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    Fluid Mechanics

    Flashcards for topic Fluid Mechanics

    Intermediate59 cardsGeneral

    Preview Cards

    Card 1

    Front

    Define pressure in a fluid and explain how it differs from force (include the mathematical expression).

    Back

    Pressure is defined as the limit of force per unit area as the area approaches zero: P=lim⁡ΔS→0FΔSP = \lim_{\Delta S \rightarrow 0} \frac{F}{\Delta S}P=limΔS→0​ΔSF​

    Key properties:

    • It's a scalar quantity (magnitude only)
    • For homogeneous and nonviscous fluids, pressure at a point is independent of the orientation of the area
    • Measured in pascals (Pa) or N/m²
    • Differs from force by being intensive (independent of size) while force is extensive
    • Pressure acts perpendicular to any surface in contact with the fluid

    Example: 1 Pa equals the pressure exerted by a force of 1 N spread over 1 m²

    Card 2

    Front

    Derive Archimedes' principle from first principles, and explain why it might not be valid in certain situations.

    Back

    Derivation of Archimedes' principle:

    1. Consider a body submerged in a fluid
    2. Imagine replacing the body with the same fluid of equal volume
    3. This substituted fluid would be in equilibrium
    4. Forces on this substitute fluid:
      • Weight mg downward
      • Buoyant force B upward from surrounding fluid
    5. For equilibrium: B = mg (where m is mass of displaced fluid)
    6. Since the surrounding fluid exerts the same pressure on the original body as on the substituted fluid, the buoyant force on the body equals the weight of the displaced fluid

    Archimedes' principle: A body partially or fully immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.

    Limitations/invalid scenarios:

    • Non-inertial reference frames: If the fluid is accelerating, the principle requires modification
    • Compressible fluids: When density varies significantly with depth
    • Surface tension effects: Dominant at very small scales
    • Non-homogeneous fluids: When density varies throughout the fluid
    • Highly viscous flows: Where viscous forces are significant
    Card 3

    Front

    Distinguish between steady and turbulent fluid flow, and explain what a streamline represents. Why can't two streamlines cross?

    Back

    Steady vs. Turbulent Flow:

    Steady (Streamline) Flow:

    • Fluid particles passing through any point follow the same path over time
    • Velocity at each point remains constant with time
    • Predictable, smooth flow patterns
    • Occurs at lower velocities
    • Can be analyzed with simpler mathematics

    Turbulent Flow:

    • Chaotic, irregular fluid motion
    • Velocity at any point changes erratically with time
    • Contains eddies and vortices
    • Occurs at higher velocities
    • Requires statistical methods for analysis

    Streamline:

    • A curve whose tangent at any point gives the direction of fluid velocity at that point
    • All fluid particles passing through any point on a streamline follow that streamline
    • Represents the path a fluid particle would take in steady flow

    Why streamlines cannot cross:

    • If two streamlines crossed, a fluid particle at the intersection would have two different velocity directions simultaneously
    • This would create a logical contradiction since a particle can only move in one direction at a time
    • At an intersection point, the fluid would need to follow two different paths, which is physically impossible
    Card 4

    Front

    How does the hydrostatic pressure equation apply when comparing two points at different depths in water? Calculate the pressure difference between points 5 meters apart vertically.

    Back

    For two points at different depths in water:

    1. Hydrostatic pressure equation: P2−P1=ρghP_2 - P_1 = \rho g hP2​−P1​=ρgh

    2. For water with density 1000 kg/m³ and points with 5 m vertical separation: P2−P1=(1000 kg/m3)(9.8 m/s2)(5 m)P_2 - P_1 = (1000 \text{ kg/m}^3)(9.8 \text{ m/s}^2)(5 \text{ m})P2​−P1​=(1000 kg/m3)(9.8 m/s2)(5 m) P2−P1=49,000 Pa=49 kPaP_2 - P_1 = 49,000 \text{ Pa} = 49 \text{ kPa}P2​−P1​=49,000 Pa=49 kPa

    3. Practical applications:

      • This is why your ears feel pressure when diving underwater
      • Water pressure increases by about 10 kPa per meter of depth
      • Deep-sea submersibles must withstand tremendous pressure differences
      • Dams are designed thicker at the base to withstand greater pressure at greater depths

    Note: This equation assumes a fluid with constant density, which is a good approximation for liquids like water under normal conditions.

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    Card 5

    Front

    Given two points in a fluid with different cross-sectional areas, velocities, and heights, how does the work-energy theorem apply to derive the relationship between pressure, height, and velocity?

    Back

    The work-energy theorem applies to fluid flow as follows:

    1. For fluid moving from point 1 to point 2:

      • Work is done by pressure forces at both ends: W1=P1A1v1ΔtW_1 = P_1A_1v_1\Delta tW1​=P1​A1​v1​Δt and W2=−P2A2v2ΔtW_2 = -P_2A_2v_2\Delta tW2​=−P2​A2​v2​Δt
      • Work is done by gravity: Wg=Δmg(h1−h2)W_g = \Delta m g(h_1 - h_2)Wg​=Δmg(h1​−h2​)
    2. This work causes a change in kinetic energy:

      • ΔKE=12Δm(v22−v12)\Delta KE = \frac{1}{2}\Delta m(v_2^2 - v_1^2)ΔKE=21​Δm(v22​−v12​)
    3. By conservation of mass (continuity equation):

      • Δm=ρA1v1Δt=ρA2v2Δt\Delta m = \rho A_1v_1\Delta t = \rho A_2v_2\Delta tΔm=ρA1​v1​Δt=ρA2​v2​Δt
    4. Applying work-energy theorem (Wtotal=ΔKEW_{total} = \Delta KEWtotal​=ΔKE):

      • P1Δmρ−P2Δmρ+Δmg(h1−h2)=12Δm(v22−v12)P_1\frac{\Delta m}{\rho} - P_2\frac{\Delta m}{\rho} + \Delta m g(h_1 - h_2) = \frac{1}{2}\Delta m(v_2^2 - v_1^2)P1​ρΔm​−P2​ρΔm​+Δmg(h1​−h2​)=21​Δm(v22​−v12​)
    5. Simplifying yields Bernoulli's equation:

      • P1ρ+gh1+12v12=P2ρ+gh2+12v22\frac{P_1}{\rho} + gh_1 + \frac{1}{2}v_1^2 = \frac{P_2}{\rho} + gh_2 + \frac{1}{2}v_2^2ρP1​​+gh1​+21​v12​=ρP2​​+gh2​+21​v22​
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    Card 6

    Front

    How does pressure change in a Venturi tube as fluid flows through the constriction, and what physical principle explains this phenomenon?

    Back

    In a Venturi tube:

    • Pressure decreases at the constriction (P₂ < P₁)
    • This follows directly from Bernoulli's principle
    • The mathematical relationship is: P₁ + ½ρv₁² = P₂ + ½ρv₂²
    • Since v₂ > v₁ (velocity increases at constriction), P₂ must decrease
    • The pressure difference can be expressed as: P₁ - P₂ = ½ρ(v₂² - v₁²)

    This demonstrates energy conservation: as kinetic energy increases at the constriction, pressure energy must decrease to maintain constant total energy along a streamline.

    Real-world application: Carburetors use this principle to draw fuel into an air stream.

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    Card 7

    Front

    In a tank with a small hole at the bottom, why does the equation of continuity (A₁v₁ = A₂v₂) combined with Bernoulli's principle lead to Torricelli's theorem?

    Back

    The derivation combines two key fluid principles:

    1. Bernoulli's equation between the free surface (1) and the hole (2): P0+12ρv12+ρgh=P0+12ρv22P_0 + \frac{1}{2}ρv_1^2 + ρgh = P_0 + \frac{1}{2}ρv_2^2P0​+21​ρv12​+ρgh=P0​+21​ρv22​

    2. The equation of continuity states that A₁v₁ = A₂v₂, which means v₁ = v₂(A₂/A₁)

    When these are combined:

    • The atmospheric pressure P₀ cancels out on both sides
    • Substituting v₁ in terms of v₂ gives: 12ρ(A2A1)2v22+ρgh=12ρv22\frac{1}{2}ρ\left(\frac{A_2}{A_1}\right)^2v_2^2 + ρgh = \frac{1}{2}ρv_2^221​ρ(A1​A2​​)2v22​+ρgh=21​ρv22​

    This simplifies to: [1−(A2A1)2]v22=2gh\left[1-\left(\frac{A_2}{A_1}\right)^2\right]v_2^2 = 2gh[1−(A1​A2​​)2]v22​=2gh

    When A₂ << A₁ (the hole is much smaller than the tank), the fraction becomes negligible, resulting in: v2=2ghv_2 = \sqrt{2gh}v2​=2gh​

    This explains why the exit velocity depends only on the height of the fluid column and not on the density of the fluid.

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    Card 8

    Front

    What is the Magnus effect and what causes a spinning ball to deviate from its expected path of motion?

    Back

    The Magnus effect is a physical phenomenon where a spinning object moving through a fluid experiences a force perpendicular to both its direction of motion and axis of rotation.

    Key mechanisms:

    • When a ball spins while moving through air, it creates asymmetric airflow patterns
    • On the side where the surface motion matches airflow direction, the relative air speed increases
    • On the opposite side where surface motion opposes airflow, the relative air speed decreases
    • According to Bernoulli's principle, faster-moving air creates lower pressure
    • This pressure differential creates a force perpendicular to both the flight path and spin axis

    Example: A cricket ball with vertical spin axis moving horizontally will experience a sideways force, causing it to "swing" left or right depending on spin direction.

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    Card 9

    Front

    How does Bernoulli's principle explain the deviation in a spinning ball's trajectory, and what specific relationship exists between spin direction and the resulting force?

    Back

    Bernoulli's principle states that as fluid velocity increases, pressure decreases. For a spinning ball:

    1. The spin creates different relative air velocities on opposite sides of the ball:

      • Higher velocity where ball surface and air move in the same direction
      • Lower velocity where ball surface and air move in opposite directions
    2. This velocity difference creates a pressure gradient:

      • Lower pressure on the side with faster relative airflow
      • Higher pressure on the side with slower relative airflow
    3. The resulting force (F) is:

      • Perpendicular to both the spin axis and velocity vector
      • Directed from the high-pressure side toward the low-pressure side
      • Proportional to both spin rate and forward velocity

    The force direction follows the right-hand rule: if your fingers curl in the direction of spin, your thumb points in the approximate direction of the force.

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    Card 10

    Front

    What practical applications utilize the pressure reduction that occurs when a fluid passes through a constriction?

    Back

    Applications utilizing pressure reduction in fluid constrictions:

    1. Aspirator pumps: Used for spraying liquids by drawing them up through a tube connected to a constriction
    2. Venturi meters: Measure fluid flow rates by calculating velocity from pressure differences
    3. Carburetors: Mix air and fuel in engines by drawing fuel into the airstream through a constriction
    4. Atomizers: Create fine mists for perfumes, medical inhalers, and paint sprayers
    5. Bunsen burners: Draw air into the gas flow to create a hotter flame
    6. Vacuum systems: Laboratory vacuum pumps use water flow through constrictions to create suction
    7. Injectors: Used in various industrial processes to introduce one fluid into another

    All these applications work because when fluid velocity increases in the constriction, pressure decreases according to Bernoulli's principle.

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