A venturi meter with a 175.0 mm approach pipe and a throat diameter ratio β = 0.50 registers a differential pressure of 71 kPa on water. With a meter coefficient of 0.98, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.50(175.0) = 87.5 mm, A_{2} = 0.00601 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.50^{4} = 0.9375$$

4. Velocity term

   $$\sqrt[2(71\times1000)/1000] = 11.92$$

5. Substituting

   $$Q = 0.98(0.00601)(11.92)/\sqrt0.9375 = 0.0725 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 12.06 m/s$$

Why the other options are there

• 0.0702 m³/s (β⁴ correction omitted)
• 0.0740 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters

A venturi meter with a 200.0 mm approach pipe and a throat diameter ratio β = 0.45 registers a differential pressure of 41 kPa on water. With a meter coefficient of 0.99, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.45(200.0) = 90.0 mm, A_{2} = 0.00636 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.45^{4} = 0.9590$$

4. Velocity term

   $$\sqrt[2(41\times1000)/1000] = 9.06$$

5. Substituting

   $$Q = 0.99(0.00636)(9.06)/\sqrt0.9590 = 0.0582 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 9.15 m/s$$

Why the other options are there

• 0.0570 m³/s (β⁴ correction omitted)
• 0.0588 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters

A venturi meter with a 225.0 mm approach pipe and a throat diameter ratio β = 0.50 registers a differential pressure of 16 kPa on water. With a meter coefficient of 0.98, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.50(225.0) = 112.5 mm, A_{2} = 0.00994 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.50^{4} = 0.9375$$

4. Velocity term

   $$\sqrt[2(16\times1000)/1000] = 5.66$$

5. Substituting

   $$Q = 0.98(0.00994)(5.66)/\sqrt0.9375 = 0.0569 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 5.73 m/s$$

Why the other options are there

• 0.0551 m³/s (β⁴ correction omitted)
• 0.0581 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters

A venturi meter with a 175.0 mm approach pipe and a throat diameter ratio β = 0.50 registers a differential pressure of 43 kPa on water. With a meter coefficient of 0.98, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.50(175.0) = 87.5 mm, A_{2} = 0.00601 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.50^{4} = 0.9375$$

4. Velocity term

   $$\sqrt[2(43\times1000)/1000] = 9.27$$

5. Substituting

   $$Q = 0.98(0.00601)(9.27)/\sqrt0.9375 = 0.0564 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 9.39 m/s$$

Why the other options are there

• 0.0546 m³/s (β⁴ correction omitted)
• 0.0576 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters

A venturi meter with a 250.0 mm approach pipe and a throat diameter ratio β = 0.45 registers a differential pressure of 13 kPa on water. With a meter coefficient of 0.97, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.45(250.0) = 112.5 mm, A_{2} = 0.00994 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.45^{4} = 0.9590$$

4. Velocity term

   $$\sqrt[2(13\times1000)/1000] = 5.10$$

5. Substituting

   $$Q = 0.97(0.00994)(5.10)/\sqrt0.9590 = 0.0502 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 5.05 m/s$$

Why the other options are there

• 0.0492 m³/s (β⁴ correction omitted)
• 0.0518 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters

A venturi meter with a 200.0 mm approach pipe and a throat diameter ratio β = 0.50 registers a differential pressure of 80 kPa on water. With a meter coefficient of 0.97, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.50(200.0) = 100.0 mm, A_{2} = 0.00785 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.50^{4} = 0.9375$$

4. Velocity term

   $$\sqrt[2(80\times1000)/1000] = 12.65$$

5. Substituting

   $$Q = 0.97(0.00785)(12.65)/\sqrt0.9375 = 0.0995 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 12.67 m/s$$

Why the other options are there

• 0.0964 m³/s (β⁴ correction omitted)
• 0.1026 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters

A venturi meter with a 225.0 mm approach pipe and a throat diameter ratio β = 0.40 registers a differential pressure of 11 kPa on water. With a meter coefficient of 0.97, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.40(225.0) = 90.0 mm, A_{2} = 0.00636 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.40^{4} = 0.9744$$

4. Velocity term

   $$\sqrt[2(11\times1000)/1000] = 4.69$$

5. Substituting

   $$Q = 0.97(0.00636)(4.69)/\sqrt0.9744 = 0.0293 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 4.61 m/s$$

Why the other options are there

• 0.0289 m³/s (β⁴ correction omitted)
• 0.0302 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters

A venturi meter with a 200.0 mm approach pipe and a throat diameter ratio β = 0.45 registers a differential pressure of 25 kPa on water. With a meter coefficient of 0.98, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.45(200.0) = 90.0 mm, A_{2} = 0.00636 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.45^{4} = 0.9590$$

4. Velocity term

   $$\sqrt[2(25\times1000)/1000] = 7.07$$

5. Substituting

   $$Q = 0.98(0.00636)(7.07)/\sqrt0.9590 = 0.0450 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 7.08 m/s$$

Why the other options are there

• 0.0441 m³/s (β⁴ correction omitted)
• 0.0459 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters

A venturi meter with a 250.0 mm approach pipe and a throat diameter ratio β = 0.70 registers a differential pressure of 75 kPa on water. With a meter coefficient of 0.98, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.70(250.0) = 175.0 mm, A_{2} = 0.02405 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.70^{4} = 0.7599$$

4. Velocity term

   $$\sqrt[2(75\times1000)/1000] = 12.25$$

5. Substituting

   $$Q = 0.98(0.02405)(12.25)/\sqrt0.7599 = 0.3312 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 13.77 m/s$$

Why the other options are there

• 0.2887 m³/s (β⁴ correction omitted)
• 0.3379 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters

A venturi meter with a 150.0 mm approach pipe and a throat diameter ratio β = 0.40 registers a differential pressure of 31 kPa on water. With a meter coefficient of 0.98, compute the throat diameter, throat velocity and the discharge.

Start with the thinking

• The (1 − β⁴) term corrects for the approach velocity — dropping it overestimates flow.
• A venturi recovers most of the pressure drop, unlike an orifice plate.

Step-by-step solution

1. Throat

   $$D_{2} = 0.40(150.0) = 60.0 mm, A_{2} = 0.00283 m^{2}$$

2. Formula

   $$Q = \dfrac{C_v A_2}{\sqrt{1-\beta^4}}\sqrt{\dfrac{2\Delta p}{\rho}}$$

3. β⁴ term

   $$1 - 0.40^{4} = 0.9744$$

4. Velocity term

   $$\sqrt[2(31\times1000)/1000] = 7.87$$

5. Substituting

   $$Q = 0.98(0.00283)(7.87)/\sqrt0.9744 = 0.0221 m^{3}/s$$

6. Throat velocity

   $$V_{2} = Q/A_{2} = 7.82 m/s$$

Why the other options are there

• 0.0218 m³/s (β⁴ correction omitted)
• 0.0226 m³/s (coefficient omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Venturi Meters