cyclone collector sized for a 50% collection efficiency cut diameter Given gas viscosity (mu) = 0.0000 Pa*s; inlet width (W) = 0.7200 m; effective turns (N_e) = 6.0000; inlet gas velocity (V_i) = 12.0000 m/s; particle density (rho_p) = 2,970 kg/m^3; gas density (rho) = 1.0900 kg/m^3, determine the cut diameter (50% efficiency) (d_p50) in m.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except d_p50 is given, so isolate d_p50 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for d_p50:

   $$d_{p50} = \sqrt{\dfrac{9\mu W}{2\pi N_e V_i(\rho_p-\rho)}}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, inlet width (W) = 0.7200 m, effective turns (N_e) = 6.0000, inlet gas velocity (V_i) = 12.0000 m/s, particle density (rho_p) = 2,970 kg/m^3, gas density (rho) = 1.0900 kg/m^3.

4. Step 4 — Substitute the given values:

   $$d_{p50} = \sqrt{\dfrac{90.0000 0.7200}{2\pi N_e V_i(\rho_p-1.0900)}}$$

5. Step 5 — Evaluate:

   $$d_{p50} = 0.0000\ \text{m}$$

6. Step 6 — Check: returning d_p50 = 0.0000 m to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0000 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency

cyclone dust collector 50% efficiency particle size determination Given gas viscosity (mu) = 0.0000 Pa*s; inlet width (W) = 0.1400 m; effective turns (N_e) = 9.5000; particle density (rho_p) = 2,900 kg/m^3; gas density (rho) = 1.2100 kg/m^3; cut diameter (50% efficiency) (d_p50) = 0.0000 m, determine the inlet gas velocity (V_i) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except V_i is given, so isolate V_i symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for V_i:

   $$V_{i} = \dfrac{9\mu W}{2\pi N_e d_{p50}^2(\rho_p-\rho)}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, inlet width (W) = 0.1400 m, effective turns (N_e) = 9.5000, particle density (rho_p) = 2,900 kg/m^3, gas density (rho) = 1.2100 kg/m^3, cut diameter (50% efficiency) (d_p50) = 0.0000 m.

4. Step 4 — Substitute the given values:

   $$V_{i} = \dfrac{90.0000 0.1400}{2\pi N_e 0.0000^2(\rho_p-1.2100)}$$

5. Step 5 — Evaluate:

   $$V_{i} = 138.4\ \text{m/s}$$

6. Step 6 — Check: returning V_i = 138.4 m/s to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 276.7 — kept a factor of two that cancels in the correct rearrangement.
• 69.1789 — dropped that same factor in the other direction.
• 152.2 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency

cyclone separator design targeting 50 percent collection efficiency of fly ash Given gas viscosity (mu) = 0.0000 Pa*s; effective turns (N_e) = 3.5000; inlet gas velocity (V_i) = 14.5000 m/s; particle density (rho_p) = 2,170 kg/m^3; gas density (rho) = 1.1000 kg/m^3; cut diameter (50% efficiency) (d_p50) = 0.0000 m, determine the inlet width (W) in m.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for W:

   $$W = \dfrac{2\pi N_e V_i (\rho_p-\rho) d_{p50}^2}{9\mu}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, effective turns (N_e) = 3.5000, inlet gas velocity (V_i) = 14.5000 m/s, particle density (rho_p) = 2,170 kg/m^3, gas density (rho) = 1.1000 kg/m^3, cut diameter (50% efficiency) (d_p50) = 0.0000 m.

4. Step 4 — Substitute the given values:

   $$W = \dfrac{2\pi N_e V_i (\rho_p-1.1000) 0.0000^2}{90.0000}$$

5. Step 5 — Evaluate:

   $$W = 1.3064\ \text{m}$$

6. Step 6 — Check: returning W = 1.3064 m to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 2.6127 — kept a factor of two that cancels in the correct rearrangement.
• 0.6532 — dropped that same factor in the other direction.
• 1.4370 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency

cyclone collector sized for a 50% collection efficiency cut diameter Given gas viscosity (mu) = 0.0000 Pa*s; inlet width (W) = 0.4700 m; effective turns (N_e) = 5.5000; inlet gas velocity (V_i) = 21.0000 m/s; particle density (rho_p) = 2,990 kg/m^3; gas density (rho) = 1.0700 kg/m^3, determine the cut diameter (50% efficiency) (d_p50) in m.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except d_p50 is given, so isolate d_p50 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for d_p50:

   $$d_{p50} = \sqrt{\dfrac{9\mu W}{2\pi N_e V_i(\rho_p-\rho)}}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, inlet width (W) = 0.4700 m, effective turns (N_e) = 5.5000, inlet gas velocity (V_i) = 21.0000 m/s, particle density (rho_p) = 2,990 kg/m^3, gas density (rho) = 1.0700 kg/m^3.

4. Step 4 — Substitute the given values:

   $$d_{p50} = \sqrt{\dfrac{90.0000 0.4700}{2\pi N_e V_i(\rho_p-1.0700)}}$$

5. Step 5 — Evaluate:

   $$d_{p50} = 0.0000\ \text{m}$$

6. Step 6 — Check: returning d_p50 = 0.0000 m to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0000 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency

cyclone dust collector 50% efficiency particle size determination Given gas viscosity (mu) = 0.0000 Pa*s; inlet width (W) = 0.5600 m; effective turns (N_e) = 7.0000; particle density (rho_p) = 2,000 kg/m^3; gas density (rho) = 1.0700 kg/m^3; cut diameter (50% efficiency) (d_p50) = 0.0000 m, determine the inlet gas velocity (V_i) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except V_i is given, so isolate V_i symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for V_i:

   $$V_{i} = \dfrac{9\mu W}{2\pi N_e d_{p50}^2(\rho_p-\rho)}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, inlet width (W) = 0.5600 m, effective turns (N_e) = 7.0000, particle density (rho_p) = 2,000 kg/m^3, gas density (rho) = 1.0700 kg/m^3, cut diameter (50% efficiency) (d_p50) = 0.0000 m.

4. Step 4 — Substitute the given values:

   $$V_{i} = \dfrac{90.0000 0.5600}{2\pi N_e 0.0000^2(\rho_p-1.0700)}$$

5. Step 5 — Evaluate:

   $$V_{i} = 0.7962\ \text{m/s}$$

6. Step 6 — Check: returning V_i = 0.7962 m/s to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1.5924 — kept a factor of two that cancels in the correct rearrangement.
• 0.3981 — dropped that same factor in the other direction.
• 0.8758 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency

cyclone separator design targeting 50 percent collection efficiency of fly ash Given gas viscosity (mu) = 0.0000 Pa*s; effective turns (N_e) = 7.5000; inlet gas velocity (V_i) = 14.5000 m/s; particle density (rho_p) = 1,630 kg/m^3; gas density (rho) = 1.1000 kg/m^3; cut diameter (50% efficiency) (d_p50) = 0.0000 m, determine the inlet width (W) in m.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for W:

   $$W = \dfrac{2\pi N_e V_i (\rho_p-\rho) d_{p50}^2}{9\mu}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, effective turns (N_e) = 7.5000, inlet gas velocity (V_i) = 14.5000 m/s, particle density (rho_p) = 1,630 kg/m^3, gas density (rho) = 1.1000 kg/m^3, cut diameter (50% efficiency) (d_p50) = 0.0000 m.

4. Step 4 — Substitute the given values:

   $$W = \dfrac{2\pi N_e V_i (\rho_p-1.1000) 0.0000^2}{90.0000}$$

5. Step 5 — Evaluate:

   $$W = 4.5467\ \text{m}$$

6. Step 6 — Check: returning W = 4.5467 m to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 9.0933 — kept a factor of two that cancels in the correct rearrangement.
• 2.2733 — dropped that same factor in the other direction.
• 5.0013 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency

cyclone collector sized for a 50% collection efficiency cut diameter Given gas viscosity (mu) = 0.0000 Pa*s; inlet width (W) = 0.6300 m; effective turns (N_e) = 4.0000; inlet gas velocity (V_i) = 23.5000 m/s; particle density (rho_p) = 2,600 kg/m^3; gas density (rho) = 1.2400 kg/m^3, determine the cut diameter (50% efficiency) (d_p50) in m.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except d_p50 is given, so isolate d_p50 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for d_p50:

   $$d_{p50} = \sqrt{\dfrac{9\mu W}{2\pi N_e V_i(\rho_p-\rho)}}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, inlet width (W) = 0.6300 m, effective turns (N_e) = 4.0000, inlet gas velocity (V_i) = 23.5000 m/s, particle density (rho_p) = 2,600 kg/m^3, gas density (rho) = 1.2400 kg/m^3.

4. Step 4 — Substitute the given values:

   $$d_{p50} = \sqrt{\dfrac{90.0000 0.6300}{2\pi N_e V_i(\rho_p-1.2400)}}$$

5. Step 5 — Evaluate:

   $$d_{p50} = 0.0000\ \text{m}$$

6. Step 6 — Check: returning d_p50 = 0.0000 m to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0000 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency

cyclone dust collector 50% efficiency particle size determination Given gas viscosity (mu) = 0.0000 Pa*s; inlet width (W) = 0.3700 m; effective turns (N_e) = 5.0000; particle density (rho_p) = 1,640 kg/m^3; gas density (rho) = 1.2800 kg/m^3; cut diameter (50% efficiency) (d_p50) = 0.0000 m, determine the inlet gas velocity (V_i) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except V_i is given, so isolate V_i symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for V_i:

   $$V_{i} = \dfrac{9\mu W}{2\pi N_e d_{p50}^2(\rho_p-\rho)}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, inlet width (W) = 0.3700 m, effective turns (N_e) = 5.0000, particle density (rho_p) = 1,640 kg/m^3, gas density (rho) = 1.2800 kg/m^3, cut diameter (50% efficiency) (d_p50) = 0.0000 m.

4. Step 4 — Substitute the given values:

   $$V_{i} = \dfrac{90.0000 0.3700}{2\pi N_e 0.0000^2(\rho_p-1.2800)}$$

5. Step 5 — Evaluate:

   $$V_{i} = 0.6600\ \text{m/s}$$

6. Step 6 — Check: returning V_i = 0.6600 m/s to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1.3201 — kept a factor of two that cancels in the correct rearrangement.
• 0.3300 — dropped that same factor in the other direction.
• 0.7260 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency

cyclone separator design targeting 50 percent collection efficiency of fly ash Given gas viscosity (mu) = 0.0000 Pa*s; effective turns (N_e) = 3.5000; inlet gas velocity (V_i) = 9.5000 m/s; particle density (rho_p) = 2,460 kg/m^3; gas density (rho) = 1.0400 kg/m^3; cut diameter (50% efficiency) (d_p50) = 0.0000 m, determine the inlet width (W) in m.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for W:

   $$W = \dfrac{2\pi N_e V_i (\rho_p-\rho) d_{p50}^2}{9\mu}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, effective turns (N_e) = 3.5000, inlet gas velocity (V_i) = 9.5000 m/s, particle density (rho_p) = 2,460 kg/m^3, gas density (rho) = 1.0400 kg/m^3, cut diameter (50% efficiency) (d_p50) = 0.0000 m.

4. Step 4 — Substitute the given values:

   $$W = \dfrac{2\pi N_e V_i (\rho_p-1.0400) 0.0000^2}{90.0000}$$

5. Step 5 — Evaluate:

   $$W = 6.3569\ \text{m}$$

6. Step 6 — Check: returning W = 6.3569 m to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 12.7137 — kept a factor of two that cancels in the correct rearrangement.
• 3.1784 — dropped that same factor in the other direction.
• 6.9925 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency

cyclone collector sized for a 50% collection efficiency cut diameter Given gas viscosity (mu) = 0.0000 Pa*s; inlet width (W) = 0.8600 m; effective turns (N_e) = 3.5000; inlet gas velocity (V_i) = 10.5000 m/s; particle density (rho_p) = 2,260 kg/m^3; gas density (rho) = 1.0000 kg/m^3, determine the cut diameter (50% efficiency) (d_p50) in m.

Start with the thinking

• The governing relation printed in this handbook section is Cyclone 50% Collection Efficiency.
• Everything except d_p50 is given, so isolate d_p50 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Cyclone 50% collection efficiency defines the cut diameter at which the cyclone collects 50% of particles of that size.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

2. Step 2 — Rearrange symbolically for d_p50:

   $$d_{p50} = \sqrt{\dfrac{9\mu W}{2\pi N_e V_i(\rho_p-\rho)}}$$

3. Step 3 — List the givens: gas viscosity (mu) = 0.0000 Pa*s, inlet width (W) = 0.8600 m, effective turns (N_e) = 3.5000, inlet gas velocity (V_i) = 10.5000 m/s, particle density (rho_p) = 2,260 kg/m^3, gas density (rho) = 1.0000 kg/m^3.

4. Step 4 — Substitute the given values:

   $$d_{p50} = \sqrt{\dfrac{90.0000 0.8600}{2\pi N_e V_i(\rho_p-1.0000)}}$$

5. Step 5 — Evaluate:

   $$d_{p50} = 0.0000\ \text{m}$$

6. Step 6 — Check: returning d_p50 = 0.0000 m to

   $$d_{p50} = \sqrt{\dfrac{9 \mu W}{2\pi N_e V_i (\rho_p - \rho)}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0000 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Cyclone 50% Collection Efficiency