A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 19.4000 ft/day; hydraulic gradient (i) = 0.0800 ft/ft; cross-sectional area (A) = 340.0 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) = 19.4000 ft/day, hydraulic gradient (i) = 0.0800 ft/ft, cross-sectional area (A) = 340.0 ft².

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

5. Step 5 — Evaluate:

   $$Q = 527.7\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 527.7 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

• 1,055 — kept a factor of two that cancels in the correct rearrangement.
• 263.8 — dropped that same factor in the other direction.
• 580.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0870 ft/ft; cross-sectional area (A) = 378.0 ft²; flow rate (Q) = 3,546 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) = 0.0870 ft/ft, cross-sectional area (A) = 378.0 ft², flow rate (Q) = 3,546 ft³/day.

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

5. Step 5 — Evaluate:

   $$K = 107.8\ \text{ft/day}$$

6. Step 6 — Check: returning K = 107.8 ft/day to

   $$Q = K i A$$

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

Why the other options are there

• 215.6 — kept a factor of two that cancels in the correct rearrangement.
• 53.9074 — dropped that same factor in the other direction.
• 118.6 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 48.6000 ft/day; hydraulic gradient (i) = 0.0520 ft/ft; flow rate (Q) = 606.8 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) = 48.6000 ft/day, hydraulic gradient (i) = 0.0520 ft/ft, flow rate (Q) = 606.8 ft³/day.

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

5. Step 5 — Evaluate:

   $$A = 240.1\ \text{ft²}$$

6. Step 6 — Check: returning A = 240.1 ft² to

   $$Q = K i A$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 6.3000 ft/day; hydraulic gradient (i) = 0.0370 ft/ft; cross-sectional area (A) = 48.0000 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) = 6.3000 ft/day, hydraulic gradient (i) = 0.0370 ft/ft, cross-sectional area (A) = 48.0000 ft².

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

5. Step 5 — Evaluate:

   $$Q = 11.1888\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 11.1888 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

• 22.3776 — kept a factor of two that cancels in the correct rearrangement.
• 5.5944 — dropped that same factor in the other direction.
• 12.3077 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0360 ft/ft; cross-sectional area (A) = 432.0 ft²; flow rate (Q) = 4,813 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) = 0.0360 ft/ft, cross-sectional area (A) = 432.0 ft², flow rate (Q) = 4,813 ft³/day.

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

5. Step 5 — Evaluate:

   $$K = 309.5\ \text{ft/day}$$

6. Step 6 — Check: returning K = 309.5 ft/day to

   $$Q = K i A$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 86.1000 ft/day; hydraulic gradient (i) = 0.0170 ft/ft; flow rate (Q) = 2,149 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) = 86.1000 ft/day, hydraulic gradient (i) = 0.0170 ft/ft, flow rate (Q) = 2,149 ft³/day.

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

5. Step 5 — Evaluate:

   $$A = 1468\ \text{ft²}$$

6. Step 6 — Check: returning A = 1,468 ft² to

   $$Q = K i A$$

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

Why the other options are there

• 2,936 — kept a factor of two that cancels in the correct rearrangement.
• 734.1 — dropped that same factor in the other direction.
• 1,615 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 54.2000 ft/day; hydraulic gradient (i) = 0.0850 ft/ft; cross-sectional area (A) = 297.0 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) = 54.2000 ft/day, hydraulic gradient (i) = 0.0850 ft/ft, cross-sectional area (A) = 297.0 ft².

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

5. Step 5 — Evaluate:

   $$Q = 1368\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 1,368 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

• 2,737 — kept a factor of two that cancels in the correct rearrangement.
• 684.1 — dropped that same factor in the other direction.
• 1,505 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0300 ft/ft; cross-sectional area (A) = 420.0 ft²; flow rate (Q) = 2,422 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) = 0.0300 ft/ft, cross-sectional area (A) = 420.0 ft², flow rate (Q) = 2,422 ft³/day.

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

5. Step 5 — Evaluate:

   $$K = 192.2\ \text{ft/day}$$

6. Step 6 — Check: returning K = 192.2 ft/day to

   $$Q = K i A$$

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

Why the other options are there

• 384.4 — kept a factor of two that cancels in the correct rearrangement.
• 96.1032 — dropped that same factor in the other direction.
• 211.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 47.7000 ft/day; hydraulic gradient (i) = 0.0060 ft/ft; flow rate (Q) = 3,908 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) = 47.7000 ft/day, hydraulic gradient (i) = 0.0060 ft/ft, flow rate (Q) = 3,908 ft³/day.

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

5. Step 5 — Evaluate:

   $$A = 13655\ \text{ft²}$$

6. Step 6 — Check: returning A = 13,655 ft² to

   $$Q = K i A$$

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

Why the other options are there

• 27,311 — kept a factor of two that cancels in the correct rearrangement.
• 6,828 — dropped that same factor in the other direction.
• 15,021 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 63.1000 ft/day; hydraulic gradient (i) = 0.0320 ft/ft; cross-sectional area (A) = 403.0 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) = 63.1000 ft/day, hydraulic gradient (i) = 0.0320 ft/ft, cross-sectional area (A) = 403.0 ft².

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

5. Step 5 — Evaluate:

   $$Q = 813.7\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 813.7 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

• 1,627 — kept a factor of two that cancels in the correct rearrangement.
• 406.9 — dropped that same factor in the other direction.
• 895.1 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined aquifer