Single-Phase Voltage Drop Calculation Method Examples

Electrical design requires careful consideration of voltage drop. To illustrate, we’ll go over a single-phase voltage drop calculation.

However, it’s important to note there are the following different methods available to calculate voltage drop, each with varying degrees of accuracy:

Approximate
Estimated
Exact

I’ll walk you through each of them step by step. By the end, you’ll have a solid understanding of the theory behind voltage drop calculations and be able to choose the best method for your specific problem. So let’s get started!

What is voltage drop?

Voltage drop refers to the loss of voltage or reduction in voltage, which occurs as current moves through an electrical circuit. This happens because of something called impedance.

Think of impedance as a circuit’s resistance to the flow of electric current, kind of like when you try to walk through deep water. But impedance is more complicated because it’s a combination of both resistance and reactance.

When a conductor carries an electrical current, the voltage drop is proportional to the impedance of the conductor and the magnitude of the current. So, the more current flowing through the conductor, the greater the voltage drop.

And if you increase the length of the conductor, the impedance also increases. That’s because all conductors have some level of impedance spread evenly across their entire length.

Issues in power systems from voltage drop

Voltage drop can lead to several issues in power systems, kind of like how wasted food can lead to a lot of problems in a kitchen. The drop in voltage happens when there’s a difference between the measured voltage at the source and the load.

One major issue caused by voltage drop is you end up paying for electricity you don’t actually use. When a utility company provides power to your home or business, you pay for all the electricity provided, even if you don’t use it.

However, an even bigger issue is voltage drop can cause your equipment to operate at a lower voltage than it’s supposed to. This is especially problematic for equipment like motors, which need a specific voltage to function properly. A low operating voltage can cause motors to draw more current, which can lead to overheating.

To address these issues, the National Electric Code (NEC) recommends the maximum combined voltage drop for both feeder and branch circuits should not exceed 5%. Additionally, the maximum voltage drop for either the feeder or branch circuit should not exceed 3%. While these are just recommendations, following them can help ensure efficient power consumption and increase the life and performance of equipment.

If you want to know more, check out the following important NEC sections on voltage drop:

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

Also, take a look at the graphic below to see the NEC’s 3% and 5% recommendations in action.

Real-world voltage drop calculation problem

Let’s start by looking at a real-world problem.

Imagine there’s an empty 1-inch PVC conduit, which runs 550-feet, connecting a low-voltage power panel to an abandoned water well. This well used to have a 1 HP pump installed. But now, the client wants to expand and fix up the site, constructing a new water well 350 feet away from the existing one. This means the total conduit run from the power panel to the new well will now be 900 feet.

To make matters more complicated, the client also wants to upgrade the well pump to a 3 HP motor. The motor’s specs are 3 HP, 240 volts, 1⌀, with a power factor of 0.90.

Per NEC 430.151(A), the Locked Rotor Current (LRC) is 102 amps (for motor starting conditions)
Per NEC 430.148, the Full Load Amps (FLA) is 17 amps (for motor running conditions)

Our goal is to determine the conduit size and conductor ratings to feed this new motor. Due to the long circuit run, we also need to calculate the voltage drop. This will help us determine if we can reuse the existing 1-inch PVC conduit already installed. Reusing the existing conduit can save a lot of money and avoid headaches.

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

Since we’re dealing with a single-phase problem, it’s important to understand how this will impact our calculations.

Take a look at the schematic below. It’s worth noting that I’ve conveniently used the resistor symbol to represent the conductor impedance.

As you can see, the current flows from the power source to the load and then returns from the load to the power source. The impedance in both segments is equal, meaning the line current to the load is the same as the neutral current to the power source.

Due to this, you’ll notice all the voltage drop equations are multiplied by two. This is because the voltage drop is a round-trip calculated value in the circuit.

Important Note: In most cases, the scenario depicted above is true. It’s important to note though, both the line and neutral wires need to be of the same length and type to have equal impedances.

However, if the current returns through a different path, the voltage drop will differ on the return path. In such cases, we would need to calculate the voltage drop separately for each wire. We can’t simply multiply by 2 in these situations.

Voltage drop calculation assumptions

Before we get into the nitty-gritty of the calculations, let’s take a look at the following assumptions we’re making:

The conductor temperatures are between 30°C and 40°C
Termination temperatures are 75°C
The load is balanced, so there’s no neutral current
All power factors are lagging
All conductors are stranded

These assumptions are essential to consider as we move forward with our single-phase voltage drop calculation. They set the stage for all our calculations and ensure we have a clear picture of what we’re dealing with.

Approximate voltage drop calculation for motor starting

To kick things off, let’s take a quick look at fire pumps.

According to NEC 695.7, 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 an important consideration because the torque output of a motor is directly proportional to the voltage, with torque scaling as the square of the voltage.

Let’s say the nominal locked rotor torque of a motor is 100-foot pounds at 240V. With reduced voltage, the motor may only create 50-foot pounds of torque. If this torque output is less than the starting torque of the motor, the motor won’t start.

Although our pump isn’t a fire pump, the 15% check is still helpful in ensuring efficient power consumption with long conductor runs. It’s a good rule of thumb to follow. For more precision, you can review the torque-speed curve of the motor.

Thus, the total voltage drop allowed in our circuit is 240 volts x 0.15 = 36 volts, meaning the voltage at the load end of our circuit can’t be less than 204 volts.

Now, we can move on to calculating 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

Next, 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, where 240 volts x 0.03 = 7.2 volts. 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 a constant value that represents the electrical resistivity of the type of conductor you’re using. Different materials have different K values, which are set and don’t change.

To calculate K, you multiply the electrical resistance of the material by its cross-sectional area, and then divide that by its longitudinal length. For copper, the K constant is 12.9 ohms-cmil per foot.

Speaking of cross-sectional area, cmil (or circular mil) is a unit of measurement that’s used to define the size of circular objects – like wires, for instance. To calculate cmil, you multiply the wire’s diameter (in decimal inches) by 1000, and then square the result. Thankfully, NEC Table 8 of Chapter 9 provides a handy table of cmil values for us to use.

Now, we 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

We’re going to use the estimated method for our voltage drop calculation, which is more accurate than the approximate method we discussed earlier.

The approximate method has a few shortcomings, though. For one thing, the K constant – which represents the average DC resistance of a given material – can lead to inaccurate AC voltage drop calculations. Additionally, the approximation equation doesn’t take into account a load’s power factor, but rather applies the K constant to any load power factor.

This means the actual AC characteristics of conductors are not considered in the approximation method.

The estimated method, on the other hand, uses the Effective Z at 0.85 power factor. This replaces the DC copper resistance value used in the approximate method, resulting in a more precise calculation.

Important Note: The Effective Z of 0.85 P.F. is typically a conservative estimate.

In the case of large conductors, it almost always results in the worst-case value.

Furthermore, in my experience, the value is almost always on the conservative side for most applications.

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 nearly 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 higher resistance than reactance, whereas for larger conductors, the reactance is greater than the resistance. Therefore, when working with larger conductors, it’s important to factor in power factors for voltage drop calculations.

It’s also important to note that the power factor differs for motor starting and running. The power factor for starting is typically much lower than for running, around 0.2 for modern motors. Additionally, when calculating for motor starting, use the motor’s Locked Rotor Amps (LRA) instead of the Full Load Amps (FLA).

It’s clear that the power factor and current draw are closely related for a given motor. As the power factor drops, the motor draws 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 rating 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)

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

According to 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.

So for our new 350-foot conduit run, we need to upgrade to 1-1/4-inch PVC conduit, and sadly, our existing 550-foot conduit is too small to reuse.

But here’s the kicker, for such a long conduit run, why not upsize to 1-1/2-inch or even 2-inch conduit? It might cost a bit more, but it’ll make pulling the conduit much easier and save headaches down the line if the client wants to upgrade the pump again.

Single-phase voltage drop calculation wrap up

In our example, the client’s project site is expanding and they need to upsize their existing 1-inch PVC conduit to at least 1-1/4-inch and pull (3) #2 Cu. THHN/THWN & (1) #6 Cu. ground conductors. It’s a pricey addition, but it’s crucial to ensuring their system properly operates.

When it comes to voltage drop calculations, the key is choosing the right approach. My suggestion is to use the exact method when you have all the data at your fingertips. It provides more accurate results than the approximate method, and you can avoid costly mistakes.

But remember, be ready to defend 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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Engineer Calcs

Koosha started Engineer Calcs in 2020 to help people better understand the engineering and construction industry, and to discuss various science and engineering-related topics to make people think. He has been working in the engineering and tech industry in California for over 15 years now and is a licensed professional electrical engineer, and also has various entrepreneurial pursuits.

Koosha has an extensive background in the design and specification of electrical systems with areas of expertise including power generation, transmission, distribution, instrumentation and controls, and water distribution and pumping as well as alternative energy (wind, solar, geothermal, and storage).

Koosha is most interested in engineering innovations, the cosmos, our history and future, sports, and fitness.