Fluids and Elasticity - Chapter 14

Chapter 14 Fluids and Elasticity

Overview

• This chapter discusses systems that flow (fluids) or deform (elasticity).

Fluids

• A fluid is a substance that flows; gases and liquids are fluids.

• Gases

  • Compressible: molecules move freely with few interactions.

  • Gases flow and exert pressure.

• Liquids

  • Incompressible: molecules are weakly bound.

  • Liquids flow and exert pressure.

Volume

• Volume (V) is an important parameter of a macroscopic system.

• SI unit of volume is m^3.

• Unit conversions:

  • 1 m^3 = 1000 L

  • 1 L = 1000 cm^3

  • 1 m^3 = 10^6 cm^3

Density

• Mass density (\rho) is the ratio of mass to volume.

  • \rho = \frac{m}{V}

• SI units of mass density are kg/m^3.

Densities of Various Fluids (at standard temperature and pressure)

• Helium gas: 0.18 kg/m^3

• Air: 1.29 kg/m^3

• Gasoline: 680 kg/m^3

• Ethyl alcohol: 790 kg/m^3

• Benzene: 880 kg/m^3

• Oil (typical): 900 kg/m^3

• Water: 1000 kg/m^3

• Seawater: 1030 kg/m^3

• Glycerin: 1260 kg/m^3

• Mercury: 13,600 kg/m^3

Pressure

• Pressure is the force-to-area ratio.

• larger v_x as you add more depth (means the pressure is greater.)

• pressure is more for more mass andd more depth

  • p = \frac{F}{A}

• SI units of pressure are N/m^2, also defined as the pascal (Pa).

  • 1 Pa = 1 N/m^2

• Fluids exert forces on the walls of their containers.

• if A is constant , then the pressure acting on the walls is uniformly distributed, as defined by the equation P = F/A, where P is pressure, F is the force exerted, and A is the area over which the force is applied. This relationship highlights the importance of understanding how variations in force or area can significantly impact pressure levels within a fluid system.

• Additionally, it is essential to note that pressure in a fluid increases with depth due to the weight of the fluid above. This relationship can be expressed by the hydrostatic pressure equation, P = P₀ + ρgh, where P₀ is the pressure at the surface, ρ is the fluid density, g is the acceleration due to gravity, and h is the depth measured from the surface.

• as pressure goes up the tempature goes up if the constant

• Contributions to pressure in a fluid:

  1. Gravitational: due to gravity pulling down on the liquid or gas (hydrostatic pressure). It effects because more molecules involve

  2. Thermal: due to collisions of gas molecules with the walls, depends on temperature. gravity pulling on gas doesnt really effect pressure. as pressure

Atmospheric Pressure

• The global average sea-level pressure is 101,300 Pa.

• 1 standard atmosphere (atm) is defined as:

  • 1 atm = 101,300 Pa = 101.3 kPa

• Fluids exert pressure forces in all directions.

• less pressure higher up

• water needs to boil longer at lower tempertures high altitudes (air thinner less oxygen).

• This phenomenon is known as the effect of altitude on boiling point, which occurs due to reduced atmospheric pressure affecting the energy required for water molecules to escape into the gas phase.

Example: Suction Cup

• A suction cup with a diameter of 10.0 cm is pushed against a smooth ceiling. Determine the maximum mass that can be suspended without detaching it from the ceiling.

• Model: Assume a perfect vacuum (p = 0 Pa) between the suction cup and the ceiling, and the room pressure is 1 atm.
  *Free-body diagram of the suction cup stuck to the ceiling.

  

• Solution:

  • The suction cup is in equilibrium: F{air} = mg+n

  • Upward force exerted by the air: F_{air}=pA=p\pi r^2=\pi(0.050m)^2(101,300Pa)=796N

  • Maximum weight when the normal force n is reduced to zero: FG = mg = F{air}

  • m = \frac{F_{air}}{g} = \frac{796 N}{9.8 m/s^2} = 81 kg

• Review: A suction cup can support a mass of up to 81 kg with a perfect vacuum inside.

• P0+A=F

• Fair+fm=pA at the bottom

• PA=P0 A+mg+P0A+pAdg

  

Pressure in Liquids

• For a liquid in static equilibrium, balancing forces in a free-body diagram:

  • pA = p_0 A + mg

  • Volume of the cylinder: V = Ad

  • Mass of the cylinder: m = \rho Ad

  • Hydrostatic pressure at depth d:

    p = p_0 + \rho g d

Example: Pressure on a Submarine

• A submarine cruises at a depth of 300 m. Calculate the pressure at this depth in pascals and atmospheres.

• Solution:

  • Density of seawater: \rho = 1030 kg/m^3

  • Pressure at depth d = 300 m:

    p = p_0 + \rho g d = 101,300 Pa + (1030 kg/m^3)(9.80 m/s^2)(300 m) = 3.13 \times 10^6 Pa

  • Converting to atmospheres:

    p = \frac{3.13 \times 10^6 Pa}{1.013 \times 10^5 Pa/atm} = 30.9 atm

Liquids in Hydrostatic Equilibrium

• A connected liquid in hydrostatic equilibrium rises to the same height in all open regions of the container.

• The pressure is the same at all points on a horizontal line through a connected liquid in hydrostatic equilibrium. P1=P2

Example: Pressure in a Closed Tube

• Water fills a tube. Determine the pressure at the top of the closed tube, assuming p_0 = 1.00 atm.

• Model: The pressure is the same at all points on a horizontal line.

• Solution:

  • At a point 40 cm above the bottom of the open tube (depth of 60 cm):

    p = p_0 + \rho g d = 1.013 \times 10^5 Pa + (1000 kg/m^3)(9.80 m/s^2)(0.60 m) = 1.072 \times 10^5 Pa = 1.06 atm

• The pressure at the top of the closed tube is 1.06 atm.

Gauge Pressure

• Many pressure gauges measure gauge pressure (p_g), which is the difference between the absolute pressure (p) and atmospheric pressure (1 atm).

• The pressure taken in the tire(example) is measured in gage pressure p-1atm=Pg Pg+1atm=P

  • p_g = p - 1 atm

Example: Underwater Pressure Gauge

• An underwater pressure gauge reads 60 kPa. What is its depth?

• Model: The gauge reads gauge pressure.

• Solution:

  • p = 1 atm + \rho g d

  • p_g = p - 1 atm = \rho g d

  • d = \frac{p_g}{\rho g} = \frac{60,000 Pa}{(1000 kg/m^3)(9.80 m/s^2)} = 6.1 m

Barometers

• A barometer measures atmospheric pressure by measuring the height (h) of a liquid column in a tube.

• p_{atmos} = \rho g h

• P=p0+pgh

• P1+p0+ O

• P2=O+pgh

• Pgh=p0

Pressure Units

Hydraulic Lift

• A hydraulic lift uses fluid pressure to lift heavy objects, such as cars.

• Areax distance V1=V2+A1d1=A2d2

• In static equilibrium:

  • \frac{F1}{A1} = \frac{F2}{A2} - \rho g h

• If piston 1 is pushed down a distance d1, the car is lifted by distance d2:

  • \frac{d1}{d2} = \frac{A2}{A1}(not tested on)

Example: Lifting a Car

• A hydraulic lift filled with oil has a 25-cm-diameter piston supporting a car. Compressed air pushes on a 6.0-cm-diameter piston. What pressure does the gauge read, so a 1300kg car is 2.0 m above the compressed-air piston?

• Model: Incompressible oil with density 900 kg/m^3.

• Solution:

  • Weight of the car: F_2 = mg = (1300 kg)(9.8 m/s^2) = 12,700 N

  • Piston areas: A1 = \pi (0.030 m)^2 = 0.00283m^2 and A2 = \pi (0.125 m)^2 = 0.0491 m^2

  • Force required to hold the car at height h = 2.0 m:

    \frac{F1}{A1} = \frac{F2}{A2} + \rho g h

    F1 = \frac{A1}{A2} \left( F2 + \rho g h A_2 \right) = \frac{0.00283 m^2}{0.0491 m^2} \left( 12,700 N + (900 kg/m^3)(9.8 m/s^2)(2.0 m)(0.0491 m^2) \right) = 782 N

  • Pressure applied by the compressed-air piston:

    p = \frac{F1}{A1} = \frac{782 N}{0.00283 m^2} = 276 \times 10^3 Pa = 276 kPa = 2.7 atm

Buoyancy

• Buoyancy is the upward force exerted by a fluid on an object.

• Fup>Fdown

• Fb=Fwater+Fg

• The pressure in the liquid exerts a net upward force on the cylinder: {net} = F{up} - F_{down}

• The buoyant force on an object is the same as the buoyant force on the fluid it displaces.

• Archimedes’ Principle: A fluid exerts an upward buoyant force on an object immersed in or floating on the fluid. The magnitude of the buoyant force equals the weight of the fluid displaced by the object.

• Fb=Wd (FB=AVfg)

  • FB = \rhof V_f g

    • \rho_f is the density of the fluid.

    • V_f is the volume of the displaced fluid. equivalent to the volume of the portion of the object that is immersed in the fluid.

Example: Holding a Block of Wood Underwater

• A 10 cm × 10 cm × 10 cm block of wood with density 700 kg/m^3 is held underwater by a string. What is the tension in the string?
  *Free-body diagram of the forces acting on the wood.

  

• Solution:

  • The block is in equilibrium: \sum Fy = FB - T - m_o g = 0

  • T = FB - mo g

    mo = \rhoo V_o

    FB = \rhof V_f g

    T = (\rhof - \rhoo) V_o g

    V_o = (0.1 m)^3 = 1.0 \times 10^{-3} m^3

    T = ((1000 kg/m^3) - (700 kg/m^3)) (1.0 \times 10^{-3} m^3) (9.8 m/s^2) = 2.9 N

Floating Object

• The volume of fluid displaced by a floating object of uniform density is:

  • Vf = \frac{mo}{\rhof} = \frac{\rhoo}{\rhof} Vo

• The volume of the displaced fluid is less than the volume of the uniform-density object:

  • \frac{Vf}{Vo} = \frac{\rhoo}{\rhof} < 1

• For icebergs:

  • \rho_{ice} = 917 kg/m^3

  • \rho_{seawater} = 1030 kg/m^3

  • \frac{Vf}{Vo} = \frac{917 kg/m^3}{1030 kg/m^3} \approx 0.89

  • About 90% of the volume of an iceberg is underwater.

Boats

• The boat will float if the weight of the displaced water equals the weight of the boat.

• The minimum height of the sides:

  • h = \frac{mo}{\rhof A}

Fluid Dynamics

• Ideal-Fluid Model Assumptions:

  1. Incompressible: More like a liquid than a gas.

  2. Nonviscous: More like water than syrup.

  3. Steady flow: Laminar flow rather than turbulent flow.

• Laminar flow: Smooth, streamline flow.

• Turbulent flow: Irregular, chaotic flow.

• Equation of Continuity for two points in a flow tube:

  • A1 v1 = A2 v2

• Volume flow rate Q is constant:

  • Q = vA

Dx1/dt A1=dx2/dtA2

Bernoulli’s Equation

• The energy equation for fluid in a flow tube:

  • p1 + \frac{1}{2} \rho v1^2 + \rho g y1 = p2 + \frac{1}{2} \rho v2^2 + \rho g y2

• Alternative form of Bernoulli’s equation:

  • p + \frac{1}{2} \rho v^2 + \rho g y = constant

  • Restatement of conservation of energy

  • Work done by pressure has to go down if potential and kinetic energy go up for everything to remain constant

Example : An Irrigation System

• Water flows through pipes. The water’s speed through the lower pipe is 5.0 m/s and a pressure gauge reads 75 kPa. What is the gauge reading of the pressure gauge on the upper pipe?
  *Treat the water as an ideal fluid obeying Bernoulli’s equation.

• Solution:

  • Bernoulli’s equation between points 1 and 2:

    p2 = p1 + \frac{1}{2} \rho (v1^2 - v2^2) + \rho g (y1 - y2)

  • The cross-section areas and water speeds at points 1 and 2 are related by

    v2 = \frac{A1}{A2} v1 = \frac{r1^2}{r2^2} v_1 = \frac{(0.030 m)^2}{(0.020 m)^2} (5.0 m/s) = 11.25 m/s

  • The pressure at point 1 is p_1 = 75 kPa + 1 atm = 176,300 Pa.

    p_2 = 176,300 Pa + \frac{1}{2} (1000 kg/m^3) ((5.0 m/s)^2 - (11.25 m/s)^2) + (1000 kg/m^3) (9.8 m/s^2) (-3.0 m) = 105,900 Pa

  • This is the absolute pressure; the pressure gauge on the upper pipe will read

    p = 105,990Pa - 1atm = 4.6kPa

Lift

• Airflow over a wing generates lift by creating unequal pressures above and below.

• If kinetic energy goes up pre sure goes down

• Pressure goes up potential energy goes down

  

Average Flow Speed

• Average flow speed where Q is the volume flow rate and A is the cross-section area of the tube.
  v_{avg} = \frac{Q}{A}

Elasticity

• Elasticity describes how objects deform under stress.

• In the elastic region, the force needed to stretch a solid rod is proportional to the change in length:

  • F = k \Delta L

• \frac{F}{A} is proportional to \frac{\Delta L}{L}.

• Young’s Modulus (Y):

  • \frac{F}{A} = Y \frac{\Delta L}{L}

  • \frac{F}{A} = tensile stress

  • \frac{\Delta L}{L} = strain

Elastic Properties of Various Materials

Example: Stretching a Wire

• A 2.0-m-long, 1.0-mm-diameter wire is suspended from the ceiling. Hanging a 4.5 kg mass from the wire stretches the wire’s length by 1.0 mm. What is Young’s modulus for this wire? Can you identify the material?

• Solution:

  • F = mg = (4.5 kg)(9.80 m/s^2)

    A = \pi r^2 = \pi (0.5 \times 10^{-3} m)^2 = 7.85 \times 10^{-7} m^2

    \Delta L = 1.0 \times 10^{-3} m

    Y = \frac{F/A}{\Delta L / L} = \frac{5.6 \times 10^7 N/m^2}{5.0 \times 10^{-4}} = 1.1 \times 10^{11} N/m^2

Volume Stress and the Bulk Modulus

• A volume stress applied to an object compresses its volume slightly.

• The volume strain is defined as \frac{\Delta V}{V}, and is negative when the volume decreases.

• Volume stress is the same as the pressure.

• \frac{\Delta V}{V} = - \frac{p}{B}, where B is the bulk modulus.

Bernoulli’s Equation

• The energy equation for fluid in a flow tube:

• p1+1/Pv1²+Pgy1=p2+1/2Pv2²+Pgy2

• Alternative form of Bernoulli’s equation:

• p + \frac{1}{2} \rho v^2 + \rho g y = constant

• Key concepts:

  • Pressure (p): Represents the force per unit area exerted by the fluid. In Bernoulli's equation, pressure accounts for the potential energy due to the compression of the fluid.

  • Kinetic Energy Density ($\frac{1}{2} \rho v^2$): Represents the kinetic energy per unit volume of the fluid, where \rho is the density and v is the velocity of the fluid. This term accounts for the energy associated with the motion of the fluid.

  • Gravitational Potential Energy Density ($\rho g y$): Represents the gravitational potential energy per unit volume of the fluid, where g is the acceleration due to gravity and y is the height above a reference point. This term accounts for the energy associated with the fluid's vertical position.

• Implications and applications:

  • Horizontal Pipe: If a pipe is horizontal (y is constant), then regions with higher velocity will have lower pressure, and vice versa.

  • Static Fluid: If the fluid is static (v = 0), Bernoulli's equation simplifies to the hydrostatic pressure equation, p + \rho g y = constant.

  • Flow through Constrictions: When a fluid flows through a constriction, its velocity increases, leading to a decrease in pressure. This principle is utilized in venturi meters and carburetors.

• Derivations and special cases:

  • Torricelli's Theorem:

  • Describes the speed of fluid flowing out of an opening at the bottom of a tank.

  • Derived from Bernoulli's equation by comparing a point at the surface of the fluid (1) and a point at the opening (2).

  • p1 + \frac{1}{2} \rho v1^2 + \rho g y1 = p2 + \frac{1}{2} \rho v2^2 + \rho g y2

  • Assuming p1 = p2 (both at atmospheric pressure), and v1 \approx 0 (large tank), the equation simplifies to: \rho g y1 = \frac{1}{2} \rho v2^2 + \rho g y2

  • Solving for v2 gives: v2 = \sqrt{2g(y1 - y2)} = \sqrt{2gh}, where h is the height of the fluid above the opening.

  • Venturi Effect:

  • Describes the reduction in fluid pressure that results when a fluid flows through a constricted section of a pipe.

  • From the continuity equation, A1 v1 = A2 v2, where A is the cross-sectional area and v is the velocity.

  • From Bernoulli's equation, p1 + \frac{1}{2} \rho v1^2 = p2 + \frac{1}{2} \rho v2^2$$

  • Combining these equations allows for the calculation of pressure differences based on changes in flow velocity and cross-sectional area.

  • Pitot Tube:

  • Measures the velocity of a fluid by comparing the stagnation pressure (where the fluid is brought to rest) to the static pressure.

  • Applying