A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 62.2000 ft/day; hydraulic gradient (i) = 0.0400 ft/ft; cross-sectional area (A) = 249.0 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) = 62.2000 ft/day, hydraulic gradient (i) = 0.0400 ft/ft, cross-sectional area (A) = 249.0 ft².

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

5. Step 5 — Evaluate:

   $$Q = 619.5\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 619.5 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

• 1,239 — kept a factor of two that cancels in the correct rearrangement.
• 309.8 — dropped that same factor in the other direction.
• 681.5 — rounded an intermediate value before the final step.

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

a municipal supply well in a sand aquifer Given hydraulic conductivity (K) = 0.0004 m/s; head at the far observation well (h_2) = 16.0000 m; head at the near observation well (h_1) = 11.0000 m; ln(r_2/r_1) (L) = 3.6000, determine the well discharge (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's formula for an unconfined aquifer well.
• 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.
• A fully penetrating well pumps steadily from an unconfined aquifer with two observation wells.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{L}$$

3. Step 3 — List the givens: hydraulic conductivity (K) = 0.0004 m/s, head at the far observation well (h_2) = 16.0000 m, head at the near observation well (h_1) = 11.0000 m, ln(r_2/r_1) (L) = 3.6000.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{\pi 0.0004 (h_2^2 - h_1^2)}{3.6000}$$

5. Step 5 — Evaluate:

   Q = 0.0424\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.0424 m^3/s to

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

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

Why the other options are there

• 0.0848 — kept a factor of two that cancels in the correct rearrangement.
• 0.0212 — dropped that same factor in the other direction.
• 0.0467 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0870 ft/ft; cross-sectional area (A) = 395.0 ft²; flow rate (Q) = 4,628 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) = 395.0 ft², flow rate (Q) = 4,628 ft³/day.

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

5. Step 5 — Evaluate:

   $$K = 134.7\ \text{ft/day}$$

6. Step 6 — Check: returning K = 134.7 ft/day to

   $$Q = K i A$$

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

Why the other options are there

• 269.3 — kept a factor of two that cancels in the correct rearrangement.
• 67.3345 — dropped that same factor in the other direction.
• 148.1 — rounded an intermediate value before the final step.

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

a test well pumped during an aquifer performance test Given well discharge (Q) = 0.0700 m^3/s; head at the far observation well (h_2) = 17.6000 m; head at the near observation well (h_1) = 12.7000 m; ln(r_2/r_1) (L) = 3.1000, determine the hydraulic conductivity (K) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's formula for an unconfined aquifer well.
• 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.
• A fully penetrating well pumps steadily from an unconfined aquifer with two observation wells.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

2. Step 2 — Rearrange symbolically for K:

   $$K = \dfrac{Q L}{\pi (h_2^2 - h_1^2)}$$

3. Step 3 — List the givens: well discharge (Q) = 0.0700 m^3/s, head at the far observation well (h_2) = 17.6000 m, head at the near observation well (h_1) = 12.7000 m, ln(r_2/r_1) (L) = 3.1000.

4. Step 4 — Substitute the given values:

   $$K = \dfrac{0.0700 3.1000}{\pi (h_2^2 - h_1^2)}$$

5. Step 5 — Evaluate:

   $$K = 0.0005\ \text{m/s}$$

6. Step 6 — Check: returning K = 0.0005 m/s to

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

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

Why the other options are there

• 0.0009 — kept a factor of two that cancels in the correct rearrangement.
• 0.0002 — dropped that same factor in the other direction.
• 0.0005 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 82.2000 ft/day; hydraulic gradient (i) = 0.0670 ft/ft; flow rate (Q) = 87.3000 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) = 82.2000 ft/day, hydraulic gradient (i) = 0.0670 ft/ft, flow rate (Q) = 87.3000 ft³/day.

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

5. Step 5 — Evaluate:

   $$A = 15.8514\ \text{ft²}$$

6. Step 6 — Check: returning A = 15.8514 ft² to

   $$Q = K i A$$

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

Why the other options are there

• 31.7028 — kept a factor of two that cancels in the correct rearrangement.
• 7.9257 — dropped that same factor in the other direction.
• 17.4365 — rounded an intermediate value before the final step.

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

a dewatering well on an excavation site Given hydraulic conductivity (K) = 0.0005 m/s; head at the far observation well (h_2) = 29.1000 m; head at the near observation well (h_1) = 8.8000 m; ln(r_2/r_1) (L) = 3.9000, determine the well discharge (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's formula for an unconfined aquifer well.
• 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.
• A fully penetrating well pumps steadily from an unconfined aquifer with two observation wells.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{L}$$

3. Step 3 — List the givens: hydraulic conductivity (K) = 0.0005 m/s, head at the far observation well (h_2) = 29.1000 m, head at the near observation well (h_1) = 8.8000 m, ln(r_2/r_1) (L) = 3.9000.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{\pi 0.0005 (h_2^2 - h_1^2)}{3.9000}$$

5. Step 5 — Evaluate:

   Q = 0.3285\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.3285 m^3/s to

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

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

Why the other options are there

• 0.6569 — kept a factor of two that cancels in the correct rearrangement.
• 0.1642 — dropped that same factor in the other direction.
• 0.3613 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 21.4000 ft/day; hydraulic gradient (i) = 0.0310 ft/ft; cross-sectional area (A) = 170.0 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) = 21.4000 ft/day, hydraulic gradient (i) = 0.0310 ft/ft, cross-sectional area (A) = 170.0 ft².

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

5. Step 5 — Evaluate:

   $$Q = 112.8\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 112.8 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

• 225.6 — kept a factor of two that cancels in the correct rearrangement.
• 56.3890 — dropped that same factor in the other direction.
• 124.1 — rounded an intermediate value before the final step.

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

a municipal supply well in a sand aquifer Given well discharge (Q) = 0.0100 m^3/s; head at the far observation well (h_2) = 24.4000 m; head at the near observation well (h_1) = 11.4000 m; ln(r_2/r_1) (L) = 3.1000, determine the hydraulic conductivity (K) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's formula for an unconfined aquifer well.
• 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.
• A fully penetrating well pumps steadily from an unconfined aquifer with two observation wells.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

2. Step 2 — Rearrange symbolically for K:

   $$K = \dfrac{Q L}{\pi (h_2^2 - h_1^2)}$$

3. Step 3 — List the givens: well discharge (Q) = 0.0100 m^3/s, head at the far observation well (h_2) = 24.4000 m, head at the near observation well (h_1) = 11.4000 m, ln(r_2/r_1) (L) = 3.1000.

4. Step 4 — Substitute the given values:

   $$K = \dfrac{0.0100 3.1000}{\pi (h_2^2 - h_1^2)}$$

5. Step 5 — Evaluate:

   $$K = 0.0000\ \text{m/s}$$

6. Step 6 — Check: returning K = 0.0000 m/s to

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

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

Why the other options are there

• 0.0000 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0770 ft/ft; cross-sectional area (A) = 120.0 ft²; flow rate (Q) = 2,942 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) = 0.0770 ft/ft, cross-sectional area (A) = 120.0 ft², flow rate (Q) = 2,942 ft³/day.

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

5. Step 5 — Evaluate:

   $$K = 318.4\ \text{ft/day}$$

6. Step 6 — Check: returning K = 318.4 ft/day to

   $$Q = K i A$$

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

Why the other options are there

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

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

a test well pumped during an aquifer performance test Given hydraulic conductivity (K) = 0.0006 m/s; head at the far observation well (h_2) = 19.9000 m; head at the near observation well (h_1) = 8.2000 m; ln(r_2/r_1) (L) = 4.1000, determine the well discharge (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's formula for an unconfined aquifer well.
• 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.
• A fully penetrating well pumps steadily from an unconfined aquifer with two observation wells.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{L}$$

3. Step 3 — List the givens: hydraulic conductivity (K) = 0.0006 m/s, head at the far observation well (h_2) = 19.9000 m, head at the near observation well (h_1) = 8.2000 m, ln(r_2/r_1) (L) = 4.1000.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{\pi 0.0006 (h_2^2 - h_1^2)}{4.1000}$$

5. Step 5 — Evaluate:

   Q = 0.1411\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.1411 m^3/s to

   $$Q = \dfrac{\pi K (h_2^2 - h_1^2)}{\ln(r_2 / r_1)}$$

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

Why the other options are there

• 0.2821 — kept a factor of two that cancels in the correct rearrangement.
• 0.0705 — dropped that same factor in the other direction.
• 0.1552 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula