A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 15.0000 ft/day; hydraulic gradient (i) = 0.0010 ft/ft; cross-sectional area (A) = 79.0000 ft², determine the flow rate (Q) in ft³/day.

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• Everything except Q is given, so isolate Q 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 15.0000 ft/day, hydraulic gradient (i) = 0.0010 ft/ft, cross-sectional area (A) = 79.0000 ft².

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$Q = 1.1850\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 1.1850 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

• 2.3700 — kept a factor of two that cancels in the correct rearrangement.
• 0.5925 — dropped that same factor in the other direction.
• 1.3035 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0900 ft/ft; cross-sectional area (A) = 349.0 ft²; flow rate (Q) = 4,527 ft³/day, determine the hydraulic conductivity (K) in ft/day.

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that K stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic gradient (i) = 0.0900 ft/ft, cross-sectional area (A) = 349.0 ft², flow rate (Q) = 4,527 ft³/day.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$K = 144.1\ \text{ft/day}$$

6. Step 6 — Check: returning K = 144.1 ft/day to

   $$Q = K i A$$

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

Why the other options are there

• 288.2 — kept a factor of two that cancels in the correct rearrangement.
• 72.0551 — dropped that same factor in the other direction.
• 158.5 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 24.6000 ft/day; hydraulic gradient (i) = 0.0740 ft/ft; flow rate (Q) = 3,005 ft³/day, determine the cross-sectional area (A) in ft².

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• Everything except A is given, so isolate 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 24.6000 ft/day, hydraulic gradient (i) = 0.0740 ft/ft, flow rate (Q) = 3,005 ft³/day.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$A = 1651\ \text{ft²}$$

6. Step 6 — Check: returning A = 1,651 ft² to

   $$Q = K i A$$

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

Why the other options are there

• 3,302 — kept a factor of two that cancels in the correct rearrangement.
• 825.5 — dropped that same factor in the other direction.
• 1,816 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 51.3000 ft/day; hydraulic gradient (i) = 0.0810 ft/ft; cross-sectional area (A) = 67.0000 ft², determine the flow rate (Q) in ft³/day.

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• Everything except Q is given, so isolate Q 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 51.3000 ft/day, hydraulic gradient (i) = 0.0810 ft/ft, cross-sectional area (A) = 67.0000 ft².

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$Q = 278.4\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 278.4 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0340 ft/ft; cross-sectional area (A) = 74.0000 ft²; flow rate (Q) = 2,060 ft³/day, determine the hydraulic conductivity (K) in ft/day.

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that K stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic gradient (i) = 0.0340 ft/ft, cross-sectional area (A) = 74.0000 ft², flow rate (Q) = 2,060 ft³/day.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$K = 818.8\ \text{ft/day}$$

6. Step 6 — Check: returning K = 818.8 ft/day to

   $$Q = K i A$$

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

Why the other options are there

• 1,638 — kept a factor of two that cancels in the correct rearrangement.
• 409.4 — dropped that same factor in the other direction.
• 900.6 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 84.1000 ft/day; hydraulic gradient (i) = 0.0240 ft/ft; flow rate (Q) = 4,848 ft³/day, determine the cross-sectional area (A) in ft².

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• Everything except A is given, so isolate 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 84.1000 ft/day, hydraulic gradient (i) = 0.0240 ft/ft, flow rate (Q) = 4,848 ft³/day.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$A = 2402\ \text{ft²}$$

6. Step 6 — Check: returning A = 2,402 ft² to

   $$Q = K i A$$

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

Why the other options are there

• 4,804 — kept a factor of two that cancels in the correct rearrangement.
• 1,201 — dropped that same factor in the other direction.
• 2,642 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 68.6000 ft/day; hydraulic gradient (i) = 0.0630 ft/ft; cross-sectional area (A) = 95.0000 ft², determine the flow rate (Q) in ft³/day.

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• Everything except Q is given, so isolate Q 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 68.6000 ft/day, hydraulic gradient (i) = 0.0630 ft/ft, cross-sectional area (A) = 95.0000 ft².

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$Q = 410.6\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 410.6 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0890 ft/ft; cross-sectional area (A) = 349.0 ft²; flow rate (Q) = 3,147 ft³/day, determine the hydraulic conductivity (K) in ft/day.

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that K stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic gradient (i) = 0.0890 ft/ft, cross-sectional area (A) = 349.0 ft², flow rate (Q) = 3,147 ft³/day.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$K = 101.3\ \text{ft/day}$$

6. Step 6 — Check: returning K = 101.3 ft/day to

   $$Q = K i A$$

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

Why the other options are there

• 202.6 — kept a factor of two that cancels in the correct rearrangement.
• 50.6519 — dropped that same factor in the other direction.
• 111.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 40.6000 ft/day; hydraulic gradient (i) = 0.0590 ft/ft; flow rate (Q) = 911.0 ft³/day, determine the cross-sectional area (A) in ft².

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• Everything except A is given, so isolate 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that A stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 40.6000 ft/day, hydraulic gradient (i) = 0.0590 ft/ft, flow rate (Q) = 911.0 ft³/day.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$A = 380.3\ \text{ft²}$$

6. Step 6 — Check: returning A = 380.3 ft² to

   $$Q = K i A$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 52.3000 ft/day; hydraulic gradient (i) = 0.0640 ft/ft; cross-sectional area (A) = 127.0 ft², determine the flow rate (Q) in ft³/day.

Start with the thinking

• The governing relation printed in this handbook section is Darcy's law.
• Everything except Q is given, so isolate Q 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.
• Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = K i A$$

2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 52.3000 ft/day, hydraulic gradient (i) = 0.0640 ft/ft, cross-sectional area (A) = 127.0 ft².

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$Q = 425.1\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 425.1 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Hydrology and Water Resources → Surface Water System Hydrologic Budget