 ohm/km.

Calculating the voltage drop for the 1/0 AWG (70mm) cable results in:

VD = 1.732 x L x R x I / 1000  
= 1.732 x 150 x 0.399 x 100 / 1000 = 10.36 V 


So, Percentage Voltage Drop = 10.36 x 100 / 380 = 2.73 %


Therefore, in order to transmit a 3 phase current of 100A per phase over a length of 150 m, with a total voltage drop equal to or less than the maximum 11.4 volts, a 1/0 AWG (70 mm2) cable is needed.




In the next Article, I will continue explaining Chapter (9) Tables Method for Voltage Drop Calculations. Please keep following. 







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