A stack gas stream of 8 m³/s carries 20 g/m³ of particulate. It passes a cyclone at 84% efficiency followed by a fabric filter at 97.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 20(1 - 0.84) = 3.200 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 3.200(1 - 0.970) = 0.0960 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.160 \times 0.0300 = 0.99520 = 99.520$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0960 g/m^{3} \times 8 m^{3}/s \times 3600 s/h \div 1000 = 2.765 kg/h$$

Why the other options are there

• η = 181.0% (efficiencies added)
• 576.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency

A stack gas stream of 33 m³/s carries 19 g/m³ of particulate. It passes a cyclone at 87% efficiency followed by a fabric filter at 90.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 19(1 - 0.87) = 2.470 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 2.470(1 - 0.905) = 0.2346 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.130 \times 0.0950 = 0.98765 = 98.765$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.2346 g/m^{3} \times 33 m^{3}/s \times 3600 s/h \div 1000 = 27.876 kg/h$$

Why the other options are there

• η = 177.5% (efficiencies added)
• 2,257 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency

A stack gas stream of 46 m³/s carries 21 g/m³ of particulate. It passes a cyclone at 84% efficiency followed by a fabric filter at 94.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 21(1 - 0.84) = 3.360 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 3.360(1 - 0.940) = 0.2016 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.160 \times 0.0600 = 0.99040 = 99.040$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.2016 g/m^{3} \times 46 m^{3}/s \times 3600 s/h \div 1000 = 33.385 kg/h$$

Why the other options are there

• η = 178.0% (efficiencies added)
• 3,478 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency

A stack gas stream of 44 m³/s carries 17 g/m³ of particulate. It passes a cyclone at 89% efficiency followed by a fabric filter at 93.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 17(1 - 0.89) = 1.870 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 1.870(1 - 0.930) = 0.1309 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.110 \times 0.0700 = 0.99230 = 99.230$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.1309 g/m^{3} \times 44 m^{3}/s \times 3600 s/h \div 1000 = 20.735 kg/h$$

Why the other options are there

• η = 182.0% (efficiencies added)
• 2,693 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency

A stack gas stream of 46 m³/s carries 21 g/m³ of particulate. It passes a cyclone at 91% efficiency followed by a fabric filter at 90.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 21(1 - 0.91) = 1.890 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 1.890(1 - 0.905) = 0.1795 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.090 \times 0.0950 = 0.99145 = 99.145$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.1795 g/m^{3} \times 46 m^{3}/s \times 3600 s/h \div 1000 = 29.733 kg/h$$

Why the other options are there

• η = 181.5% (efficiencies added)
• 3,478 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency

A stack gas stream of 55 m³/s carries 5 g/m³ of particulate. It passes a cyclone at 83% efficiency followed by a fabric filter at 92.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 5(1 - 0.83) = 0.850 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 0.850(1 - 0.925) = 0.0637 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.170 \times 0.0750 = 0.98725 = 98.725$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0637 g/m^{3} \times 55 m^{3}/s \times 3600 s/h \div 1000 = 12.622 kg/h$$

Why the other options are there

• η = 175.5% (efficiencies added)
• 990.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency

A stack gas stream of 11 m³/s carries 22 g/m³ of particulate. It passes a cyclone at 79% efficiency followed by a fabric filter at 96.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 22(1 - 0.79) = 4.620 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 4.620(1 - 0.960) = 0.1848 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.210 \times 0.0400 = 0.99160 = 99.160$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.1848 g/m^{3} \times 11 m^{3}/s \times 3600 s/h \div 1000 = 7.318 kg/h$$

Why the other options are there

• η = 175.0% (efficiencies added)
• 871.2 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency

A stack gas stream of 27 m³/s carries 21 g/m³ of particulate. It passes a cyclone at 77% efficiency followed by a fabric filter at 92.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 21(1 - 0.77) = 4.830 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 4.830(1 - 0.925) = 0.3622 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.230 \times 0.0750 = 0.98275 = 98.275$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.3622 g/m^{3} \times 27 m^{3}/s \times 3600 s/h \div 1000 = 35.211 kg/h$$

Why the other options are there

• η = 169.5% (efficiencies added)
• 2,041 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency

A stack gas stream of 60 m³/s carries 3 g/m³ of particulate. It passes a cyclone at 79% efficiency followed by a fabric filter at 98.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 3(1 - 0.79) = 0.630 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 0.630(1 - 0.985) = 0.0095 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.210 \times 0.0150 = 0.99685 = 99.685$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0095 g/m^{3} \times 60 m^{3}/s \times 3600 s/h \div 1000 = 2.041 kg/h$$

Why the other options are there

• η = 177.5% (efficiencies added)
• 648.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency

A stack gas stream of 22 m³/s carries 25 g/m³ of particulate. It passes a cyclone at 86% efficiency followed by a fabric filter at 96.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_{1} = C_in(1 - \eta_{1})$$

2. Substituting

   $$C_{1} = 25(1 - 0.86) = 3.500 g/m^{3}$$

3. Formula

   $$C_{2} = C_{1}(1 - \eta_{2})$$

4. Substituting

   $$C_{2} = 3.500(1 - 0.960) = 0.1400 g/m^{3}$$

5. Formula

   $$\eta_overall = 1 - (1 - \eta_{1})(1 - \eta_{2})$$

6. Substituting

   $$\eta = 1 - 0.140 \times 0.0400 = 0.99440 = 99.440$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.1400 g/m^{3} \times 22 m^{3}/s \times 3600 s/h \div 1000 = 11.088 kg/h$$

Why the other options are there

• η = 182.0% (efficiencies added)
• 1,980 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Cyclone Collection (Particle Removal) Efficiency