A 75 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 2.7 m. With a discharge coefficient of 0.60, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times2.7) = 7.28 m/s$$

3. Orifice area

   $$A = \pi(0.075)^{2}/4 = 0.00442 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.60(0.00442)(7.28) = 0.0193 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0193(60) = 1.16 m^{3}$$

Why the other options are there

• 0.0322 m³/s (C_d ignored)
• 0.1404 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

A 75 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.5 m. With a discharge coefficient of 0.59, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times4.5) = 9.40 m/s$$

3. Orifice area

   $$A = \pi(0.075)^{2}/4 = 0.00442 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.59(0.00442)(9.40) = 0.0245 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0245(60) = 1.47 m^{3}$$

Why the other options are there

• 0.0415 m³/s (C_d ignored)
• 0.2301 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

A 35 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 1.6 m. With a discharge coefficient of 0.61, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times1.6) = 5.60 m/s$$

3. Orifice area

   $$A = \pi(0.035)^{2}/4 = 0.00096 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.61(0.00096)(5.60) = 0.0033 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0033(60) = 0.20 m^{3}$$

Why the other options are there

• 0.0054 m³/s (C_d ignored)
• 0.0184 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

A 35 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.9 m. With a discharge coefficient of 0.62, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times4.9) = 9.80 m/s$$

3. Orifice area

   $$A = \pi(0.035)^{2}/4 = 0.00096 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.62(0.00096)(9.80) = 0.0058 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0058(60) = 0.35 m^{3}$$

Why the other options are there

• 0.0094 m³/s (C_d ignored)
• 0.0573 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

A 70 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 6.2 m. With a discharge coefficient of 0.59, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times6.2) = 11.03 m/s$$

3. Orifice area

   $$A = \pi(0.070)^{2}/4 = 0.00385 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.59(0.00385)(11.03) = 0.0250 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0250(60) = 1.50 m^{3}$$

Why the other options are there

• 0.0424 m³/s (C_d ignored)
• 0.2762 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

A 30 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.0 m. With a discharge coefficient of 0.65, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times4.0) = 8.86 m/s$$

3. Orifice area

   $$A = \pi(0.030)^{2}/4 = 0.00071 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.65(0.00071)(8.86) = 0.0041 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0041(60) = 0.24 m^{3}$$

Why the other options are there

• 0.0063 m³/s (C_d ignored)
• 0.0361 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

A 25 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 3.8 m. With a discharge coefficient of 0.65, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times3.8) = 8.63 m/s$$

3. Orifice area

   $$A = \pi(0.025)^{2}/4 = 0.00049 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.65(0.00049)(8.63) = 0.0028 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0028(60) = 0.17 m^{3}$$

Why the other options are there

• 0.0042 m³/s (C_d ignored)
• 0.0238 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

A 95 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 8.3 m. With a discharge coefficient of 0.64, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times8.3) = 12.76 m/s$$

3. Orifice area

   $$A = \pi(0.095)^{2}/4 = 0.00709 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.64(0.00709)(12.76) = 0.0579 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0579(60) = 3.47 m^{3}$$

Why the other options are there

• 0.0905 m³/s (C_d ignored)
• 0.7387 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

A 60 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.1 m. With a discharge coefficient of 0.64, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times4.1) = 8.97 m/s$$

3. Orifice area

   $$A = \pi(0.060)^{2}/4 = 0.00283 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.64(0.00283)(8.97) = 0.0162 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0162(60) = 0.97 m^{3}$$

Why the other options are there

• 0.0254 m³/s (C_d ignored)
• 0.1456 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

A 95 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.0 m. With a discharge coefficient of 0.65, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Start with the thinking

• Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
• The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

1. Formula

   $$V = \sqrt{2gH}$$

2. Substituting

   $$V = \sqrt(2\times9.81\times4.0) = 8.86 m/s$$

3. Orifice area

   $$A = \pi(0.095)^{2}/4 = 0.00709 m^{2}$$

4. Formula

   $$Q = C_d A\sqrt{2gH}$$

5. Substituting

   $$Q = 0.65(0.00709)(8.86) = 0.0408 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0408(60) = 2.45 m^{3}$$

Why the other options are there

• 0.0628 m³/s (C_d ignored)
• 0.3616 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices