Fluid Mechanics Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 6

Question 101Question →

A heavy transport vehicle uses a pneumatic (air) suspension system to maintain a constant ride height regardless of cargo weight. Place the following events in the operation of this pneumatic leveling system in the correct chronological order, starting from the moment a heavy load is placed on the vehicle.

Drag items to arrange them in the correct order

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Answer

The correct sequence is: load compresses springs, valve detects drop, air flows into springs, increased pressure raises vehicle.

The correct sequence follows the logical physical response of a pneumatic system to an external load. First, the load mechanically compresses the air springs. This downward movement triggers the control valve. The open valve then allows high-pressure air to flow from the reservoir into the springs. Finally, the newly added air increases the internal pressure, expanding the springs to lift the load and reset the system.

Step-by-Step Solution

1

Identify the initial triggering event.

The addition of heavy cargo forces the vehicle down, compressing the pneumatic springs.

Air is a compressible fluid, so adding weight decreases the volume of the air in the springs until the pressure balances the new weight.

2

Determine the system's mechanical response.

The leveling valve detects this drop in height and opens.

Pneumatic systems require a mechanical or electrical trigger to allow high-pressure air to enter the working components.

3

Trace the flow of the compressed gas.

Air moves from the storage reservoir into the springs.

Gas naturally flows from an area of higher pressure (reservoir) to an area of lower pressure (compressed springs).

4

Identify the final mechanical work performed by the gas.

The added air increases the pressure inside the springs, forcing them to expand and lift the vehicle.

As more gas is introduced into the flexible space, pressure increases, generating enough mechanical force to lift the load.

Key Concept

Operation of pneumatic control systems and the relationship between air pressure, volume, and mechanical force.

Question 102Question →

A vertical standpipe in a building's fire suppression system is completely filled with fresh water. The top of the water column is exactly $25\text{ meters}$ above a closed test valve on the ground floor. What is the hydrostatic pressure exerted by the water directly on this valve?

(Assume the density of water is $1000\text{ kg/m}^3$ and the acceleration due to gravity is $10\text{ m/s}^2$ . Provide your answer in kilopascals, $\text{kPa}$ .)

Show answer & explanation

Answer: 250

Answer

250

The hydrostatic pressure at the bottom of the water column is found using the equation

P = \rho g h

. Substituting the given values gives

1000 \times 10 \times 25 = 250,000\text{ Pascals}

. To express this in kilopascals, you divide by 1,000, resulting in exactly

250\text{ kPa}

.

Step-by-Step Solution

1

Identify the relevant formula for hydrostatic pressure.

P = \rho g h

Hydrostatic pressure at a given depth depends strictly on the fluid's density, gravity, and the depth of the fluid column.

2

Substitute the given values into the formula.

P = 1000\text{ kg/m}^3 \times 10\text{ m/s}^2 \times 25\text{ m}

These are the precise physical parameters provided in the scenario.

3

Calculate the total pressure in standard SI units (Pascals).

P = 250,000\text{ Pa}

Multiplying the standard SI units yields the pressure in Pascals, since

1\text{ Pa} = 1\text{ kg}/(\text{m}\cdot\text{s}^2)

.

4

Convert the pressure from Pascals to kilopascals.

250,000 / 1000 = 250\text{ kPa}

The question specifically requests the final numerical answer in kilopascals.

Key Concept

Hydrostatic pressure is directly proportional to the density of the fluid, the acceleration due to gravity, and the vertical depth of the fluid column (

P = \rho g h

), regardless of the total fluid volume.

Estimated Time:1m 0s

Question 103Question →

A pneumatic lifting system uses compressed air to operate a main lifting cylinder. Initially, the cylinder contains $2.0 \text{ m}^3$ of air at an absolute pressure of $100 \text{ kPa}$ . To prepare for a heavy lift, the system's compressor forces the piston inward, reducing the trapped air's volume to $0.5 \text{ m}^3$ . Assuming the temperature of the air remains constant during this process, what is the new absolute pressure inside the cylinder?

100 \text{ kPa}

25 \text{ kPa}

400 \text{ kPa}

200 \text{ kPa}

Show answer & explanation

Answer: $400 \text{ kPa}$

Answer

The new absolute pressure is

400 \text{ kPa}

.

According to Boyle's Law, the pressure of a fixed mass of gas is inversely proportional to its volume, provided the temperature remains constant. The equation is

P_1V_1 = P_2V_2

. By substituting the given values (

100 \text{ kPa} \times 2.0 \text{ m}^3 = P_2 \times 0.5 \text{ m}^3

), we find that reducing the volume to one-quarter of its original size forces the pressure to quadruple, resulting in a final pressure of

400 \text{ kPa}

.

Step-by-Step Solution

1

Identify the relevant gas law for a situation with changing pressure and volume at a constant temperature.

Boyle's Law (

P_1V_1 = P_2V_2

) applies to this scenario.

The problem specifies a trapped gas undergoing volume compression while maintaining a constant temperature.

2

Extract the known initial and final values from the problem statement.

Initial pressure (

P_1

) =

100 \text{ kPa}

. Initial volume (

V_1

) =

2.0 \text{ m}^3

. Final volume (

V_2

) =

0.5 \text{ m}^3

.

Organizing known variables is necessary before attempting to solve for the unknown final pressure (

P_2

).

3

Substitute the known values into Boyle's Law equation and solve for the unknown final pressure.

100 \times 2.0 = P_2 \times 0.5

. This simplifies to

200 = 0.5P_2

, which yields

P_2 = 400 \text{ kPa}

.

Calculating the algebraic solution provides the final absolute pressure required by the question.

Key Concept

Boyle's Law (Inverse Relationship of Pressure and Volume)

Question 104Question →

An open-ended U-shaped tube contains two immiscible static liquids, Liquid A and Liquid B, which meet at a distinct separation line. Above this horizontal separation line, the column of Liquid A on the left side is exactly $20\text{ cm}$ tall. The column of Liquid B on the right side is exactly $30\text{ cm}$ tall. If the density of Liquid A is $1.5\text{ g/cm}^3$ , what must be the density of Liquid B to maintain this equilibrium?

1.0\text{ g/cm}^3

2.25\text{ g/cm}^3

0.67\text{ g/cm}^3

1.5\text{ g/cm}^3

Show answer & explanation

Answer: $1.0\text{ g/cm}^3$

Answer

The density of Liquid B is

1.0\text{ g/cm}^3

.

In an open U-tube manometer at equilibrium, the hydrostatic pressure exerted by both fluid columns at their meeting point must be equal. Since pressure equals density times gravity times height (

P = \rho g h

), we can drop the constant gravity and set

\rho_A \times h_A = \rho_B \times h_B

. Multiplying Liquid A's density (

1.5

) by its height (

20

) gives

30

. To balance this, Liquid B must have a density that, when multiplied by its height (

30

), also equals

30

. Therefore, Liquid B's density is

1.0\text{ g/cm}^3

.

Step-by-Step Solution

1

Identify the relevant physics principle for a static fluid system.

The hydrostatic pressure at the horizontal interface must be equal on both sides (

P_A = P_B

).

If the pressures were not equal at the interface, the fluid would move until equilibrium is reached.

2

Set up the pressure equivalence equation.

\rho_A \times g \times h_A = \rho_B \times g \times h_B

Hydrostatic pressure is the product of fluid density, gravitational acceleration, and depth.

3

Cancel out gravity (

g

) and substitute the given values into the equation.

1.5 \times 20 = \rho_B \times 30

Gravity is constant on both sides, and the values provided are density (

1.5\text{ g/cm}^3

) and heights (

20\text{ cm}

and

30\text{ cm}

).

4

Solve for the unknown density (

\rho_B

).

30 = 30 \times \rho_B \Rightarrow \rho_B = 1.0\text{ g/cm}^3

Isolating the variable gives the exact density required to balance the pressure.

Key Concept

Hydrostatic pressure depends on the product of a fluid's density and its vertical depth.

Question 105Question →

A municipal water network is supplied by a central water tower. The free surface of the water inside the tower is maintained at a constant elevation of $50\text{ meters}$ above sea level. The plumbing network is a closed, interconnected system, and the water is completely static.

Arrange the following four locations within this network in order from the LOWEST hydrostatic water pressure to the HIGHEST hydrostatic water pressure:

Drag items to arrange them in the correct order

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Answer

The correct order from lowest pressure to highest pressure is: Hilltop valve, Second-floor residential tap, Street-level fire hydrant, and Basement water main.

In a closed, static fluid system, hydrostatic pressure is determined entirely by the depth of the fluid relative to its highest free surface. By calculating the depth of each location from the 50-meter water tower surface, we find that locations at higher absolute elevations have shallower water columns above them, resulting in lower pressure. Conversely, locations at lower elevations have deeper water columns and thus higher pressure. Therefore, ranking the locations from highest elevation to lowest elevation correctly orders them from lowest hydrostatic pressure to highest hydrostatic pressure.

Step-by-Step Solution

1

Identify the reference point for hydrostatic pressure.

The reference point is the highest free surface of the interconnected water system, which is the water level in the tower at an elevation of 50 meters above sea level.

Hydrostatic pressure in a connected, static fluid system is driven by the depth of the fluid column below its highest free surface.

2

Calculate the depth of each location relative to the free surface.

Hilltop valve depth:

50 - 40 = 10\text{ m}

. Residential tap depth:

50 - 20 = 30\text{ m}

. Fire hydrant depth:

50 - 10 = 40\text{ m}

. Basement main depth:

50 - (-5) = 55\text{ m}

.

Depth (

h

) is found by subtracting the elevation of the specific point from the elevation of the fluid surface.

3

Relate the calculated depths to hydrostatic pressure and order them.

Since pressure increases proportionally with depth, the order from lowest to highest depth (and thus pressure) is

10\text{ m}

,

30\text{ m}

,

40\text{ m}

, and

55\text{ m}

.

Hydrostatic pressure (

P

) is directly proportional to the fluid depth (

h

) according to the formula

P = \rho g h

.

Key Concept

Hydrostatic pressure in communicating vessels depends solely on the depth of the fluid from the highest free surface, independent of the shape of the pipes.

Question 106Question →

A heavy-duty vehicle's pneumatic braking system relies on an accumulator tank to store potential energy. Initially, a volume of $150 \text{ in}^3$ of air is compressed at an absolute pressure of $90 \text{ psi}$ inside the accumulator. When the operator engages the brakes, the air expands isothermally (at a constant temperature) into the connected brake chambers, causing the system's absolute pressure to drop to $30 \text{ psi}$ . What is the total volume that the air now occupies, in cubic inches?

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Answer: 450

Answer

450

According to Boyle's Law, the pressure and volume of a confined gas are inversely proportional when the temperature is held constant. The initial state is

90 \text{ psi}

and

150 \text{ in}^3

. The final pressure drops to

30 \text{ psi}

, which is one-third of the original pressure. Therefore, to balance the equation and conserve the quantity of gas, the volume must expand to three times its original size.

150 \times 3 = 450 \text{ in}^3

.

Step-by-Step Solution

1

Identify the known variables from the scenario.

P_1 = 90 \text{ psi}

,

V_1 = 150 \text{ in}^3

,

P_2 = 30 \text{ psi}

.

Establishing the initial and final states of the gas is necessary to select the correct formula.

2

Select the appropriate gas law.

Use Boyle's Law:

P_1V_1 = P_2V_2

.

The problem states the expansion happens isothermally (constant temperature) and involves inversely proportional pressure and volume changes.

3

Substitute the values and solve for

V_2

.

90 \times 150 = 30 \times V_2

, so

V_2 = 13500 / 30 = 450 \text{ in}^3

.

Mathematical calculation is required to find the unknown final volume.

Key Concept

Boyle's Law (Pressure-Volume relationship at constant temperature)

Alternative Method

Instead of using the formula

P_1V_1 = P_2V_2

algebraically, you can use proportional reasoning. The pressure drops from

90 \text{ psi}

to

30 \text{ psi}

, which is a reduction by a factor of 3 (

90 / 30 = 3

). Because pressure and volume are inversely proportional at a constant temperature, the volume must increase by that exact same factor. Therefore, simply multiply the initial volume by 3:

150 \times 3 = 450 \text{ in}^3

.

Estimated Time:1m 0s

Question 107Question →

A specialized deep-well chemical injection pump is designed to operate safely up to a specific maximum hydrostatic pressure limit. During factory calibration in a test well filled with fresh water (density of $1,000\text{ kg/m}^3$ ), the pump reaches its maximum safe pressure rating at exactly $50\text{ meters}$ of depth. For its field deployment, the pump is lowered into a vertical shaft filled with heavy drilling mud (density of $1,250\text{ kg/m}^3$ ). Assuming the pressure rating remains constant, at what depth will the pump reach its maximum hydrostatic pressure limit in the drilling mud?

40\text{ meters}

62.5\text{ meters}

50\text{ meters}

25\text{ meters}

Show answer & explanation

Answer: $40\text{ meters}$

Answer

The pump will reach its maximum safe pressure limit at

40\text{ meters}

of depth in the drilling mud.

Hydrostatic pressure is determined by the formula

P = \rho g h

. Because the drilling mud is denser (

1,250\text{ kg/m}^3

) than the fresh water (

1,000\text{ kg/m}^3

), it exerts more pressure for every meter of depth. Since the mud is

1.25

times as dense as the water (

1,250 / 1,000 = 1.25

), the pump will reach the maximum pressure limit at

1/1.25

(or

4/5

) of the original depth. Multiplying

50\text{ meters}

by

0.8

yields exactly

40\text{ meters}

.

Step-by-Step Solution

1

Identify the relationship between hydrostatic pressure, fluid density, and depth.

P = \rho \cdot g \cdot h

Hydrostatic pressure is directly proportional to both the fluid's density and the depth.

2

Set up an equation equating the pressure in both fluids.

\rho_{\text{water}} \cdot h_{\text{water}} = \rho_{\text{mud}} \cdot h_{\text{mud}}

Since the maximum pressure (

P

) and gravity (

g

) are identical in both scenarios, the product of density and depth must remain constant.

3

Substitute the known values and solve for the target depth.

1,000 \cdot 50 = 1,250 \cdot h_{\text{mud}}

, which simplifies to

50,000 = 1,250 \cdot h_{\text{mud}}

, resulting in

h_{\text{mud}} = 40

.

Dividing the constant pressure factor by the new, higher density yields the shallower depth required.

Key Concept

Hydrostatic pressure at a given depth is directly proportional to the density of the fluid. A denser fluid exerts more pressure per meter, requiring less depth to reach the same total pressure.

Question 108Question →

An open industrial mixing vat contains a liquid chemical with a specific gravity of $1.25$ . The liquid in the vat reaches a depth of $4\text{ meters}$ . Assuming the density of fresh water is $1,000\text{ kg/m}^3$ and the acceleration due to gravity is $10\text{ m/s}^2$ , what is the hydrostatic pressure exerted by the liquid at the bottom of the vat? (Enter your answer in kilopascals)

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Answer: 50

Answer

50

The hydrostatic pressure at the bottom of a fluid column is determined by the formula

P = \rho g h

. First, the specific gravity of

1.25

indicates the fluid is

1.25

times as dense as water, giving a density of

1,250\text{ kg/m}^3

. Multiplying this density by the acceleration due to gravity (

10\text{ m/s}^2

) and the depth (

4\text{ meters}

) yields

50,000\text{ Pascals}

. Since

1,000\text{ Pa}

equals

1\text{ kPa}

, this is equivalent to

50\text{ kPa}

.

Step-by-Step Solution

1

Determine the density of the fluid.

Density = 1,250 kg/m³

Specific gravity is the ratio of the fluid's density to the density of water. Multiplying 1.25 by 1,000 kg/m³ gives the actual density.

2

Calculate the hydrostatic pressure in Pascals.

Pressure = 50,000 Pa

Hydrostatic pressure is the product of density, gravity, and depth (1,250 × 10 × 4 = 50,000).

3

Convert the pressure to kilopascals (kPa).

50 kPa

The question asks for the answer in kilopascals, so divide the Pascal value by 1,000.

Key Concept

Hydrostatic Pressure

Question 109Question →

A sealed pneumatic cylinder in a robotic assembly arm contains air at an absolute pressure of $20 \text{ psi}$ . The cylinder has an initial internal volume of $60 \text{ cm}^3$ . During a rapid compression stroke at a constant temperature, the piston reduces the volume to $15 \text{ cm}^3$ . What is the new absolute pressure inside the cylinder?

80 \text{ psi}

5 \text{ psi}

15 \text{ psi}

60 \text{ psi}

Show answer & explanation

Answer: $80 \text{ psi}$

Answer

The absolute pressure increases to

80 \text{ psi}

because reducing the volume by a factor of 4 causes the pressure to multiply by 4.

According to Boyle's Law, pressure and volume are inversely proportional when temperature is constant (

P_1V_1 = P_2V_2

). The initial pressure-volume product is

20 \times 60 = 1200

. To maintain this constant with a new volume of

15 \text{ cm}^3

, the new pressure must be

1200 \div 15 = 80 \text{ psi}

. Alternatively, recognizing that the volume is reduced by a factor of 4 (

60 \div 15 = 4

) means the pressure must increase by a factor of 4 (

20 \times 4 = 80 \text{ psi}

).

Step-by-Step Solution

1

Identify the initial conditions and the relevant gas law.

Initial pressure (

P_1

) is

20 \text{ psi}

and initial volume (

V_1

) is

60 \text{ cm}^3

. Since temperature is constant, use Boyle's Law:

P_1V_1 = P_2V_2

.

Boyle's Law defines the inversely proportional relationship between the pressure and volume of a gas when temperature remains unchanged.

2

Set up the equation with the known values.

20 \text{ psi} \times 60 \text{ cm}^3 = P_2 \times 15 \text{ cm}^3

Substituting the given parameters prepares the equation to isolate the unknown final pressure (

P_2

).

3

Solve for the final pressure (

P_2

).

1200 = P_2 \times 15

, which simplifies to

P_2 = \frac{1200}{15} = 80 \text{ psi}

.

Isolating

P_2

reveals the new absolute pressure after the compression stroke.

Key Concept

Boyle's Law (Pressure-Volume Inverse Relationship)

Question 110Question →

A mechanical pressure relief valve is installed on a pneumatic air receiver tank to prevent over-pressurization. Arrange the following events in the correct chronological sequence, from the moment the tank begins to over-pressurize until the system stabilizes.

Drag items to arrange them in the correct order

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Answer

The sequence begins with system pressure exceeding the setpoint. This creates a pneumatic force that overcomes the spring tension, unseating the valve. The compressed air then vents, lowering system pressure until the spring reseats the valve.

The correct operational sequence of a pneumatic relief valve relies on the balance between gas pressure and mechanical spring force. Initially, an over-pressure condition occurs in the tank. Because force equals pressure multiplied by area, this high pressure exerts an upward force on the valve plug. When this pneumatic force exceeds the downward force of the calibrated spring, the plug lifts. This physical unseating creates a pathway, allowing the highly compressed gas to rapidly expand and vent to the lower-pressure atmosphere. As the gas escapes, the internal tank pressure drops until the pneumatic force is weaker than the spring tension, allowing the spring to snap the valve closed again.

Step-by-Step Solution

1

Identify the initial condition that triggers the safety system.

The air pressure inside the tank rises above the safe setpoint limit.

A pneumatic relief sequence must begin with an over-pressure event.

2

Determine the mechanical interaction between the compressed air and the valve.

The pneumatic force overcomes the mechanical spring tension.

For the valve to operate, fluid force (Pressure × Area) must defeat the physical mechanism keeping it closed.

3

Establish the physical movement resulting from the force imbalance.

The valve plug lifts off its seat to expose the exhaust port.

The net upward pneumatic force pushes the valve open.

4

Identify the behavior of the gas once the pathway is open.

Compressed air escapes into the atmosphere, reducing tank pressure.

Gas naturally expands and flows from areas of high pressure to low pressure.

5

Determine the condition that ends the sequence.

The spring pushes the valve closed as the pneumatic pressure decreases.

When tank pressure drops sufficiently, the pneumatic force is no longer strong enough to resist the spring.

Key Concept

Pneumatic System Control and Pressure Relief Operations

Question 111Question →

A remote-controlled submersible is conducting an inspection in a freshwater lake. It initially hovers at a static depth of $10\text{ meters}$ below the surface. Later, the operator commands the submersible to descend to a depth of $3,000\text{ centimeters}$ . Assuming the water density remains constant, how does the hydrostatic pressure on the submersible at the new depth compare to the pressure at the initial depth?

The hydrostatic pressure is exactly

3\text{ times}

as great.

The hydrostatic pressure is exactly

300\text{ times}

as great.

The hydrostatic pressure remains the same because the volume of the submersible displacing the water has not changed.

The hydrostatic pressure increases exponentially because the deeper water is highly compressed to a greater density.

Show answer & explanation

Answer: The hydrostatic pressure is exactly $3\text{ times}$ as great.

Answer

The hydrostatic pressure is exactly 3 times as great.

Hydrostatic pressure is calculated using the formula

P = \rho \cdot g \cdot h

. Because the fluid (freshwater) and gravity remain constant, the pressure is directly proportional to the depth (

h

). To accurately compare the depths, they must be in the same unit. Converting the new depth to meters (

3,000\text{ cm} = 30\text{ m}

) reveals that the submersible has gone from

10\text{ meters}

to

30\text{ meters}

. Since the depth has increased by a factor of 3, the hydrostatic pressure also increases by a factor of 3.

Step-by-Step Solution

1

Identify the formula for hydrostatic pressure.

P = \rho \cdot g \cdot h

, meaning pressure is directly proportional to depth (

h

) when density (

\rho

) and gravity (

g

) are constant.

Establishing the mathematical relationship is necessary to compare the two states.

2

Convert the second depth into meters to match the initial depth's units.

3,000\text{ cm} \div 100\text{ cm/m} = 30\text{ meters}

.

Physical quantities must be in the same units before they can be accurately compared or divided.

3

Compare the new depth to the initial depth.

30\text{ meters} \div 10\text{ meters} = 3

.

Because the depth has tripled, the directly proportional hydrostatic pressure also triples.

Key Concept

Hydrostatic pressure in an incompressible fluid is directly proportional to its depth.

Estimated Time:1m 0s

Question 112Question →

An oceanarium features a massive main saltwater tank with several viewing windows and maintenance hatches of varying sizes. Arrange the following locations in order of the hydrostatic pressure exerted by the seawater at their center points, from the lowest pressure to the highest pressure.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from lowest to highest pressure corresponds strictly to depth: the 5-foot deep panoramic panel, the 10-foot deep porthole, the 15-foot deep observation window, and the 20-foot deep maintenance hatch.

Hydrostatic pressure is calculated based on the vertical depth below the surface. Because all windows are in the same saltwater tank, fluid density and gravity are constant. Therefore, the pressure increases linearly with depth, making the 5-foot depth the lowest pressure and the 20-foot depth the highest, regardless of the physical size of each window.

Step-by-Step Solution

1

Recall the principles determining hydrostatic pressure in a static fluid.

Hydrostatic pressure is determined by the equation

P = \rho gh

, meaning it depends only on fluid density (

\rho

), gravity (

g

), and depth (

h

).

Establishing the governing physical rule prevents confusion from irrelevant details like window size or shape.

2

Identify the relevant variable for each location.

The relevant variable is the depth of the center point for each window: 5 ft, 10 ft, 15 ft, and 20 ft.

Extracting the depth isolates the only factor that changes hydrostatic pressure in this single, uniform tank.

3

Sort the locations by depth to determine pressure order.

Ranking from lowest depth to highest depth gives: 5 ft < 10 ft < 15 ft < 20 ft.

Since pressure is directly proportional to depth, ordering from shallowest to deepest correctly ranks the pressure from lowest to highest.

Key Concept

Hydrostatic pressure in a static fluid is directly proportional to depth and is completely independent of the total fluid volume, container shape, or the area of the surface it acts upon.

Estimated Time:1m 0s

Question 113Question →

At a commercial fuel distribution center, two open-top vertical storage tanks hold the exact same grade of liquid diesel. Tank X is a wide cylinder with a $12\text{-meter}$ base diameter. Tank Y is a tall, narrow cylinder with a $2\text{-meter}$ base diameter. Both tanks are currently filled with diesel to an exact depth of $8\text{ meters}$ . Which of the following statements correctly compares the hydrostatic pressure exerted by the fuel on the bottom surface of each tank?

The pressure is identical in both tanks, because hydrostatic pressure depends only on fluid depth and density.

The pressure in Tank X is greater, because it contains a much larger total weight and volume of fuel.

The pressure in Tank Y is greater, because the narrower base restricts the fuel and concentrates its force.

The pressure cannot be compared without first calculating the total surface area of the tank walls.

Show answer & explanation

Answer: The pressure is identical in both tanks, because hydrostatic pressure depends only on fluid depth and density.

Answer

The pressure is identical in both tanks, because hydrostatic pressure depends only on fluid depth and density.

Hydrostatic pressure is determined by the formula

P = \rho gh

(where

\rho

is fluid density,

g

is gravity, and

h

is vertical fluid depth). Because both tanks hold the exact same fluid at the exact same depth, the pressure at the bottom is identical. The overall width, shape, or total volume of the tank does not affect hydrostatic pressure.

Step-by-Step Solution

1

Identify the standard formula for calculating hydrostatic pressure.

P = \rho \cdot g \cdot h

(Pressure = density

\times

gravity

\times

depth).

To determine which physical variables actually affect the pressure at the bottom of the fluid column.

2

Analyze the variables provided for both storage tanks.

Both tanks contain the exact same diesel fuel (identical density,

\rho

) and are filled to the same

8\text{-meter}

level (identical depth,

h

).

To compare the factors that influence hydrostatic pressure in this specific scenario.

3

Evaluate the effect of the different tank diameters (

12\text{ meters}

vs.

2\text{ meters}

).

The diameters and total fluid volumes do not appear in the hydrostatic pressure formula and therefore have no effect on the bottom pressure.

To confirm that the shape and width of the container are irrelevant to hydrostatic pressure at a given depth.

Key Concept

Hydrostatic pressure at a given depth depends strictly on the density of the fluid and the vertical depth itself, regardless of the container's shape, base width, or total fluid volume.

Question 114Question →

A hydrostatic pressure sensor at the bottom of an upright agricultural water cistern reads $45\text{ psi}$ when the water depth is $30\text{ feet}$ . During a dry season, the water level drops to a depth of $18\text{ feet}$ . Assuming the fluid density remains constant, what is the new hydrostatic pressure reading in $\text{psi}$ at the sensor?

Show answer & explanation

Answer: 27

Answer

27

Since hydrostatic pressure is directly proportional to depth (

P = \rho \cdot g \cdot h

), the ratio of pressure to depth remains constant. The initial ratio is

45\text{ psi} / 30\text{ feet} = 1.5\text{ psi/ft}

. At a new depth of

18\text{ feet}

, the pressure is

18\text{ feet} \times 1.5\text{ psi/ft} = 27\text{ psi}

.

Step-by-Step Solution

1

Identify the relationship between hydrostatic pressure and fluid depth.

Hydrostatic pressure is directly proportional to the depth of the fluid.

The formula for hydrostatic pressure is

P = \rho \cdot g \cdot h

, meaning if density (

\rho

) and gravity (

g

) are constant, the pressure changes linearly with depth (

h

).

2

Determine the pressure per foot of water depth.

45\text{ psi} \div 30\text{ feet} = 1.5\text{ psi}

per foot.

This establishes the constant of proportionality for the fluid in this specific cistern.

3

Calculate the new pressure at the reduced depth.

1.5\text{ psi/foot} \times 18\text{ feet} = 27\text{ psi}

.

Multiplying the pressure per foot by the new depth yields the corresponding hydrostatic pressure.

Key Concept

Hydrostatic pressure is directly proportional to fluid depth.

Question 115Question →

A structural engineer is evaluating two submerged sensor housings within a continuous freshwater facility. Housing X is mounted at the bottom of a narrow vertical standpipe, exactly $12\text{ meters}$ below the water's surface. Housing Y is mounted on the open floor of the main reservoir, also exactly $12\text{ meters}$ below the water's surface. Both locations share the same body of water. Which of the following statements correctly compares the hydrostatic pressure acting on the two housings?

The hydrostatic pressure is equal on both housings because they are situated at the identical vertical depth.

The hydrostatic pressure is greater on Housing Y because the expansive reservoir holds a much larger total weight of water.

The hydrostatic pressure is greater on Housing X because the narrow standpipe concentrates the fluid's downward force into a smaller area.

The hydrostatic pressure cannot be compared directly unless the depth measurements are mathematically converted into volumetric flow rates.

Show answer & explanation

Answer: The hydrostatic pressure is equal on both housings because they are situated at the identical vertical depth.

Answer

The hydrostatic pressure is equal on both housings because they are situated at the identical vertical depth.

The correct answer accurately states that hydrostatic pressure depends only on fluid depth. Because both housings are situated exactly

12\text{ meters}

below the surface of the same continuous body of water, the pressure exerted upon them is identical, regardless of the surrounding container geometry.

Step-by-Step Solution

1

Identify the factors that determine hydrostatic pressure.

Hydrostatic pressure (

P

) depends on fluid density (

\rho

), gravity (

g

), and vertical depth (

h

) according to the formula

P = \rho \cdot g \cdot h

.

This establishes the physical law governing the scenario.

2

Evaluate the fluid conditions for both housings.

Both housings are in the same continuous freshwater system, meaning the fluid density (

\rho

) is identical.

Eliminates fluid density as a variable between the two locations.

3

Compare the depth (

h

) and container shapes for both housings.

Housing X is in a narrow pipe at a

12\text{ m}

depth, and Housing Y is in a wide reservoir at a

12\text{ m}

depth. Container width and total fluid volume are not variables in the hydrostatic pressure formula.

Demonstrates that despite the drastically different container shapes, the equal depth guarantees equal pressure.

Key Concept

Hydrostatic pressure relies strictly on the vertical depth and density of the fluid, and is completely unaffected by the shape of the container or the total volume of the fluid.

Question 116Question →

A maintenance crew is inspecting a closed, static water pipeline fed by a mountain reservoir. They monitor static pressure sensors located at different points along the pipeline's path.

Arrange the following sensor locations in order from the LOWEST hydrostatic pressure to the HIGHEST hydrostatic pressure.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest pressure is: the surface sensor (0 feet), the sensor in the 24-inch pipe (30 feet), the sensor in the 12-inch pipe (60 feet), and the sensor at the bend (90 feet).

Hydrostatic pressure in a continuous, static fluid is directly proportional to its vertical depth below the free surface (

P = \rho g h

). Factors like pipe diameter, pipe orientation, or the total volume of fluid in a specific section do not influence the pressure at a specific point. Therefore, ranking from lowest to highest pressure simply requires ordering the depths from shallowest to deepest: 0 feet, 30 feet, 60 feet, and finally 90 feet.

Step-by-Step Solution

1

Identify the principle governing hydrostatic pressure in a static fluid.

Hydrostatic pressure is determined by the formula

P = \rho g h

, where

h

is the vertical depth from the free surface of the fluid.

Establishing the core rule is necessary to evaluate which variables provided in the scenario actually affect the pressure.

2

Evaluate the effect of pipe diameter and shape on the pressure.

The 12-inch diameter, 24-inch diameter, and horizontal pipe bend do not affect static pressure.

According to Pascal's principle and fluid statics, pressure at a given depth is independent of the container's volume, shape, or width.

3

Rank the locations based solely on their vertical depth below the reservoir surface.

Surface (0 ft) < 30 ft < 60 ft < 90 ft.

Since vertical depth is the only changing variable that affects the pressure, the ranking strictly follows the depth from shallowest to deepest.

Key Concept

Hydrostatic pressure in a continuous fluid depends exclusively on fluid density, gravity, and vertical depth, remaining entirely independent of the container's volume or shape.

Question 117Question →

A hydroelectric dam holds back a massive artificial lake. Along the submerged face of the dam, there is a small diagnostic sensor port and a massive turbine water intake. Both the sensor port and the center of the turbine intake are located exactly $45\text{ meters}$ below the surface of the lake. The diameter of the turbine intake is twenty times larger than the diagnostic port. When the water in the lake is perfectly still, how does the hydrostatic pressure at the diagnostic port compare to the pressure at the center of the turbine intake?

The pressure is identical at both locations because they are positioned at the exact same depth.

The pressure is twenty times greater at the turbine intake because its larger diameter captures more force from the water.

The pressure is significantly higher at the diagnostic port because the fluid is compressed into a narrower opening.

The pressure at both locations is determined by the total mass of the reservoir, so the depth of the specific openings is not the primary factor.

Show answer & explanation

Answer: The pressure is identical at both locations because they are positioned at the exact same depth.

Answer

The pressure is identical at both locations because they are positioned at the exact same depth.

Hydrostatic pressure in a static fluid is calculated using the formula

P = \rho g h

(Pressure = density × gravity × depth). Because both the diagnostic port and the center of the turbine intake are located exactly

45\text{ meters}

below the surface, they experience the exact same water pressure. The size, diameter, or area of the opening affects the total force exerted on that opening, but it does not change the pressure, which is a measure of force per unit area.

Step-by-Step Solution

1

Identify the factors that determine hydrostatic pressure in a static fluid.

Hydrostatic pressure (

P

) is determined by the formula

P = \rho g h

, where

\rho

is fluid density,

g

is gravity, and

h

is depth.

This establishes the physical rule governing the scenario, showing that only depth matters when the fluid is consistent.

2

Compare the depth of the two locations provided in the scenario.

Both the diagnostic port and the turbine intake center are located at a depth of

45\text{ meters}

.

Since the

h

value is identical for both locations, the pressure calculated from

P = \rho g h

will be equal.

3

Evaluate the effect of the opening sizes on the hydrostatic pressure.

The diameter of the openings has no effect on the hydrostatic pressure, though it does affect the total force (

F = P \times A

).

It separates the concept of pressure (force per unit area) from the concept of total force, confirming that the pressure remains identical.

Key Concept

Hydrostatic pressure in a liquid depends only on the depth of the liquid and its density, and is entirely independent of the shape or size of the container or opening.

Question 118Question →

A vertical ballast column on a maritime vessel is completely filled with fresh water. A pressure gauge is mounted precisely $60\text{ feet}$ below the open surface of the water. If the specific weight of the fresh water is $62.4\text{ lb/ft}^3$ , what is the hydrostatic gauge pressure at this depth in pounds per square inch (psi)?

Show answer & explanation

Answer: 26

Answer

The hydrostatic pressure is 26 psi.

The correct calculation requires two phases: first, determining the pressure based on depth and specific weight, which yields

3744\text{ psf}

(

60 \times 62.4

). Second, converting area units from square feet to square inches by dividing by 144, resulting in exactly

26\text{ psi}

.

Step-by-Step Solution

1

Calculate the hydrostatic pressure in pounds per square foot (psf).

62.4\text{ lb/ft}^3 \times 60\text{ ft} = 3744\text{ lb/ft}^2

(psf)

Hydrostatic pressure is the product of the fluid's specific weight and the depth of the fluid column.

2

Convert the pressure from psf to pounds per square inch (psi).

3744\text{ psf} \div 144\text{ in}^2/\text{ft}^2 = 26\text{ psi}

There are

144\text{ square inches}

in

1\text{ square foot}

(

12\text{ inches} \times 12\text{ inches}

). Dividing the psf value by 144 converts the units appropriately.

Key Concept

Hydrostatic Pressure Calculation and Unit Conversion

Fluid Mechanics

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