Duct Size worksheet

1. Equivalent round diameter: 13.06 in

1. Use the Huebscher formula to find the round duct that loses the same pressure as this rectangle (a and b are its sides)

   $\mathrm{Dₑ}$$=\frac{1.3\times {\left(11\times 13\right)}^{0.625}}{{\left(11+13\right)}^{0.25}}$$=13.060974\text{\hspace{0.17em}in}$

2. Compare the areas. The rectangle needs more area than the round duct to lose the same pressure, because it has more wall

   $11\times 13$$=143\text{\hspace{0.17em}in²}$

2. Round diameter (estimate): 11.16 in

1. Change the friction rate to pascals per metre, the units the friction formula uses (1 in. w.c. = 249.08891 Pa, 100 ft = 30.48 m)

   $\frac{0.05\times 249.08891}{30.48}$$=0.40861\text{\hspace{0.17em}Pa/m}$

2. Change the airflow to cubic metres per second for the same reason (1 CFM = 0.00047194745 m³/s)

   $400\text{\hspace{0.17em}CFM}$$=0.18877898\text{\hspace{0.17em}m³/s}$

3. Search for the duct size whose pressure loss matches your rate. Bigger ducts lose less, so the calculator keeps halving its guess range until the loss matches. It uses the Darcy–Weisbach friction formula for standard air (ρ = 1.204 kg/m³, ν = 1.516 × 10⁻⁵ m²/s, roughness 0.09 mm)

   $d$$=0.283555\text{\hspace{0.17em}m}$

4. Check that size: air speed V = 2.9894 m/s, Reynolds number Re = 55,915 (how the air flows), friction factor f = 0.02154

   $\frac{0.021537}{0.283555}\times \frac{1.204\times {2.9894}^2}{2}$$=0.40861\text{\hspace{0.17em}Pa/m}$

5. Change the diameter to inches (1 in = 0.0254 m)

   $\frac{0.283555}{0.0254}$$=11.163599\text{\hspace{0.17em}in}$

6. Find the rectangle that loses the same pressure: with one side of 6 in, find the width whose Huebscher match is 11.1636 in

   $b$$=18.778207\text{\hspace{0.17em}in}$

3. Other rectangular side (width): 5.87 in

1. Find the width b that gives a Huebscher match of 7 in with a 7 in side. There's no direct way to solve for b, so the calculator narrows it down by halving its guess range again and again

   $b$$=5.869054\text{\hspace{0.17em}in}$

2. Check the answer: putting that width back into the Huebscher formula gives the round size you started with

   $\frac{1.3\times {\left(7\times 5.869054\right)}^{0.625}}{{\left(7+5.869054\right)}^{0.25}}$$=7\text{\hspace{0.17em}in}$

3. Compare the areas. The rectangle needs more area than the round duct to lose the same pressure, because it has more wall

   $7\times 5.869054$$=41.083378\text{\hspace{0.17em}in²}$

4. Equivalent round diameter: 290.2 mm

1. Use the Huebscher formula to find the round duct that loses the same pressure as this rectangle (a and b are its sides)

   $\mathrm{Dₑ}$$=\frac{1.3\times {\left(200\times 360\right)}^{0.625}}{{\left(200+360\right)}^{0.25}}$$=290.221215\text{\hspace{0.17em}mm}$

2. Compare the areas. The rectangle needs more area than the round duct to lose the same pressure, because it has more wall

   $200\times 360$$=\mathrm{72,000}\text{\hspace{0.17em}mm²}$

5. Round diameter (estimate): 5.78 in

1. Change the friction rate to pascals per metre, the units the friction formula uses (1 in. w.c. = 249.08891 Pa, 100 ft = 30.48 m)

   $\frac{0.1\times 249.08891}{30.48}$$=0.817221\text{\hspace{0.17em}Pa/m}$

2. Change the airflow to cubic metres per second for the same reason (1 CFM = 0.00047194745 m³/s)

   $100\text{\hspace{0.17em}CFM}$$=0.04719474\text{\hspace{0.17em}m³/s}$

3. Search for the duct size whose pressure loss matches your rate. Bigger ducts lose less, so the calculator keeps halving its guess range until the loss matches. It uses the Darcy–Weisbach friction formula for standard air (ρ = 1.204 kg/m³, ν = 1.516 × 10⁻⁵ m²/s, roughness 0.09 mm)

   $d$$=0.146731\text{\hspace{0.17em}m}$

4. Check that size: air speed V = 2.791 m/s, Reynolds number Re = 27,014 (how the air flows), friction factor f = 0.02557

   $\frac{0.025571}{0.146731}\times \frac{1.204\times {2.791}^2}{2}$$=0.817221\text{\hspace{0.17em}Pa/m}$

5. Change the diameter to inches (1 in = 0.0254 m)

   $\frac{0.146731}{0.0254}$$=5.776814\text{\hspace{0.17em}in}$

6. Find the rectangle that loses the same pressure: with one side of 7 in, find the width whose Huebscher match is 5.7768 in

   $b$$=4.06296\text{\hspace{0.17em}in}$

6. Equivalent round diameter: 315.4 mm

1. Use the Huebscher formula to find the round duct that loses the same pressure as this rectangle (a and b are its sides)

   $\mathrm{Dₑ}$$=\frac{1.3\times {\left(240\times 350\right)}^{0.625}}{{\left(240+350\right)}^{0.25}}$$=315.431385\text{\hspace{0.17em}mm}$

2. Compare the areas. The rectangle needs more area than the round duct to lose the same pressure, because it has more wall

   $240\times 350$$=\mathrm{84,000}\text{\hspace{0.17em}mm²}$

7. Equivalent round diameter: 312.6 mm

1. Use the Huebscher formula to find the round duct that loses the same pressure as this rectangle (a and b are its sides)

   $\mathrm{Dₑ}$$=\frac{1.3\times {\left(230\times 360\right)}^{0.625}}{{\left(230+360\right)}^{0.25}}$$=312.60744\text{\hspace{0.17em}mm}$

2. Compare the areas. The rectangle needs more area than the round duct to lose the same pressure, because it has more wall

   $230\times 360$$=\mathrm{82,800}\text{\hspace{0.17em}mm²}$

8. Other rectangular side (width): 333.4 mm

1. Find the width b that gives a Huebscher match of 280 mm with a 200 mm side. There's no direct way to solve for b, so the calculator narrows it down by halving its guess range again and again

   $b$$=333.368372\text{\hspace{0.17em}mm}$

2. Check the answer: putting that width back into the Huebscher formula gives the round size you started with

   $\frac{1.3\times {\left(200\times 333.368372\right)}^{0.625}}{{\left(200+333.368372\right)}^{0.25}}$$=280\text{\hspace{0.17em}mm}$

3. Compare the areas. The rectangle needs more area than the round duct to lose the same pressure, because it has more wall

   $200\times 333.368372$$=\mathrm{66,673.674449}\text{\hspace{0.17em}mm²}$

9. Round diameter (estimate): 9.76 in

1. Change the friction rate to pascals per metre, the units the friction formula uses (1 in. w.c. = 249.08891 Pa, 100 ft = 30.48 m)

   $\frac{0.1\times 249.08891}{30.48}$$=0.817221\text{\hspace{0.17em}Pa/m}$

2. Change the airflow to cubic metres per second for the same reason (1 CFM = 0.00047194745 m³/s)

   $400\text{\hspace{0.17em}CFM}$$=0.18877898\text{\hspace{0.17em}m³/s}$

3. Search for the duct size whose pressure loss matches your rate. Bigger ducts lose less, so the calculator keeps halving its guess range until the loss matches. It uses the Darcy–Weisbach friction formula for standard air (ρ = 1.204 kg/m³, ν = 1.516 × 10⁻⁵ m²/s, roughness 0.15 mm)

   $d$$=0.247974\text{\hspace{0.17em}m}$

4. Check that size: air speed V = 3.9089 m/s, Reynolds number Re = 63,938 (how the air flows), friction factor f = 0.02203

   $\frac{0.022031}{0.247974}\times \frac{1.204\times {3.9089}^2}{2}$$=0.817221\text{\hspace{0.17em}Pa/m}$

5. Change the diameter to inches (1 in = 0.0254 m)

   $\frac{0.247974}{0.0254}$$=9.762745\text{\hspace{0.17em}in}$

6. Find the rectangle that loses the same pressure: with one side of 9 in, find the width whose Huebscher match is 9.7627 in

   $b$$=8.862091\text{\hspace{0.17em}in}$

10. Minimum round diameter: 19.15 in

1. Find the area the air needs: airflow ÷ air speed, because airflow is speed times area

   $\frac{\mathrm{1,200}}{600}$$=2\text{\hspace{0.17em}ft²}$

2. Find the diameter of a circle with that area, then multiply by 12 to turn feet into inches

   $12\times \sqrt{\frac{4\times 2}{\pi }}$$=19.149229\text{\hspace{0.17em}in}$

3. Find the rectangle that loses the same pressure: with one side of 8 in, find the width whose Huebscher match is 19.1492 in

   $b$$=45.400536\text{\hspace{0.17em}in}$