A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 17.8000 ft/day; hydraulic gradient (i) = 0.0340 ft/ft; cross-sectional area (A) = 272.0 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) = 17.8000 ft/day, hydraulic gradient (i) = 0.0340 ft/ft, cross-sectional area (A) = 272.0 ft².

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

5. Step 5 — Evaluate:

   $$Q = 164.6\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 164.6 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

• 329.2 — kept a factor of two that cancels in the correct rearrangement.
• 82.3072 — dropped that same factor in the other direction.
• 181.1 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

a municipal supply well in a sand aquifer Given hydraulic conductivity (K) = 0.0006 m/s; head at the far observation well (h_2) = 17.9000 m; head at the near observation well (h_1) = 9.8000 m; ln(r_2/r_1) (L) = 2.6000, 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) = 0.0006 m/s, head at the far observation well (h_2) = 17.9000 m, head at the near observation well (h_1) = 9.8000 m, ln(r_2/r_1) (L) = 2.6000.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{\pi 0.0006 (h_2^2 - h_1^2)}{2.6000}$$

5. Step 5 — Evaluate:

   Q = 0.1600\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.1600 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

• 0.3199 — kept a factor of two that cancels in the correct rearrangement.
• 0.0800 — dropped that same factor in the other direction.
• 0.1759 — 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) = 0.0240 ft/ft; cross-sectional area (A) = 249.0 ft²; flow rate (Q) = 3,891 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) = 0.0240 ft/ft, cross-sectional area (A) = 249.0 ft², flow rate (Q) = 3,891 ft³/day.

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

5. Step 5 — Evaluate:

   $$K = 651.1\ \text{ft/day}$$

6. Step 6 — Check: returning K = 651.1 ft/day to

   $$Q = K i A$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

a test well pumped during an aquifer performance test Given well discharge (Q) = 0.0930 m^3/s; head at the far observation well (h_2) = 18.6000 m; head at the near observation well (h_1) = 11.0000 m; ln(r_2/r_1) (L) = 2.3000, 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) = 0.0930 m^3/s, head at the far observation well (h_2) = 18.6000 m, head at the near observation well (h_1) = 11.0000 m, ln(r_2/r_1) (L) = 2.3000.

4. Step 4 — Substitute the given values:

   $$K = \dfrac{0.0930 2.3000}{\pi (h_2^2 - h_1^2)}$$

5. Step 5 — Evaluate:

   $$K = 0.0003\ \text{m/s}$$

6. Step 6 — Check: returning K = 0.0003 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

• 0.0006 — kept a factor of two that cancels in the correct rearrangement.
• 0.0002 — dropped that same factor in the other direction.
• 0.0003 — 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) = 77.1000 ft/day; hydraulic gradient (i) = 0.0120 ft/ft; flow rate (Q) = 1,544 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) = 77.1000 ft/day, hydraulic gradient (i) = 0.0120 ft/ft, flow rate (Q) = 1,544 ft³/day.

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

5. Step 5 — Evaluate:

   $$A = 1669\ \text{ft²}$$

6. Step 6 — Check: returning A = 1,669 ft² to

   $$Q = K i A$$

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

Why the other options are there

• 3,337 — kept a factor of two that cancels in the correct rearrangement.
• 834.3 — dropped that same factor in the other direction.
• 1,835 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

a dewatering well on an excavation site Given hydraulic conductivity (K) = 0.0008 m/s; head at the far observation well (h_2) = 25.6000 m; head at the near observation well (h_1) = 9.5000 m; ln(r_2/r_1) (L) = 2.7000, 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) = 0.0008 m/s, head at the far observation well (h_2) = 25.6000 m, head at the near observation well (h_1) = 9.5000 m, ln(r_2/r_1) (L) = 2.7000.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{\pi 0.0008 (h_2^2 - h_1^2)}{2.7000}$$

5. Step 5 — Evaluate:

   Q = 0.5523\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.5523 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

• 1.1047 — kept a factor of two that cancels in the correct rearrangement.
• 0.2762 — dropped that same factor in the other direction.
• 0.6076 — 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) = 41.9000 ft/day; hydraulic gradient (i) = 0.0950 ft/ft; cross-sectional area (A) = 389.0 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) = 41.9000 ft/day, hydraulic gradient (i) = 0.0950 ft/ft, cross-sectional area (A) = 389.0 ft².

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

5. Step 5 — Evaluate:

   $$Q = 1548\ \text{ft³/day}$$

6. Step 6 — Check: returning Q = 1,548 ft³/day to

   $$Q = K i A$$

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

Why the other options are there

• 3,097 — kept a factor of two that cancels in the correct rearrangement.
• 774.2 — dropped that same factor in the other direction.
• 1,703 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

a municipal supply well in a sand aquifer Given well discharge (Q) = 0.0500 m^3/s; head at the far observation well (h_2) = 15.2000 m; head at the near observation well (h_1) = 8.5000 m; ln(r_2/r_1) (L) = 2.7000, 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) = 0.0500 m^3/s, head at the far observation well (h_2) = 15.2000 m, head at the near observation well (h_1) = 8.5000 m, ln(r_2/r_1) (L) = 2.7000.

4. Step 4 — Substitute the given values:

   $$K = \dfrac{0.0500 2.7000}{\pi (h_2^2 - h_1^2)}$$

5. Step 5 — Evaluate:

   $$K = 0.0003\ \text{m/s}$$

6. Step 6 — Check: returning K = 0.0003 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

• 0.0005 — kept a factor of two that cancels in the correct rearrangement.
• 0.0001 — dropped that same factor in the other direction.
• 0.0003 — 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) = 0.0580 ft/ft; cross-sectional area (A) = 440.0 ft²; flow rate (Q) = 1,712 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) = 0.0580 ft/ft, cross-sectional area (A) = 440.0 ft², flow rate (Q) = 1,712 ft³/day.

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

5. Step 5 — Evaluate:

   $$K = 67.0768\ \text{ft/day}$$

6. Step 6 — Check: returning K = 67.0768 ft/day to

   $$Q = K i A$$

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

Why the other options are there

• 134.2 — kept a factor of two that cancels in the correct rearrangement.
• 33.5384 — dropped that same factor in the other direction.
• 73.7845 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

a test well pumped during an aquifer performance test Given hydraulic conductivity (K) = 0.0009 m/s; head at the far observation well (h_2) = 21.1000 m; head at the near observation well (h_1) = 11.7000 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) = 0.0009 m/s, head at the far observation well (h_2) = 21.1000 m, head at the near observation well (h_1) = 11.7000 m, ln(r_2/r_1) (L) = 3.9000.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{\pi 0.0009 (h_2^2 - h_1^2)}{3.9000}$$

5. Step 5 — Evaluate:

   Q = 0.2285\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.2285 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

• 0.4570 — kept a factor of two that cancels in the correct rearrangement.
• 0.1142 — dropped that same factor in the other direction.
• 0.2513 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula