Fluid Mechanics

118 soru

Tüm türler118 Çoktan seçmeli60 Doğru / Yanlış4 Sayısal22 Eşleştirme12 Sıralama20

An underwater research drone is deployed in a freshwater lake. It records a hydrostatic pressure reading at a depth of 10 meters. The drone then descends vertically and records a second pressure reading at a depth of 40 meters. Assuming the water density is uniform, how does the second pressure reading compare to the first?

The second reading is four times as large as the first reading.

The second reading is one-fourth the magnitude of the first reading.

Both readings display equivalent values since they were taken in the same body of water.

The pressure difference depends on the weight of the drone rather than the change in depth.

Cevabı ve açıklamayı göster

Cevap: The second reading is four times as large as the first reading.

Cevap

The second reading is four times as large as the first reading.

Hydrostatic pressure is calculated using the equation

P = \rho g h

. Because the density of the water (

\rho

) and the acceleration due to gravity (

g

) remain constant in this scenario, the pressure (

P

) is directly proportional to the depth (

h

). When the drone descends from 10 meters to 40 meters, the depth increases by a factor of 4. Therefore, the hydrostatic pressure also increases by exactly a factor of 4.

Adım Adım Çözüm

Identify the mathematical formula for hydrostatic pressure.

P = \rho g h

, where

P

is pressure,

\rho

is density,

g

is gravity, and

h

is depth.

This formula establishes the mathematical relationship between fluid pressure and depth.

Determine the proportional relationship between the two recorded depths.

The second depth (40 meters) is exactly four times the first depth (10 meters).

Calculating the ratio of the depths allows for a direct comparison of the resulting pressures.

Apply the depth ratio to the pressure formula.

Since density and gravity are constant, increasing

h

by a factor of 4 increases

P

by a factor of 4.

Hydrostatic pressure is directly proportional to fluid depth.

Anahtar Kavram

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

Soru 82Soru →

A heavy steel transport barge leaves the salty water of an ocean bay and enters a freshwater river. The total weight of the barge and its cargo remains exactly the same. Given that saltwater has a significantly higher density than freshwater, how will the depth that the barge sinks into the water (its draft) change, and why?

It will sink deeper into the water, because it must displace a larger volume of the less-dense freshwater to equal its own weight.

Its draft will remain exactly the same, because the required buoyant force depends entirely on the barge's total weight, which has not changed.

It will float higher in the water, because the lighter freshwater pushes upward against the hull with a greater pressure.

It will sink deeper into the water, because the lack of salt in the river water causes the barge's overall mass to increase.

Cevabı ve açıklamayı göster

Cevap: It will sink deeper into the water, because it must displace a larger volume of the less-dense freshwater to equal its own weight.

Cevap

The barge will sink deeper into the water because it must displace a larger volume of the less-dense freshwater to equal its own weight.

According to Archimedes' principle, a floating object displaces a weight of fluid equal to its own true weight. Because the barge's weight is constant, it must displace the exact same *weight* of water in both locations to stay afloat. Since freshwater is less dense (weighs less per unit of volume) than saltwater, a larger physical *volume* of freshwater must be displaced to equal the barge's massive weight. To displace more volume, the barge must sink deeper into the water, increasing its draft.

Adım Adım Çözüm

Identify the conditions for a floating object.

A floating object is in equilibrium, meaning the upward buoyant force (

F_b

) must exactly equal the object's downward true weight (

W

This is the fundamental principle of flotation derived from Archimedes' principle.

Relate the buoyant force to fluid density and displaced volume.

Buoyant force equals the weight of the displaced fluid, expressed as

F_b = \rho \times V \times g

, where

\rho

is fluid density,

V

is displaced volume, and

g

is gravity.

This formula shows how fluid properties generate the upward force.

Apply the relationship to the barge moving from saltwater to freshwater.

Since

F_b

must remain constant (because the barge's weight is constant), and the fluid density (

\rho

) decreases in freshwater, the displaced volume (

V

) must increase to compensate.

Displacing a larger volume of water means the hull of the barge must sink deeper (increasing its draft).

Anahtar Kavram

Buoyancy and floating equilibrium (Archimedes' Principle)

Tahmini Süre:1m 0s

Soru 83Soru →

During a hydro-engineering project, a completely sealed hollow steel sphere is used as an underwater sensor housing. The sphere has an external volume of $2.5 \text{ m}^3$ and a dry weight of $35,000 \text{ N}$ in the air. When completely submerged in a testing pool where the water's weight density is $10,000 \text{ N/m}^3$ , what is the net downward force (apparent weight) of the housing in Newtons?

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Cevap: 10000

Cevap

10000

The net downward force, or apparent weight, is found by subtracting the upward buoyant force from the object's true dry weight. Since the object displaces

2.5 \text{ m}^3

of water with a weight density of

10,000 \text{ N/m}^3

, the buoyant force is

25,000 \text{ N}

. Subtracting this upward force from the

35,000 \text{ N}

downward dry weight leaves a net downward force of

10,000 \text{ N}

Adım Adım Çözüm

Calculate the buoyant force acting on the submerged housing.

25,000 N

According to Archimedes' principle, the buoyant force equals the weight of the fluid displaced by the object's external volume (

2.5 \text{ m}^3 \times 10,000 \text{ N/m}^3 = 25,000 \text{ N}

Calculate the apparent weight (net downward force).

10,000 N

The apparent weight is the actual weight in air minus the upward buoyant force (

35,000 \text{ N} - 25,000 \text{ N} = 10,000 \text{ N}

Anahtar Kavram

Archimedes' Principle and Apparent Weight

Soru 84Soru →

Two outdoor swimming pools are filled with fresh water to an exact depth of $3\text{ meters}$ . The first is a small backyard wading pool, while the second is a massive Olympic-sized pool holding a much larger total volume of water. How does the hydrostatic pressure at the very bottom of the backyard pool compare to the hydrostatic pressure at the bottom of the Olympic-sized pool?

The pressure is exactly equal in both pools because they share the same fluid density and water depth.

The pressure is significantly greater in the Olympic-sized pool because it contains a larger total mass of water.

The pressure at the bottom of both pools is exactly

3\text{ Pascals}

because the depth is exactly

3\text{ meters}

The pressure depends entirely on the buoyant force, which requires a heavy object to be placed in the water.

Cevabı ve açıklamayı göster

Cevap: The pressure is exactly equal in both pools because they share the same fluid density and water depth.

Cevap

The pressure is exactly equal in both pools because they share the same fluid density and water depth.

The correct answer accurately states that hydrostatic pressure relies solely on depth and fluid density. Because both pools contain fresh water and are filled to the exact same depth of

3\text{ meters}

, the pressure at the bottom of both pools is identical, despite their vast difference in total volume.

Adım Adım Çözüm

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

Hydrostatic pressure is determined by the fluid's density, the acceleration due to gravity, and the depth of the fluid.

To evaluate whether factors like total volume or container shape influence the final pressure reading.

Compare the relevant factors between the two distinct swimming pools.

Both pools contain fresh water (same density) and have an exact depth of

3\text{ meters}

To determine if there are any differences in the variables that actually calculate pressure.

Deduce the relative pressure at the bottom of each pool based on the identified factors.

Since both fluid density and depth are identical, the hydrostatic pressure at the bottom must be identical.

According to the principles of fluid mechanics, hydrostatic pressure is independent of the container's width or the fluid's total mass.

Anahtar Kavram

Hydrostatic pressure is dependent only on fluid density and depth, not on the container's shape or the total volume of fluid.

Soru 85Soru →

A heavy-duty diving bell is being tested in a deep saltwater facility. At a depth of $10\text{ meters}$ , the hydrostatic pressure exerted on its exterior is measured to be $100\text{ kilopascals (kPa)}$ . If the diving bell is lowered further to a depth of $30\text{ meters}$ , what will the new hydrostatic pressure be in $\text{kPa}$ , assuming the fluid density remains constant?

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Cevap: 300

Cevap

The new hydrostatic pressure will be 300 kPa.

Because hydrostatic pressure is directly proportional to depth in a fluid of constant density, tripling the depth from 10 meters to 30 meters results in exactly three times the pressure. Thus, 100 kPa multiplied by 3 equals 300 kPa.

Adım Adım Çözüm

Identify the relationship between hydrostatic pressure and depth.

Pressure is directly proportional to depth.

This is a fundamental principle of hydrostatic pressure for a fluid with constant density.

Calculate the ratio of the new depth to the initial depth.

30\text{ m} / 10\text{ m} = 3

This ratio tells us how much the pressure will increase.

Calculate the new pressure.

100\text{ kPa} \times 3 = 300\text{ kPa}

Applying the direct proportion to find the final pressure value.

Anahtar Kavram

Hydrostatic pressure increases in direct proportion to the depth of the fluid, assuming fluid density remains constant.

Soru 86Soru →

Two solid machine components, Component X and Component Y, have the exact same volume but are cast from different metals. Component X weighs $15 \text{ N}$ in air, and Component Y weighs $45 \text{ N}$ in air.

Both components are separately lowered into identical displacement tanks filled to the brim with water. When Component X is completely submerged, exactly $5 \text{ N}$ of water spills out of the tank. When Component Y is completely submerged, what is the weight of the water that spills out?

5 \text{ N}

15 \text{ N}

40 \text{ N}

45 \text{ N}

Cevabı ve açıklamayı göster

Cevap: $5 \text{ N}$

Cevap

The weight of the water that spills out is

5 \text{ N}

The correct answer is

5 \text{ N}

. When an object is fully submerged in a fluid, it displaces a volume of fluid equal to its own volume. According to Archimedes' Principle, the buoyant force acting on the object is equal to the weight of this displaced fluid. Because Component Y has the exact same volume as Component X, it must displace the exact same amount of water, meaning the weight of the overflowing water remains

5 \text{ N}

. The object's weight in air does not affect how much fluid it displaces once it is completely submerged.

Adım Adım Çözüm

Identify the relationship between the two components.

Both Component X and Component Y share the exact same volume, despite having different weights in air.

The physical volume of an object dictates how much fluid it can displace when fully submerged.

Apply Archimedes' principle of displacement.

Because both components have identical volumes, completely submerging either one will displace the exact same volume of water.

A fully submerged object always displaces a volume of fluid equal to its own total volume, regardless of its density or mass.

Determine the weight of the displaced water for Component Y.

Since Component X displaced

5 \text{ N}

of water, Component Y must also displace

5 \text{ N}

of water.

Equal volumes of the same fluid (water) will naturally have the same weight.

Anahtar Kavram

Buoyancy and Archimedes' Principle

Tahmini Süre:1m 0s

Soru 87Soru →

A naval submarine uses internal ballast tanks to take in or pump out seawater, allowing it to change its overall weight while its total external volume remains exactly the same. Arrange the following operational states of the submarine in order of the submarine's OVERALL AVERAGE DENSITY, from LEAST dense to MOST dense.

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Cevap

The correct sequence from least dense to most dense is: Floating on the surface, followed by Hovering motionless, and finally Resting on the ocean floor.

Density directly dictates buoyancy states. An object with a density lower than seawater will float on the surface. An object with a density equal to seawater will hover neutrally. An object with a density greater than seawater will sink to the bottom. Therefore, the order from least to most dense is: floating, hovering, sinking.

Adım Adım Çözüm

Analyze the condition required for an object to float.

The state of 'Floating on the surface' represents the lowest density.

According to Archimedes' principle, an object floats (positive buoyancy) when its overall density is less than the density of the fluid it is in.

Analyze the condition required for an object to hover underwater.

The state of 'Hovering motionless' represents a medium density.

An object achieves neutral buoyancy and hovers when its weight perfectly matches the buoyant force of the fully displaced fluid, meaning its density is exactly equal to the fluid's density.

Analyze the condition required for an object to sink.

The state of 'Resting on the solid ocean floor' represents the highest density.

An object sinks (negative buoyancy) when its weight exceeds the maximum possible buoyant force, meaning its overall density is greater than the fluid's density.

Anahtar Kavram

Buoyancy and Density Relationships (Positive, Neutral, and Negative Buoyancy)

Tahmini Süre:45s

Soru 88Soru →

A laboratory container is filled to the very brim with water. A 3-kilogram solid wood block is gently placed into the water, where it floats, and the water that spills over is collected and weighed. The wood is then removed, the container is completely refilled, and a 3-kilogram solid steel weight is lowered into the water until it rests on the bottom. The water that spills over from the steel is also collected and weighed. Based on Archimedes' principle, which of the following statements correctly compares the weight of the water displaced by the two objects?

The wood block displaces a greater weight of water than the steel weight.

The steel weight displaces a greater weight of water because it completely submerges beneath the surface.

Both objects displace the exact same weight of water because their initial masses are identical.

The steel weight displaces a greater weight of water because steel has a higher density than wood.

Cevabı ve açıklamayı göster

Cevap: The wood block displaces a greater weight of water than the steel weight.

Cevap

The wood block displaces a greater weight of water than the steel weight.

According to Archimedes' principle, a floating object (like the wood) displaces a weight of fluid equal to its own weight, meaning the wood displaces 3 kg of water. A sinking object (like the steel) displaces a volume of fluid equal to its own volume. Since steel is very dense, a 3 kg piece of steel takes up very little volume and therefore displaces much less than 3 kg of water. Thus, the floating wood displaces more water than the sinking steel.

Adım Adım Çözüm

Analyze the displaced water for the floating wood block.

According to Archimedes' principle, a floating object displaces a weight of fluid equal to its own weight. Therefore, the 3-kilogram wood block displaces exactly 3 kilograms of water.

This establishes the buoyant force required to keep the wood in equilibrium on the surface of the water.

Analyze the displaced water for the sinking steel weight.

A sinking object displaces a volume of fluid equal to its own total volume. Because steel is much denser than water, a 3-kilogram piece of steel has a very small volume.

This determines how much water is physically pushed out of the container by the fully submerged steel.

Compare the weight of the water displaced by both objects.

The wood displaces 3 kilograms of water. The steel displaces a volume of water equal to its own small volume, which weighs much less than 3 kilograms. Therefore, the floating wood displaces a greater weight of water.

Directly answering the question by comparing the calculated fluid displacements.

Anahtar Kavram

Buoyancy and Archimedes' Principle

Soru 89Soru →

Two ships are traveling parallel to each other at high speed, leaving a narrow channel of water between their hulls. The water flowing into this narrow space moves at a significantly higher relative velocity than the water on the outer sides of the ships. According to Bernoulli's principle, what happens to the water pressure in the gap, and what is the resulting effect on the two ships?

The water pressure in the gap decreases, creating a net inward force that pulls the ships toward each other.

The water pressure in the gap increases due to the higher velocity, creating a net outward force that pushes the ships apart.

The water compresses as it enters the narrow gap, causing the pressure to increase and violently push the ships apart.

The water pressure remains unchanged, but the increased fluid friction in the gap creates a drag force that pushes the ships apart.

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Cevap: The water pressure in the gap decreases, creating a net inward force that pulls the ships toward each other.

Cevap

The water pressure in the gap decreases, creating a net inward force that pulls the ships toward each other.

According to Bernoulli's principle, an increase in fluid velocity is accompanied by a decrease in static pressure. Because the water between the parallel ships is forced to flow faster through the narrow gap, the pressure in that specific region drops. The relatively slower-moving water on the outer sides of the ships maintains a higher pressure. This pressure differential creates a net inward force, pushing the two vessels toward each other.

Adım Adım Çözüm

Identify the relationship between fluid velocity and static pressure.

Bernoulli's principle states that as the speed of a moving fluid increases, its static pressure decreases.

This is the fundamental concept governing fluid dynamics in constricted flow areas.

Apply the principle to the given maritime scenario.

The water channeled between the ships moves faster relative to the hulls than the water on the outside. Therefore, the static pressure between the hulls becomes lower than the static pressure on the outer sides of the ships.

The narrowing of the channel forces the water to speed up to maintain continuous flow, which directly affects the localized pressure.

Determine the resulting physical force acting on the vessels.

The higher pressure on the outside of the ships pushes inward toward the lower pressure zone between them, effectively drawing the ships together.

Fluids exert a net force moving from high-pressure areas toward low-pressure areas.

Anahtar Kavram

Bernoulli's Principle and Pressure Differentials

Soru 90Soru →

When a steady flow of water moves from a wide horizontal pipe into a narrower section, the water's velocity increases, causing the static pressure against the walls of the narrow section to be higher than in the wider section.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

False

The statement is false because Bernoulli's principle dictates that as the velocity of a horizontal fluid flow increases (which it must do when entering a narrower pipe to maintain flow rate), its static pressure decreases. Therefore, the pressure against the walls is lower, not higher, in the narrow section.

Adım Adım Çözüm

Determine the change in fluid velocity.

The water's velocity increases as it enters the narrow section.

The principle of continuity states that for an incompressible fluid, the flow rate must remain constant; thus, a smaller cross-sectional area requires a higher velocity.

Relate the change in velocity to static pressure.

The static pressure in the narrow section decreases.

Bernoulli's principle states that in a steady, horizontal fluid flow, an increase in fluid velocity results in a decrease in static pressure because total energy is conserved.

Evaluate the original statement.

The statement is false.

The statement incorrectly claims that the static pressure would be higher in the narrow section, when in fact it is lower.

Anahtar Kavram

Bernoulli's Principle and Fluid Continuity

Soru 91Soru →

A mechanical eductor (or jet pump) uses a primary fluid flowing through a converging nozzle to extract a secondary fluid from a lower tank. As the primary fluid accelerates through the narrowest section of the nozzle, how does its static pressure change, and what is the resulting effect on the secondary fluid?

The static pressure rises proportionally with the fluid's velocity, creating a force that pushes the secondary fluid forward.

The increased friction against the narrow nozzle walls creates a high-pressure zone that drives the secondary fluid upward.

The static pressure drops below the surrounding pressure, creating a suction effect that draws the secondary fluid into the main stream.

The static pressure remains constant because liquids are incompressible, and the fluids mix solely due to physical contact.

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Cevap: The static pressure drops below the surrounding pressure, creating a suction effect that draws the secondary fluid into the main stream.

Cevap

The static pressure drops below the surrounding pressure, creating a suction effect that draws the secondary fluid into the main stream.

The correct answer accurately applies Bernoulli's principle, which states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure. In an eductor, the narrow nozzle accelerates the primary fluid, causing a significant pressure drop. This localized low-pressure zone acts as a vacuum, drawing the secondary fluid from the tank into the main flow.

Adım Adım Çözüm

Analyze the fluid flow through the converging nozzle.

The cross-sectional area of the nozzle decreases, which forces the fluid velocity to increase in order to maintain a constant flow rate (continuity principle).

Understanding how area affects velocity is the first step in applying fluid dynamics.

Apply Bernoulli's principle to the high-velocity region.

According to Bernoulli's equation, an increase in fluid velocity (kinetic energy) must be accompanied by a decrease in static pressure to conserve total mechanical energy.

This establishes the inverse relationship between velocity and static pressure.

Determine the physical consequence of the lowered static pressure.

The decreased static pressure in the narrow section becomes lower than the atmospheric pressure acting on the secondary fluid, creating a pressure differential (suction) that draws the secondary fluid up into the eductor.

This links the theoretical physics concept directly to the mechanical action described in the problem.

Anahtar Kavram

As a fluid's velocity increases, its static pressure decreases, which can be utilized to create suction in mechanical systems.

Tahmini Süre:1m 0s

Soru 92Soru →

A steady stream of water flows through a horizontal pipeline of varying diameter. Four vertical piezometer tubes (standpipes open to the atmosphere) are installed along the pipe to measure the local static pressure. The height of the water column in each tube is observed as follows:

- Station 1: $1.2$ meters
- Station 2: $2.5$ meters
- Station 3: $0.8$ meters
- Station 4: $1.8$ meters

Based on Bernoulli's principle, arrange the stations in order from the fastest fluid velocity to the slowest fluid velocity.

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Cevap

The correct order from fastest to slowest velocity is: Station 3, Station 1, Station 4, Station 2.

According to Bernoulli's principle, for an incompressible fluid flowing through a horizontal pipe, an increase in fluid velocity occurs simultaneously with a decrease in static pressure. The height of the water in a vertical piezometer tube is a direct measure of the static pressure at that location. Therefore, the station with the lowest water column (Station 3,

0.8

m) has the lowest static pressure and the fastest velocity. The station with the highest water column (Station 2,

2.5

m) has the highest static pressure and the slowest velocity. Sorting from lowest to highest water column gives the correct sequence from fastest to slowest velocity.

Adım Adım Çözüm

Understand the relationship between water column height and static pressure.

The height of a fluid column in a piezometer tube directly indicates the static pressure at that point (

P = \rho g h

Establishing what the given measurements represent.

Apply Bernoulli's principle to relate static pressure and fluid velocity.

In a horizontal pipe, regions with higher static pressure correspond to slower fluid velocities, and regions with lower static pressure correspond to faster fluid velocities.

This inverse relationship is the core concept of Bernoulli's principle for steady, horizontal flow.

Rank the static pressures based on the water column heights.

Lowest pressure to highest pressure: Station 3 (

0.8

m), Station 1 (

1.2

m), Station 4 (

1.8

m), Station 2 (

2.5

m).

To find the order of fluid velocities, we first need to order the static pressures.

Determine the final order of fluid velocities.

Since the fastest velocity occurs at the lowest pressure, the order from fastest to slowest is: Station 3, Station 1, Station 4, Station 2.

Translating the pressure ranking into the final velocity ranking required by the question.

Anahtar Kavram

Fluid Dynamics and Bernoulli's Principle

Tahmini Süre:1m 30s

Soru 93Soru →

During routine maintenance of a municipal water line, engineers observe that a thick layer of mineral scale has formed inside a section of a horizontal supply pipe. This scale acts as a partial blockage, significantly reducing the pipe's interior cross-sectional area in that specific location. Assuming the flow rate of water through the system remains steady, how are the water's velocity and static pressure affected as it passes through this narrowed section?

The water's velocity increases, and its static pressure also increases due to the confined space.

The water's velocity decreases due to the restricted opening, causing the static pressure to drop.

The water's velocity increases, and its static pressure decreases.

The water's velocity remains constant, but its static pressure increases because the liquid is compressed.

Cevabı ve açıklamayı göster

Cevap: The water's velocity increases, and its static pressure decreases.

Cevap

The water's velocity increases, and its static pressure decreases.

In a closed, continuous fluid system, an incompressible liquid like water must maintain a constant flow rate. When it encounters a constriction (like the mineral scale buildup), the only way the same volume of water can get through the smaller opening in the same amount of time is to speed up (principle of continuity). As the velocity increases, Bernoulli's principle dictates that the static pressure must decrease so that the total energy of the fluid flow remains constant.

Adım Adım Çözüm

Determine the change in velocity through the narrowed pipe section.

The velocity of the water increases.

Because water is an incompressible fluid, the same volume of water must pass through the restricted area in the same amount of time. This relies on the principle of continuity (

A_1V_1 = A_2V_2

). To maintain the flow rate through a smaller cross-sectional area (

A

), the velocity (

V

) must increase.

Determine the change in static pressure based on the new velocity.

The static pressure of the water decreases.

According to Bernoulli's principle, the total mechanical energy in a steady horizontal fluid flow remains constant. When the water's kinetic energy increases (due to higher velocity), its potential energy—manifested as static pressure against the pipe walls—must decrease proportionally to conserve energy.

Anahtar Kavram

Fluid Dynamics and Bernoulli's Principle

Tahmini Süre:45s

Soru 94Soru →

As a race car travels at high speeds, the air flowing through the narrow clearance between the car's flat underbody and the track surface accelerates. According to the principles of fluid dynamics, this increase in air velocity creates a region of elevated static pressure that pushes upward against the bottom of the car.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

False

The statement is false. According to Bernoulli's principle, an increase in fluid velocity results in a decrease in static pressure. Therefore, the fast-moving air under the race car creates a low-pressure zone that produces downforce, rather than an upward push.

Adım Adım Çözüm

Analyze the change in air velocity under the vehicle.

The narrow gap between the car's underbody and the track restricts the airflow area, causing the air velocity to increase in order to maintain a continuous flow.

The continuity equation dictates that fluid velocity must increase when the cross-sectional area of the flow path decreases.

Relate the increased velocity to static pressure using Bernoulli's principle.

According to Bernoulli's principle, the higher kinetic energy of the fast-moving air results in a simultaneous decrease in static pressure.

In a flowing fluid system, an increase in velocity is accompanied by a drop in static pressure to conserve total energy.

Determine the direction of the resulting force on the vehicle.

Because the static pressure under the car decreases, the relatively higher ambient pressure above the car pushes it downward, creating downforce rather than upward lift.

Objects are naturally pushed toward regions of lower pressure.

Anahtar Kavram

Bernoulli's Principle and the Continuity Equation

Soru 95Soru →

An aircraft's emergency pneumatic blow-down bottle contains nitrogen gas at an absolute pressure of $3000 \text{ psi}$ with an internal volume of $200 \text{ in}^3$ . During an emergency landing gear extension, a valve opens and the gas expands into the actuator lines and cylinders, increasing the total volume occupied by the gas to $600 \text{ in}^3$ . Assuming the temperature remains constant during this rapid expansion, what is the new absolute pressure in the system?

1500 \text{ psi}

1000 \text{ psi}

4500 \text{ psi}

9000 \text{ psi}

Cevabı ve açıklamayı göster

Cevap: $1000 \text{ psi}$

Cevap

The new absolute pressure in the system is

1000 \text{ psi}

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

200 \text{ in}^3

to a total final volume of

600 \text{ in}^3

. Because the volume increased by a factor of 3 (

600 / 200 = 3

), the pressure must decrease by a factor of 3. Dividing the initial

3000 \text{ psi}

by 3 results in

1000 \text{ psi}

Adım Adım Çözüm

Identify the appropriate gas law for a scenario with changing pressure and volume at a constant temperature.

Boyle's Law applies, which is expressed as

P_1 \times V_1 = P_2 \times V_2

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

Identify the known initial and final variables from the problem statement.

Initial pressure (

P_1

) =

3000 \text{ psi}

, Initial volume (

V_1

) =

200 \text{ in}^3

, and Final volume (

V_2

) =

600 \text{ in}^3

These are the critical parameters needed to solve for the unknown final pressure (

P_2

Substitute the known values into Boyle's Law and solve for the final pressure (

P_2

3000 \times 200 = P_2 \times 600 \rightarrow 600000 = P_2 \times 600 \rightarrow P_2 = 1000 \text{ psi}

Isolating

P_2

algebraically provides the exact pressure after the gas has fully expanded into the larger volume.

Anahtar Kavram

Boyle's Law (Inverse relationship between pressure and volume)

Soru 96Soru →

A mechanic pours four common automotive fluids into a clear, tall recovery cylinder. Assuming they do not mix and are allowed to settle completely into distinct layers, arrange the fluids in order from the TOP layer to the BOTTOM layer.

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Cevap

The correct order from the top layer to the bottom layer is: Gasoline, SAE 30 Engine Oil, Fresh Water, and finally Ethylene Glycol (Antifreeze).

When immiscible fluids are mixed and allowed to settle, they separate into distinct layers based on their specific gravity. The fluid with the lowest specific gravity (gasoline, ~0.73) will float on top. Engine oil (~0.88) comes next, floating on the fresh water (1.0). Ethylene glycol antifreeze (~1.11) is the densest fluid of the group and will sink to the very bottom.

Adım Adım Çözüm

Identify the physical property that determines how fluids layer.

Fluids layer based on their density or specific gravity. The least dense fluid floats on top, and the most dense fluid sinks to the bottom.

Buoyancy dictates that lighter fluids will displace heavier ones downwards.

Recall the relative specific gravities of the listed automotive fluids.

Gasoline is the lightest (SG ~0.73), followed by engine oil (SG ~0.88), fresh water (SG 1.0), and ethylene glycol antifreeze (SG ~1.11).

Understanding the density of common mechanical fluids is necessary for diagnosing leaks and fluid separation.

Arrange the fluids according to their specific gravity from lowest (top) to highest (bottom).

Gasoline, Engine Oil, Water, Antifreeze.

This matches the instruction to order from the top layer to the bottom layer.

Anahtar Kavram

Density and Specific Gravity

Soru 97Soru →

During the operation of a pneumatic nailer, its internal firing reservoir holds a compressed air volume of $20 \text{ in}^3$ at an absolute pressure of $120 \text{ psi}$ . When the trigger is pulled, the air expands into the driver chamber at a constant temperature until the absolute pressure drops to $30 \text{ psi}$ . Based on Boyle's Law, what is the new total volume of the expanded air in cubic inches?

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Cevap: 80

Cevap

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

P_1 \times V_1 = P_2 \times V_2

). The pressure drops from 120 psi to 30 psi, which is a factor of 4 decrease. Because the relationship is inversely proportional, the volume must increase by a factor of 4. Therefore, the initial volume of 20 cubic inches is multiplied by 4, resulting in a final volume of 80 cubic inches.

Adım Adım Çözüm

Identify the initial and final states of the gas.

Initial pressure

P_1 = 120 \text{ psi}

, Initial volume

V_1 = 20 \text{ in}^3

. Final pressure

P_2 = 30 \text{ psi}

Establishing the known parameters is required before applying the appropriate gas law.

Apply Boyle's Law formula.

P_1 \times V_1 = P_2 \times V_2

Since the temperature is constant, Boyle's Law relates the pressure and volume of the trapped air.

Substitute values and solve for the unknown final volume (

V_2

120 \times 20 = 30 \times V_2 \rightarrow 2400 = 30 \times V_2 \rightarrow V_2 = 80

Isolating

V_2

provides the final expanded volume of the air.

Anahtar Kavram

Boyle's Law states that the absolute pressure of a confined gas is inversely proportional to its volume, provided the temperature remains constant.

Alternatif Yöntem

Use proportional reasoning: Observe that the final pressure (30 psi) is exactly one-quarter (1/4) of the initial pressure (120 psi). Because pressure and volume are inversely proportional under constant temperature, the final volume must be four times (4x) the initial volume. 4 × 20 = 80.

Tahmini Süre:1m 0s

Soru 98Soru →

A hydraulic accumulator is a sealed, rigid tank that contains both hydraulic fluid (a liquid) and pressurized nitrogen (a gas). When a pump forces additional hydraulic fluid into the tank and significantly raises the internal pressure, which of the following best describes the volumetric changes of the original contents?

Both the original hydraulic fluid and the nitrogen gas compress by an equal percentage to absorb the increased pressure.

The volume of the nitrogen gas decreases significantly, while the volume of the original hydraulic fluid remains virtually unchanged.

The original hydraulic fluid compresses substantially to make room, while the nitrogen gas volume remains constant.

The nitrogen gas expands to increase the pressure, while the hydraulic fluid decreases in density.

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Cevap: The volume of the nitrogen gas decreases significantly, while the volume of the original hydraulic fluid remains virtually unchanged.

Cevap

The volume of the nitrogen gas decreases significantly, while the volume of the original hydraulic fluid remains virtually unchanged.

Liquids, such as hydraulic fluid, are practically incompressible, meaning their volume does not significantly change when subjected to high pressure. Gases, such as nitrogen, are highly compressible. When the pump forces more liquid into the rigid tank, the internal pressure rises. The gas acts like a spring, compressing into a smaller volume to accommodate the new liquid, while the original liquid retains its volume.

Adım Adım Çözüm

Identify the states of matter of the materials inside the accumulator.

The nitrogen is a gas, and the hydraulic fluid is a liquid.

Different states of matter respond differently to applied pressure.

Recall the compressibility property of liquids.

Liquids are practically incompressible.

This means adding pressure will not noticeably change the volume of the original hydraulic fluid.

Recall the compressibility property of gases.

Gases are highly compressible.

This means the nitrogen gas will be forced into a smaller volume as the pressure inside the rigid tank increases.

Combine these principles to determine the outcome.

The added fluid forces the gas to compress into a smaller space, while the original liquid maintains its volume.

This allows the rigid tank to accommodate the additional hydraulic fluid.

Anahtar Kavram

Compressibility of Liquids vs. Gases

Tahmini Süre:1m 0s

Soru 99Soru →

A marine engineer is filling a sealed buoyancy tank with a specialized flotation fluid. The system requires a fluid with a specific gravity of $0.85$ to provide the correct lift. If the internal volume of the tank is $0.4 \text{ m}^3$ , what must the total mass of the fluid be? (Assume the density of water is $1000 \text{ kg/m}^3$ )

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Cevap: 340

Cevap

The correct mass is

340 \text{ kg}

The correct answer is found by first converting the unitless specific gravity into an actual density by multiplying it by the density of water (

0.85 \times 1000 = 850 \text{ kg/m}^3

). The total mass is then calculated by multiplying this density by the given volume (

850 \times 0.4 = 340 \text{ kg}

Adım Adım Çözüm

Calculate the density of the fluid using its specific gravity.

Density =

850 \text{ kg/m}^3

Specific gravity is defined as the ratio of a fluid's density to the density of water (

1000 \text{ kg/m}^3

). Therefore, density equals

0.85 \times 1000

Calculate the mass of the fluid using its density and volume.

Mass =

340 \text{ kg}

Mass is the product of density and volume.

850 \text{ kg/m}^3 \times 0.4 \text{ m}^3 = 340 \text{ kg}

Anahtar Kavram

The relationship between specific gravity, density, and volume to determine the mass of a fluid.

Soru 100Soru →

An outdoor hydraulic lift operates noticeably slower and requires more effort from the pump on a freezing winter morning compared to a hot summer afternoon. Assuming the mechanical components are functioning normally, which of the following fluid properties best explains this sluggish performance?

The viscosity of the hydraulic fluid increases in cold temperatures, creating greater internal resistance to flow.

The liquid hydraulic fluid becomes highly compressible in freezing conditions, causing the pump's energy to compress the fluid rather than move the cylinder.

The fluid expands as it cools down, significantly lowering its specific gravity and reducing its ability to push the cylinder.

The colder fluid provides less buoyant force against the internal piston, making the external load effectively heavier to lift.

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Cevap: The viscosity of the hydraulic fluid increases in cold temperatures, creating greater internal resistance to flow.

Cevap

The viscosity of the hydraulic fluid increases in cold temperatures, creating greater internal resistance to flow.

Viscosity measures a fluid's internal resistance to flow. In virtually all lubricating and hydraulic liquids, viscosity increases as the temperature drops, causing the fluid to become thicker. This increased thickness creates drag inside the hydraulic lines and valves, resulting in slower mechanical operation.

Adım Adım Çözüm

Analyze the mechanical symptom described in the scenario.

The hydraulic lift is operating slowly (sluggishly) in cold temperatures.

Identifying the symptom helps isolate which physical property of the fluid is changing with the temperature drop.

Evaluate how temperature affects the physical properties of liquids.

As temperature decreases, the internal friction of a liquid increases.

Understanding the inverse relationship between temperature and liquid viscosity is a core concept in mechanical fluid maintenance.

Connect the changing fluid property to the mechanical outcome.

Higher internal friction (higher viscosity) means the fluid resists flowing through hoses and valves, slowing down the machine's operation.

This confirms that increased viscosity is the direct cause of the sluggish mechanical performance.

Anahtar Kavram

Viscosity and Temperature Relationship

Tahmini Süre:45s

ÖncekiSayfa 5 / 6Sonraki

Fluid Mechanics

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