A permeameter has k = 0.0018 ft/s, cross-sectional area A = 13.5 ft², sample length L = 7.0 ft and head difference Δh = 9.5 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 9.5/7.0 = 1.357

2. Darcy velocity

   $$v = ki = 0.0018(1.357) = 2.443e-3 ft/s$$

3. Discharge

   $$Q = vA = 2.443e-3(13.5) = 3.298e-2 ft^{3}/s$$

4. Per day

   $$Q = 2,849 ft^{3}/day$$

Why the other options are there

• Q = 2.31e-1 ft³/s (length omitted from the gradient)
• Q = 2.44e-3 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)

A permeameter has k = 0.0022 ft/s, cross-sectional area A = 12.0 ft², sample length L = 4.0 ft and head difference Δh = 4.0 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 4.0/4.0 = 1.000

2. Darcy velocity

   $$v = ki = 0.0022(1.000) = 2.200e-3 ft/s$$

3. Discharge

   $$Q = vA = 2.200e-3(12.0) = 2.640e-2 ft^{3}/s$$

4. Per day

   $$Q = 2,281 ft^{3}/day$$

Why the other options are there

• Q = 1.06e-1 ft³/s (length omitted from the gradient)
• Q = 2.20e-3 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)

A permeameter has k = 0.0039 ft/s, cross-sectional area A = 14.0 ft², sample length L = 12.5 ft and head difference Δh = 11.5 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 11.5/12.5 = 0.920

2. Darcy velocity

   $$v = ki = 0.0039(0.920) = 3.588e-3 ft/s$$

3. Discharge

   $$Q = vA = 3.588e-3(14.0) = 5.023e-2 ft^{3}/s$$

4. Per day

   $$Q = 4,340 ft^{3}/day$$

Why the other options are there

• Q = 6.28e-1 ft³/s (length omitted from the gradient)
• Q = 3.59e-3 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)

A permeameter has k = 0.0037 ft/s, cross-sectional area A = 7.0 ft², sample length L = 13.5 ft and head difference Δh = 3.0 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 3.0/13.5 = 0.222

2. Darcy velocity

   $$v = ki = 0.0037(0.222) = 8.222e-4 ft/s$$

3. Discharge

   $$Q = vA = 8.222e-4(7.0) = 5.756e-3 ft^{3}/s$$

4. Per day

   $$Q = 497.3 ft^{3}/day$$

Why the other options are there

• Q = 7.77e-2 ft³/s (length omitted from the gradient)
• Q = 8.22e-4 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)

A permeameter has k = 0.0022 ft/s, cross-sectional area A = 5.0 ft², sample length L = 6.0 ft and head difference Δh = 11.0 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 11.0/6.0 = 1.833

2. Darcy velocity

   $$v = ki = 0.0022(1.833) = 4.033e-3 ft/s$$

3. Discharge

   $$Q = vA = 4.033e-3(5.0) = 2.017e-2 ft^{3}/s$$

4. Per day

   $$Q = 1,742 ft^{3}/day$$

Why the other options are there

• Q = 1.21e-1 ft³/s (length omitted from the gradient)
• Q = 4.03e-3 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)

A permeameter has k = 0.0033 ft/s, cross-sectional area A = 19.5 ft², sample length L = 9.0 ft and head difference Δh = 8.5 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 8.5/9.0 = 0.944

2. Darcy velocity

   $$v = ki = 0.0033(0.944) = 3.117e-3 ft/s$$

3. Discharge

   $$Q = vA = 3.117e-3(19.5) = 6.077e-2 ft^{3}/s$$

4. Per day

   $$Q = 5,251 ft^{3}/day$$

Why the other options are there

• Q = 5.47e-1 ft³/s (length omitted from the gradient)
• Q = 3.12e-3 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)

A permeameter has k = 0.0012 ft/s, cross-sectional area A = 15.5 ft², sample length L = 9.5 ft and head difference Δh = 7.5 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 7.5/9.5 = 0.789

2. Darcy velocity

   $$v = ki = 0.0012(0.789) = 9.474e-4 ft/s$$

3. Discharge

   $$Q = vA = 9.474e-4(15.5) = 1.468e-2 ft^{3}/s$$

4. Per day

   $$Q = 1,269 ft^{3}/day$$

Why the other options are there

• Q = 1.39e-1 ft³/s (length omitted from the gradient)
• Q = 9.47e-4 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)

A permeameter has k = 0.0038 ft/s, cross-sectional area A = 14.5 ft², sample length L = 19.0 ft and head difference Δh = 11.0 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 11.0/19.0 = 0.579

2. Darcy velocity

   $$v = ki = 0.0038(0.579) = 2.200e-3 ft/s$$

3. Discharge

   $$Q = vA = 2.200e-3(14.5) = 3.190e-2 ft^{3}/s$$

4. Per day

   $$Q = 2,756 ft^{3}/day$$

Why the other options are there

• Q = 6.06e-1 ft³/s (length omitted from the gradient)
• Q = 2.20e-3 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)

A permeameter has k = 0.0007 ft/s, cross-sectional area A = 15.0 ft², sample length L = 17.0 ft and head difference Δh = 11.5 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 11.5/17.0 = 0.676

2. Darcy velocity

   $$v = ki = 0.0007(0.676) = 4.735e-4 ft/s$$

3. Discharge

   $$Q = vA = 4.735e-4(15.0) = 7.103e-3 ft^{3}/s$$

4. Per day

   $$Q = 613.7 ft^{3}/day$$

Why the other options are there

• Q = 1.21e-1 ft³/s (length omitted from the gradient)
• Q = 4.74e-4 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)

A permeameter has k = 0.0003 ft/s, cross-sectional area A = 10.5 ft², sample length L = 7.5 ft and head difference Δh = 9.0 ft. Find the seepage discharge.

Start with the thinking

• Gradient is dimensionless: head lost divided by the flow path length.
• Darcy velocity is a superficial velocity, not the pore velocity.

Step-by-step solution

1. Gradient — i = Δh/L = 9.0/7.5 = 1.200

2. Darcy velocity

   $$v = ki = 0.0003(1.200) = 3.600e-4 ft/s$$

3. Discharge

   $$Q = vA = 3.600e-4(10.5) = 3.780e-3 ft^{3}/s$$

4. Per day

   $$Q = 326.6 ft^{3}/day$$

Why the other options are there

• Q = 2.83e-2 ft³/s (length omitted from the gradient)
• Q = 3.60e-4 ft³/s (area omitted)

Reference: FE Reference Handbook — Geotechnical → Hydraulic conductivity (also coefficient of permeability)