A stack gas stream of 59 m³/s carries 7 g/m³ of particulate. It passes a cyclone at 90% 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

   $$C_{1} = 7(1 - 0.90) = 0.700 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 0.700(1 - 0.940) = 0.0420 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.100 \times 0.0600 = 0.99400 = 99.400$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0420 g/m^{3} \times 59 m^{3}/s \times 3600 s/h \div 1000 = 8.921 kg/h$$

Why the other options are there

• η = 184.0% (efficiencies added)
• 1,487 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency

A stack gas stream of 41 m³/s carries 4 g/m³ of particulate. It passes a cyclone at 91% 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

   $$C_{1} = 4(1 - 0.91) = 0.360 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 0.360(1 - 0.940) = 0.0216 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.090 \times 0.0600 = 0.99460 = 99.460$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0216 g/m^{3} \times 41 m^{3}/s \times 3600 s/h \div 1000 = 3.188 kg/h$$

Why the other options are there

• η = 185.0% (efficiencies added)
• 590.4 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency

A stack gas stream of 21 m³/s carries 20 g/m³ of particulate. It passes a cyclone at 88% efficiency followed by a fabric filter at 95.5% 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

   $$C_{1} = 20(1 - 0.88) = 2.400 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 2.400(1 - 0.955) = 0.1080 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.120 \times 0.0450 = 0.99460 = 99.460$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.1080 g/m^{3} \times 21 m^{3}/s \times 3600 s/h \div 1000 = 8.165 kg/h$$

Why the other options are there

• η = 183.5% (efficiencies added)
• 1,512 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency

A stack gas stream of 15 m³/s carries 14 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

   $$C_{1} = 14(1 - 0.77) = 3.220 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 3.220(1 - 0.925) = 0.2415 g/m^{3}$$

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

   $$0.2415 g/m^{3} \times 15 m^{3}/s \times 3600 s/h \div 1000 = 13.041 kg/h$$

Why the other options are there

• η = 169.5% (efficiencies added)
• 756.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency

A stack gas stream of 28 m³/s carries 6 g/m³ of particulate. It passes a cyclone at 87% 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

   $$C_{1} = 6(1 - 0.87) = 0.780 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 0.780(1 - 0.925) = 0.0585 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.130 \times 0.0750 = 0.99025 = 99.025$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0585 g/m^{3} \times 28 m^{3}/s \times 3600 s/h \div 1000 = 5.897 kg/h$$

Why the other options are there

• η = 179.5% (efficiencies added)
• 604.8 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency

A stack gas stream of 55 m³/s carries 4 g/m³ of particulate. It passes a cyclone at 77% 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

   $$C_{1} = 4(1 - 0.77) = 0.920 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 0.920(1 - 0.990) = 0.0092 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.230 \times 0.0100 = 0.99770 = 99.770$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.0092 g/m^{3} \times 55 m^{3}/s \times 3600 s/h \div 1000 = 1.822 kg/h$$

Why the other options are there

• η = 176.0% (efficiencies added)
• 792.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency

A stack gas stream of 18 m³/s carries 25 g/m³ of particulate. It passes a cyclone at 78% efficiency followed by a fabric filter at 95.5% 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

   $$C_{1} = 25(1 - 0.78) = 5.500 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 5.500(1 - 0.955) = 0.2475 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.220 \times 0.0450 = 0.99010 = 99.010$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.2475 g/m^{3} \times 18 m^{3}/s \times 3600 s/h \div 1000 = 16.038 kg/h$$

Why the other options are there

• η = 173.5% (efficiencies added)
• 1,620 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency

A stack gas stream of 29 m³/s carries 25 g/m³ of particulate. It passes a cyclone at 83% 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

   $$C_{1} = 25(1 - 0.83) = 4.250 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 4.250(1 - 0.945) = 0.2338 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.170 \times 0.0550 = 0.99065 = 99.065$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.2338 g/m^{3} \times 29 m^{3}/s \times 3600 s/h \div 1000 = 24.404 kg/h$$

Why the other options are there

• η = 177.5% (efficiencies added)
• 2,610 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency

A stack gas stream of 55 m³/s carries 18 g/m³ of particulate. It passes a cyclone at 75% efficiency followed by a fabric filter at 92.0% 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

   $$C_{1} = 18(1 - 0.75) = 4.500 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 4.500(1 - 0.920) = 0.3600 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.250 \times 0.0800 = 0.98000 = 98.000$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.3600 g/m^{3} \times 55 m^{3}/s \times 3600 s/h \div 1000 = 71.280 kg/h$$

Why the other options are there

• η = 167.0% (efficiencies added)
• 3,564 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency

A stack gas stream of 25 m³/s carries 20 g/m³ of particulate. It passes a cyclone at 78% 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

   $$C_{1} = 20(1 - 0.78) = 4.400 g/m^{3}$$

3. Formula

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

4. Substituting

   $$C_{2} = 4.400(1 - 0.935) = 0.2860 g/m^{3}$$

5. Formula

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

6. Substituting

   $$\eta = 1 - 0.220 \times 0.0650 = 0.98570 = 98.570$$

7. Formula

   $$emission = C_{2} Q$$

8. Substituting

   $$0.2860 g/m^{3} \times 25 m^{3}/s \times 3600 s/h \div 1000 = 25.740 kg/h$$

Why the other options are there

• η = 171.5% (efficiencies added)
• 1,800 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Electrostatic Precipitator Efficiency