Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 1.5 m/s through a 0.75 m × 0.15 m rectangular conduit 88 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.75(0.15) = 0.1125 m^{2}, P = 2(0.75 + 0.15) = 1.800 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.1125)/1.800 = 0.2500 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 1.5(0.2500)/1.0\times10^{-6} = 3.75e+5$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(3.75e+5)^{-0.25} = 0.0128$$

8. Head loss

   $$h_f = 0.0128(88/0.2500)(0.1147) = 0.515 m$$

Why the other options are there

• D_h = 0.0625 m (factor of 4 omitted)
• D_h = 0.450 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits

Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 1.5 m/s through a 0.35 m × 0.45 m rectangular conduit 70 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.35(0.45) = 0.1575 m^{2}, P = 2(0.35 + 0.45) = 1.600 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.1575)/1.600 = 0.3938 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 1.5(0.3938)/1.0\times10^{-6} = 5.91e+5$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(5.91e+5)^{-0.25} = 0.0114$$

8. Head loss

   $$h_f = 0.0114(70/0.3938)(0.1147) = 0.232 m$$

Why the other options are there

• D_h = 0.0984 m (factor of 4 omitted)
• D_h = 0.400 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits

Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 4.5 m/s through a 0.80 m × 0.15 m rectangular conduit 74 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.80(0.15) = 0.1200 m^{2}, P = 2(0.80 + 0.15) = 1.900 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.1200)/1.900 = 0.2526 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 4.5(0.2526)/1.0\times10^{-6} = 1.14e+6$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(1.14e+6)^{-0.25} = 0.0097$$

8. Head loss

   $$h_f = 0.0097(74/0.2526)(1.0321) = 2.926 m$$

Why the other options are there

• D_h = 0.0632 m (factor of 4 omitted)
• D_h = 0.475 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits

Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 3.0 m/s through a 0.80 m × 0.30 m rectangular conduit 23 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.80(0.30) = 0.2400 m^{2}, P = 2(0.80 + 0.30) = 2.200 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.2400)/2.200 = 0.4364 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 3.0(0.4364)/1.0\times10^{-6} = 1.31e+6$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(1.31e+6)^{-0.25} = 0.0093$$

8. Head loss

   $$h_f = 0.0093(23/0.4364)(0.4587) = 0.226 m$$

Why the other options are there

• D_h = 0.1091 m (factor of 4 omitted)
• D_h = 0.550 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits

Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 6.5 m/s through a 0.85 m × 0.40 m rectangular conduit 57 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.85(0.40) = 0.3400 m^{2}, P = 2(0.85 + 0.40) = 2.500 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.3400)/2.500 = 0.5440 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 6.5(0.5440)/1.0\times10^{-6} = 3.54e+6$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(3.54e+6)^{-0.25} = 0.0073$$

8. Head loss

   $$h_f = 0.0073(57/0.5440)(2.1534) = 1.644 m$$

Why the other options are there

• D_h = 0.1360 m (factor of 4 omitted)
• D_h = 0.625 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits

Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 3.5 m/s through a 0.75 m × 0.35 m rectangular conduit 24 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.75(0.35) = 0.2625 m^{2}, P = 2(0.75 + 0.35) = 2.200 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.2625)/2.200 = 0.4773 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 3.5(0.4773)/1.0\times10^{-6} = 1.67e+6$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(1.67e+6)^{-0.25} = 0.0088$$

8. Head loss

   $$h_f = 0.0088(24/0.4773)(0.6244) = 0.276 m$$

Why the other options are there

• D_h = 0.1193 m (factor of 4 omitted)
• D_h = 0.550 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits

Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 1.5 m/s through a 0.25 m × 0.30 m rectangular conduit 81 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.25(0.30) = 0.0750 m^{2}, P = 2(0.25 + 0.30) = 1.100 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.0750)/1.100 = 0.2727 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 1.5(0.2727)/1.0\times10^{-6} = 4.09e+5$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(4.09e+5)^{-0.25} = 0.0125$$

8. Head loss

   $$h_f = 0.0125(81/0.2727)(0.1147) = 0.426 m$$

Why the other options are there

• D_h = 0.0682 m (factor of 4 omitted)
• D_h = 0.275 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits

Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 5.0 m/s through a 0.30 m × 0.55 m rectangular conduit 57 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.30(0.55) = 0.1650 m^{2}, P = 2(0.30 + 0.55) = 1.700 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.1650)/1.700 = 0.3882 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 5.0(0.3882)/1.0\times10^{-6} = 1.94e+6$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(1.94e+6)^{-0.25} = 0.0085$$

8. Head loss

   $$h_f = 0.0085(57/0.3882)(1.2742) = 1.584 m$$

Why the other options are there

• D_h = 0.0971 m (factor of 4 omitted)
• D_h = 0.425 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits

Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 8.0 m/s through a 0.65 m × 0.20 m rectangular conduit 68 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.65(0.20) = 0.1300 m^{2}, P = 2(0.65 + 0.20) = 1.700 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.1300)/1.700 = 0.3059 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 8.0(0.3059)/1.0\times10^{-6} = 2.45e+6$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(2.45e+6)^{-0.25} = 0.0080$$

8. Head loss

   $$h_f = 0.0080(68/0.3059)(3.2620) = 5.794 m$$

Why the other options are there

• D_h = 0.0765 m (factor of 4 omitted)
• D_h = 0.425 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits

Water at 20 °C (ν = 1.0 × 10⁻⁶ m²/s) flows at 3.0 m/s through a 0.50 m × 0.10 m rectangular conduit 16 m long. Compute the hydraulic diameter, Reynolds number, Blasius friction factor and head loss.

Start with the thinking

• Noncircular conduits reuse every circular-pipe relation once the hydraulic diameter replaces D.
• D_h = 4A/P — the factor of four is what makes it reduce to D for a full circular pipe.

Step-by-step solution

1. Area and perimeter

   $$A = 0.50(0.10) = 0.0500 m^{2}, P = 2(0.50 + 0.10) = 1.200 m$$

2. Formula

   $$D_h = \dfrac{4A}{P}$$

3. Substituting

   $$D_h = 4(0.0500)/1.200 = 0.1667 m$$

4. Formula

   $$Re = \dfrac{V D_h}{\nu}$$

5. Substituting

   $$Re = 3.0(0.1667)/1.0\times10^{-6} = 5.00e+5$$

6. Formula

   $$f = 0.316\,Re^{-1/4}$$

7. Substituting

   $$f = 0.316(5.00e+5)^{-0.25} = 0.0119$$

8. Head loss

   $$h_f = 0.0119(16/0.1667)(0.4587) = 0.523 m$$

Why the other options are there

• D_h = 0.0417 m (factor of 4 omitted)
• D_h = 0.300 m (mean side used)

Reference: FE Reference Handbook — Fluid Mechanics → Flow in Noncircular Conduits