A rectangular lined conduit 2.6 m wide runs 0.8 m deep on a slope of 0.0016 m/m with Manning n = 0.015. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 2.6(0.8) = 2.080 m^{2}, P = 2.6 + 2(0.8) = 4.20 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 2.080/4.20 = 0.495 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.015)(0.495)^{2/3}(0.0016)^{1/2} = 1.67 m/s$$

6. Discharge

   $$Q = AV = 2.080(1.67) = 3.472 m^{3}/s$$

Why the other options are there

• V = 2.48 m/s (US constant used in SI)
• R = 0.306 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation

A rectangular lined conduit 1.6 m wide runs 0.5 m deep on a slope of 0.0052 m/m with Manning n = 0.015. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 1.6(0.5) = 0.800 m^{2}, P = 1.6 + 2(0.5) = 2.60 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 0.800/2.60 = 0.308 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.015)(0.308)^{2/3}(0.0052)^{1/2} = 2.19 m/s$$

6. Discharge

   $$Q = AV = 0.800(2.19) = 1.753 m^{3}/s$$

Why the other options are there

• V = 3.26 m/s (US constant used in SI)
• R = 0.190 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation

A rectangular lined conduit 2.2 m wide runs 0.8 m deep on a slope of 0.0054 m/m with Manning n = 0.013. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 2.2(0.8) = 1.760 m^{2}, P = 2.2 + 2(0.8) = 3.80 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 1.760/3.80 = 0.463 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.013)(0.463)^{2/3}(0.0054)^{1/2} = 3.38 m/s$$

6. Discharge

   $$Q = AV = 1.760(3.38) = 5.956 m^{3}/s$$

Why the other options are there

• V = 5.03 m/s (US constant used in SI)
• R = 0.293 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation

A rectangular lined conduit 1.2 m wide runs 0.4 m deep on a slope of 0.0020 m/m with Manning n = 0.012. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 1.2(0.4) = 0.480 m^{2}, P = 1.2 + 2(0.4) = 2.00 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 0.480/2.00 = 0.240 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.012)(0.240)^{2/3}(0.0020)^{1/2} = 1.44 m/s$$

6. Discharge

   $$Q = AV = 0.480(1.44) = 0.691 m^{3}/s$$

Why the other options are there

• V = 2.14 m/s (US constant used in SI)
• R = 0.150 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation

A rectangular lined conduit 2.0 m wide runs 1.1 m deep on a slope of 0.0036 m/m with Manning n = 0.022. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 2.0(1.1) = 2.200 m^{2}, P = 2.0 + 2(1.1) = 4.20 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 2.200/4.20 = 0.524 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.022)(0.524)^{2/3}(0.0036)^{1/2} = 1.77 m/s$$

6. Discharge

   $$Q = AV = 2.200(1.77) = 3.899 m^{3}/s$$

Why the other options are there

• V = 2.63 m/s (US constant used in SI)
• R = 0.355 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation

A rectangular lined conduit 2.6 m wide runs 0.5 m deep on a slope of 0.0042 m/m with Manning n = 0.022. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 2.6(0.5) = 1.300 m^{2}, P = 2.6 + 2(0.5) = 3.60 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 1.300/3.60 = 0.361 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.022)(0.361)^{2/3}(0.0042)^{1/2} = 1.49 m/s$$

6. Discharge

   $$Q = AV = 1.300(1.49) = 1.942 m^{3}/s$$

Why the other options are there

• V = 2.22 m/s (US constant used in SI)
• R = 0.210 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation

A rectangular lined conduit 1.6 m wide runs 0.6 m deep on a slope of 0.0058 m/m with Manning n = 0.015. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 1.6(0.6) = 0.960 m^{2}, P = 1.6 + 2(0.6) = 2.80 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 0.960/2.80 = 0.343 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.015)(0.343)^{2/3}(0.0058)^{1/2} = 2.49 m/s$$

6. Discharge

   $$Q = AV = 0.960(2.49) = 2.388 m^{3}/s$$

Why the other options are there

• V = 3.70 m/s (US constant used in SI)
• R = 0.218 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation

A rectangular lined conduit 2.0 m wide runs 0.6 m deep on a slope of 0.0034 m/m with Manning n = 0.012. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 2.0(0.6) = 1.200 m^{2}, P = 2.0 + 2(0.6) = 3.20 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 1.200/3.20 = 0.375 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.012)(0.375)^{2/3}(0.0034)^{1/2} = 2.53 m/s$$

6. Discharge

   $$Q = AV = 1.200(2.53) = 3.032 m^{3}/s$$

Why the other options are there

• V = 3.75 m/s (US constant used in SI)
• R = 0.231 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation

A rectangular lined conduit 2.4 m wide runs 1.3 m deep on a slope of 0.0026 m/m with Manning n = 0.015. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 2.4(1.3) = 3.120 m^{2}, P = 2.4 + 2(1.3) = 5.00 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 3.120/5.00 = 0.624 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.015)(0.624)^{2/3}(0.0026)^{1/2} = 2.48 m/s$$

6. Discharge

   $$Q = AV = 3.120(2.48) = 7.745 m^{3}/s$$

Why the other options are there

• V = 3.69 m/s (US constant used in SI)
• R = 0.422 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation

A rectangular lined conduit 2.2 m wide runs 1.2 m deep on a slope of 0.0028 m/m with Manning n = 0.013. Compute the hydraulic radius, mean velocity and discharge in SI units.

Start with the thinking

• In SI the Manning coefficient is 1.000; in US units it is 1.486 — mixing them is the classic error.
• The wetted perimeter of an open rectangular section excludes the free surface.

Step-by-step solution

1. Geometry

   $$A = 2.2(1.2) = 2.640 m^{2}, P = 2.2 + 2(1.2) = 4.60 m$$

2. Formula

   $$R = A/P$$

3. Substituting

   $$R = 2.640/4.60 = 0.574 m$$

4. Formula

   $$V = \dfrac{1}{n}R^{2/3}S^{1/2}$$

5. Substituting

   $$V = (1/0.013)(0.574)^{2/3}(0.0028)^{1/2} = 2.81 m/s$$

6. Discharge

   $$Q = AV = 2.640(2.81) = 7.421 m^{3}/s$$

Why the other options are there

• V = 4.18 m/s (US constant used in SI)
• R = 0.388 m (free surface added to the perimeter)

Reference: FE Reference Handbook — Fluid Mechanics → Manning's Equation