A 2.5 m diameter cylindrical tank filled to 6.0 m drains through a 90 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(2.5)^{2}/4 = 4.909 m^{2}, A_o = 0.00636 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(4.909)(\sqrt6.0)/[0.62(0.00636)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(4.909)(2.449) = 24.048$$

5. Denominator

   $$0.62(0.00636)(4.429) = 0.017471$$

6. Result

   $$t = 1,376 s = 22.9 \min$$

Why the other options are there

• 688.2 s (factor of 2 dropped)
• 688.2 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank

A 4.0 m diameter cylindrical tank filled to 2.5 m drains through a 50 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(4.0)^{2}/4 = 12.566 m^{2}, A_o = 0.00196 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(12.566)(\sqrt2.5)/[0.62(0.00196)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(12.566)(1.581) = 39.738$$

5. Denominator

   $$0.62(0.00196)(4.429) = 0.005392$$

6. Result

   $$t = 7,370 s = 122.8 \min$$

Why the other options are there

• 3,685 s (factor of 2 dropped)
• 3,685 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank

A 4.5 m diameter cylindrical tank filled to 3.0 m drains through a 60 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(4.5)^{2}/4 = 15.904 m^{2}, A_o = 0.00283 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(15.904)(\sqrt3.0)/[0.62(0.00283)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(15.904)(1.732) = 55.094$$

5. Denominator

   $$0.62(0.00283)(4.429) = 0.007765$$

6. Result

   $$t = 7,095 s = 118.3 \min$$

Why the other options are there

• 3,548 s (factor of 2 dropped)
• 3,548 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank

A 4.5 m diameter cylindrical tank filled to 5.0 m drains through a 100.0 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(4.5)^{2}/4 = 15.904 m^{2}, A_o = 0.00785 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(15.904)(\sqrt5.0)/[0.62(0.00785)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(15.904)(2.236) = 71.126$$

5. Denominator

   $$0.62(0.00785)(4.429) = 0.021569$$

6. Result

   $$t = 3,298 s = 55.0 \min$$

Why the other options are there

• 1,649 s (factor of 2 dropped)
• 1,649 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank

A 4.5 m diameter cylindrical tank filled to 5.5 m drains through a 110.0 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(4.5)^{2}/4 = 15.904 m^{2}, A_o = 0.00950 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(15.904)(\sqrt5.5)/[0.62(0.00950)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(15.904)(2.345) = 74.598$$

5. Denominator

   $$0.62(0.00950)(4.429) = 0.026099$$

6. Result

   $$t = 2,858 s = 47.6 \min$$

Why the other options are there

• 1,429 s (factor of 2 dropped)
• 1,429 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank

A 1.5 m diameter cylindrical tank filled to 2.5 m drains through a 50 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(1.5)^{2}/4 = 1.767 m^{2}, A_o = 0.00196 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(1.767)(\sqrt2.5)/[0.62(0.00196)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(1.767)(1.581) = 5.588$$

5. Denominator

   $$0.62(0.00196)(4.429) = 0.005392$$

6. Result

   $$t = 1,036 s = 17.3 \min$$

Why the other options are there

• 518.2 s (factor of 2 dropped)
• 518.2 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank

A 2.0 m diameter cylindrical tank filled to 2.5 m drains through a 60 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(2.0)^{2}/4 = 3.142 m^{2}, A_o = 0.00283 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(3.142)(\sqrt2.5)/[0.62(0.00283)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(3.142)(1.581) = 9.935$$

5. Denominator

   $$0.62(0.00283)(4.429) = 0.007765$$

6. Result

   $$t = 1,279 s = 21.3 \min$$

Why the other options are there

• 639.7 s (factor of 2 dropped)
• 639.7 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank

A 3.5 m diameter cylindrical tank filled to 2.5 m drains through a 100.0 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(3.5)^{2}/4 = 9.621 m^{2}, A_o = 0.00785 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(9.621)(\sqrt2.5)/[0.62(0.00785)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(9.621)(1.581) = 30.425$$

5. Denominator

   $$0.62(0.00785)(4.429) = 0.021569$$

6. Result

   $$t = 1,411 s = 23.5 \min$$

Why the other options are there

• 705.3 s (factor of 2 dropped)
• 705.3 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank

A 1.5 m diameter cylindrical tank filled to 4.5 m drains through a 120.0 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(1.5)^{2}/4 = 1.767 m^{2}, A_o = 0.01131 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(1.767)(\sqrt4.5)/[0.62(0.01131)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(1.767)(2.121) = 7.497$$

5. Denominator

   $$0.62(0.01131)(4.429) = 0.031059$$

6. Result

   $$t = 241.4 s = 4.0 \min$$

Why the other options are there

• 120.7 s (factor of 2 dropped)
• 120.7 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank

A 3.5 m diameter cylindrical tank filled to 3.0 m drains through a 70 mm bottom orifice with C_d = 0.62. Compute the time required to drain the tank completely.

Start with the thinking

• The head falls as the tank empties, so the discharge is unsteady — integrate rather than use Q = C_dA√(2gH) once.
• Integrating dt = −A_t dh /(C_dA_o√(2gh)) from h₁ to 0 gives the 2A_t√h₁ form.

Step-by-step solution

1. Areas

   $$A_t = \pi(3.5)^{2}/4 = 9.621 m^{2}, A_o = 0.00385 m^{2}$$

2. Formula

   $$t = \dfrac{2A_t\left(\sqrt{h_1}-\sqrt{h_2}\right)}{C_d A_o\sqrt{2g}}$$

3. Substituting

   $$t = 2(9.621)(\sqrt3.0)/[0.62(0.00385)\sqrt(2\times9.81)]$$

4. Numerator

   $$2(9.621)(1.732) = 33.329$$

5. Denominator

   $$0.62(0.00385)(4.429) = 0.010569$$

6. Result

   $$t = 3,153 s = 52.6 \min$$

Why the other options are there

• 1,577 s (factor of 2 dropped)
• 1,577 s (steady discharge assumed)

Reference: FE Reference Handbook — Fluid Mechanics → Time required to drain a tank