238 lines
5.5 KiB
Markdown
238 lines
5.5 KiB
Markdown
---
|
|
id:
|
|
aliases: []
|
|
tags:
|
|
- destiny/fleeting
|
|
- status/incomplete
|
|
- topic/construction/electrical
|
|
- type/encyclopedia
|
|
- authorship/original
|
|
title: Conductor Sizing
|
|
---
|
|
# Conductor Sizing
|
|
|
|
Conductors are sized to be suitable for the load,
|
|
overcurrent protection is sized to protect the conductors.
|
|
|
|
## "The 80% Rule"
|
|
|
|
"The 80% Rule" is a rule of thumb
|
|
referring to a common convention of several articles
|
|
including:
|
|
|
|
* [[nfpa-70_210_branch-circuits#210.19(A)(1) General.]]
|
|
* [[nfpa-70_215_feeders#215.2(A)(1) General.]]
|
|
* [[nfpa-70_430_motors#430.22 Single Motor.]]
|
|
|
|
which paraphrased states:
|
|
|
|
> ... the minimum conductor size shall have an ampacity
|
|
> not less than the noncontinuous load
|
|
> plus 125 percent of the continuous load
|
|
|
|
When the rule is repeated,
|
|
the noncontinuous load is ignored
|
|
and it is stated that conductors are suitable
|
|
for 80% their listed rating,
|
|
since 80% is the reciprocal of 125%.
|
|
|
|
## Branch Circuits
|
|
|
|
### Receptacle Branch
|
|
|
|
> [!important]
|
|
> There is no maximum number of receptacles per circuit _in any occupancy_.
|
|
|
|
It is a common misconception that the limit
|
|
can be calculated with a formula like
|
|
|
|
$$
|
|
\frac{1.25(180VA)}{120V} = 1.875A, \quad \frac{20A}{1.875A} = 10.\bar{6}
|
|
$$
|
|
|
|
but the 180VA per yoke load specified in
|
|
[[nfpa-70_220_load-calculations#220.14(I) Receptacle Outlets.|220.14(I)]]
|
|
is specifically for calculating service and feeder sizing.
|
|
|
|
Per [[nfpa-70_210_branch-circuits#210.19(A)(1) General.|210.19(A)(1)]]
|
|
a receptacle branch circuit's load
|
|
is the load of the equipment intended to be served by it.
|
|
|
|
Where general-use receptacles are provided
|
|
without specific equipment in mind,
|
|
circuits will be engineered at the minimum load.
|
|
If a receptacle circuit's load is a whole multiple of 180VA
|
|
there's a good chance that's the number of devices,
|
|
or at least was at some point in the design.
|
|
|
|
## Feeders
|
|
|
|
> [!cite] NEC Article 250 (emphasis added)
|
|
> ### 250.122 Size of Equipment Grounding Conductors
|
|
> #### (A) General.
|
|
> Copper, aluminum, or copper-clad aluminum
|
|
> equipment grounding conductors of the wire type
|
|
> shall not be smaller than shown in Table 250.122,
|
|
> but in no case shall they be required to be larger
|
|
> than the circuit conductors supplying the equipment...
|
|
|
|
Apparently in the 2026 NEC First Draft Meetings,
|
|
Code Making Panel 5 clarified that the equipment grounding conductor (EGC)
|
|
never needs to be larger than the largest ungrounded conductor in any raceway
|
|
when installed in parallel.
|
|
I can not find a source to verify this.
|
|
Statements from other reputable sources including Mike Holt
|
|
are in contradiction to this idea.
|
|
|
|
***
|
|
|
|
Given a minimum ampacity, find all valid configurations.
|
|
|
|
> [!cite] NEC Article 310 (emphasis added)
|
|
> #### 310.10(H) Conductors in Parallel.
|
|
> ##### (1) General.
|
|
> Aluminum, copper-clad aluminum, or copper conductors,
|
|
> for each phase, polarity, neutral, or grounded circuit
|
|
> shall be permitted to be connected in parallel
|
|
> (electrically joined at both ends)
|
|
> _only in sizes 1/0 AWG and larger_
|
|
> where installed in accordance with 310.10(H)(2) through (H)(6).
|
|
|
|
Rank by total cost of install.
|
|
|
|
### Complexity to Ignore
|
|
|
|
#### Conductor Material
|
|
|
|
Tinned copper and copper-clad aluminum conductors
|
|
can be assumed out of scope.
|
|
|
|
### Complexity to Respect
|
|
|
|
#### Equipment Grounding Conductor Material
|
|
|
|
Wire and EGC conductors are usually assumed to match,
|
|
but it is sometimes necessary to use a copper EGC with aluminum wires,
|
|
either for spec requirements or conduit fill considerations.
|
|
|
|
## Voltage Drop
|
|
|
|
> [!info] Ohm's Law
|
|
>
|
|
> $$
|
|
> V = I \times R, \quad R = \frac{V}{I}, \quad I = \frac{V}{R}
|
|
> $$
|
|
|
|
### Step 1: Effective Impedance $Z$
|
|
|
|
$$
|
|
Z = R \cos(\theta) + X \sin(\theta)
|
|
$$
|
|
|
|
where
|
|
* $Z$ = Effective impedance
|
|
* $R$ = AC resistance
|
|
* $X$ = Reactance
|
|
* $\theta$ = Power factor angle = $\arccos(PF)$
|
|
|
|
#### Parallel Runs
|
|
|
|
The equivalent resistance of parallel resistances is given by
|
|
|
|
$$
|
|
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}
|
|
$$
|
|
|
|
For $n$ parallel resistances of value $R$
|
|
|
|
$$
|
|
\begin{align*}
|
|
\frac{1}{R_{\text{eq}}} &= n \times \left(\frac{1}{R}\right) \\
|
|
&= \frac{n}{R} \\
|
|
R_{\text{eq}} &= \frac{R}{n}
|
|
\end{align*}
|
|
$$
|
|
|
|
### Step 2: Voltage Drop
|
|
|
|
> [!important]
|
|
> This section assumes a 3-phase
|
|
> 208Y/120V or 480Y/277V voltage system
|
|
|
|
> [!info] 3-Phase Voltage
|
|
>
|
|
> $$
|
|
> V_{LL} = \sqrt{3} \times V_{LN}, \quad V_{LN} = \frac{V_{LL}}{\sqrt{3}}
|
|
> $$
|
|
|
|
3% allowable voltage drop for a 120V line-to-neutral load:
|
|
|
|
$$
|
|
\text{Max} \Delta V = 0.03 \times 120 \text{V}_{LN} = 3.6 \text{V}_{LN}
|
|
$$
|
|
|
|
3% allowable voltage drop for a 208V line-to-line load:
|
|
|
|
$$
|
|
\text{Max} \Delta V = 0.03 \times 208 \text{V}_{LL} = 6.24 \text{V}_{LL}
|
|
$$
|
|
|
|
#### Line to Neutral Loads
|
|
|
|
$$
|
|
\Delta V_{LN} = I \times Z \times 2L
|
|
$$
|
|
|
|
#### Line to Line Loads
|
|
|
|
$$
|
|
\Delta V_{LL} = \sqrt{3} \times \left( I \times Z \times 2L \right)
|
|
$$
|
|
|
|
#### 3-Phase Loads
|
|
|
|
$$
|
|
\Delta V_{3\phi} = \sqrt{3} \times \left( I \times Z \times L \right)
|
|
$$
|
|
|
|
where
|
|
* $\Delta V$ = Voltage drop in volts ($V$)
|
|
* $I$ = Current in amperes ($A$)
|
|
* $L$ = Length of wire one way in feet ($\text{ft}$)
|
|
|
|
> [!important]
|
|
> "Current" is not the OCPD rating,
|
|
> but the actual load.
|
|
|
|
***
|
|
|
|
It is often more useful to know the maximum length
|
|
a certain wiring configuration is suitable for.
|
|
|
|
$$
|
|
L = \frac{ \Delta V }{ I \times M } \times \frac{1}{Z}
|
|
$$
|
|
|
|
## Transformers
|
|
|
|
$$
|
|
I = \frac{S}{ \sqrt{3} \times V \times E }
|
|
$$
|
|
|
|
* $I$ = nameplate current rating
|
|
* $S$ = nameplate kVA rating
|
|
* $V$ = feeder voltage
|
|
* $E$ = efficiency
|
|
|
|
## Motors
|
|
|
|
1 electric horsepower = 746 watts
|
|
|
|
full-load current (FLC) / full-load amperes (FLA)
|
|
|
|
minimum circuit ampacity (MCA)
|
|
|
|
$$
|
|
\text{MCA} = 1.25 \times \text{FLC}
|
|
$$
|