A reservoir with a surface area of 110 ha receives 31 mm of rainfall and loses 158 mm to evaporation in a 30-day month while a stream delivers 1.45 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$1.45(86,400)(30) = 3,758,400 m^{3}$$

3. Rainfall

   $$(31/1000)(1,100,000) = 34,100 m^{3}$$

4. Evaporation

   $$(158/1000)(1,100,000) = 173,800 m^{3}$$

5. Substituting — ΔS = 3,758,400 + 34,100 − 173,800 = 3,618,700 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 3,618,700/1,100,000 = 3.290 m

Why the other options are there

• -139,700 m³ (stream inflow ignored)
• -127.0 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

A reservoir with a surface area of 256 ha receives 30 mm of rainfall and loses 57 mm to evaporation in a 30-day month while a stream delivers 0.80 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$0.80(86,400)(30) = 2,073,600 m^{3}$$

3. Rainfall

   $$(30/1000)(2,560,000) = 76,800 m^{3}$$

4. Evaporation

   $$(57/1000)(2,560,000) = 145,920 m^{3}$$

5. Substituting — ΔS = 2,073,600 + 76,800 − 145,920 = 2,004,480 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 2,004,480/2,560,000 = 0.783 m

Why the other options are there

• -69,120 m³ (stream inflow ignored)
• -27 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

A reservoir with a surface area of 601 ha receives 53 mm of rainfall and loses 107 mm to evaporation in a 30-day month while a stream delivers 0.95 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$0.95(86,400)(30) = 2,462,400 m^{3}$$

3. Rainfall

   $$(53/1000)(6,010,000) = 318,530 m^{3}$$

4. Evaporation

   $$(107/1000)(6,010,000) = 643,070 m^{3}$$

5. Substituting — ΔS = 2,462,400 + 318,530 − 643,070 = 2,137,860 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 2,137,860/6,010,000 = 0.356 m

Why the other options are there

• -324,540 m³ (stream inflow ignored)
• -54 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

A reservoir with a surface area of 643 ha receives 156 mm of rainfall and loses 94 mm to evaporation in a 30-day month while a stream delivers 0.30 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$0.30(86,400)(30) = 777,600 m^{3}$$

3. Rainfall

   $$(156/1000)(6,430,000) = 1,003,080 m^{3}$$

4. Evaporation

   $$(94/1000)(6,430,000) = 604,420 m^{3}$$

5. Substituting — ΔS = 777,600 + 1,003,080 − 604,420 = 1,176,260 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 1,176,260/6,430,000 = 0.183 m

Why the other options are there

• 398,660 m³ (stream inflow ignored)
• 62 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

A reservoir with a surface area of 832 ha receives 153 mm of rainfall and loses 177 mm to evaporation in a 30-day month while a stream delivers 0.20 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$0.20(86,400)(30) = 518,400 m^{3}$$

3. Rainfall

   $$(153/1000)(8,320,000) = 1,272,960 m^{3}$$

4. Evaporation

   $$(177/1000)(8,320,000) = 1,472,640 m^{3}$$

5. Substituting — ΔS = 518,400 + 1,272,960 − 1,472,640 = 318,720 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 318,720/8,320,000 = 0.038 m

Why the other options are there

• -199,680 m³ (stream inflow ignored)
• -24 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

A reservoir with a surface area of 895 ha receives 163 mm of rainfall and loses 135 mm to evaporation in a 30-day month while a stream delivers 1.50 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$1.50(86,400)(30) = 3,888,000 m^{3}$$

3. Rainfall

   $$(163/1000)(8,950,000) = 1,458,850 m^{3}$$

4. Evaporation

   $$(135/1000)(8,950,000) = 1,208,250 m^{3}$$

5. Substituting — ΔS = 3,888,000 + 1,458,850 − 1,208,250 = 4,138,600 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 4,138,600/8,950,000 = 0.462 m

Why the other options are there

• 250,600 m³ (stream inflow ignored)
• 28 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

A reservoir with a surface area of 350 ha receives 120 mm of rainfall and loses 140 mm to evaporation in a 30-day month while a stream delivers 0.45 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$0.45(86,400)(30) = 1,166,400 m^{3}$$

3. Rainfall

   $$(120/1000)(3,500,000) = 420,000 m^{3}$$

4. Evaporation

   $$(140/1000)(3,500,000) = 490,000 m^{3}$$

5. Substituting — ΔS = 1,166,400 + 420,000 − 490,000 = 1,096,400 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 1,096,400/3,500,000 = 0.313 m

Why the other options are there

• -70,000 m³ (stream inflow ignored)
• -20 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

A reservoir with a surface area of 164 ha receives 175 mm of rainfall and loses 88 mm to evaporation in a 30-day month while a stream delivers 1.40 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$1.40(86,400)(30) = 3,628,800 m^{3}$$

3. Rainfall

   $$(175/1000)(1,640,000) = 287,000 m^{3}$$

4. Evaporation

   $$(88/1000)(1,640,000) = 144,320 m^{3}$$

5. Substituting — ΔS = 3,628,800 + 287,000 − 144,320 = 3,771,480 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 3,771,480/1,640,000 = 2.300 m

Why the other options are there

• 142,680 m³ (stream inflow ignored)
• 87 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

A reservoir with a surface area of 824 ha receives 30 mm of rainfall and loses 75 mm to evaporation in a 30-day month while a stream delivers 1.40 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$1.40(86,400)(30) = 3,628,800 m^{3}$$

3. Rainfall

   $$(30/1000)(8,240,000) = 247,200 m^{3}$$

4. Evaporation

   $$(75/1000)(8,240,000) = 618,000 m^{3}$$

5. Substituting — ΔS = 3,628,800 + 247,200 − 618,000 = 3,258,000 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 3,258,000/8,240,000 = 0.395 m

Why the other options are there

• -370,800 m³ (stream inflow ignored)
• -45 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses

A reservoir with a surface area of 485 ha receives 75 mm of rainfall and loses 81 mm to evaporation in a 30-day month while a stream delivers 0.65 m³/s. Compute the net storage change in volume and the corresponding change in water-surface elevation.

Start with the thinking

• The water budget is simply inflow + precipitation − evaporation − outflow = ΔS.
• Depths in millimetres must be converted to metres before multiplying by the surface area.

Step-by-step solution

1. Formula — ΔS = Q_in Δt + P·A − E·A

2. Stream inflow

   $$0.65(86,400)(30) = 1,684,800 m^{3}$$

3. Rainfall

   $$(75/1000)(4,850,000) = 363,750 m^{3}$$

4. Evaporation

   $$(81/1000)(4,850,000) = 392,850 m^{3}$$

5. Substituting — ΔS = 1,684,800 + 363,750 − 392,850 = 1,655,700 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 1,655,700/4,850,000 = 0.341 m

Why the other options are there

• -29,100 m³ (stream inflow ignored)
• -6 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Evapotranspiration Rates for Grasses