fixed-film equation with recycle for a recirculating trickling filter Given combined feed BOD (with recycle) (S_a) = 123.0 mg/L; treatability constant (k) = 0.0190; filter depth (D) = 1.0000 m; hydraulic loading (with recycle) (Q) = 3.5000 m^3/m^2/day; depth exponent (n) = 0.7300; flow exponent (m) = 0.6800, determine the effluent BOD (S_e) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except S_e is given, so isolate S_e 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_e:

   $$S_{e} = S_a e^{-kD^n/Q^m}$$

3. Step 3 — List the givens: combined feed BOD (with recycle) (S_a) = 123.0 mg/L, treatability constant (k) = 0.0190, filter depth (D) = 1.0000 m, hydraulic loading (with recycle) (Q) = 3.5000 m^3/m^2/day, depth exponent (n) = 0.7300, flow exponent (m) = 0.6800.

4. Step 4 — Substitute the given values:

   $$S_{e} = S_a e^{-k1.0000^0.7300/3.5000^0.6800}$$

5. Step 5 — Evaluate:

   $$S_{e} = 122.0\ \text{mg/L}$$

6. Step 6 — Check: returning S_e = 122.0 mg/L to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 244.0 — kept a factor of two that cancels in the correct rearrangement.
• 61.0035 — dropped that same factor in the other direction.
• 134.2 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)

trickling filter fixed-film equation with recycle around the filter Given treatability constant (k) = 0.0270; filter depth (D) = 1.0000 m; hydraulic loading (with recycle) (Q) = 4.5000 m^3/m^2/day; depth exponent (n) = 0.8800; flow exponent (m) = 0.6200; effluent BOD (S_e) = 33.5000 mg/L, determine the combined feed BOD (with recycle) (S_a) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except S_a is given, so isolate S_a 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_a:

   $$S_{a} = \dfrac{S_e}{e^{-kD^n/Q^m}}$$

3. Step 3 — List the givens: treatability constant (k) = 0.0270, filter depth (D) = 1.0000 m, hydraulic loading (with recycle) (Q) = 4.5000 m^3/m^2/day, depth exponent (n) = 0.8800, flow exponent (m) = 0.6200, effluent BOD (S_e) = 33.5000 mg/L.

4. Step 4 — Substitute the given values:

   $$S_{a} = \dfrac{S_e}{e^{-k1.0000^0.8800/4.5000^0.6200}}$$

5. Step 5 — Evaluate:

   $$S_{a} = 33.8579\ \text{mg/L}$$

6. Step 6 — Check: returning S_a = 33.8579 mg/L to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 67.7157 — kept a factor of two that cancels in the correct rearrangement.
• 16.9289 — dropped that same factor in the other direction.
• 37.2437 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)

fixed-film equation with recycle design of a two-stage filter Given combined feed BOD (with recycle) (S_a) = 178.0 mg/L; filter depth (D) = 1.8000 m; hydraulic loading (with recycle) (Q) = 50.0000 m^3/m^2/day; depth exponent (n) = 0.7600; flow exponent (m) = 0.5800; effluent BOD (S_e) = 148.5 mg/L, determine the treatability constant (k).

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except k is given, so isolate k 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for k:

   $$k = -\ln\left(\dfrac{S_e}{S_a}\right)\dfrac{Q^m}{D^n}$$

3. Step 3 — List the givens: combined feed BOD (with recycle) (S_a) = 178.0 mg/L, filter depth (D) = 1.8000 m, hydraulic loading (with recycle) (Q) = 50.0000 m^3/m^2/day, depth exponent (n) = 0.7600, flow exponent (m) = 0.5800, effluent BOD (S_e) = 148.5 mg/L.

4. Step 4 — Substitute the given values:

   $$k = -\l0.7600\left(\dfrac{S_e}{S_a}\right)\dfrac{50.0000^0.5800}{1.8000^0.7600}$$

5. Step 5 — Evaluate:

   $$k = 1.1209$$

6. Step 6 — Check: returning k = 1.1209 to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 2.2417 — kept a factor of two that cancels in the correct rearrangement.
• 0.5604 — dropped that same factor in the other direction.
• 1.2329 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)

fixed-film equation with recycle for a recirculating trickling filter Given combined feed BOD (with recycle) (S_a) = 155.0 mg/L; treatability constant (k) = 0.0850; filter depth (D) = 2.1000 m; hydraulic loading (with recycle) (Q) = 46.0000 m^3/m^2/day; depth exponent (n) = 0.6800; flow exponent (m) = 0.6100, determine the effluent BOD (S_e) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except S_e is given, so isolate S_e 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_e:

   $$S_{e} = S_a e^{-kD^n/Q^m}$$

3. Step 3 — List the givens: combined feed BOD (with recycle) (S_a) = 155.0 mg/L, treatability constant (k) = 0.0850, filter depth (D) = 2.1000 m, hydraulic loading (with recycle) (Q) = 46.0000 m^3/m^2/day, depth exponent (n) = 0.6800, flow exponent (m) = 0.6100.

4. Step 4 — Substitute the given values:

   $$S_{e} = S_a e^{-k2.1000^0.6800/46.0000^0.6100}$$

5. Step 5 — Evaluate:

   $$S_{e} = 152.9\ \text{mg/L}$$

6. Step 6 — Check: returning S_e = 152.9 mg/L to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 305.8 — kept a factor of two that cancels in the correct rearrangement.
• 76.4514 — dropped that same factor in the other direction.
• 168.2 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)

trickling filter fixed-film equation with recycle around the filter Given treatability constant (k) = 0.0930; filter depth (D) = 2.9000 m; hydraulic loading (with recycle) (Q) = 7.0000 m^3/m^2/day; depth exponent (n) = 0.7600; flow exponent (m) = 0.5900; effluent BOD (S_e) = 23.5000 mg/L, determine the combined feed BOD (with recycle) (S_a) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except S_a is given, so isolate S_a 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_a:

   $$S_{a} = \dfrac{S_e}{e^{-kD^n/Q^m}}$$

3. Step 3 — List the givens: treatability constant (k) = 0.0930, filter depth (D) = 2.9000 m, hydraulic loading (with recycle) (Q) = 7.0000 m^3/m^2/day, depth exponent (n) = 0.7600, flow exponent (m) = 0.5900, effluent BOD (S_e) = 23.5000 mg/L.

4. Step 4 — Substitute the given values:

   $$S_{a} = \dfrac{S_e}{e^{-k2.9000^0.7600/7.0000^0.5900}}$$

5. Step 5 — Evaluate:

   $$S_{a} = 25.1100\ \text{mg/L}$$

6. Step 6 — Check: returning S_a = 25.1100 mg/L to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 50.2201 — kept a factor of two that cancels in the correct rearrangement.
• 12.5550 — dropped that same factor in the other direction.
• 27.6210 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)

fixed-film equation with recycle design of a two-stage filter Given combined feed BOD (with recycle) (S_a) = 235.0 mg/L; filter depth (D) = 2.0000 m; hydraulic loading (with recycle) (Q) = 41.0000 m^3/m^2/day; depth exponent (n) = 0.7100; flow exponent (m) = 0.6800; effluent BOD (S_e) = 206.0 mg/L, determine the treatability constant (k).

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except k is given, so isolate k 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for k:

   $$k = -\ln\left(\dfrac{S_e}{S_a}\right)\dfrac{Q^m}{D^n}$$

3. Step 3 — List the givens: combined feed BOD (with recycle) (S_a) = 235.0 mg/L, filter depth (D) = 2.0000 m, hydraulic loading (with recycle) (Q) = 41.0000 m^3/m^2/day, depth exponent (n) = 0.7100, flow exponent (m) = 0.6800, effluent BOD (S_e) = 206.0 mg/L.

4. Step 4 — Substitute the given values:

   $$k = -\l0.7100\left(\dfrac{S_e}{S_a}\right)\dfrac{41.0000^0.6800}{2.0000^0.7100}$$

5. Step 5 — Evaluate:

   $$k = 1.0060$$

6. Step 6 — Check: returning k = 1.0060 to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 2.0119 — kept a factor of two that cancels in the correct rearrangement.
• 0.5030 — dropped that same factor in the other direction.
• 1.1065 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)

fixed-film equation with recycle for a recirculating trickling filter Given combined feed BOD (with recycle) (S_a) = 268.0 mg/L; treatability constant (k) = 0.0290; filter depth (D) = 1.0000 m; hydraulic loading (with recycle) (Q) = 42.5000 m^3/m^2/day; depth exponent (n) = 0.7900; flow exponent (m) = 0.4700, determine the effluent BOD (S_e) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except S_e is given, so isolate S_e 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_e:

   $$S_{e} = S_a e^{-kD^n/Q^m}$$

3. Step 3 — List the givens: combined feed BOD (with recycle) (S_a) = 268.0 mg/L, treatability constant (k) = 0.0290, filter depth (D) = 1.0000 m, hydraulic loading (with recycle) (Q) = 42.5000 m^3/m^2/day, depth exponent (n) = 0.7900, flow exponent (m) = 0.4700.

4. Step 4 — Substitute the given values:

   $$S_{e} = S_a e^{-k1.0000^0.7900/42.5000^0.4700}$$

5. Step 5 — Evaluate:

   $$S_{e} = 266.7\ \text{mg/L}$$

6. Step 6 — Check: returning S_e = 266.7 mg/L to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

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

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)

trickling filter fixed-film equation with recycle around the filter Given treatability constant (k) = 0.0940; filter depth (D) = 1.9000 m; hydraulic loading (with recycle) (Q) = 48.5000 m^3/m^2/day; depth exponent (n) = 0.8800; flow exponent (m) = 0.4900; effluent BOD (S_e) = 70.5000 mg/L, determine the combined feed BOD (with recycle) (S_a) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except S_a is given, so isolate S_a 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_a:

   $$S_{a} = \dfrac{S_e}{e^{-kD^n/Q^m}}$$

3. Step 3 — List the givens: treatability constant (k) = 0.0940, filter depth (D) = 1.9000 m, hydraulic loading (with recycle) (Q) = 48.5000 m^3/m^2/day, depth exponent (n) = 0.8800, flow exponent (m) = 0.4900, effluent BOD (S_e) = 70.5000 mg/L.

4. Step 4 — Substitute the given values:

   $$S_{a} = \dfrac{S_e}{e^{-k1.9000^0.8800/48.5000^0.4900}}$$

5. Step 5 — Evaluate:

   $$S_{a} = 72.2619\ \text{mg/L}$$

6. Step 6 — Check: returning S_a = 72.2619 mg/L to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 144.5 — kept a factor of two that cancels in the correct rearrangement.
• 36.1309 — dropped that same factor in the other direction.
• 79.4881 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)

fixed-film equation with recycle design of a two-stage filter Given combined feed BOD (with recycle) (S_a) = 203.0 mg/L; filter depth (D) = 2.3000 m; hydraulic loading (with recycle) (Q) = 43.5000 m^3/m^2/day; depth exponent (n) = 0.6900; flow exponent (m) = 0.6200; effluent BOD (S_e) = 146.5 mg/L, determine the treatability constant (k).

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except k is given, so isolate k 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for k:

   $$k = -\ln\left(\dfrac{S_e}{S_a}\right)\dfrac{Q^m}{D^n}$$

3. Step 3 — List the givens: combined feed BOD (with recycle) (S_a) = 203.0 mg/L, filter depth (D) = 2.3000 m, hydraulic loading (with recycle) (Q) = 43.5000 m^3/m^2/day, depth exponent (n) = 0.6900, flow exponent (m) = 0.6200, effluent BOD (S_e) = 146.5 mg/L.

4. Step 4 — Substitute the given values:

   $$k = -\l0.6900\left(\dfrac{S_e}{S_a}\right)\dfrac{43.5000^0.6200}{2.3000^0.6900}$$

5. Step 5 — Evaluate:

   $$k = 1.9043$$

6. Step 6 — Check: returning k = 1.9043 to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 3.8085 — kept a factor of two that cancels in the correct rearrangement.
• 0.9521 — dropped that same factor in the other direction.
• 2.0947 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)

fixed-film equation with recycle for a recirculating trickling filter Given combined feed BOD (with recycle) (S_a) = 96.0000 mg/L; treatability constant (k) = 0.0260; filter depth (D) = 2.3000 m; hydraulic loading (with recycle) (Q) = 17.5000 m^3/m^2/day; depth exponent (n) = 0.8700; flow exponent (m) = 0.4100, determine the effluent BOD (S_e) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation with Recycle.
• Everything except S_e is given, so isolate S_e 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.
• The fixed-film equation with recycle accounts for recirculation in trickling filter effluent BOD prediction.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_e:

   $$S_{e} = S_a e^{-kD^n/Q^m}$$

3. Step 3 — List the givens: combined feed BOD (with recycle) (S_a) = 96.0000 mg/L, treatability constant (k) = 0.0260, filter depth (D) = 2.3000 m, hydraulic loading (with recycle) (Q) = 17.5000 m^3/m^2/day, depth exponent (n) = 0.8700, flow exponent (m) = 0.4100.

4. Step 4 — Substitute the given values:

   $$S_{e} = S_a e^{-k2.3000^0.8700/17.5000^0.4100}$$

5. Step 5 — Evaluate:

   $$S_{e} = 94.4198\ \text{mg/L}$$

6. Step 6 — Check: returning S_e = 94.4198 mg/L to

   $$\dfrac{S_e}{S_a} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 188.8 — kept a factor of two that cancels in the correct rearrangement.
• 47.2099 — dropped that same factor in the other direction.
• 103.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation with Recycle (NRC formula)