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