A stack gas stream of 23 m³/s carries 19 g/m³ of particulate. It passes a cyclone at 91% efficiency followed by a fabric filter at 94.5% 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

   $$C_{1} = 19(1 - 0.91) = 1.710 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 1.710(1 - 0.945) = 0.0941 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.090 \times 0.0550 = 0.99505 = 99.505$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0941 g/m^{3} \times 23 m^{3}/s \times 3600 s/h \div 1000 = 7.787 kg/h$$

Why the other options are there

• η = 185.5% (efficiencies added)
• 1,573 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

A stack gas stream of 19 m³/s carries 23 g/m³ of particulate. It passes a cyclone at 75% 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

   $$C_{1} = 23(1 - 0.75) = 5.750 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 5.750(1 - 0.970) = 0.1725 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.250 \times 0.0300 = 0.99250 = 99.250$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.1725 g/m^{3} \times 19 m^{3}/s \times 3600 s/h \div 1000 = 11.799 kg/h$$

Why the other options are there

• η = 172.0% (efficiencies added)
• 1,573 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

A stack gas stream of 48 m³/s carries 9 g/m³ of particulate. It passes a cyclone at 85% efficiency followed by a fabric filter at 97.5% 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

   $$C_{1} = 9(1 - 0.85) = 1.350 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 1.350(1 - 0.975) = 0.0338 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.150 \times 0.0250 = 0.99625 = 99.625$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0338 g/m^{3} \times 48 m^{3}/s \times 3600 s/h \div 1000 = 5.832 kg/h$$

Why the other options are there

• η = 182.5% (efficiencies added)
• 1,555 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

A stack gas stream of 41 m³/s carries 17 g/m³ of particulate. It passes a cyclone at 81% 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

   $$C_{1} = 17(1 - 0.81) = 3.230 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 3.230(1 - 0.970) = 0.0969 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.190 \times 0.0300 = 0.99430 = 99.430$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0969 g/m^{3} \times 41 m^{3}/s \times 3600 s/h \div 1000 = 14.302 kg/h$$

Why the other options are there

• η = 178.0% (efficiencies added)
• 2,509 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

A stack gas stream of 10 m³/s carries 16 g/m³ of particulate. It passes a cyclone at 78% efficiency followed by a fabric filter at 99.0% 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

   $$C_{1} = 16(1 - 0.78) = 3.520 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 3.520(1 - 0.990) = 0.0352 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.220 \times 0.0100 = 0.99780 = 99.780$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0352 g/m^{3} \times 10 m^{3}/s \times 3600 s/h \div 1000 = 1.267 kg/h$$

Why the other options are there

• η = 177.0% (efficiencies added)
• 576.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

A stack gas stream of 46 m³/s carries 22 g/m³ of particulate. It passes a cyclone at 84% efficiency followed by a fabric filter at 91.0% 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

   $$C_{1} = 22(1 - 0.84) = 3.520 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 3.520(1 - 0.910) = 0.3168 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.160 \times 0.0900 = 0.98560 = 98.560$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.3168 g/m^{3} \times 46 m^{3}/s \times 3600 s/h \div 1000 = 52.462 kg/h$$

Why the other options are there

• η = 175.0% (efficiencies added)
• 3,643 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

A stack gas stream of 40 m³/s carries 26 g/m³ of particulate. It passes a cyclone at 75% 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

   $$C_{1} = 26(1 - 0.75) = 6.500 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 6.500(1 - 0.985) = 0.0975 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.250 \times 0.0150 = 0.99625 = 99.625$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0975 g/m^{3} \times 40 m^{3}/s \times 3600 s/h \div 1000 = 14.040 kg/h$$

Why the other options are there

• η = 173.5% (efficiencies added)
• 3,744 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

A stack gas stream of 50 m³/s carries 15 g/m³ of particulate. It passes a cyclone at 85% efficiency followed by a fabric filter at 91.0% 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

   $$C_{1} = 15(1 - 0.85) = 2.250 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 2.250(1 - 0.910) = 0.2025 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.150 \times 0.0900 = 0.98650 = 98.650$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.2025 g/m^{3} \times 50 m^{3}/s \times 3600 s/h \div 1000 = 36.450 kg/h$$

Why the other options are there

• η = 176.0% (efficiencies added)
• 2,700 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

A stack gas stream of 46 m³/s carries 30 g/m³ of particulate. It passes a cyclone at 86% efficiency followed by a fabric filter at 93.5% 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

   $$C_{1} = 30(1 - 0.86) = 4.200 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 4.200(1 - 0.935) = 0.2730 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.140 \times 0.0650 = 0.99090 = 99.090$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.2730 g/m^{3} \times 46 m^{3}/s \times 3600 s/h \div 1000 = 45.209 kg/h$$

Why the other options are there

• η = 179.5% (efficiencies added)
• 4,968 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

A stack gas stream of 49 m³/s carries 26 g/m³ of particulate. It passes a cyclone at 91% efficiency followed by a fabric filter at 90.0% 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

   $$C_{1} = 26(1 - 0.91) = 2.340 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 2.340(1 - 0.900) = 0.2340 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.090 \times 0.1000 = 0.99100 = 99.100$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.2340 g/m^{3} \times 49 m^{3}/s \times 3600 s/h \div 1000 = 41.278 kg/h$$

Why the other options are there

• η = 181.0% (efficiencies added)
• 4,586 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse