Conduit Fill worksheet

1. Conduit fill: 333.06%

1. Find the open space inside the 1/2 in Schedule 40 PVC. Treat it as a circle with inside diameter d = 0.578 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {0.578}^2$$=0.262389\text{\hspace{0.17em}in²}$

2. Find the space one 2/0 AWG THHN takes up (0.524 in across, including insulation), then multiply by 4 wires

   $4\times \frac{\pi }{4}\times {0.524}^2$$=0.862606\text{\hspace{0.17em}in²}$

3. Find the space one 0.12 in wire takes up (0.12 in across, including insulation), then multiply by 1 wires

   $1\times \frac{\pi }{4}\times {0.12}^2$$=0.01131\text{\hspace{0.17em}in²}$

4. Add up the space taken by all the wires

   $0.862606+0.01131$$=0.873916\text{\hspace{0.17em}in²}$

5. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{0.873916}{0.262389}\times 100$$=\mathrm{333.0611\%}$

6. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $0.262389\times \frac{40}{100}$$=0.104956\text{\hspace{0.17em}in²}\text{allowed}$

7. Work out how many 2/0 AWG THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.215651\text{≤ allowed area: n =}0$

2. Conduit fill: 7.52%

1. Find the open space inside the 3/4 in EMT. Treat it as a circle with inside diameter d = 0.824 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {0.824}^2$$=0.533267\text{\hspace{0.17em}in²}$

2. Find the space one 14 AWG THHN takes up (0.113 in across, including insulation), then multiply by 4 wires

   $4\times \frac{\pi }{4}\times {0.113}^2$$=0.040115\text{\hspace{0.17em}in²}$

3. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{0.040115}{0.533267}\times 100$$=\mathrm{7.5225\%}$

4. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $0.533267\times \frac{40}{100}$$=0.213307\text{\hspace{0.17em}in²}\text{allowed}$

5. Work out how many 14 AWG THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.010029\text{≤ allowed area: n =}21$

3. Conduit fill: 7.11%

1. Find the open space inside the 3 in EMT. Treat it as a circle with inside diameter d = 3.356 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {3.356}^2$$=8.845732\text{\hspace{0.17em}in²}$

2. Find the space one 1 AWG THHN takes up (0.441 in across, including insulation), then multiply by 4 wires

   $4\times \frac{\pi }{4}\times {0.441}^2$$=0.61098\text{\hspace{0.17em}in²}$

3. Find the space one 0.15 in wire takes up (0.15 in across, including insulation), then multiply by 1 wires

   $1\times \frac{\pi }{4}\times {0.15}^2$$=0.017671\text{\hspace{0.17em}in²}$

4. Add up the space taken by all the wires

   $0.61098+0.017671$$=0.628652\text{\hspace{0.17em}in²}$

5. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{0.628652}{8.845732}\times 100$$=\mathrm{7.1068\%}$

6. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $8.845732\times \frac{40}{100}$$=3.538293\text{\hspace{0.17em}in²}\text{allowed}$

7. Work out how many 1 AWG THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.152745\text{≤ allowed area: n =}23$

4. Conduit fill: 36.68%

1. Find the open space inside the 2 1/2 in EMT. Treat it as a circle with inside diameter d = 2.731 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {2.731}^2$$=5.857783\text{\hspace{0.17em}in²}$

2. Find the space one 400 kcmil THHN takes up (0.827 in across, including insulation), then multiply by 4 wires

   $4\times \frac{\pi }{4}\times {0.827}^2$$=2.148626\text{\hspace{0.17em}in²}$

3. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{2.148626}{5.857783}\times 100$$=\mathrm{36.6799\%}$

4. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $5.857783\times \frac{40}{100}$$=2.343113\text{\hspace{0.17em}in²}\text{allowed}$

5. Work out how many 400 kcmil THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.537157\text{≤ allowed area: n =}4$

5. Conduit fill: 1.44%

1. Find the open space inside the 4 in EMT. Treat it as a circle with inside diameter d = 4.334 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {4.334}^2$$=14.75257\text{\hspace{0.17em}in²}$

2. Find the space one 8 AWG THHN takes up (0.217 in across, including insulation), then multiply by 4 wires

   $4\times \frac{\pi }{4}\times {0.217}^2$$=0.147934\text{\hspace{0.17em}in²}$

3. Find the space one 10 AWG THHN takes up (0.166 in across, including insulation), then multiply by 3 wires

   $3\times \frac{\pi }{4}\times {0.166}^2$$=0.064927\text{\hspace{0.17em}in²}$

4. Add up the space taken by all the wires

   $0.147934+0.064927$$=0.212862\text{\hspace{0.17em}in²}$

5. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{0.212862}{14.75257}\times 100$$=\mathrm{1.4429\%}$

6. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $14.75257\times \frac{40}{100}$$=5.901028\text{\hspace{0.17em}in²}\text{allowed}$

7. Work out how many 8 AWG THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.036984\text{≤ allowed area: n =}159$

6. Conduit fill: 100.56%

1. Find the open space inside the conduit with a 0.6 in bore. Treat it as a circle with inside diameter d = 0.6 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {0.6}^2$$=0.282743\text{\hspace{0.17em}in²}$

2. Find the space one 4 AWG THHN takes up (0.324 in across, including insulation), then multiply by 3 wires

   $3\times \frac{\pi }{4}\times {0.324}^2$$=0.247344\text{\hspace{0.17em}in²}$

3. Find the space one 8 AWG THHN takes up (0.217 in across, including insulation), then multiply by 1 wires

   $1\times \frac{\pi }{4}\times {0.217}^2$$=0.036984\text{\hspace{0.17em}in²}$

4. Add up the space taken by all the wires

   $0.247344+0.036984$$=0.284327\text{\hspace{0.17em}in²}$

5. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{0.284327}{0.282743}\times 100$$=\mathrm{100.5603\%}$

6. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $0.282743\times \frac{40}{100}$$=0.113097\text{\hspace{0.17em}in²}\text{allowed}$

7. Work out how many 4 AWG THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.082448\text{≤ allowed area: n =}1$

7. Conduit fill: 199.87%

1. Find the open space inside the 3/4 in Schedule 40 PVC. Treat it as a circle with inside diameter d = 0.78 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {0.78}^2$$=0.477836\text{\hspace{0.17em}in²}$

2. Find the space one 1/0 AWG THHN takes up (0.48 in across, including insulation), then multiply by 5 wires

   $5\times \frac{\pi }{4}\times {0.48}^2$$=0.904779\text{\hspace{0.17em}in²}$

3. Find the space one 6 AWG THHN takes up (0.253 in across, including insulation), then multiply by 1 wires

   $1\times \frac{\pi }{4}\times {0.253}^2$$=0.050273\text{\hspace{0.17em}in²}$

4. Add up the space taken by all the wires

   $0.904779+0.050273$$=0.955051\text{\hspace{0.17em}in²}$

5. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{0.955051}{0.477836}\times 100$$=\mathrm{199.87\%}$

6. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $0.477836\times \frac{40}{100}$$=0.191134\text{\hspace{0.17em}in²}\text{allowed}$

7. Work out how many 1/0 AWG THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.180956\text{≤ allowed area: n =}1$

8. Conduit fill: 6.25%

1. Find the open space inside the 4 in EMT. Treat it as a circle with inside diameter d = 4.334 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {4.334}^2$$=14.75257\text{\hspace{0.17em}in²}$

2. Find the space one 3 AWG THHN takes up (0.35 in across, including insulation), then multiply by 4 wires

   $4\times \frac{\pi }{4}\times {0.35}^2$$=0.384845\text{\hspace{0.17em}in²}$

3. Find the space one 400 kcmil THHN takes up (0.827 in across, including insulation), then multiply by 1 wires

   $1\times \frac{\pi }{4}\times {0.827}^2$$=0.537157\text{\hspace{0.17em}in²}$

4. Add up the space taken by all the wires

   $0.384845+0.537157$$=0.922002\text{\hspace{0.17em}in²}$

5. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{0.922002}{14.75257}\times 100$$=\mathrm{6.2498\%}$

6. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $14.75257\times \frac{40}{100}$$=5.901028\text{\hspace{0.17em}in²}\text{allowed}$

7. Work out how many 3 AWG THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.096211\text{≤ allowed area: n =}61$

9. Conduit fill: 105.73%

1. Find the open space inside the 1 1/2 in EMT. Treat it as a circle with inside diameter d = 1.61 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {1.61}^2$$=2.035831\text{\hspace{0.17em}in²}$

2. Find the space one 2/0 AWG THHN takes up (0.524 in across, including insulation), then multiply by 5 wires

   $5\times \frac{\pi }{4}\times {0.524}^2$$=1.078257\text{\hspace{0.17em}in²}$

3. Find the space one 400 kcmil THHN takes up (0.827 in across, including insulation), then multiply by 2 wires

   $2\times \frac{\pi }{4}\times {0.827}^2$$=1.074313\text{\hspace{0.17em}in²}$

4. Add up the space taken by all the wires

   $1.078257+1.074313$$=2.152571\text{\hspace{0.17em}in²}$

5. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{2.152571}{2.035831}\times 100$$=\mathrm{105.7343\%}$

6. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $2.035831\times \frac{40}{100}$$=0.814332\text{\hspace{0.17em}in²}\text{allowed}$

7. Work out how many 2/0 AWG THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.215651\text{≤ allowed area: n =}3$

10. Conduit fill: 28.62%

1. Find the open space inside the 1 in EMT. Treat it as a circle with inside diameter d = 1.049 in, so its area A = π ÷ 4 × d²

   $A$$=\frac{\pi }{4}\times {1.049}^2$$=0.864253\text{\hspace{0.17em}in²}$

2. Find the space one 4 AWG THHN takes up (0.324 in across, including insulation), then multiply by 3 wires

   $3\times \frac{\pi }{4}\times {0.324}^2$$=0.247344\text{\hspace{0.17em}in²}$

3. Find the percent fill: the wires' area divided by the conduit's open area, times 100

   $\frac{0.247344}{0.864253}\times 100$$=\mathrm{28.6194\%}$

4. Look up the fill limit for more than two conductors (40%) in NEC Chapter 9, Table 1, then find how much of the area the wires may use

   $0.864253\times \frac{40}{100}$$=0.345701\text{\hspace{0.17em}in²}\text{allowed}$

5. Work out how many 4 AWG THHN wires would fit on their own (each count is checked against its own limit)

   $\text{largest n with}n\times 0.082448\text{≤ allowed area: n =}4$