A reservoir with a 80 acre surface receives 97 ac-ft of stream inflow in a month, plus 5.5 in of direct rainfall, while losing 2.0 in to evaporation and releasing 82 ac-ft to demand. Find the change in storage.

Start with the thinking

• Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
• The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

1. Rain volume

   $$V_P = (P/12)A_s$$

2. Substituting

   $$V_P = (5.5/12) \times 80 = 36.67 ac-ft$$

3. Evaporation volume

   $$V_E = (E/12)A_s$$

4. Substituting

   $$V_E = (2.0/12) \times 80 = 13.33 ac-ft$$

5. Balance — ΔS = I + V_P − V_E − D

6. Substituting — ΔS = 97 + 36.67 − 13.33 − 82

7. Evaluate — ΔS = 38.33 ac-ft (storage gained)

Why the other options are there

• 15.0 ac-ft (rain and evaporation ignored)
• 18.5 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

A reservoir with a surface area of 847 ha receives 181 mm of rainfall and loses 167 mm to evaporation in a 30-day month while a stream delivers 0.05 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

   $$0.05(86,400)(30) = 129,600 m^{3}$$

3. Rainfall

   $$(181/1000)(8,470,000) = 1,533,070 m^{3}$$

4. Evaporation

   $$(167/1000)(8,470,000) = 1,414,490 m^{3}$$

5. Substituting — ΔS = 129,600 + 1,533,070 − 1,414,490 = 248,180 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 248,180/8,470,000 = 0.029 m

Why the other options are there

• 118,580 m³ (stream inflow ignored)
• 14 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

A reservoir with a 330.0 acre surface receives 85 ac-ft of stream inflow in a month, plus 3.0 in of direct rainfall, while losing 4.0 in to evaporation and releasing 65 ac-ft to demand. Find the change in storage.

Start with the thinking

• Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
• The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

1. Rain volume

   $$V_P = (P/12)A_s$$

2. Substituting

   $$V_P = (3.0/12) \times 330.0 = 82.50 ac-ft$$

3. Evaporation volume

   $$V_E = (E/12)A_s$$

4. Substituting

   $$V_E = (4.0/12) \times 330.0 = 110.0 ac-ft$$

5. Balance — ΔS = I + V_P − V_E − D

6. Substituting — ΔS = 85 + 82.50 − 110.0 − 65

7. Evaluate — ΔS = -7.50 ac-ft (storage drawn down)

Why the other options are there

• 20.0 ac-ft (rain and evaporation ignored)
• 19.0 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

A reservoir with a surface area of 260 ha receives 102 mm of rainfall and loses 96 mm to evaporation in a 30-day month while a stream delivers 0.15 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

   $$0.15(86,400)(30) = 388,800 m^{3}$$

3. Rainfall

   $$(102/1000)(2,600,000) = 265,200 m^{3}$$

4. Evaporation

   $$(96/1000)(2,600,000) = 249,600 m^{3}$$

5. Substituting — ΔS = 388,800 + 265,200 − 249,600 = 404,400 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 404,400/2,600,000 = 0.156 m

Why the other options are there

• 15,600 m³ (stream inflow ignored)
• 6 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

A reservoir with a 220.0 acre surface receives 44 ac-ft of stream inflow in a month, plus 6.0 in of direct rainfall, while losing 7.0 in to evaporation and releasing 17 ac-ft to demand. Find the change in storage.

Start with the thinking

• Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
• The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

1. Rain volume

   $$V_P = (P/12)A_s$$

2. Substituting

   $$V_P = (6.0/12) \times 220.0 = 110.0 ac-ft$$

3. Evaporation volume

   $$V_E = (E/12)A_s$$

4. Substituting

   $$V_E = (7.0/12) \times 220.0 = 128.3 ac-ft$$

5. Balance — ΔS = I + V_P − V_E − D

6. Substituting — ΔS = 44 + 110.0 − 128.3 − 17

7. Evaluate — ΔS = 8.67 ac-ft (storage gained)

Why the other options are there

• 27.0 ac-ft (rain and evaporation ignored)
• 26.0 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

A reservoir with a surface area of 440 ha receives 98 mm of rainfall and loses 93 mm to evaporation in a 30-day month while a stream delivers 0.50 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

   $$0.50(86,400)(30) = 1,296,000 m^{3}$$

3. Rainfall

   $$(98/1000)(4,400,000) = 431,200 m^{3}$$

4. Evaporation

   $$(93/1000)(4,400,000) = 409,200 m^{3}$$

5. Substituting — ΔS = 1,296,000 + 431,200 − 409,200 = 1,318,000 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 1,318,000/4,400,000 = 0.300 m

Why the other options are there

• 22,000 m³ (stream inflow ignored)
• 5 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

A reservoir with a 40 acre surface receives 55 ac-ft of stream inflow in a month, plus 4.0 in of direct rainfall, while losing 4.0 in to evaporation and releasing 51 ac-ft to demand. Find the change in storage.

Start with the thinking

• Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
• The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

1. Rain volume

   $$V_P = (P/12)A_s$$

2. Substituting

   $$V_P = (4.0/12) \times 40 = 13.33 ac-ft$$

3. Evaporation volume

   $$V_E = (E/12)A_s$$

4. Substituting

   $$V_E = (4.0/12) \times 40 = 13.33 ac-ft$$

5. Balance — ΔS = I + V_P − V_E − D

6. Substituting — ΔS = 55 + 13.33 − 13.33 − 51

7. Evaluate — ΔS = 4.00 ac-ft (storage gained)

Why the other options are there

• 4.0 ac-ft (rain and evaporation ignored)
• 4.0 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

A reservoir with a surface area of 149 ha receives 130 mm of rainfall and loses 100 mm to evaporation in a 30-day month while a stream delivers 0.85 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

   $$0.85(86,400)(30) = 2,203,200 m^{3}$$

3. Rainfall

   $$(130/1000)(1,490,000) = 193,700 m^{3}$$

4. Evaporation

   $$(100/1000)(1,490,000) = 149,000 m^{3}$$

5. Substituting — ΔS = 2,203,200 + 193,700 − 149,000 = 2,247,900 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 2,247,900/1,490,000 = 1.509 m

Why the other options are there

• 44,700 m³ (stream inflow ignored)
• 30 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

A reservoir with a 140.0 acre surface receives 111.0 ac-ft of stream inflow in a month, plus 5.0 in of direct rainfall, while losing 5.0 in to evaporation and releasing 23 ac-ft to demand. Find the change in storage.

Start with the thinking

• Depths over the water surface convert to volume with V = (depth in inches / 12) × area in acres.
• The continuity budget is ΔS = inflows − outflows.

Step-by-step solution

1. Rain volume

   $$V_P = (P/12)A_s$$

2. Substituting

   $$V_P = (5.0/12) \times 140.0 = 58.33 ac-ft$$

3. Evaporation volume

   $$V_E = (E/12)A_s$$

4. Substituting

   $$V_E = (5.0/12) \times 140.0 = 58.33 ac-ft$$

5. Balance — ΔS = I + V_P − V_E − D

6. Substituting — ΔS = 111.0 + 58.33 − 58.33 − 23

7. Evaluate — ΔS = 88.00 ac-ft (storage gained)

Why the other options are there

• 88.0 ac-ft (rain and evaporation ignored)
• 88.0 ac-ft (inches added as volumes)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation

A reservoir with a surface area of 208 ha receives 38 mm of rainfall and loses 161 mm to evaporation in a 30-day month while a stream delivers 1.05 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

   $$1.05(86,400)(30) = 2,721,600 m^{3}$$

3. Rainfall

   $$(38/1000)(2,080,000) = 79,040 m^{3}$$

4. Evaporation

   $$(161/1000)(2,080,000) = 334,880 m^{3}$$

5. Substituting — ΔS = 2,721,600 + 79,040 − 334,880 = 2,465,760 m³

6. Formula — Δz = ΔS/A

7. Substituting — Δz = 2,465,760/2,080,000 = 1.185 m

Why the other options are there

• -255,840 m³ (stream inflow ignored)
• -123.0 m³ (millimetres treated as volume)

Reference: FE Reference Handbook — Hydrology and Water Resources → Pan Evaporation