A fluid mechanics problem uses Hydrostatic pressure. Given unit weight (gamma) = 56.0000 lb/ft³; depth (h) = 12.5000 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

   $$List the givens: unit weight (gamma) = 56.0000 lb/ft^{3}, depth (h) = 12.5000 ft$$

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

5. Step 5 — Evaluate:

   $$p = 700.0\ \text{psf}$$

6. Step 6 — Check: returning p = 700.0 psf to

   $$p = \gamma h$$

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

Why the other options are there

• 1,400 — kept a factor of two that cancels in the correct rearrangement.
• 350.0 — dropped that same factor in the other direction.
• 770.0 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Bernoulli Equation

the Bernoulli equation across a pipe contraction Given pressure at 1 (p_1) = 164,800 Pa; velocity at 1 (V_1) = 2.5000 m/s; elevation at 1 (z_1) = 4.7000 m; velocity at 2 (V_2) = 1.7000 m/s; elevation at 2 (z_2) = 0.8000 m; specific weight (gamma) = 9,240 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) = 164,800 Pa, velocity at 1 (V_1) = 2.5000 m/s, elevation at 1 (z_1) = 4.7000 m, velocity at 2 (V_2) = 1.7000 m/s, elevation at 2 (z_2) = 0.8000 m, specific weight (gamma) = 9,240 N/m^3.

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

5. Step 5 — Evaluate:

   $$p_{2} = 202418\ \text{Pa}$$

6. Step 6 — Check: returning p_2 = 202,418 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

• 404,837 — kept a factor of two that cancels in the correct rearrangement.
• 101,209 — dropped that same factor in the other direction.
• 222,660 — rounded an intermediate value before the final step.

Reference: FE Handbook — Bernoulli Equation

A fluid mechanics problem uses Hydrostatic pressure. Given depth (h) = 93.0000 ft; pressure (p) = 4,915 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

   $$List the givens: depth (h) = 93.0000 ft, pressure (p) = 4,915 psf$$

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

5. Step 5 — Evaluate:

   $$\gamma = 52.8495\ \text{lb/ft³}$$

6. Step 6 — Check: returning gamma = 52.8495 lb/ft³ to

   $$p = \gamma h$$

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

Why the other options are there

• 105.7 — kept a factor of two that cancels in the correct rearrangement.
• 26.4247 — dropped that same factor in the other direction.
• 58.1344 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Bernoulli Equation

the Bernoulli equation for flow discharging from a tank nozzle Given pressure at 1 (p_1) = 336,900 Pa; velocity at 1 (V_1) = 0.8000 m/s; elevation at 1 (z_1) = 4.8000 m; pressure at 2 (p_2) = 203,500 Pa; elevation at 2 (z_2) = 3.2000 m; specific weight (gamma) = 9,320 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) = 336,900 Pa, velocity at 1 (V_1) = 0.8000 m/s, elevation at 1 (z_1) = 4.8000 m, pressure at 2 (p_2) = 203,500 Pa, elevation at 2 (z_2) = 3.2000 m, specific weight (gamma) = 9,320 N/m^3.

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

5. Step 5 — Evaluate:

   $$V_{2} = 17.6878\ \text{m/s}$$

6. Step 6 — Check: returning V_2 = 17.6878 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

• 35.3756 — kept a factor of two that cancels in the correct rearrangement.
• 8.8439 — dropped that same factor in the other direction.
• 19.4566 — rounded an intermediate value before the final step.

Reference: FE Handbook — Bernoulli Equation

A fluid mechanics problem uses Hydrostatic pressure. Given unit weight (gamma) = 56.0000 lb/ft³; pressure (p) = 5,282 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

   $$List the givens: unit weight (gamma) = 56.0000 lb/ft^{3}, pressure (p) = 5,282 psf$$

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

5. Step 5 — Evaluate:

   $$h = 94.3214\ \text{ft}$$

6. Step 6 — Check: returning h = 94.3214 ft to

   $$p = \gamma h$$

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

Why the other options are there

• 188.6 — kept a factor of two that cancels in the correct rearrangement.
• 47.1607 — dropped that same factor in the other direction.
• 103.8 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Bernoulli Equation

the Bernoulli equation applied to a venturi meter Given pressure at 1 (p_1) = 148,600 Pa; velocity at 1 (V_1) = 2.8000 m/s; elevation at 1 (z_1) = 1.0000 m; pressure at 2 (p_2) = 319,000 Pa; velocity at 2 (V_2) = 3.5000 m/s; specific weight (gamma) = 9,140 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) = 148,600 Pa, velocity at 1 (V_1) = 2.8000 m/s, elevation at 1 (z_1) = 1.0000 m, pressure at 2 (p_2) = 319,000 Pa, velocity at 2 (V_2) = 3.5000 m/s, specific weight (gamma) = 9,140 N/m^3.

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

5. Step 5 — Evaluate:

   $$z_{2} = -17.8681\ \text{m}$$

6. Step 6 — Check: returning z_2 = -17.8681 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

• -35.7362 — kept a factor of two that cancels in the correct rearrangement.
• -8.9340 — dropped that same factor in the other direction.
• -19.6549 — rounded an intermediate value before the final step.

Reference: FE Handbook — Bernoulli Equation

A fluid mechanics problem uses Hydrostatic pressure. Given unit weight (gamma) = 59.0000 lb/ft³; depth (h) = 39.0000 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

   $$List the givens: unit weight (gamma) = 59.0000 lb/ft^{3}, depth (h) = 39.0000 ft$$

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

5. Step 5 — Evaluate:

   $$p = 2301\ \text{psf}$$

6. Step 6 — Check: returning p = 2,301 psf to

   $$p = \gamma h$$

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

Why the other options are there

• 4,602 — kept a factor of two that cancels in the correct rearrangement.
• 1,151 — dropped that same factor in the other direction.
• 2,531 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Bernoulli Equation

the Bernoulli equation across a pipe contraction Given pressure at 1 (p_1) = 236,100 Pa; velocity at 1 (V_1) = 1.4000 m/s; elevation at 1 (z_1) = 2.3000 m; velocity at 2 (V_2) = 5.3000 m/s; elevation at 2 (z_2) = 0.1000 m; specific weight (gamma) = 9,720 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) = 236,100 Pa, velocity at 1 (V_1) = 1.4000 m/s, elevation at 1 (z_1) = 2.3000 m, velocity at 2 (V_2) = 5.3000 m/s, elevation at 2 (z_2) = 0.1000 m, specific weight (gamma) = 9,720 N/m^3.

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

5. Step 5 — Evaluate:

   $$p_{2} = 244539\ \text{Pa}$$

6. Step 6 — Check: returning p_2 = 244,539 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

• 489,078 — kept a factor of two that cancels in the correct rearrangement.
• 122,269 — dropped that same factor in the other direction.
• 268,993 — rounded an intermediate value before the final step.

Reference: FE Handbook — Bernoulli Equation

A fluid mechanics problem uses Hydrostatic pressure. Given depth (h) = 94.0000 ft; pressure (p) = 1,964 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

   $$List the givens: depth (h) = 94.0000 ft, pressure (p) = 1,964 psf$$

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

5. Step 5 — Evaluate:

   $$\gamma = 20.8936\ \text{lb/ft³}$$

6. Step 6 — Check: returning gamma = 20.8936 lb/ft³ to

   $$p = \gamma h$$

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

Why the other options are there

• 41.7872 — kept a factor of two that cancels in the correct rearrangement.
• 10.4468 — dropped that same factor in the other direction.
• 22.9830 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Fluid Mechanics → Bernoulli Equation

the Bernoulli equation for flow discharging from a tank nozzle Given pressure at 1 (p_1) = 157,400 Pa; velocity at 1 (V_1) = 1.7000 m/s; elevation at 1 (z_1) = 0.3000 m; pressure at 2 (p_2) = 266,200 Pa; elevation at 2 (z_2) = 2.0000 m; specific weight (gamma) = 9,630 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) = 157,400 Pa, velocity at 1 (V_1) = 1.7000 m/s, elevation at 1 (z_1) = 0.3000 m, pressure at 2 (p_2) = 266,200 Pa, elevation at 2 (z_2) = 2.0000 m, specific weight (gamma) = 9,630 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