A 35 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 5.1 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

   $$V = \sqrt(2\times9.81\times5.1) = 10.00 m/s$$

3. Orifice area

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

4. Formula

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

5. Substituting

   $$Q = 0.61(0.00096)(10.00) = 0.0059 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0059(60) = 0.35 m^{3}$$

Why the other options are there

• 0.0096 m³/s (C_d ignored)
• 0.0587 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere

A 95 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 1.7 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

   $$V = \sqrt(2\times9.81\times1.7) = 5.78 m/s$$

3. Orifice area

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

4. Formula

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

5. Substituting

   $$Q = 0.60(0.00709)(5.78) = 0.0246 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0246(60) = 1.47 m^{3}$$

Why the other options are there

• 0.0409 m³/s (C_d ignored)
• 0.1419 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere

A 55 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 7.9 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

   $$V = \sqrt(2\times9.81\times7.9) = 12.45 m/s$$

3. Orifice area

   $$A = \pi(0.055)^{2}/4 = 0.00238 m^{2}$$

4. Formula

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

5. Substituting

   $$Q = 0.61(0.00238)(12.45) = 0.0180 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0180(60) = 1.08 m^{3}$$

Why the other options are there

• 0.0296 m³/s (C_d ignored)
• 0.2246 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere

A 90 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 8.0 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

   $$V = \sqrt(2\times9.81\times8.0) = 12.53 m/s$$

3. Orifice area

   $$A = \pi(0.090)^{2}/4 = 0.00636 m^{2}$$

4. Formula

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

5. Substituting

   $$Q = 0.64(0.00636)(12.53) = 0.0510 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0510(60) = 3.06 m^{3}$$

Why the other options are there

• 0.0797 m³/s (C_d ignored)
• 0.6391 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere

A 80 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 2.3 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

   $$V = \sqrt(2\times9.81\times2.3) = 6.72 m/s$$

3. Orifice area

   $$A = \pi(0.080)^{2}/4 = 0.00503 m^{2}$$

4. Formula

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

5. Substituting

   $$Q = 0.61(0.00503)(6.72) = 0.0206 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0206(60) = 1.24 m^{3}$$

Why the other options are there

• 0.0338 m³/s (C_d ignored)
• 0.1384 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere

A 40 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 7.4 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

   $$V = \sqrt(2\times9.81\times7.4) = 12.05 m/s$$

3. Orifice area

   $$A = \pi(0.040)^{2}/4 = 0.00126 m^{2}$$

4. Formula

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

5. Substituting

   $$Q = 0.60(0.00126)(12.05) = 0.0091 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0091(60) = 0.55 m^{3}$$

Why the other options are there

• 0.0151 m³/s (C_d ignored)
• 0.1095 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere

A 50 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.2 m. With a discharge coefficient of 0.63, 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

   $$V = \sqrt(2\times9.81\times4.2) = 9.08 m/s$$

3. Orifice area

   $$A = \pi(0.050)^{2}/4 = 0.00196 m^{2}$$

4. Formula

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

5. Substituting

   $$Q = 0.63(0.00196)(9.08) = 0.0112 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0112(60) = 0.67 m^{3}$$

Why the other options are there

• 0.0178 m³/s (C_d ignored)
• 0.1019 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere

A 25 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 7.4 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

   $$V = \sqrt(2\times9.81\times7.4) = 12.05 m/s$$

3. Orifice area

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

4. Formula

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

5. Substituting

   $$Q = 0.62(0.00049)(12.05) = 0.0037 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0037(60) = 0.22 m^{3}$$

Why the other options are there

• 0.0059 m³/s (C_d ignored)
• 0.0442 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere

A 110.0 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 8.4 m. With a discharge coefficient of 0.63, 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

   $$V = \sqrt(2\times9.81\times8.4) = 12.84 m/s$$

3. Orifice area

   $$A = \pi(0.110)^{2}/4 = 0.00950 m^{2}$$

4. Formula

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

5. Substituting

   $$Q = 0.63(0.00950)(12.84) = 0.0769 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0769(60) = 4.61 m^{3}$$

Why the other options are there

• 0.1220 m³/s (C_d ignored)
• 0.9867 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere

A 105.0 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 8.2 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

   $$V = \sqrt(2\times9.81\times8.2) = 12.68 m/s$$

3. Orifice area

   $$A = \pi(0.105)^{2}/4 = 0.00866 m^{2}$$

4. Formula

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

5. Substituting

   $$Q = 0.64(0.00866)(12.68) = 0.0703 m^{3}/s$$

6. Volume in 60 s

   $$\forall = 0.0703(60) = 4.22 m^{3}$$

Why the other options are there

• 0.1098 m³/s (C_d ignored)
• 0.8916 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifice Discharging Freely into Atmosphere