A fluid mechanics problem uses Hydrostatic pressure. Given unit weight (gamma) = 59.0000 lb/ft³; depth (h) = 82.0000 ft, determine the pressure (p) in psf.

Start with the thinking

• The governing relation printed in this handbook section is Hydrostatic pressure.
• Everything except p is given, so isolate p symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Fluid Mechanics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$p = \gamma h$$

2. Step 2 — Rearrange the relation so that p stands alone on the left-hand side.

3. Step 3

   $$List the givens: unit weight (gamma) = 59.0000 lb/ft^{3}, depth (h) = 82.0000 ft$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$p = 4838\ \text{psf}$$

6. Step 6 — Check: returning p = 4,838 psf to

   $$p = \gamma h$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 9,676 — kept a factor of two that cancels in the correct rearrangement.
• 2,419 — dropped that same factor in the other direction.
• 5,322 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

the Bernoulli equation across a pipe contraction Given pressure at 1 (p_1) = 262,600 Pa; velocity at 1 (V_1) = 2.1000 m/s; elevation at 1 (z_1) = 2.1000 m; velocity at 2 (V_2) = 5.2000 m/s; elevation at 2 (z_2) = 4.1000 m; specific weight (gamma) = 9,090 N/m^3, determine the pressure at 2 (p_2) in Pa.

Start with the thinking

• The governing relation printed in this handbook section is Bernoulli equation.
• Everything except p_2 is given, so isolate p_2 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Bernoulli equation relates pressure, velocity, and elevation along a pipeline streamline.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange the relation so that p_2 stands alone on the left-hand side.

3. Step 3 — List the givens: pressure at 1 (p_1) = 262,600 Pa, velocity at 1 (V_1) = 2.1000 m/s, elevation at 1 (z_1) = 2.1000 m, velocity at 2 (V_2) = 5.2000 m/s, elevation at 2 (z_2) = 4.1000 m, specific weight (gamma) = 9,090 N/m^3.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$p_{2} = 233935\ \text{Pa}$$

6. Step 6 — Check: returning p_2 = 233,935 Pa to

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

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 467,871 — kept a factor of two that cancels in the correct rearrangement.
• 116,968 — dropped that same factor in the other direction.
• 257,329 — rounded an intermediate value before the final step.

Reference: FE Handbook — Bernoulli Equation

A fluid mechanics problem uses Hydrostatic pressure. Given depth (h) = 98.0000 ft; pressure (p) = 4,271 psf, determine the unit weight (gamma) in lb/ft³.

Start with the thinking

• The governing relation printed in this handbook section is Hydrostatic pressure.
• Everything except gamma is given, so isolate gamma symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Fluid Mechanics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$p = \gamma h$$

2. Step 2 — Rearrange the relation so that gamma stands alone on the left-hand side.

3. Step 3

   $$List the givens: depth (h) = 98.0000 ft, pressure (p) = 4,271 psf$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\gamma = 43.5816\ \text{lb/ft³}$$

6. Step 6 — Check: returning gamma = 43.5816 lb/ft³ to

   $$p = \gamma h$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 87.1633 — kept a factor of two that cancels in the correct rearrangement.
• 21.7908 — dropped that same factor in the other direction.
• 47.9398 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

the Bernoulli equation for flow discharging from a tank nozzle Given pressure at 1 (p_1) = 230,500 Pa; velocity at 1 (V_1) = 2.2000 m/s; elevation at 1 (z_1) = 3.2000 m; pressure at 2 (p_2) = 166,100 Pa; elevation at 2 (z_2) = 4.1000 m; specific weight (gamma) = 9,500 N/m^3, determine the velocity at 2 (V_2) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Bernoulli equation.
• Everything except V_2 is given, so isolate V_2 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Bernoulli equation relates pressure, velocity, and elevation along a pipeline streamline.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange the relation so that V_2 stands alone on the left-hand side.

3. Step 3 — List the givens: pressure at 1 (p_1) = 230,500 Pa, velocity at 1 (V_1) = 2.2000 m/s, elevation at 1 (z_1) = 3.2000 m, pressure at 2 (p_2) = 166,100 Pa, elevation at 2 (z_2) = 4.1000 m, specific weight (gamma) = 9,500 N/m^3.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$V_{2} = 10.9629\ \text{m/s}$$

6. Step 6 — Check: returning V_2 = 10.9629 m/s to

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

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 21.9258 — kept a factor of two that cancels in the correct rearrangement.
• 5.4814 — dropped that same factor in the other direction.
• 12.0592 — rounded an intermediate value before the final step.

Reference: FE Handbook — Bernoulli Equation

A fluid mechanics problem uses Hydrostatic pressure. Given unit weight (gamma) = 64.0000 lb/ft³; pressure (p) = 3,982 psf, determine the depth (h) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Hydrostatic pressure.
• Everything except h is given, so isolate h symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Fluid Mechanics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$p = \gamma h$$

2. Step 2 — Rearrange the relation so that h stands alone on the left-hand side.

3. Step 3

   $$List the givens: unit weight (gamma) = 64.0000 lb/ft^{3}, pressure (p) = 3,982 psf$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$h = 62.2188\ \text{ft}$$

6. Step 6 — Check: returning h = 62.2188 ft to

   $$p = \gamma h$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 124.4 — kept a factor of two that cancels in the correct rearrangement.
• 31.1094 — dropped that same factor in the other direction.
• 68.4406 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

the Bernoulli equation applied to a venturi meter Given pressure at 1 (p_1) = 106,100 Pa; velocity at 1 (V_1) = 2.3000 m/s; elevation at 1 (z_1) = 2.7000 m; pressure at 2 (p_2) = 226,800 Pa; velocity at 2 (V_2) = 5.8000 m/s; specific weight (gamma) = 9,020 N/m^3, determine the elevation at 2 (z_2) in m.

Start with the thinking

• The governing relation printed in this handbook section is Bernoulli equation.
• Everything except z_2 is given, so isolate z_2 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Bernoulli equation relates pressure, velocity, and elevation along a pipeline streamline.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange the relation so that z_2 stands alone on the left-hand side.

3. Step 3 — List the givens: pressure at 1 (p_1) = 106,100 Pa, velocity at 1 (V_1) = 2.3000 m/s, elevation at 1 (z_1) = 2.7000 m, pressure at 2 (p_2) = 226,800 Pa, velocity at 2 (V_2) = 5.8000 m/s, specific weight (gamma) = 9,020 N/m^3.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$z_{2} = -12.1263\ \text{m}$$

6. Step 6 — Check: returning z_2 = -12.1263 m to

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

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• -24.2527 — kept a factor of two that cancels in the correct rearrangement.
• -6.0632 — dropped that same factor in the other direction.
• -13.3390 — rounded an intermediate value before the final step.

Reference: FE Handbook — Bernoulli Equation

A fluid mechanics problem uses Hydrostatic pressure. Given unit weight (gamma) = 64.0000 lb/ft³; depth (h) = 58.5000 ft, determine the pressure (p) in psf.

Start with the thinking

• The governing relation printed in this handbook section is Hydrostatic pressure.
• Everything except p is given, so isolate p symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Fluid Mechanics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$p = \gamma h$$

2. Step 2 — Rearrange the relation so that p stands alone on the left-hand side.

3. Step 3

   $$List the givens: unit weight (gamma) = 64.0000 lb/ft^{3}, depth (h) = 58.5000 ft$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$p = 3744\ \text{psf}$$

6. Step 6 — Check: returning p = 3,744 psf to

   $$p = \gamma h$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 7,488 — kept a factor of two that cancels in the correct rearrangement.
• 1,872 — dropped that same factor in the other direction.
• 4,118 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

the Bernoulli equation across a pipe contraction Given pressure at 1 (p_1) = 283,900 Pa; velocity at 1 (V_1) = 1.1000 m/s; elevation at 1 (z_1) = 3.9000 m; velocity at 2 (V_2) = 4.5000 m/s; elevation at 2 (z_2) = 0.4000 m; specific weight (gamma) = 9,430 N/m^3, determine the pressure at 2 (p_2) in Pa.

Start with the thinking

• The governing relation printed in this handbook section is Bernoulli equation.
• Everything except p_2 is given, so isolate p_2 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Bernoulli equation relates pressure, velocity, and elevation along a pipeline streamline.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange the relation so that p_2 stands alone on the left-hand side.

3. Step 3 — List the givens: pressure at 1 (p_1) = 283,900 Pa, velocity at 1 (V_1) = 1.1000 m/s, elevation at 1 (z_1) = 3.9000 m, velocity at 2 (V_2) = 4.5000 m/s, elevation at 2 (z_2) = 0.4000 m, specific weight (gamma) = 9,430 N/m^3.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$p_{2} = 307754\ \text{Pa}$$

6. Step 6 — Check: returning p_2 = 307,754 Pa to

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

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 615,508 — kept a factor of two that cancels in the correct rearrangement.
• 153,877 — dropped that same factor in the other direction.
• 338,529 — rounded an intermediate value before the final step.

Reference: FE Handbook — Bernoulli Equation

A fluid mechanics problem uses Hydrostatic pressure. Given depth (h) = 75.0000 ft; pressure (p) = 4,983 psf, determine the unit weight (gamma) in lb/ft³.

Start with the thinking

• The governing relation printed in this handbook section is Hydrostatic pressure.
• Everything except gamma is given, so isolate gamma symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Fluid Mechanics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$p = \gamma h$$

2. Step 2 — Rearrange the relation so that gamma stands alone on the left-hand side.

3. Step 3

   $$List the givens: depth (h) = 75.0000 ft, pressure (p) = 4,983 psf$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\gamma = 66.4400\ \text{lb/ft³}$$

6. Step 6 — Check: returning gamma = 66.4400 lb/ft³ to

   $$p = \gamma h$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 132.9 — kept a factor of two that cancels in the correct rearrangement.
• 33.2200 — dropped that same factor in the other direction.
• 73.0840 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

the Bernoulli equation for flow discharging from a tank nozzle Given pressure at 1 (p_1) = 139,000 Pa; velocity at 1 (V_1) = 1.7000 m/s; elevation at 1 (z_1) = 2.4000 m; pressure at 2 (p_2) = 235,100 Pa; elevation at 2 (z_2) = 3.6000 m; specific weight (gamma) = 9,610 N/m^3, determine the velocity at 2 (V_2) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Bernoulli equation.
• Everything except V_2 is given, so isolate V_2 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The Bernoulli equation relates pressure, velocity, and elevation along a pipeline streamline.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange the relation so that V_2 stands alone on the left-hand side.

3. Step 3 — List the givens: pressure at 1 (p_1) = 139,000 Pa, velocity at 1 (V_1) = 1.7000 m/s, elevation at 1 (z_1) = 2.4000 m, pressure at 2 (p_2) = 235,100 Pa, elevation at 2 (z_2) = 3.6000 m, specific weight (gamma) = 9,610 N/m^3.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$V_{2} = 0.1000\ \text{m/s}$$

6. Step 6 — Check: returning V_2 = 0.1000 m/s to

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

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.2000 — kept a factor of two that cancels in the correct rearrangement.
• 0.0500 — dropped that same factor in the other direction.
• 0.1100 — rounded an intermediate value before the final step.

Reference: FE Handbook — Bernoulli Equation