fixed-film trickling filter equation without recycle for effluent BOD Given influent BOD (S_0) = 307.0 mg/L; treatability constant (k) = 0.0160; filter depth (D) = 1.2000 m; hydraulic loading (Q) = 14.5000 m^3/m^2/day; depth exponent (n) = 0.7800; flow exponent (m) = 0.6500, 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 without 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_e:

   $$S_{e} = S_0 e^{-kD^n/Q^m}$$

3. Step 3 — List the givens: influent BOD (S_0) = 307.0 mg/L, treatability constant (k) = 0.0160, filter depth (D) = 1.2000 m, hydraulic loading (Q) = 14.5000 m^3/m^2/day, depth exponent (n) = 0.7800, flow exponent (m) = 0.6500.

4. Step 4 — Substitute the given values:

   $$S_{e} = S_0 e^{-k1.2000^0.7800/14.5000^0.6500}$$

5. Step 5 — Evaluate:

   $$S_{e} = 306.0\ \text{mg/L}$$

6. Step 6 — Check: returning S_e = 306.0 mg/L to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

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

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)

fixed-film equation without recycle applied to a single-stage filter Given treatability constant (k) = 0.0390; filter depth (D) = 1.1000 m; hydraulic loading (Q) = 38.5000 m^3/m^2/day; depth exponent (n) = 0.6500; flow exponent (m) = 0.5500; effluent BOD (S_e) = 19.0000 mg/L, determine the influent BOD (S_0) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation without Recycle.
• Everything except S_0 is given, so isolate S_0 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_0:

   $$S_{0} = \dfrac{S_e}{e^{-kD^n/Q^m}}$$

3. Step 3 — List the givens: treatability constant (k) = 0.0390, filter depth (D) = 1.1000 m, hydraulic loading (Q) = 38.5000 m^3/m^2/day, depth exponent (n) = 0.6500, flow exponent (m) = 0.5500, effluent BOD (S_e) = 19.0000 mg/L.

4. Step 4 — Substitute the given values:

   $$S_{0} = \dfrac{S_e}{e^{-k1.1000^0.6500/38.5000^0.5500}}$$

5. Step 5 — Evaluate:

   $$S_{0} = 19.1062\ \text{mg/L}$$

6. Step 6 — Check: returning S_0 = 19.1062 mg/L to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 38.2123 — kept a factor of two that cancels in the correct rearrangement.
• 9.5531 — dropped that same factor in the other direction.
• 21.0168 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)

trickling filter fixed-film equation without recycle design check Given influent BOD (S_0) = 351.0 mg/L; filter depth (D) = 2.8000 m; hydraulic loading (Q) = 14.5000 m^3/m^2/day; depth exponent (n) = 0.8500; flow exponent (m) = 0.4700; effluent BOD (S_e) = 183.5 mg/L, determine the treatability constant (k).

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation without 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for k:

   $$k = -\ln\left(\dfrac{S_e}{S_0}\right)\dfrac{Q^m}{D^n}$$

3. Step 3 — List the givens: influent BOD (S_0) = 351.0 mg/L, filter depth (D) = 2.8000 m, hydraulic loading (Q) = 14.5000 m^3/m^2/day, depth exponent (n) = 0.8500, flow exponent (m) = 0.4700, effluent BOD (S_e) = 183.5 mg/L.

4. Step 4 — Substitute the given values:

   $$k = -\l0.8500\left(\dfrac{S_e}{S_0}\right)\dfrac{14.5000^0.4700}{2.8000^0.8500}$$

5. Step 5 — Evaluate:

   $$k = 0.9500$$

6. Step 6 — Check: returning k = 0.9500 to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 1.9000 — kept a factor of two that cancels in the correct rearrangement.
• 0.4750 — dropped that same factor in the other direction.
• 1.0450 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)

fixed-film trickling filter equation without recycle for effluent BOD Given influent BOD (S_0) = 187.0 mg/L; treatability constant (k) = 0.0990; filter depth (D) = 2.5000 m; hydraulic loading (Q) = 3.5000 m^3/m^2/day; depth exponent (n) = 0.6700; flow exponent (m) = 0.3400, 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 without 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_e:

   $$S_{e} = S_0 e^{-kD^n/Q^m}$$

3. Step 3 — List the givens: influent BOD (S_0) = 187.0 mg/L, treatability constant (k) = 0.0990, filter depth (D) = 2.5000 m, hydraulic loading (Q) = 3.5000 m^3/m^2/day, depth exponent (n) = 0.6700, flow exponent (m) = 0.3400.

4. Step 4 — Substitute the given values:

   $$S_{e} = S_0 e^{-k2.5000^0.6700/3.5000^0.3400}$$

5. Step 5 — Evaluate:

   $$S_{e} = 165.9\ \text{mg/L}$$

6. Step 6 — Check: returning S_e = 165.9 mg/L to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 331.9 — kept a factor of two that cancels in the correct rearrangement.
• 82.9707 — dropped that same factor in the other direction.
• 182.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)

fixed-film equation without recycle applied to a single-stage filter Given treatability constant (k) = 0.0470; filter depth (D) = 2.5000 m; hydraulic loading (Q) = 40.0000 m^3/m^2/day; depth exponent (n) = 0.8700; flow exponent (m) = 0.6400; effluent BOD (S_e) = 87.5000 mg/L, determine the influent BOD (S_0) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation without Recycle.
• Everything except S_0 is given, so isolate S_0 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_0:

   $$S_{0} = \dfrac{S_e}{e^{-kD^n/Q^m}}$$

3. Step 3 — List the givens: treatability constant (k) = 0.0470, filter depth (D) = 2.5000 m, hydraulic loading (Q) = 40.0000 m^3/m^2/day, depth exponent (n) = 0.8700, flow exponent (m) = 0.6400, effluent BOD (S_e) = 87.5000 mg/L.

4. Step 4 — Substitute the given values:

   $$S_{0} = \dfrac{S_e}{e^{-k2.5000^0.8700/40.0000^0.6400}}$$

5. Step 5 — Evaluate:

   $$S_{0} = 88.3652\ \text{mg/L}$$

6. Step 6 — Check: returning S_0 = 88.3652 mg/L to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 176.7 — kept a factor of two that cancels in the correct rearrangement.
• 44.1826 — dropped that same factor in the other direction.
• 97.2018 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)

trickling filter fixed-film equation without recycle design check Given influent BOD (S_0) = 275.0 mg/L; filter depth (D) = 2.4000 m; hydraulic loading (Q) = 33.5000 m^3/m^2/day; depth exponent (n) = 0.5800; flow exponent (m) = 0.5100; effluent BOD (S_e) = 59.0000 mg/L, determine the treatability constant (k).

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation without 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for k:

   $$k = -\ln\left(\dfrac{S_e}{S_0}\right)\dfrac{Q^m}{D^n}$$

3. Step 3 — List the givens: influent BOD (S_0) = 275.0 mg/L, filter depth (D) = 2.4000 m, hydraulic loading (Q) = 33.5000 m^3/m^2/day, depth exponent (n) = 0.5800, flow exponent (m) = 0.5100, effluent BOD (S_e) = 59.0000 mg/L.

4. Step 4 — Substitute the given values:

   $$k = -\l0.5800\left(\dfrac{S_e}{S_0}\right)\dfrac{33.5000^0.5100}{2.4000^0.5800}$$

5. Step 5 — Evaluate:

   $$k = 5.5533$$

6. Step 6 — Check: returning k = 5.5533 to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 11.1067 — kept a factor of two that cancels in the correct rearrangement.
• 2.7767 — dropped that same factor in the other direction.
• 6.1087 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)

fixed-film trickling filter equation without recycle for effluent BOD Given influent BOD (S_0) = 362.0 mg/L; treatability constant (k) = 0.0920; filter depth (D) = 3.0000 m; hydraulic loading (Q) = 2.0000 m^3/m^2/day; depth exponent (n) = 0.5100; flow exponent (m) = 0.5300, 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 without 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_e:

   $$S_{e} = S_0 e^{-kD^n/Q^m}$$

3. Step 3 — List the givens: influent BOD (S_0) = 362.0 mg/L, treatability constant (k) = 0.0920, filter depth (D) = 3.0000 m, hydraulic loading (Q) = 2.0000 m^3/m^2/day, depth exponent (n) = 0.5100, flow exponent (m) = 0.5300.

4. Step 4 — Substitute the given values:

   $$S_{e} = S_0 e^{-k3.0000^0.5100/2.0000^0.5300}$$

5. Step 5 — Evaluate:

   $$S_{e} = 323.8\ \text{mg/L}$$

6. Step 6 — Check: returning S_e = 323.8 mg/L to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

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

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)

fixed-film equation without recycle applied to a single-stage filter Given treatability constant (k) = 0.0510; filter depth (D) = 1.5000 m; hydraulic loading (Q) = 31.5000 m^3/m^2/day; depth exponent (n) = 0.6200; flow exponent (m) = 0.6700; effluent BOD (S_e) = 230.0 mg/L, determine the influent BOD (S_0) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation without Recycle.
• Everything except S_0 is given, so isolate S_0 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_0:

   $$S_{0} = \dfrac{S_e}{e^{-kD^n/Q^m}}$$

3. Step 3 — List the givens: treatability constant (k) = 0.0510, filter depth (D) = 1.5000 m, hydraulic loading (Q) = 31.5000 m^3/m^2/day, depth exponent (n) = 0.6200, flow exponent (m) = 0.6700, effluent BOD (S_e) = 230.0 mg/L.

4. Step 4 — Substitute the given values:

   $$S_{0} = \dfrac{S_e}{e^{-k1.5000^0.6200/31.5000^0.6700}}$$

5. Step 5 — Evaluate:

   $$S_{0} = 231.5\ \text{mg/L}$$

6. Step 6 — Check: returning S_0 = 231.5 mg/L to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

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

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)

trickling filter fixed-film equation without recycle design check Given influent BOD (S_0) = 234.0 mg/L; filter depth (D) = 2.2000 m; hydraulic loading (Q) = 15.5000 m^3/m^2/day; depth exponent (n) = 0.5900; flow exponent (m) = 0.4300; effluent BOD (S_e) = 38.0000 mg/L, determine the treatability constant (k).

Start with the thinking

• The governing relation printed in this handbook section is Fixed-Film Equation without 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for k:

   $$k = -\ln\left(\dfrac{S_e}{S_0}\right)\dfrac{Q^m}{D^n}$$

3. Step 3 — List the givens: influent BOD (S_0) = 234.0 mg/L, filter depth (D) = 2.2000 m, hydraulic loading (Q) = 15.5000 m^3/m^2/day, depth exponent (n) = 0.5900, flow exponent (m) = 0.4300, effluent BOD (S_e) = 38.0000 mg/L.

4. Step 4 — Substitute the given values:

   $$k = -\l0.5900\left(\dfrac{S_e}{S_0}\right)\dfrac{15.5000^0.4300}{2.2000^0.5900}$$

5. Step 5 — Evaluate:

   $$k = 3.7097$$

6. Step 6 — Check: returning k = 3.7097 to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

• 7.4195 — kept a factor of two that cancels in the correct rearrangement.
• 1.8549 — dropped that same factor in the other direction.
• 4.0807 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)

fixed-film trickling filter equation without recycle for effluent BOD Given influent BOD (S_0) = 233.0 mg/L; treatability constant (k) = 0.0710; filter depth (D) = 1.5000 m; hydraulic loading (Q) = 8.0000 m^3/m^2/day; depth exponent (n) = 0.6100; flow exponent (m) = 0.4300, 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 without 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 without recycle predicts trickling filter effluent BOD as a function of depth and hydraulic loading.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

2. Step 2 — Rearrange symbolically for S_e:

   $$S_{e} = S_0 e^{-kD^n/Q^m}$$

3. Step 3 — List the givens: influent BOD (S_0) = 233.0 mg/L, treatability constant (k) = 0.0710, filter depth (D) = 1.5000 m, hydraulic loading (Q) = 8.0000 m^3/m^2/day, depth exponent (n) = 0.6100, flow exponent (m) = 0.4300.

4. Step 4 — Substitute the given values:

   $$S_{e} = S_0 e^{-k1.5000^0.6100/8.0000^0.4300}$$

5. Step 5 — Evaluate:

   $$S_{e} = 224.5\ \text{mg/L}$$

6. Step 6 — Check: returning S_e = 224.5 mg/L to

   $$\dfrac{S_e}{S_0} = e^{-k D^n / Q^m}$$

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

Why the other options are there

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

Reference: FE Handbook — Fixed-Film Equation without Recycle (NRC formula)