Single-Phase Voltage Drop Calculation Method Examples

Voltage drop is an issue to consider in electrical design. To better understand, we’ll go over a real-world single-phase voltage drop calculation example.

I’m going to focus on three different voltage drop calculation methods. The methods will output the following calculated voltage drop accuracy levels:

  • Approximate
  • Estimated
  • Exact

Depending on your voltage drop problem, you need to choose the best fit method.

That said, in my discussion, I’m going to closely outline all parts of the calculations. This way, you learn the theory of a single-phase voltage drop calculation too. You’ll then better know which method to use.

What is voltage drop? 

Voltage drop is the reduction in voltage or voltage loss. As current moves through a circuit, there will be a loss of voltage due to an impedance.

Think of impedance as a circuit’s total opposition to electric current flow. Like resistance. But impedance is a combination of resistance and reactance.

With a current-carrying conductor, the voltage drop is directly proportional to the conductor’s impedance. And also, the current magnitude. So as the current increases, the voltage drop across the conductor increases.

Now, if you increase the conductor length, the impedance increases as well. Because all conductors have an impedance. Also, the impedance of a conductor is uniformly distributed along its entire length.

To illustrate voltage drop, take a look at the image below.

Our load rating of 15 amps is the same in all three scenarios. But look what happens as the conductor length increases from 10, to 100, and finally to 500-feet.

The voltage drop increases across the conductor shown in orange. As a result, the voltage at our load drops.

voltage drop circuit schematic

Issues in power systems from voltage drop

Think of voltage drop like wasted electricity.

This happens when there’s a difference in measured voltage at your source and load.

Because of this difference, several issues come up.

The first is you pay for the electricity you don’t use. When a utility supplies you with power, you pay for all the provided electricity. This includes the electricity you never use.

But an even bigger issue is the operating voltage of your equipment. A voltage drop causes your equipment to operate at lower than rated voltage.

For example, think of an inductive load, like a motor. The low operating voltage forces the motor to draw more current so it can meet its power demand. In return, the motor windings can overheat.

National Electric Code (NEC) notes on voltage drop

I want to point out the following important NEC sections on voltage drop:

  • 210.19(A)
  • 215.2(A)(4)
  • 310.15(A)(1)

The NEC recommends the max combined voltage drop for both the feeder and branch circuit to not exceed 5%. Also, the max voltage drop on either the feeder or branch circuit shouldn’t exceed 3%.

This is only a suggestion by the NEC.

The suggestion is to help ensure efficient power consumption. Also, it’ll increase equipment life and performance.

In the graphic below, you can see this NEC 3% and 5% recommendation illustrated.

nec voltage drop recommendations

Real-world voltage drop calculation problem 

Let’s outline our real-world problem.

An existing empty 1-inch PVC conduit runs 550-feet. It connects a low voltage power panel to an abandoned water well. This water well once had a 1 HP well pump inside.

The client now wants to fix up and expand the project site. The scope is to construct a new water well 350 feet away from the existing well. Thus, the total conduit run from the low voltage power panel to the new well now totals 900-feet.

What’s more, the client wants to upsize the well pump to 3 HP.

The motor specs are 3 HP, 240-volt, 1⌀ with a power factor of 0.90

  • Per NEC 430.151(A): Locked Rotor Current (LRC) = 102 amps (for motor starting conditions)
  • Per NEC 430.148: Full Load Amps (FLA) = 17 amps (for motor running conditions)

Our goal is to determine the conduit size and conductor ratings to feed this new motor. And because of the long circuit run, we need to do voltage drop calculations.

This way we’ll know if we can reuse the existing installed 1-inch PVC conduit. By reusing this conduit, we’d save a lot of money and have fewer headaches.

The theory behind single-phase, two-wire voltage drop

Since we have a single-phase problem, let’s find out how this will impact our calculations.

Take a look at the below schematic. While noting, I’ve conveniently used the resistor symbol for the conductor impedance.

That said, notice the current path is from the power source to the load. Then the return path is from the load to the power source.

There’s an equal impedance to the current in both segments. In other words, the line current to the load is equal to the neutral current to the power source.

For this reason, you’ll see how all the voltage drop equations are multiplied by two. As again, the voltage drop is a round trip calculated value in the circuit.

single phase two wire voltage drop schematic

Important Note: this depicted scenario is true in most instances. More specifically, both the line and neutral wires need to be the same length and type. Only then, will their impedances be equal. 

To point out, if the current returns some other way, the voltage drop will differ on the return path. We’d then need to calculate the voltage drop individually for each wire. We can’t simply multiply by 2.   

Voltage drop calculation assumptions in the analysis

Before we crunch any numbers, the following are our study assumptions:

  • Conductor temperatures are between 30°C and 40°C
  • Termination temperatures are 75°C
  • The load is balanced and thus there’s no neutral current
  • Power factors are all lagging
  • Conductors are all stranded

These assumptions will set the stage for each single-phase voltage drop calculation.

Approximate voltage drop calculation for motor starting 

To get started, let’s take a quick look at fire pumps.

The voltage at the line terminals of a fire pump can’t drop more than 15% of the rated voltage when the motor starts. This is according to NEC 695.7.

To dig deeper, the torque output of a motor typically scales as the square of the voltage. So the torque output can substantially drop with reduced voltage.

Now imagine the nominal locked rotor torque of a motor is 100-foot pounds at 240V.

With reduced voltage, let’s say the motor creates 50-foot pounds of torque. If this torque is greater than the starting torque of the motor, the motor can start. But if not, the motor won’t start.

Now, let’s jump back to our real-world problem.

I know our pump isn’t a fire pump. Regardless though, this 15% check is helpful with long conductor runs.

I’ve found using 15% is typically a good enough check across the board. For greater precision checking, review the torque-speed curve of a motor.

Thus, the total voltage drop allowed in our circuit is 240 volts x 0.15 = 36 volts. Where the voltage at the load end of our circuit can be no less than 204 volts.

Now, we calculate the wire size.

cmil = \dfrac{2 \times K \times I \times L}{V_{D}}

Where,
K constant = ohms-cmil per foot (copper has a K value of 12.9 ohms-cmil/ft)
I = current (or amperes) going to the load and returning from the load
L = length of conductor in feet (one-way)
cmil = circular mil area of the conductor (using Table 8 of Chapter 9 from the NEC)

cmil = \dfrac{2 \times K \times I \times L}{V_{D}} = \dfrac{2 \times 12.9 \times 102 \times 900}{36}} = 65,790 cmils

Then to calculate the voltage drop, we rearrange the equation.

V_{Drop} = \dfrac{2\times K \times I \times L}{A} =  \dfrac{2\times 12.9 \times 102 \times 900}{66,360} = 35.69 volts

\Rightarrow (35.69V/240V) \times 100 = 14.87% voltage drop

Looking at the NEC Table 8 of Chapter 9, a #2 conductor has a cross-sectional area of 66,360 cmils. This is greater than the calculated 65,790 cmils. This makes a #2 conductor a good choice so far.

Approximate voltage drop calculation for motor running 

We first calculate the total voltage drop allowed in the branch circuit: 240 volts x 0.03 = 7.2 volts. Where the voltage at the load end of the circuit can be no less than 232.8 volts.

Next, we calculate the conductor size.

cmil = \dfrac{2 \times K \times I \times L}{V_{D}}

Where,
K constant = 12.9 ohms-cmil per foot
Current (I) = 17 amps
Length (L) = 900 feet

Important Note: K is the electrical resistivity of the conductor type you’re using. It’s a constant set value for different materials.

It’s calculated by ‘electrical resistance’ multiplied by ‘cross-sectional area’ over ‘longitudinal length.’ For copper, the K constant is 12.9 ohms-cmil per foot.

Also, cmil or circular mill is a unit area. It’s used to define the cross-sectional size of something that’s circular in shape. In our case, a wire.

You calculate cmil by multiplying wire diameter in decimal inches by 1000. Then squaring the result. Fortunately, NEC Table 8 of Chapter 9 gives us a table of cmil values to use.

We now can plug numbers into our equation.

cmil = \dfrac{2 \times K \times I \times L}{V_{D}} = \dfrac{2 \times 12.9 \times 17 \times 900}{7.2}} = 54,825 cmils

Looking at the NEC Table 8 of Chapter 9, a #2 conductor has a cross-sectional area of 66,360 cmils. This is greater than the calculated 54,825 cmils. While #3 conductor has a cross-sectional area of 52,620 cmils.

Again, to calculate the voltage drop, we rearrange the equation.

V_{Drop} = \dfrac{2\times I \times K \times L}{A} =  \dfrac{2\times 17 \times 12.9 \times 900}{66,360} = 5.95 volts

\Rightarrow (5.95V/240V) \times 100 = 2.48% voltage drop

So using the approximate method, a #2 conductor will work. A #2 conductor meets both the motor’s starting and running conditions.

Estimated voltage drop calculation for motor running 

Now, we’ll use the more accurate estimate method for our voltage drop calculation.

Let’s first go over the shortcomings of the approximate method.

The K constant is the average DC resistance value for all conductor sizes of a given material type. Thus, using the K constant can lead to inaccurate AC voltage drop calculations.

Plus, the approximation equation doesn’t consider a load’s power factor. Rather, the K constant applies to any load power factor.

In other words, the actual AC characteristics of conductors aren’t considered.

But with the estimated method, we use the Effective Z at 0.85 power factor. This will replace the DC copper resistance value from the approximation method.

Important Note: using the Effective Z of 0.85 P.F. is typically conservative.

For large conductors, it’s almost always the worst-case value. 

Also, I’ve found for most applications, the value is always on the conservative side. 

The equation is V_{drop} = 2 x I x Z x L

Using NEC Table 9 of Chapter 9, a power factor of 0.85 results in an impedance of 0.19 ohms per 1000-feet for the #2 conductor in PVC.

V_{drop} = 2 x 17 x 0.19 x \dfrac{900}{1000} = 5.81 volts

\Rightarrow (5.81V /240V) \times 100 = 2.42% voltage drop

This shows how the estimate and approximate methods output near the same results.

Exact voltage drop calculation for motor running and starting 

For our third and final calculation, we’re going to calculate the exact voltage drop.

We’ll use the following equation:

V_{D} = 2 \times (V_{S} + IRLcos(\theta) + IXLsin(\theta) - \surd[V_{S}^{2} - (IXLcos(\theta) - IRLsin(\theta))^{2}])

In our calculation, I’m going to use the resistance (R) and reactance (X) of the #2 conductor in PVC.

Where,
V_{S} (Voltage Source) = 120 volts
Current (I) = 17 amps
Resistance (R) =  0.19 ohms/1000-feet (from NEC Table 8 of Chapter 9)
Reactance (X) = 0.045 ohms/1000-feet (from NEC Table 8 of Chapter 9)
Length (L) = 900 feet
Running Power Factor = 0.90 or cos^{-1}(0.90) = 25.8\degree

Now, let’s plug our numbers into the equation.

V_{D} = 2 \times (240 + (17)(0.19)(\dfrac{900}{1000})cos(25.8\degree) + (17)(0.045)(\dfrac{900}{1000})sin(25.8\degree) - \surd[240^{2} - ((17)(0.045)(\dfrac{900}{1000})cos(25.8\degree) - (17)(0.19)(\dfrac{900}{1000})sin(25.8\degree))^{2}])

= 2 x (240 + 2.62 + 0.30 – 240) = 5.84 volts

\Rightarrow (5.84V / 240V) \times 100 = 2.43% voltage drop

So all three methods output near the same voltage drop.

Important Note: smaller wire sizes have much greater resistance than reactance. Then for larger conductors, the reactance becomes larger than the resistance. Thus, use power factors in your voltage drop calculations with large conductors. 

Also, understand how the power factor for motor starting and running differ. For motor starting, the power factor will be much less than with motor running.

Typically, for modern motors, the power factor is 0.2 for starting. At the same time, for motor starting, use the motor Locked Rotor Amps (LRA) in your calculation. Not the motor Full Load Amps (FLA). 

Clearly, the power factor and current draw go hand in hand for a given motor. As the power factor drops, a motor will draw a greater current. 

Exact voltage drop calculation for motor starting 

To wrap things up, let’s do the calculation for the motor starting using the exact method.

We’ll use a power factor of 0.20 with an LRC of 102 amps.

V_{D} = 2 \times (240 + (102)(0.19)(\dfrac{900}{1000})cos(78.5\degree) + (102)(0.045)(\dfrac{900}{1000})sin(78.5\degree) - \surd[240^{2} - ((102)(0.045)(\dfrac{900}{1000})cos(78.5\degree) - (102)(0.19)(\dfrac{900}{1000})sin(78.5\degree))^{2}])

= 2 x (240 + 3.48 + 4.05 – 239.4) = 16.26 volts

\Rightarrow (16.26V / 240V) \times 100 = 6.78% voltage drop

You can see how the voltage drop here differs from the output from the approximate method. It’s because with the exact method we used an accurate power factor value.

Conduit fill calculation

Through all of our motor running calculation methods, we’ve determined #2 conductor is the right choice.

Our conduit fill is (3) #2 Cu. THHN/THWN & (1) #6 Cu. Ground.

The following is the cross-sectional areas of our conductors

  • #2 THHN/THWN: 0.1158 in² (from NEC Table 5 of Chapter 9)
  • #6 Cu.: 0.027 in² (from NEC Table 8 of Chapter 9)

\Rightarrow (0.1158\:in^{2} \times 3) + 0.027 \: in^{2} = 0.3744 \: in^{2}

Using NEC Table 4 Article 352 of Chapter 9, rigid PVC conduit schedule 40 has an allowable 40% fill of 0.581 in² for 1-1/4-inch conduit. While the allowable 40% fill for 1-inch conduit is 0.333 in².

Thus, for our new 350-foot conduit run, we need to use a 1-1/4-inch PVC conduit. Not only that, but we can’t reuse our existing in-ground 550-foot conduit, as it’s undersized.

What’s more, for a cable run this long, I’d upsize to 1-1/2-inch or even 2-inch conduit. This is an engineering decision.

For one, the cost is trivial in the grand scheme. Plus, the conduit will be easier to pull with such a long conduit run.

But also, in the future, the client may want to upsize the pump. Especially since they’ve already once upsized, with their eyes set on further expansion. Thus, they can reuse the larger conduit in the future without re-trenching for a new conduit.

Single-phase voltage drop calculation wrap up

In our problem, the client needs to upsize their existing 1-inch PVC conduit to at least 1-1/4-inch. Then pull (3) #2 Cu. THHN/THWN & (1) #6 Cu. ground conductors.

Without a doubt, this new conduit run adds a great cost to the project. But at least now, we won’t compromise the system efficiency and motor.

In short, voltage drop calculations can be simple or complex. It depends on how you approach a problem.

But in most instances, use the exact method when you have all the data at your fingertips. Because the approximate method is sometimes simply wrong.

In the end, just be ready to justify your decision on the voltage drop calculation method you choose.

What are your thoughts on the single-phase voltage drop calculation? Which single-phase voltage drop calculation do you typically use?

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