Part 3. Electrical conductor voltage drop.
By Ed Butts, PE, CPI
There is no other term involved in electrical circuit and motor design with more confusion or misunderstanding than “voltage drop.”
Figure 1a. AC power sinewave for a pure resistive or “tuned” circuit.
This month, as a final companion to the last two months’ columns, we will attempt to dispel many of the inaccuracies associated with permissible and actual allowable values of voltage drop.
Please note the following information is based on my personal experience and application to the National Electrical Code (NEC) plus local codes. The actual values of permissible voltage drop for your locality and application may vary with jurisdictions, local customs, specific equipment requirements, and code interpretation. Please verify your application with the local authorities or equipment manufacturer, if appropriate, before proceeding with wire sizing and allowable voltage drop values.
Defining Voltage Drop
Figure 1b. AC power sinewave for an inductive circuit.
Voltage drop or loss is defined as the decrease of electrical potential (voltage) along the path of a current flowing in or through a circuit. Cumulative voltage drops from the source and across conductors, contacts, and connectors to the load are undesirable and inefficient because a portion of the energy supplied is dissipated.
The voltage drop across the load is proportional to the power available to be converted in that load to some other useful form of energy, such as a motor or transformer, and the lost energy is usually in the form of heat as watts. This relationship is described in the well-known equation, Ohm’s Law:
DC Voltage (V) =
Current in Amperes (I) × Resistance (R) in Ohms or E = I × R (V= A × O)
For DC Power, Watt’s Law: Power (in Watts) = Voltage × Current, in Amperes or P = V × A
For an AC circuit, the Ohm’s Law equation is slightly modified to reflect the change from pure resistance to the combined impact of resistance, along with capacitance or induction when present, known in combination as reactance. Reactance, when combined with the basic circuit resistance, is then called impedance (Z). Ohm’s and Watt’s Laws for an AC circuit are then modified as:
Ohm’s Law: AC Voltage (V) = Current in Amperes (I) × Impedance (Z) or V = I × Z
For 1Ǿ: Power (in Watts) = Voltage × Current × Power Factor (PF) or P = V × A × PF
For 3Ǿ: Power (in Watts) =
Voltage × Current × Power Factor (PF) × 1.732 or P = V × A × PF × 1.732
Figure 1c. AC power sinewave for a capacitive circuit.
For instance, consider a circuit for a single-phase, AC power system where a 120-volt, 60 Hz, AC voltage source delivers 2 amps of power to a purely resistive load with a power factor of 1, such as an incandescent light bulb.
In this example, the power fundamentally consumed at the load would be 240 watts (120 VAC × 2 amps). Because this load is purely resistive (i.e., without the impact of reactance due to the absence of capacitance or induction), the current is in phase with the voltage or tuned and the calculations will be the same as that obtained in an equivalent DC circuit.
This means that power is always being consumed by the resistive load and never returned to the source as it is with reactive loads (Figure 1a). A DC or purely resistive AC circuit is measured in ohms, but a reactive load on the other hand includes various types of loads such as motors (inductance) and fluorescent light fixtures (capacitance). This is a load carried by an alternating current generating system in which the current and voltage are 90° out of phase (for a purely reactive load), which is measured in volt-amperes or kilovolt-amperes.
For two quarters of each cycle, the product of voltage and current is positive, but for the other two quarters, the product is negative, indicating that on average exactly as much energy flows into the load as flows back out. This means there is no net energy flow over each half cycle and only reactive power flows. This also means there is no net transfer of energy to the load, but electrical power does flow along the wires and returns by flowing in reverse along the same wires.
The current required for this reactive power flow dissipates energy in the line resistance even if the ideal load device consumes no energy itself. Practical AC loads have resistance as well as inductance and capacitance, so both active and reactive power will flow to normal loads.
Figure 2. NEC recommended voltage drop percentages.
Impedance is a combined function of the offsetting values of capacitance (measured in farads or C) and induction (measured in henrys or L) along with basic circuit resistance with the final result (Z) also measured in ohms. A circuit with both inductive (L) and capacitive (C) influence is referred to as an LC circuit.
Reactance is the opposition to a change of electric voltage or current due to the inductance or capacitance present in a circuit. If the circuit contains inductors, then the reactance is inductive. Conversely, if the circuit contains capacitors, then the reactance is said to be capacitive.
Reactance is not constant and varies according to the frequency of the input frequency. Purely inductive circuits supply reactive power with the current waveform lagging the voltage waveform by 90° (Figure 1b), while purely capacitive circuits absorb reactive power with the current waveform leading the voltage waveform by 90° (Figure 1c).
The result of this is that equivalized capacitive and inductive circuit elements tend to cancel each other out. Capacitors are used in submersible motor control boxes to help offset the effects of induction by adding capacitance to the motor circuit. A reactive circuit in which the level of capacitance and induction cancels each other so that only circuit resistance remains is said to be a resonant or tuned circuit.
Unlike a resistive circuit, the impedance is a function of the applied frequency in a reactive load. As we change the frequency in hertz, capacitors and inductors change impedance to the frequency. In the case of voltage drop, too much loss of voltage at the motor, particularly during starting, will affect the motor’s torque, possibly to the point that the motor cannot start or accelerate adequately to overcome the inertia of the motor rotor, pump impeller stack, and opposing water head.
In this case, the motor may stall and rapidly fail. Always remember that the starting current is generally five to seven times the full load current value of a motor. A 1 HP, 230 VAC, one-phase submersible motor with a full-load amperage of 8.2 has a locked rotor current, also known as locked rotor amperage (LRA), translating to a stuck motor starting inrush current of around 42 amps.
The power factor also becomes an important factor during motor starting as it will temporarily drop to as low as 20% to 25% to develop the instantaneous magnetizing force necessary to induce rotation of the rotor. Disregarding power factor and other complex power issues, an instantaneous current value of 42 amps on a 230 VAC motor circuit already designed to allow 10% of voltage drop at 8 amps will incur an approximate voltage drop of around 120 volts, approximately 50%, at the motor.
Most motor manufacturers recommend no less than 65% of rated voltage at the motor during starting to maintain an adequate level of starting torque. This translates to a maximum voltage drop during startup of 35%. Obviously, 50% is considerably more than 35%, and in our example, the motor may stall and not even begin to turn.
Conversely, by limiting the total voltage drop to 5% based on an electric power supply of 230 volts, the designer will ensure that the total voltage drop at the motor during starting will not exceed 58 volts, or roughly 25%, ensuring there will be adequate voltage to develop the necessary magnetizing effect and torque needed to start and accelerate the motor and pump to get it running.
As far as the difference between the 230-volt motor operating voltage and the 240-volt voltage often provided from utilities, a smart designer in my opinion will not use this buffer in the design, but will retain it as an additional safety factor to protect against possible voltage drop from the utility’s transformer or transmission/feeder lines due to a high electrical load consumption on the system (common during summer months), weakened supply transformer, bad or weak connections, or within the facility itself.
In addition, the extra voltage will help stabilize the starting voltage and torque at the motor, which can be critical with sandy wells or systems with long offset runs or high operating heads.
So to summarize: When calculating the voltage drop of an electrical circuit, I suggest you use the value of the nominal voltage supply (120 volts or 240 volts); verify that each conductor has a minimum size to handle the full load motor current with a 125% NEC-required safety factor (not service factor current); and design the total length of the well drop cable plus any offset wiring for a maximum 5% voltage drop.
Electrical Code and Standard Implications
There are several electrical guidelines, standards, and recommendations applicable to allowable voltage drop. Many of these are not code enforceable but recommendations. For example, NEC Section 210.19(A)(1) Fine Print Note 4 and Section 215.2(A) Fine Print Note 2 recommend the branch circuit and feeder voltage drop should be limited to 3%, with the combined voltage drop not to exceed 5% of the supply voltage (Figure 2). This is purely to maintain electrical efficiency in the circuit.
The national standard for utility voltage tolerance in North America is ANSI C84.1-2020. Essentially, ANSI and NEMA recognize two types of normally expected voltages (Range A): service and utilization.
The service voltage is the voltage expected at the service entrance, meter, or motor controller, while the utilization voltage is the value that should be at the end-user location, such as a motor.
Basically, the ANSI standard tolerance limits overvoltage to 5% of the nominal voltage and undervoltage to 10% of the nominal voltage. However, these limits, particularly the undervoltage, may prove problematic for motor circuits with long lengths, such as a typical submersible motor installation.
This, combined with the damage that overvoltage can cause to motor windings, is the primary reason I recommend limiting both overvoltage and undervoltage values to +/–5% of the nominal voltage. The values shown in Table 1 reflect ANSI and my personal recommended Range A voltages for each typical low voltage (less than 1000 VAC) system.
NEC Section 110.3(B) states: “Listed or labeled equipment shall be installed and used in accordance with any instructions included in the listing or labeling.” This is the NEC catch-all and simply means the manufacturer can dictate the voltage value required for their equipment under all operating conditions.
Therefore, designers should consult with the relevant motor manufacturer to ensure the proper voltage is being applied to their product. This basic step can preserve a warranty and prevent headaches down the road.
Another controversial topic is the allowable or permitted voltage drop in a feeder or branch circuit conductor. This is one element of design that is usually left to the total discretion of the system designer, provided they initially satisfy a few critical NEC stipulations, although some jurisdictions limit the permitted voltage drop.
First and foremost is a code issue. The NEC specifies the minimum size of conductor based on the ampacity of the circuit. As previously stated: For a 1 HP, 230-volt, one-phase motor, we would need to use the NEC-published value for the design of our branch circuit conductors, or 8 amps.
Since NEC Article 430 requires each branch circuit conductor to be designed to provide a minimum of 125% of the full load current (in this case, 8 amps × 1.25 = 10 amps), a conductor with no less than 10 amps of ampacity must be selected.
Referring to NEC Table 310.16, a #14-gauge, copper conductor with a total of no more than three current-carrying #14-gauge conductors grouped together in a common raceway or individual conductors directly buried in earth is capable of 15 amps, easily satisfying the first requirement.
Once the NEC requirement is met, the designer can technically design the voltage drop for 3%, 5%, or even 10% of drop. Thus, the question then becomes: Which value should be used?
The voltage drop of a circuit is addressed in the NEC through use of Fine Print Notes (FPN), which are recommendations. To ensure the proper level of performance under all conditions, the FPN recommends limiting this to a conservative value of the maximum voltage drop to 3% in a circuit, while most motor manufacturers require the total voltage drop to be at or less than 5% at the motor.
In my judgment, the only viable justification for using a 3% voltage drop is if either the motor manufacturer requires this limit or a specific specification, energy code, or stipulation does.
There are even certain occasions when you can design a circuit with up to a 10% voltage drop based on the following rationale: Most single phase electrical systems now supply a nominal power source of 120/240 volts since most single-phase motors are designed to operate with a nominal supply of 230 volts This built-in difference of 10 volts or around 4% provides the designer with an automatic voltage safety factor before even starting the design.
This same concept often applies to three-phase electrical systems, as a 230- or 460-volt motor can effectively operate on a 240- or 480-volt power supply, respectively. In addition, since most motors will easily tolerate up to a 5% voltage drop while running, this means you could conceivably design a drop cable or offset wire for up to 9% to 10% voltage drop, right?
Although competitive instincts and situations may frequently tempt a system designer to undersize wire in the hopes of saving those few dollars needed to secure the job, they should resist the temptation for the following three reasons.
- Most of the pumps and motors used in current water system designs are the submersible type. These motors are built with specific and limited design characteristics. Most notable is the fact that the motor is always located at the end of a run of possibly thousands of feet of wire (conductors). Since the voltage drop in a circuit is cumulative, it’s logical to surmise that the longer the run, the higher the overall voltage drop will be.All functions of an electric motor are inherently tied to the other functions. In other words, if you vary the voltage up or down, it will show up in some other motor condition such as starting torque, starting or running amperage, or motor efficiency.
- I have also observed numerous instances where customers have decided to upsize their well pump to realize additional capacity or the well had to be deepened due to a loss of water from nearby or local well use and draft. In either case, a larger pump and motor horsepower will almost always be required.If the offset or well drop cable size was compromised or marginal during the initial installation, there will obviously be no latitude towards using the existing cable for a larger motor. This can be very embarrassing and hard to explain to a client after using their system for only a year or so following the initial installation. In many of these instances, in order to salvage the existing offset cable, I have resorted to moving the control box to the wellhead and using a boost transformer at the wellhead to increase the motor voltage back to permissible levels.However, don’t forget that you can never get something for nothing in return. The trade-off to this will be higher heat in the offset conductors. Although this increased heat will occur, this is generally allowed if the conductor’s size complies with the 125% ampacity requirement and that the insulation rating does not exceed the allowable operating temperature. Trying to squeeze more electrical power through a given area of conductor will obviously result in higher heat in the wire. Higher heat in an electrical conductor results in lower efficiency, greater risk of overload and fire, and shorter life of the insulation.Although the added heat may not be injurious, particularly with direct buried wires where the ground tends to act as a heat sink, the effect from day-to-day usage can be cumulative upon the system, and in certain cases, the extra power costs to the customer from the added heat losses in the conductors can amount to more than the difference between the initial cost of using larger or smaller wire. If you work with electrical design of any kind, always hold one thought paramount in your thinking: Heat kills electrical equipment.
- As a comparison, a 3% voltage drop on a single-phase, 120 VAC system would be 0.03 × 120V = 3.6V. Similarly, a 2% voltage drop on a three-phase, 480 VAC system would be 0.02 × 480V = 9.6V.The main concern of voltage drop calculations is to make sure the load has enough voltage to function properly. This includes ensuring adequate starting torque. For example, a motor on a 480 VAC system is rated to operate at 460 VAC, and there is even a tolerance around that lower voltage at which the motor will still start and function properly. Generally, an induction motor with full voltage starting requires at least 65% of its nameplate voltage to start and between 90% to 110% of the same nameplate voltage to operate efficiently.If the motor can be started without too many problems,typically it will also run without too many issues. This is where proper wire sizing is important. NEC 90.5(C) states that the explanatory information provided in the Code, such as Informational Notes, are not enforceable as Code requirements. Therefore, voltage drop recommendations that appear as Informational Notes in the Code are not NEC enforceable, but this does not mean that local authorities will not enforce a specific voltage drop as a code interpretation. This means they are usually required and expected to comply with the Code for maximum ampacity and on projects for conductor run lengths that merit consideration.
Calculating Wire Size and Allowable Voltage Drop
Let’s look at the general requirements for sizing conductors once the calculated load current is known. This procedure applies to all types of AC loads and induction motors as a two-step beginning process:
- The first step is to look at the temperature rating of the terminals and the permitted ampacity of the conductor that could be used at a matching temperature rating.
- The second step is to look at the effect of ambient temperature and conductor derating factors on the ampacity of the conductor that results from where and how the conductors are installed.
Because Underwriters Laboratories Inc. (UL) places certain restrictions on the size and temperature rating of conductors that can be used at its terminals, the conductor size needs to be carefully determined. When manufacturers’ tables or charts are unavailable, use of the circular mils method is the next most common and accurate method of determining voltage drop and wire sizes. The equation is:
Required Circular Mils (CM) =
R × Circuit Amperage × 1-Way Length of Circuit in Feet × C
Nominal or Supply AC Voltage × Allowable Voltage Drop (in %)
C = Phase Factor: Use 2 for Single Phase or 1.732 for Three Phase.
Voltage Drop % Entry at 5% = 0.05.
Although the skin effect is generally considered for conductor sizes of #2/0 and greater during design, for purposes of drop cable sizing, this can typically be ignored. Engineering Your Business column’s Table 3 in the WWJ December 2023 issue provides circular mil data for submersible pump cable and other conductors.
Example: Referring to Figure 3, the customer desires to replace a 20 HP, 460 VAC, 3Ǿ, pump and motor with a new 30 HP unit. Assuming the existing #6 copper 75°C rated direct buried offset cable can be reused and no derating factors are needed, what size of new drop cable is required?
Verify the #6 copper wire is Code acceptable for the load (refer to NEC Table 310.16 or Table 3 in the WWJ December 2023 column):
30 HP, 460 VAC, three-phase Motor FLA = 40 amps × 1.25 = 50 amps < 65 amps
(Okay for #6 copper at 75°C)
Step 2: (see Table 2)
Select the appropriate R (12.9 ohms/1000 feet for 75°C conductors) and determine the offset voltage drop by modifying the above equation:
Offset Voltage Drop (Δ) =
12.9 ohms/1000 feet × 40 amps × 400 feet × 1.732 = 13.6 volts
*Circular Mils for #6 conductor, from Table 3 in December 2023 WWJ
Determine remaining allowable voltage drop: 480 VAC × 0.05 (5%) = 24V – 13.6V = 10.4V (Δ)
Determine the new drop cable size by utilizing the remaining allowable voltage drop:
Required CM = 12.9 × 40 amps × 400 feet × 1.732
Δ = 10.4 volts
= 39,373 CM < #4 (41,740 CM)
(Okay to use #4)
This example obviously assumes the replacement submersible cable will also be rated for 75°C. This is an example of a typical water well pump wire sizing situation but can be applied to virtually any wire sizing need. This also underscores the importance of properly determining voltage drop in a well pump application.
This concludes this month’s—and this year’s—edition of Engineering Your Business. Next month, we will conclude this series dedicated to electrical oriented themes and this three-part miniseries on electrical wire types and sizing with an overview on determining and calculating voltage drop.
I wish all of you a safe and happy holiday season. Until next month (and next year), work safe and smart.
Ed Butts, PE, CPI, is the chief engineer at 4B Engineering & Consulting, Salem, Oregon. He has more than 40 years of experience in the water well business, specializing in engineering and business management. He can be reached at email@example.com.