A 175.0 mm pipeline carries 0.130 m³/s over 550 m with a friction factor of 0.023. At the upstream gauge the pipe centreline is at elevation 20.0 m and the pressure is 385 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.130/0.02405 = 5.40 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 5.40^{2}/(2\times9.81) = 1.489 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 20.0 + 385/9.81 = 59.25 m$$

6. EGL

   $$EGL = HGL + h_v = 59.25 + 1.489 = 60.73 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.023(550/0.175)(1.489) = 107.6 m$$

9. Gradient

   $$S = h_f/L = 107.6/550 = 0.19568 m/m$$

Why the other options are there

• EGL = 57.76 m (velocity head subtracted)
• S = 107.6 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

A 175.0 mm pipeline carries 0.055 m³/s over 883 m with a friction factor of 0.023. At the upstream gauge the pipe centreline is at elevation 19.5 m and the pressure is 275 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.055/0.02405 = 2.29 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 2.29^{2}/(2\times9.81) = 0.266 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 19.5 + 275/9.81 = 47.53 m$$

6. EGL

   $$EGL = HGL + h_v = 47.53 + 0.266 = 47.80 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.023(883/0.175)(0.266) = 30.93 m$$

9. Gradient

   $$S = h_f/L = 30.93/883 = 0.03503 m/m$$

Why the other options are there

• EGL = 47.27 m (velocity head subtracted)
• S = 30.9275 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

A 225.0 mm pipeline carries 0.095 m³/s over 551 m with a friction factor of 0.024. At the upstream gauge the pipe centreline is at elevation 27.5 m and the pressure is 237 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.095/0.03976 = 2.39 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 2.39^{2}/(2\times9.81) = 0.291 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 27.5 + 237/9.81 = 51.66 m$$

6. EGL

   $$EGL = HGL + h_v = 51.66 + 0.291 = 51.95 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.024(551/0.225)(0.291) = 17.10 m$$

9. Gradient

   $$S = h_f/L = 17.10/551 = 0.03104 m/m$$

Why the other options are there

• EGL = 51.37 m (velocity head subtracted)
• S = 17.1009 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

A 275.0 mm pipeline carries 0.040 m³/s over 186 m with a friction factor of 0.027. At the upstream gauge the pipe centreline is at elevation 23.0 m and the pressure is 260 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.040/0.05940 = 0.67 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 0.67^{2}/(2\times9.81) = 0.023 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 23.0 + 260/9.81 = 49.50 m$$

6. EGL

   $$EGL = HGL + h_v = 49.50 + 0.023 = 49.53 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.027(186/0.275)(0.023) = 0.42 m$$

9. Gradient

   $$S = h_f/L = 0.42/186 = 0.00227 m/m$$

Why the other options are there

• EGL = 49.48 m (velocity head subtracted)
• S = 0.4221 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

A 300.0 mm pipeline carries 0.085 m³/s over 686 m with a friction factor of 0.022. At the upstream gauge the pipe centreline is at elevation 17.0 m and the pressure is 280 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.085/0.07069 = 1.20 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 1.20^{2}/(2\times9.81) = 0.074 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 17.0 + 280/9.81 = 45.54 m$$

6. EGL

   $$EGL = HGL + h_v = 45.54 + 0.074 = 45.62 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.022(686/0.300)(0.074) = 3.71 m$$

9. Gradient

   $$S = h_f/L = 3.71/686 = 0.00540 m/m$$

Why the other options are there

• EGL = 45.47 m (velocity head subtracted)
• S = 3.7077 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

A 200.0 mm pipeline carries 0.065 m³/s over 665 m with a friction factor of 0.017. At the upstream gauge the pipe centreline is at elevation 14.5 m and the pressure is 296 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.065/0.03142 = 2.07 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 2.07^{2}/(2\times9.81) = 0.218 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 14.5 + 296/9.81 = 44.67 m$$

6. EGL

   $$EGL = HGL + h_v = 44.67 + 0.218 = 44.89 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.017(665/0.200)(0.218) = 12.33 m$$

9. Gradient

   $$S = h_f/L = 12.33/665 = 0.01855 m/m$$

Why the other options are there

• EGL = 44.46 m (velocity head subtracted)
• S = 12.3330 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

A 325.0 mm pipeline carries 0.140 m³/s over 882 m with a friction factor of 0.023. At the upstream gauge the pipe centreline is at elevation 22.5 m and the pressure is 391 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.140/0.08296 = 1.69 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 1.69^{2}/(2\times9.81) = 0.145 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 22.5 + 391/9.81 = 62.36 m$$

6. EGL

   $$EGL = HGL + h_v = 62.36 + 0.145 = 62.50 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.023(882/0.325)(0.145) = 9.06 m$$

9. Gradient

   $$S = h_f/L = 9.06/882 = 0.01027 m/m$$

Why the other options are there

• EGL = 62.21 m (velocity head subtracted)
• S = 9.0606 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

A 300.0 mm pipeline carries 0.085 m³/s over 827 m with a friction factor of 0.018. At the upstream gauge the pipe centreline is at elevation 17.0 m and the pressure is 243 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.085/0.07069 = 1.20 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 1.20^{2}/(2\times9.81) = 0.074 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 17.0 + 243/9.81 = 41.77 m$$

6. EGL

   $$EGL = HGL + h_v = 41.77 + 0.074 = 41.84 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.018(827/0.300)(0.074) = 3.66 m$$

9. Gradient

   $$S = h_f/L = 3.66/827 = 0.00442 m/m$$

Why the other options are there

• EGL = 41.70 m (velocity head subtracted)
• S = 3.6570 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

A 325.0 mm pipeline carries 0.080 m³/s over 573 m with a friction factor of 0.018. At the upstream gauge the pipe centreline is at elevation 25.0 m and the pressure is 336 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.080/0.08296 = 0.96 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 0.96^{2}/(2\times9.81) = 0.047 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 25.0 + 336/9.81 = 59.25 m$$

6. EGL

   $$EGL = HGL + h_v = 59.25 + 0.047 = 59.30 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.018(573/0.325)(0.047) = 1.50 m$$

9. Gradient

   $$S = h_f/L = 1.50/573 = 0.00263 m/m$$

Why the other options are there

• EGL = 59.20 m (velocity head subtracted)
• S = 1.5042 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

A 225.0 mm pipeline carries 0.145 m³/s over 899 m with a friction factor of 0.020. At the upstream gauge the pipe centreline is at elevation 14.5 m and the pressure is 206 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Start with the thinking

• The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
• The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

1. Velocity

   $$V = Q/A = 0.145/0.03976 = 3.65 m/s$$

2. Formula

   $$h_v = \dfrac{V^2}{2g}$$

3. Substituting

   $$h_v = 3.65^{2}/(2\times9.81) = 0.678 m$$

4. Formula

   $$HGL = z + p/\gamma$$

5. Substituting

   $$HGL = 14.5 + 206/9.81 = 35.50 m$$

6. EGL

   $$EGL = HGL + h_v = 35.50 + 0.678 = 36.18 m$$

7. Formula

   $$h_f = f\dfrac{L}{D}\dfrac{V^2}{2g}$$

8. Substituting

   $$h_f = 0.020(899/0.225)(0.678) = 54.17 m$$

9. Gradient

   $$S = h_f/L = 54.17/899 = 0.06025 m/m$$

Why the other options are there

• EGL = 34.82 m (velocity head subtracted)
• S = 54.1669 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)