Dupuit's formula applied to a fully penetrating well in an unconfined aquifer Given hydraulic conductivity (K) = 23.0000 m/day; head at outer radius (h_2) = 21.5000 m; head at well radius (h_1) = 8.0000 m; outer (influence) radius (r_2) = 420.0 m; well radius (r_1) = 0.6000 m, determine the well discharge (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's Formula.
• 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.
• Dupuit's formula estimates steady radial flow to a pumping well in an unconfined aquifer.

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 the relation so that Q stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 23.0000 m/day, head at outer radius (h_2) = 21.5000 m, head at well radius (h_1) = 8.0000 m, outer (influence) radius (r_2) = 420.0 m, well radius (r_1) = 0.6000 m.

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

5. Step 5 — Evaluate:

   Q = 4393\ \text{m^3/day}

6. Step 6 — Check: returning Q = 4,393 m^3/day 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

• 8,785 — kept a factor of two that cancels in the correct rearrangement.
• 2,196 — dropped that same factor in the other direction.
• 4,832 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

a municipal supply well in a sand aquifer Given hydraulic conductivity (K) = 0.0006 m/s; head at the far observation well (h_2) = 29.1000 m; head at the near observation well (h_1) = 8.0000 m; ln(r_2/r_1) (L) = 2.0000, 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) = 29.1000 m, head at the near observation well (h_1) = 8.0000 m, ln(r_2/r_1) (L) = 2.0000.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{\pi 0.0006 (h_2^2 - h_1^2)}{2.0000}$$

5. Step 5 — Evaluate:

   Q = 0.7255\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.7255 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

• 1.4510 — kept a factor of two that cancels in the correct rearrangement.
• 0.3627 — dropped that same factor in the other direction.
• 0.7980 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

Dupuit's formula used to estimate discharge from a dewatering well Given head at outer radius (h_2) = 21.5000 m; head at well radius (h_1) = 5.5000 m; outer (influence) radius (r_2) = 445.0 m; well radius (r_1) = 0.7500 m; well discharge (Q) = 135.0 m^3/day, determine the hydraulic conductivity (K) in m/day.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's Formula.
• 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.
• Dupuit's formula estimates steady radial flow to a pumping well in an unconfined aquifer.

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 the relation so that K stands alone on the left-hand side.

3. Step 3 — List the givens: head at outer radius (h_2) = 21.5000 m, head at well radius (h_1) = 5.5000 m, outer (influence) radius (r_2) = 445.0 m, well radius (r_1) = 0.7500 m, well discharge (Q) = 135.0 m^3/day.

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

5. Step 5 — Evaluate:

   $$K = 0.6352\ \text{m/day}$$

6. Step 6 — Check: returning K = 0.6352 m/day 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

• 1.2704 — kept a factor of two that cancels in the correct rearrangement.
• 0.3176 — dropped that same factor in the other direction.
• 0.6987 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

a test well pumped during an aquifer performance test Given well discharge (Q) = 0.1470 m^3/s; head at the far observation well (h_2) = 28.6000 m; head at the near observation well (h_1) = 10.8000 m; ln(r_2/r_1) (L) = 4.4000, 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) = 0.1470 m^3/s, head at the far observation well (h_2) = 28.6000 m, head at the near observation well (h_1) = 10.8000 m, ln(r_2/r_1) (L) = 4.4000.

4. Step 4 — Substitute the given values:

   $$K = \dfrac{0.1470 4.4000}{\pi (h_2^2 - h_1^2)}$$

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.
• 0.0001 — dropped that same factor in the other direction.
• 0.0003 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

Dupuit's formula for groundwater flow to a municipal water-supply well Given hydraulic conductivity (K) = 25.0000 m/day; head at outer radius (h_2) = 27.5000 m; outer (influence) radius (r_2) = 485.0 m; well radius (r_1) = 0.6500 m; well discharge (Q) = 4,920 m^3/day, determine the head at well radius (h_1) in m.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's Formula.
• Everything except h_1 is given, so isolate h_1 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.
• Dupuit's formula estimates steady radial flow to a pumping well in an unconfined aquifer.

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 the relation so that h_1 stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 25.0000 m/day, head at outer radius (h_2) = 27.5000 m, outer (influence) radius (r_2) = 485.0 m, well radius (r_1) = 0.6500 m, well discharge (Q) = 4,920 m^3/day.

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

5. Step 5 — Evaluate:

   $$h_{1} = 18.4897\ \text{m}$$

6. Step 6 — Check: returning h_1 = 18.4897 m 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

• 36.9794 — kept a factor of two that cancels in the correct rearrangement.
• 9.2448 — dropped that same factor in the other direction.
• 20.3386 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

a dewatering well on an excavation site Given hydraulic conductivity (K) = 0.0002 m/s; head at the far observation well (h_2) = 22.8000 m; head at the near observation well (h_1) = 9.8000 m; ln(r_2/r_1) (L) = 1.9000, 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) = 0.0002 m/s, head at the far observation well (h_2) = 22.8000 m, head at the near observation well (h_1) = 9.8000 m, ln(r_2/r_1) (L) = 1.9000.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{\pi 0.0002 (h_2^2 - h_1^2)}{1.9000}$$

5. Step 5 — Evaluate:

   Q = 0.1191\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.1191 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

• 0.2383 — kept a factor of two that cancels in the correct rearrangement.
• 0.0596 — dropped that same factor in the other direction.
• 0.1310 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

Dupuit's formula applied to a fully penetrating well in an unconfined aquifer Given hydraulic conductivity (K) = 15.5000 m/day; head at outer radius (h_2) = 19.5000 m; head at well radius (h_1) = 4.5000 m; outer (influence) radius (r_2) = 420.0 m; well radius (r_1) = 0.6500 m, determine the well discharge (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's Formula.
• 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.
• Dupuit's formula estimates steady radial flow to a pumping well in an unconfined aquifer.

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 the relation so that Q stands alone on the left-hand side.

3. Step 3 — List the givens: hydraulic conductivity (K) = 15.5000 m/day, head at outer radius (h_2) = 19.5000 m, head at well radius (h_1) = 4.5000 m, outer (influence) radius (r_2) = 420.0 m, well radius (r_1) = 0.6500 m.

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

5. Step 5 — Evaluate:

   Q = 2709\ \text{m^3/day}

6. Step 6 — Check: returning Q = 2,709 m^3/day 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

• 5,418 — kept a factor of two that cancels in the correct rearrangement.
• 1,355 — dropped that same factor in the other direction.
• 2,980 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

a municipal supply well in a sand aquifer Given well discharge (Q) = 0.0430 m^3/s; head at the far observation well (h_2) = 14.3000 m; head at the near observation well (h_1) = 9.8000 m; ln(r_2/r_1) (L) = 3.6000, 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) = 0.0430 m^3/s, head at the far observation well (h_2) = 14.3000 m, head at the near observation well (h_1) = 9.8000 m, ln(r_2/r_1) (L) = 3.6000.

4. Step 4 — Substitute the given values:

   $$K = \dfrac{0.0430 3.6000}{\pi (h_2^2 - h_1^2)}$$

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

Dupuit's formula used to estimate discharge from a dewatering well Given head at outer radius (h_2) = 24.5000 m; head at well radius (h_1) = 4.0000 m; outer (influence) radius (r_2) = 260.0 m; well radius (r_1) = 0.2500 m; well discharge (Q) = 1,440 m^3/day, determine the hydraulic conductivity (K) in m/day.

Start with the thinking

• The governing relation printed in this handbook section is Dupuit's Formula.
• 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.
• Dupuit's formula estimates steady radial flow to a pumping well in an unconfined aquifer.

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 the relation so that K stands alone on the left-hand side.

3. Step 3 — List the givens: head at outer radius (h_2) = 24.5000 m, head at well radius (h_1) = 4.0000 m, outer (influence) radius (r_2) = 260.0 m, well radius (r_1) = 0.2500 m, well discharge (Q) = 1,440 m^3/day.

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

5. Step 5 — Evaluate:

   $$K = 5.4502\ \text{m/day}$$

6. Step 6 — Check: returning K = 5.4502 m/day 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

• 10.9003 — kept a factor of two that cancels in the correct rearrangement.
• 2.7251 — dropped that same factor in the other direction.
• 5.9952 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula

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) = 27.4000 m; head at the near observation well (h_1) = 11.3000 m; ln(r_2/r_1) (L) = 2.1000, 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) = 0.0009 m/s, head at the far observation well (h_2) = 27.4000 m, head at the near observation well (h_1) = 11.3000 m, ln(r_2/r_1) (L) = 2.1000.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{\pi 0.0009 (h_2^2 - h_1^2)}{2.1000}$$

5. Step 5 — Evaluate:

   Q = 0.8109\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.8109 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

• 1.6219 — kept a factor of two that cancels in the correct rearrangement.
• 0.4055 — dropped that same factor in the other direction.
• 0.8920 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dupuit's Formula