Water flows steadily at 0.150 m³/s through a reducer from 360.0 mm to 100.0 mm diameter; the outlet is 2.0 m above the inlet. The inlet gauge pressure is 207 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.360)^{2}/4 = 0.10179 m^{2}, A_{2} = 0.00785 m^{2}$$

3. Substituting

   $$V_{1} = 0.150/0.10179 = 1.47 m/s, V_{2} = 0.150/0.00785 = 19.10 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(19.10^{2} - 1.47^{2})/2 = 181.3 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(2.0) = 19.6 kPa

7. Substituting

   $$p_{2} = 207 - 181.3 - 19.6 = 6.1 kPa$$

Why the other options are there

• p₂ = 388.3 kPa (sign of the velocity term reversed)
• V₂ = 0.41 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow

Water flows steadily at 0.130 m³/s through a reducer from 360.0 mm to 130.0 mm diameter; the outlet is 1.5 m above the inlet. The inlet gauge pressure is 339 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.360)^{2}/4 = 0.10179 m^{2}, A_{2} = 0.01327 m^{2}$$

3. Substituting

   $$V_{1} = 0.130/0.10179 = 1.28 m/s, V_{2} = 0.130/0.01327 = 9.79 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(9.79^{2} - 1.28^{2})/2 = 47.1 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(1.5) = 14.7 kPa

7. Substituting

   $$p_{2} = 339 - 47.1 - 14.7 = 277.1 kPa$$

Why the other options are there

• p₂ = 386.1 kPa (sign of the velocity term reversed)
• V₂ = 0.46 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow

Water flows steadily at 0.065 m³/s through a reducer from 370.0 mm to 140.0 mm diameter; the outlet is 1.0 m above the inlet. The inlet gauge pressure is 390 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.370)^{2}/4 = 0.10752 m^{2}, A_{2} = 0.01539 m^{2}$$

3. Substituting

   $$V_{1} = 0.065/0.10752 = 0.60 m/s, V_{2} = 0.065/0.01539 = 4.22 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(4.22^{2} - 0.60^{2})/2 = 8.7 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(1.0) = 9.8 kPa

7. Substituting

   $$p_{2} = 390 - 8.7 - 9.8 = 371.5 kPa$$

Why the other options are there

• p₂ = 398.7 kPa (sign of the velocity term reversed)
• V₂ = 0.23 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow

Water flows steadily at 0.075 m³/s through a reducer from 180.0 mm to 110.0 mm diameter; the outlet is 1.5 m above the inlet. The inlet gauge pressure is 516 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.180)^{2}/4 = 0.02545 m^{2}, A_{2} = 0.00950 m^{2}$$

3. Substituting

   $$V_{1} = 0.075/0.02545 = 2.95 m/s, V_{2} = 0.075/0.00950 = 7.89 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(7.89^{2} - 2.95^{2})/2 = 26.8 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(1.5) = 14.7 kPa

7. Substituting

   $$p_{2} = 516 - 26.8 - 14.7 = 474.5 kPa$$

Why the other options are there

• p₂ = 542.8 kPa (sign of the velocity term reversed)
• V₂ = 1.80 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow

Water flows steadily at 0.030 m³/s through a reducer from 340.0 mm to 90 mm diameter; the outlet is 1.5 m above the inlet. The inlet gauge pressure is 383 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.340)^{2}/4 = 0.09079 m^{2}, A_{2} = 0.00636 m^{2}$$

3. Substituting

   $$V_{1} = 0.030/0.09079 = 0.33 m/s, V_{2} = 0.030/0.00636 = 4.72 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(4.72^{2} - 0.33^{2})/2 = 11.1 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(1.5) = 14.7 kPa

7. Substituting

   $$p_{2} = 383 - 11.1 - 14.7 = 357.2 kPa$$

Why the other options are there

• p₂ = 394.1 kPa (sign of the velocity term reversed)
• V₂ = 0.09 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow

Water flows steadily at 0.045 m³/s through a reducer from 310.0 mm to 90 mm diameter; the outlet is 3.5 m above the inlet. The inlet gauge pressure is 457 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.310)^{2}/4 = 0.07548 m^{2}, A_{2} = 0.00636 m^{2}$$

3. Substituting

   $$V_{1} = 0.045/0.07548 = 0.60 m/s, V_{2} = 0.045/0.00636 = 7.07 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(7.07^{2} - 0.60^{2})/2 = 24.8 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(3.5) = 34.3 kPa

7. Substituting

   $$p_{2} = 457 - 24.8 - 34.3 = 397.8 kPa$$

Why the other options are there

• p₂ = 481.8 kPa (sign of the velocity term reversed)
• V₂ = 0.17 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow

Water flows steadily at 0.130 m³/s through a reducer from 380.0 mm to 90 mm diameter; the outlet is 3.5 m above the inlet. The inlet gauge pressure is 380 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.380)^{2}/4 = 0.11341 m^{2}, A_{2} = 0.00636 m^{2}$$

3. Substituting

   $$V_{1} = 0.130/0.11341 = 1.15 m/s, V_{2} = 0.130/0.00636 = 20.43 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(20.43^{2} - 1.15^{2})/2 = 208.1 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(3.5) = 34.3 kPa

7. Substituting

   $$p_{2} = 380 - 208.1 - 34.3 = 137.5 kPa$$

Why the other options are there

• p₂ = 588.1 kPa (sign of the velocity term reversed)
• V₂ = 0.27 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow

Water flows steadily at 0.055 m³/s through a reducer from 390.0 mm to 90 mm diameter; the outlet is 0.5 m above the inlet. The inlet gauge pressure is 398 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.390)^{2}/4 = 0.11946 m^{2}, A_{2} = 0.00636 m^{2}$$

3. Substituting

   $$V_{1} = 0.055/0.11946 = 0.46 m/s, V_{2} = 0.055/0.00636 = 8.65 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(8.65^{2} - 0.46^{2})/2 = 37.3 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(0.5) = 4.9 kPa

7. Substituting

   $$p_{2} = 398 - 37.3 - 4.9 = 355.8 kPa$$

Why the other options are there

• p₂ = 435.3 kPa (sign of the velocity term reversed)
• V₂ = 0.11 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow

Water flows steadily at 0.105 m³/s through a reducer from 180.0 mm to 80 mm diameter; the outlet is 0.0 m above the inlet. The inlet gauge pressure is 408 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.180)^{2}/4 = 0.02545 m^{2}, A_{2} = 0.00503 m^{2}$$

3. Substituting

   $$V_{1} = 0.105/0.02545 = 4.13 m/s, V_{2} = 0.105/0.00503 = 20.89 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(20.89^{2} - 4.13^{2})/2 = 209.7 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(0.0) = 0.0 kPa

7. Substituting

   $$p_{2} = 408 - 209.7 - 0.0 = 198.3 kPa$$

Why the other options are there

• p₂ = 617.7 kPa (sign of the velocity term reversed)
• V₂ = 1.83 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow

Water flows steadily at 0.090 m³/s through a reducer from 320.0 mm to 60 mm diameter; the outlet is 4.0 m above the inlet. The inlet gauge pressure is 324 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Start with the thinking

• Continuity fixes the velocities from areas alone — pressure never enters that step.
• Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

1. Formula

   $$Q = A_1 V_1 = A_2 V_2$$

2. Areas

   $$A_{1} = \pi(0.320)^{2}/4 = 0.08042 m^{2}, A_{2} = 0.00283 m^{2}$$

3. Substituting

   $$V_{1} = 0.090/0.08042 = 1.12 m/s, V_{2} = 0.090/0.00283 = 31.83 m/s$$

4. Formula

   $$\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + g z_1 = \dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + g z_2$$

5. Velocity term

   $$\rho(V_{2}^{2} - V_{1}^{2})/2 = 1000(31.83^{2} - 1.12^{2})/2 = 506.0 kPa$$

6. Elevation term — ρgΔz = 1000(9.81)(4.0) = 39.2 kPa

7. Substituting

   $$p_{2} = 324 - 506.0 - 39.2 = -221.2 kPa$$

Why the other options are there

• p₂ = 830.0 kPa (sign of the velocity term reversed)
• V₂ = 0.21 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Principles of One-Dimensional Fluid Flow