A reactor of volume 1,966 m³ treats 0.80 m³/s carrying 101 mg/L of a contaminant that decays first-order with k = 0.30 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 1966/0.80 = 2,458 s = 0.683 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 101/(1 + 0.30 \times 0.683) = 83.83 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 101 e^(-0.30 \times 0.683) = 82.30 mg/L$$

7. Comparison — plug flow removes 18.5% versus 17.0% for the CSTR

Why the other options are there

• 80.3 mg/L (linear decay assumed)
• 83.8 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

A reactor of volume 4,642 m³ treats 1.20 m³/s carrying 350 mg/L of a contaminant that decays first-order with k = 0.30 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 4642/1.20 = 3,868 s = 1.075 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 350/(1 + 0.30 \times 1.075) = 264.7 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 350 e^(-0.30 \times 1.075) = 253.6 mg/L$$

7. Comparison — plug flow removes 27.6% versus 24.4% for the CSTR

Why the other options are there

• 237.2 mg/L (linear decay assumed)
• 264.7 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

A reactor of volume 3,931 m³ treats 1.85 m³/s carrying 213 mg/L of a contaminant that decays first-order with k = 0.40 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 3931/1.85 = 2,125 s = 0.590 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 213/(1 + 0.40 \times 0.590) = 172.3 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 213 e^(-0.40 \times 0.590) = 168.2 mg/L$$

7. Comparison — plug flow removes 21.0% versus 19.1% for the CSTR

Why the other options are there

• 162.7 mg/L (linear decay assumed)
• 172.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

A reactor of volume 4,830 m³ treats 1.85 m³/s carrying 322 mg/L of a contaminant that decays first-order with k = 0.50 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 4830/1.85 = 2,611 s = 0.725 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 322/(1 + 0.50 \times 0.725) = 236.3 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 322 e^(-0.50 \times 0.725) = 224.1 mg/L$$

7. Comparison — plug flow removes 30.4% versus 26.6% for the CSTR

Why the other options are there

• 205.2 mg/L (linear decay assumed)
• 236.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

A reactor of volume 4,326 m³ treats 1.75 m³/s carrying 173 mg/L of a contaminant that decays first-order with k = 0.40 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 4326/1.75 = 2,472 s = 0.687 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 173/(1 + 0.40 \times 0.687) = 135.7 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 173 e^(-0.40 \times 0.687) = 131.4 mg/L$$

7. Comparison — plug flow removes 24.0% versus 21.5% for the CSTR

Why the other options are there

• 125.5 mg/L (linear decay assumed)
• 135.7 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

A reactor of volume 2,804 m³ treats 0.45 m³/s carrying 318 mg/L of a contaminant that decays first-order with k = 0.10 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 2804/0.45 = 6,231 s = 1.731 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 318/(1 + 0.10 \times 1.731) = 271.1 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 318 e^(-0.10 \times 1.731) = 267.5 mg/L$$

7. Comparison — plug flow removes 15.9% versus 14.8% for the CSTR

Why the other options are there

• 263.0 mg/L (linear decay assumed)
• 271.1 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

A reactor of volume 3,256 m³ treats 2.00 m³/s carrying 179 mg/L of a contaminant that decays first-order with k = 0.10 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 3256/2.00 = 1,628 s = 0.452 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 179/(1 + 0.10 \times 0.452) = 171.3 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 179 e^(-0.10 \times 0.452) = 171.1 mg/L$$

7. Comparison — plug flow removes 4.4% versus 4.3% for the CSTR

Why the other options are there

• 170.9 mg/L (linear decay assumed)
• 171.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

A reactor of volume 1,513 m³ treats 0.65 m³/s carrying 370 mg/L of a contaminant that decays first-order with k = 0.25 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 1513/0.65 = 2,328 s = 0.647 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 370/(1 + 0.25 \times 0.647) = 318.5 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 370 e^(-0.25 \times 0.647) = 314.8 mg/L$$

7. Comparison — plug flow removes 14.9% versus 13.9% for the CSTR

Why the other options are there

• 310.2 mg/L (linear decay assumed)
• 318.5 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

A reactor of volume 3,460 m³ treats 1.90 m³/s carrying 324 mg/L of a contaminant that decays first-order with k = 0.50 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 3460/1.90 = 1,821 s = 0.506 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 324/(1 + 0.50 \times 0.506) = 258.6 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 324 e^(-0.50 \times 0.506) = 251.6 mg/L$$

7. Comparison — plug flow removes 22.3% versus 20.2% for the CSTR

Why the other options are there

• 242.1 mg/L (linear decay assumed)
• 258.6 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

A reactor of volume 3,559 m³ treats 1.30 m³/s carrying 100 mg/L of a contaminant that decays first-order with k = 0.25 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau = V/Q$$

2. Substituting

   $$\tau = 3559/1.30 = 2,738 s = 0.760 h$$

3. Formula (CSTR)

   $$C = C_{0}/(1 + k \tau)$$

4. Substituting

   $$C = 100/(1 + 0.25 \times 0.760) = 84.03 mg/L$$

5. Formula (PFR)

   $$C = C_{0} e^(-k \tau)$$

6. Substituting

   $$C = 100 e^(-0.25 \times 0.760) = 82.69 mg/L$$

7. Comparison — plug flow removes 17.3% versus 16.0% for the CSTR

Why the other options are there

• 81.0 mg/L (linear decay assumed)
• 84.0 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes