A reservoir has a growing-season total phosphorus of 16 µg/L, chlorophyll-a of 3.5 µg/L and a Secchi depth of 7.7 m. Its volume is 15.5 × 10⁶ m³ with an inflow of 10.5 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (16) + 4.15 = 44.1$$

3. Classification

   $$TP = 16 µg/L \to mesotrophic (TSI 40-50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 15.5\times10^{6}/(10.5 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 3.5 µg/L and Secchi 7.7 m are consistent with a clear, low-productivity system.

Why the other options are there

• 21.5 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification

A reservoir has a growing-season total phosphorus of 24 µg/L, chlorophyll-a of 37.5 µg/L and a Secchi depth of 0.7 m. Its volume is 30.0 × 10⁶ m³ with an inflow of 8.5 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (24) + 4.15 = 50.0$$

3. Classification

   $$TP = 24 µg/L \to mesotrophic (TSI 40-50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 30.0\times10^{6}/(8.5 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 37.5 µg/L and Secchi 0.7 m are consistent with a productive system.

Why the other options are there

• 24.1 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification

A reservoir has a growing-season total phosphorus of 72 µg/L, chlorophyll-a of 21.0 µg/L and a Secchi depth of 5.3 m. Its volume is 11.0 × 10⁶ m³ with an inflow of 7.5 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (72) + 4.15 = 65.8$$

3. Classification

   $$TP = 72 µg/L \to eutrophic (TSI > 50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 11.0\times10^{6}/(7.5 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 21.0 µg/L and Secchi 5.3 m are consistent with a productive system.

Why the other options are there

• 30.9 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification

A reservoir has a growing-season total phosphorus of 15 µg/L, chlorophyll-a of 13.5 µg/L and a Secchi depth of 4.8 m. Its volume is 9.5 × 10⁶ m³ with an inflow of 4.0 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (15) + 4.15 = 43.2$$

3. Classification

   $$TP = 15 µg/L \to mesotrophic (TSI 40-50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 9.5\times10^{6}/(4.0 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 13.5 µg/L and Secchi 4.8 m are consistent with a productive system.

Why the other options are there

• 21.1 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification

A reservoir has a growing-season total phosphorus of 78 µg/L, chlorophyll-a of 35.0 µg/L and a Secchi depth of 6.2 m. Its volume is 16.5 × 10⁶ m³ with an inflow of 6.0 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (78) + 4.15 = 67.0$$

3. Classification

   $$TP = 78 µg/L \to eutrophic (TSI > 50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 16.5\times10^{6}/(6.0 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 35.0 µg/L and Secchi 6.2 m are consistent with a productive system.

Why the other options are there

• 31.4 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification

A reservoir has a growing-season total phosphorus of 71 µg/L, chlorophyll-a of 27.0 µg/L and a Secchi depth of 2.3 m. Its volume is 10.5 × 10⁶ m³ with an inflow of 3.0 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (71) + 4.15 = 65.6$$

3. Classification

   $$TP = 71 µg/L \to eutrophic (TSI > 50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 10.5\times10^{6}/(3.0 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 27.0 µg/L and Secchi 2.3 m are consistent with a productive system.

Why the other options are there

• 30.8 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification

A reservoir has a growing-season total phosphorus of 13 µg/L, chlorophyll-a of 40.0 µg/L and a Secchi depth of 2.8 m. Its volume is 18.0 × 10⁶ m³ with an inflow of 10.0 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (13) + 4.15 = 41.1$$

3. Classification

   $$TP = 13 µg/L \to mesotrophic (TSI 40-50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 18.0\times10^{6}/(10.0 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 40.0 µg/L and Secchi 2.8 m are consistent with a productive system.

Why the other options are there

• 20.2 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification

A reservoir has a growing-season total phosphorus of 35 µg/L, chlorophyll-a of 41.0 µg/L and a Secchi depth of 4.4 m. Its volume is 14.0 × 10⁶ m³ with an inflow of 8.5 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (35) + 4.15 = 55.4$$

3. Classification

   $$TP = 35 µg/L \to eutrophic (TSI > 50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 14.0\times10^{6}/(8.5 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 41.0 µg/L and Secchi 4.4 m are consistent with a productive system.

Why the other options are there

• 26.4 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification

A reservoir has a growing-season total phosphorus of 16 µg/L, chlorophyll-a of 40.0 µg/L and a Secchi depth of 6.7 m. Its volume is 21.0 × 10⁶ m³ with an inflow of 10.5 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (16) + 4.15 = 44.1$$

3. Classification

   $$TP = 16 µg/L \to mesotrophic (TSI 40-50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 21.0\times10^{6}/(10.5 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 40.0 µg/L and Secchi 6.7 m are consistent with a productive system.

Why the other options are there

• 21.5 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification

A reservoir has a growing-season total phosphorus of 30 µg/L, chlorophyll-a of 34.5 µg/L and a Secchi depth of 7.7 m. Its volume is 15.0 × 10⁶ m³ with an inflow of 1.0 m³/s. Compute the Carlson trophic state index from phosphorus, classify the lake, and find the hydraulic residence time.

Start with the thinking

• Phosphorus is normally the limiting nutrient in fresh water, so it drives classification.
• Residence time controls how long nutrients stay available for algal growth.

Step-by-step solution

1. Formula

   $$TSI(TP) = 14.42\ln(TP) + 4.15$$

2. Substituting

   $$TSI = 14.42 \ln (30) + 4.15 = 53.2$$

3. Classification

   $$TP = 30 µg/L \to eutrophic (TSI > 50)$$

4. Formula

   $$\tau = \dfrac{V}{Q}$$

5. Substituting

   $$\tau = 15.0\times10^{6}/(1.0 \times 86 400) = 0.0 days$$

6. Cross-check — chlorophyll-a of 34.5 µg/L and Secchi 7.7 m are consistent with a productive system.

Why the other options are there

• 25.5 (log base 10 instead of natural log)
• τ = 0.0 days (seconds-to-days conversion applied twice)

Reference: FE Reference Handbook — Hydrology and Water Resources → Lake Classification