Chapter 4, Problem 36R

In problem 35, it is desired to keep the voltage drop to 3% maximum. What minimum size wire would be installed to accomplish this 3% maximum voltage drop? See 210.19(A), Informational Note No. 4 and 215.2(A)(1), Informational Note No. 2.

To determine

Find the minimum size wire that should be installed to accomplish 3% maximum voltage drop.

Answer to Problem 36R

10 AWG or larger value conductor should be installed to accomplish 3% maximum voltage drop.

Explanation of Solution

Given data:

Supply voltage is 120-volt.

Current is 10 A.

The distance between panel and heater is approximately 140 ft.

14 AWG, 12 AWG, 10AWG, and 8 AWG copper conductor.

Calculation:

Write the expression for voltage drop in a single-phase circuit.

${\text{E}}_{\text{d}}=\frac{\text{K}\times \text{I}\times \text{L}\times 2}{\text{CMA}}$ (1)

Here,

${\text{E}}_{\text{d}}$ is the voltage drop,

K is the approximate resistance in ohms per circular-mil foot at $75°\text{C}$. The value of K for the uncoated copper wire is $12.9$ and for aluminum wire is $21.2$.

I is the current, and

CMA is cross-sectional area of the conductors.

Modify Equation (1) for 14 AWG copper wire.

${\text{E}}_{\text{d}\left(14\text{AWG}\right)}=\frac{\text{K}\times \text{I}\times \text{L}\times 2}{{\text{CMA}}_{\left(14\text{AWG}\right)}}$ (2)

Refer TABLE 4-7 in the textbook for circular mil area (CMA) for many of the common conductors.

The value of CMA for 14 AWG is 4110.

Substitute $12.9$ for K, 10 A for I, 140 ft for L, and 4110 for ${\text{CMA}}_{\left(14\text{AWG}\right)}$ in Equation (2) to find the voltage drop in 14 AWG copper wire.

$\begin{array}{c}{\text{E}}_{\text{d}\left(14\text{AWG}\right)}=\frac{12.9\times 10\times 140\times 2}{4110}\\ =8.788\text{Volts drop}\\ \cong 8.79\text{Volts drop}\end{array}$

Modify Equation (1) for 12 AWG copper wire.

${\text{E}}_{\text{d}\left(12\text{AWG}\right)}=\frac{\text{K}\times \text{I}\times \text{L}\times 2}{{\text{CMA}}_{\left(12\text{AWG}\right)}}$ (3)

Refer TABLE 4-7 in the textbook for circular mil area (CMA) for many of the common conductors.

The value of CMA for 12 AWG is 6530.

Substitute $12.9$ for K, 10 A for I, 140 ft for L, and 6530 for ${\text{CMA}}_{\left(12\text{AWG}\right)}$ in Equation (3) to find the voltage drop in 12 AWG copper wire.

$\begin{array}{c}{\text{E}}_{\text{d}\left(12\text{AWG}\right)}=\frac{12.9\times 10\times 140\times 2}{6530}\\ =5.531\text{Volts drop}\\ \cong 5.53\text{Volts drop}\end{array}$

Modify Equation (1) for 10 AWG copper wire.

${\text{E}}_{\text{d}\left(10\text{AWG}\right)}=\frac{\text{K}\times \text{I}\times \text{L}\times 2}{{\text{CMA}}_{\left(10\text{AWG}\right)}}$ (4)

Refer TABLE 4-7 in the textbook for circular mil area (CMA) for many of the common conductors.

The value of CMA for 10 AWG is 10,380.

Substitute $12.9$ for K, 10 A for I, 140 ft for L, and 10,380 for ${\text{CMA}}_{\left(10\text{AWG}\right)}$ in Equation (4) to find the voltage drop in 10 AWG copper wire.

$\begin{array}{c}{\text{E}}_{\text{d}\left(10\text{AWG}\right)}=\frac{12.9\times 10\times 140\times 2}{10,380}\\ =3.479\text{Volts drop}\\ \cong \text{3}\text{.48}\text{Volts drop}\end{array}$

Modify Equation (1) for 8 AWG copper wire.

${\text{E}}_{\text{d}\left(\text{8AWG}\right)}=\frac{\text{K}\times \text{I}\times \text{L}\times 2}{{\text{CMA}}_{\left(\text{8AWG}\right)}}$ (5)

Refer TABLE 4-7 in the textbook for circular mil area (CMA) for many of the common conductors.

The value of CMA for 8 AWG is 16,510.

Substitute $12.9$ for K, 10 A for I, 140 ft for L, and 16,510 for ${\text{CMA}}_{\left(\text{8AWG}\right)}$ in Equation (5) to find the voltage drop in 8 AWG copper wire.

$\begin{array}{c}{\text{E}}_{\text{d}\left(\text{8AWG}\right)}=\frac{12.9\times 10\times 140\times 2}{16,510}\\ =2.187\text{Volts drop}\\ \cong \text{2}\text{.19}\text{Volts drop}\end{array}$

Tabulate the calculated value of the voltage drop as shown in Table 1.

Table 1

Sl. No.	Conductor Size	Voltage drop
1	14 AWG	$8.79\text{Volts drop}$
2	12 AWG	$5.53\text{Volts drop}$
3	10 AWG	$\text{3}\text{.48}\text{Volts drop}$
4	8 AWG	$\text{2}\text{.19}\text{Volts drop}$

Find the value of $3\%$ voltage drop of the applied voltage.

$\begin{array}{c}{\text{Voltage drop}}_{3\%\text{ applied voltage}}=3\%\left(120\text{V}\right)\\ =\frac{3}{100}\left(120\text{V}\right)\\ =3.6\text{V}\end{array}$

Compare the obtained value of 3% voltage drop of applied voltage with voltage drops of each conductor as tabulated in Table 1.

A voltage drop of 8AWG conductor (that is 2.19 Volts drop) is less than $3\%$ voltage drop of the applied voltage (that is 3.6 Volts drop). Therefore, conductor size with 10 AWG or larger should be installed to accomplish 3% maximum voltage drop.

Conclusion:

Thus, 10 AWG or larger value conductor should be installed to accomplish 3% maximum voltage drop.