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Classical Mechanics
Topic 9: Fluid Mechanics
-Density and Pressure
-Pascal’s Law
-Archimedes Principle
-Continuity Equation
-Bernoulli’s Equation
Based from Sears and Zemansky’s University Physics
with Modern Physics 13th ed
Fluid Mechanics
Fluid – any substance that can flow (liquid and gas)
Fluid statics – study of fluids at rest / equilibrium conditions.
Fluid dynamics – study of fluids in motion.
Density (ρ) – tells you how compact the object is.
ρ =
𝑚
𝑉
m is the mass
V is the volume of space occupied by the object
For the same object, the density is constant:
𝑚1
𝑉1
=
𝑚2
𝑉2
Specific gravity – ratio of the object's density to water's density at 4.0 ºC.
density of water at 4.0 ºC. = 1000
𝑘𝑔
𝑚3
The density of a material is independent of its mass and volume.
Useful conversion: 1
𝑔
𝑐𝑚3
= 1000
𝑘𝑔
𝑚3
Density
Pressure (p) – normal force per unit area
𝑝 =
𝑑𝐹⊥
𝑑𝐴
Other units of pressure:
SI unit is Pascal:1 𝑃𝑎 = 1
𝑁
𝑚2
1 𝑎𝑡𝑚 = 1.01325 × 105 𝑃𝑎
1 𝑏𝑎𝑟 = 105 𝑃𝑎
For constant pressure: 𝐹⊥ = 𝑝𝐴
Pressure
Suppose that the pressure on the back of the swimmer's hand is 1.2
x 105 [Pa]. The surface area of the back of the hand is 8.4 x 10 –3
[m2]. Determine the magnitude of the force that acts on the hand
and its direction.
For constant pressure: 𝐹⊥ = 𝑝𝐴
𝐹⊥ = 𝑝𝐴 = 1.2 × 105
𝑁
𝑚2
8.4 × 10−3
𝑚2
= 1.0 × 103 𝑁
Direction is downward
Goal: Relationship between the pressure p at any point in a fluid at
rest and the elevation y at that point.
The pressure decreases as we move upward in the fluid.
Pressure
If the density and the acceleration due to gravity
are both constants as the elevation varies, then
𝑝2 − 𝑝1 = −ρ𝑔 𝑦2 − 𝑦1
In terms of depth h: 𝑝 = 𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 + ρ𝑔ℎ
● The pressure is the same at any two points
at the same level in the fluid.
● The shape of the container does not
affect the pressure due to the fluid.
Pressure
Absolute and Gauge Pressure
• Absolute pressure – the total pressure due to the fluid's weight and the
pressure on the fluid's surface (atmospheric pressure).
𝑝 = 𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 + ρ𝑔ℎ
• Gauge pressure – the excess pressure above the atmospheric
pressure.
𝑝𝑔𝑎𝑢𝑔𝑒 = 𝑝 − 𝑝𝑎𝑡𝑚
E2
Absolute and Gauge Pressure
Pressure
𝑝2 − 𝑝1 = −ρ𝑔 𝑦2 − 𝑦1
y is measured from the bottom of the container (y = 0). Point 1 is at
a lower altitude and point 2 is at a higher altitude inside the fluid.
Set point 2 to be the surface of the fluid and let h = y2 – y1.
𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 − 𝑝1 = −ρ𝑔ℎ
𝑝1 = 𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 + ρ𝑔ℎ
𝑝 = 𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 + ρ𝑔ℎ This is the absolute pressure.
ℎ1
ℎ = ℎ2 − ℎ1
𝑝
ℎ2
𝑝𝑎𝑡𝑚
The U-tube contains a liquid of known density ρ with the right tube open to the atmosphere.
The left end of the tube is connected to the container where the pressure p is to be
measured.
Open tube manometer
The absolute pressure at the bottom of the right arm is:
𝑝𝑟𝑖𝑔ℎ𝑡 = 𝑝𝑎𝑡𝑚 + ρ𝑔ℎ2
E3. What is the absolute pressure at the bottom of the left tube?
If the system is in equilibrium, then
𝑝𝑙𝑒𝑓𝑡 = 𝑝𝑟𝑖𝑔ℎ𝑡
E4. Show that the pressure inside the bulb is:
𝑝 = 𝑝𝑎𝑡𝑚 + ρ𝑔ℎ
Pascal's law
• Pressure applied to an enclosed fluid is transmitted undiminished to every portion of
the fluid and the walls of the containing vessel.
𝑝1 = 𝑝2
𝐹1
𝐴1
=
𝐹2
𝐴2
Example:
For the hydraulic lift shown in the figure, what must be the
ratio of the diameter of the vessel at the car to the
diameter of the vessel where the force F1 is applied so
that a 1520-kg car can be lifted with a force of F1 of just
125 [N]?
𝐹1
𝐴1
=
𝐹2
𝐴2
𝐴2
𝐴1
=
𝐹2
𝐹1
E5. Continue this.
𝑑2
𝑑1
= 10.9
Archimedes principle
• When a body is completely or partially immersed in a fluid, the fluid exerts an
upward force on the body equal to the weight of the fluid displaced by the body.
• Buoyant force – upward force that the fluid exerts on a body that is completely or
partially immersed.
𝐵 = 𝑤𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙 = 𝑚𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙𝑔
𝐵 = ρ𝑓𝑙𝑢𝑖𝑑𝑉𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙𝑔
• An object will float if it has a lower density than the fluid:
ρ𝑓𝑙𝑢𝑖𝑑 ≥ ρ𝑜𝑏𝑗𝑒𝑐𝑡
𝐵
𝑉𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙𝑔
≥
𝑤
𝑉𝑜𝑏𝑗𝑒𝑐𝑡𝑔
• The buoyant force B and the weight of the object
must cancel each other for the object to float.
𝑉𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙 ≤ 𝑉𝑜𝑏𝑗𝑒𝑐𝑡
• No need to submerge the entire volume of the
object just to float.
Fluid Flow
Steady flow - the flow pattern is not changing with time.
Streamlines/ flow lines
Streamlines in a steady flow do not intersect each other
Nonviscous fluid - no internal friction (no viscosity)
Non viscous fluid → fluid particles passing
through a given cross-section will have
the same speed.
viscous fluid → fluid particles near
the walls of the tube will have lower
speed than the one at the center.
The densities of the fluid on these regions are the same.
Incompressible fluid - the density of the fluid remains
constant on any sections of the flow
Continuity Equation
The mass of the fluid that enters a tube section should be equal to the
mass of the fluid that leaves the tube section.
ρ2𝐴2𝑣2 = ρ1𝐴1𝑣1
𝐴2𝑣2 = 𝐴1𝑣1 (Incompressible fluid)
𝑄 = 𝐴𝑣 The volume flow rate (Q) is constant
for an incompressible fluid
Bernoulli's Equation
𝑃 +
1
2
ρ𝑣2 + ρ𝑔𝑦 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
For an incompressible, non viscous fluid moving
with a steady flow:
𝑃2 +
1
2
ρ𝑣2
2
+ ρ𝑔𝑦2 = 𝑃1 +
1
2
ρ𝑣1
2
+ ρ𝑔𝑦1
For two different points:
Bernoulli's equation is derived using
the conservation of mechanical
energy.
The fluid's speeds at points 1 and 2 are
still related via:
𝐴2𝑣2 = 𝐴1𝑣1
• Uniform pipe:
𝑃2 + ρ𝑔𝑦2 = 𝑃1 + ρ𝑔𝑦1
• The pressure at point with
lower elevation is larger than
those at higher elevation.
• Result is similar to fluid at rest.
Bernoulli's Equation

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CM9_-Fluid Mechanics- v2.pdf

  • 1. Classical Mechanics Topic 9: Fluid Mechanics -Density and Pressure -Pascal’s Law -Archimedes Principle -Continuity Equation -Bernoulli’s Equation Based from Sears and Zemansky’s University Physics with Modern Physics 13th ed
  • 2. Fluid Mechanics Fluid – any substance that can flow (liquid and gas) Fluid statics – study of fluids at rest / equilibrium conditions. Fluid dynamics – study of fluids in motion. Density (ρ) – tells you how compact the object is. ρ = 𝑚 𝑉 m is the mass V is the volume of space occupied by the object For the same object, the density is constant: 𝑚1 𝑉1 = 𝑚2 𝑉2 Specific gravity – ratio of the object's density to water's density at 4.0 ºC. density of water at 4.0 ºC. = 1000 𝑘𝑔 𝑚3 The density of a material is independent of its mass and volume. Useful conversion: 1 𝑔 𝑐𝑚3 = 1000 𝑘𝑔 𝑚3
  • 4. Pressure (p) – normal force per unit area 𝑝 = 𝑑𝐹⊥ 𝑑𝐴 Other units of pressure: SI unit is Pascal:1 𝑃𝑎 = 1 𝑁 𝑚2 1 𝑎𝑡𝑚 = 1.01325 × 105 𝑃𝑎 1 𝑏𝑎𝑟 = 105 𝑃𝑎 For constant pressure: 𝐹⊥ = 𝑝𝐴 Pressure Suppose that the pressure on the back of the swimmer's hand is 1.2 x 105 [Pa]. The surface area of the back of the hand is 8.4 x 10 –3 [m2]. Determine the magnitude of the force that acts on the hand and its direction. For constant pressure: 𝐹⊥ = 𝑝𝐴 𝐹⊥ = 𝑝𝐴 = 1.2 × 105 𝑁 𝑚2 8.4 × 10−3 𝑚2 = 1.0 × 103 𝑁 Direction is downward
  • 5. Goal: Relationship between the pressure p at any point in a fluid at rest and the elevation y at that point. The pressure decreases as we move upward in the fluid. Pressure
  • 6. If the density and the acceleration due to gravity are both constants as the elevation varies, then 𝑝2 − 𝑝1 = −ρ𝑔 𝑦2 − 𝑦1 In terms of depth h: 𝑝 = 𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 + ρ𝑔ℎ ● The pressure is the same at any two points at the same level in the fluid. ● The shape of the container does not affect the pressure due to the fluid. Pressure
  • 7. Absolute and Gauge Pressure • Absolute pressure – the total pressure due to the fluid's weight and the pressure on the fluid's surface (atmospheric pressure). 𝑝 = 𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 + ρ𝑔ℎ • Gauge pressure – the excess pressure above the atmospheric pressure. 𝑝𝑔𝑎𝑢𝑔𝑒 = 𝑝 − 𝑝𝑎𝑡𝑚
  • 8. E2 Absolute and Gauge Pressure Pressure 𝑝2 − 𝑝1 = −ρ𝑔 𝑦2 − 𝑦1 y is measured from the bottom of the container (y = 0). Point 1 is at a lower altitude and point 2 is at a higher altitude inside the fluid. Set point 2 to be the surface of the fluid and let h = y2 – y1. 𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 − 𝑝1 = −ρ𝑔ℎ 𝑝1 = 𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 + ρ𝑔ℎ 𝑝 = 𝑝𝑠𝑢𝑟𝑓𝑎𝑐𝑒 + ρ𝑔ℎ This is the absolute pressure.
  • 9. ℎ1 ℎ = ℎ2 − ℎ1 𝑝 ℎ2 𝑝𝑎𝑡𝑚 The U-tube contains a liquid of known density ρ with the right tube open to the atmosphere. The left end of the tube is connected to the container where the pressure p is to be measured. Open tube manometer The absolute pressure at the bottom of the right arm is: 𝑝𝑟𝑖𝑔ℎ𝑡 = 𝑝𝑎𝑡𝑚 + ρ𝑔ℎ2 E3. What is the absolute pressure at the bottom of the left tube? If the system is in equilibrium, then 𝑝𝑙𝑒𝑓𝑡 = 𝑝𝑟𝑖𝑔ℎ𝑡 E4. Show that the pressure inside the bulb is: 𝑝 = 𝑝𝑎𝑡𝑚 + ρ𝑔ℎ
  • 10. Pascal's law • Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. 𝑝1 = 𝑝2 𝐹1 𝐴1 = 𝐹2 𝐴2 Example: For the hydraulic lift shown in the figure, what must be the ratio of the diameter of the vessel at the car to the diameter of the vessel where the force F1 is applied so that a 1520-kg car can be lifted with a force of F1 of just 125 [N]? 𝐹1 𝐴1 = 𝐹2 𝐴2 𝐴2 𝐴1 = 𝐹2 𝐹1 E5. Continue this. 𝑑2 𝑑1 = 10.9
  • 11. Archimedes principle • When a body is completely or partially immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body. • Buoyant force – upward force that the fluid exerts on a body that is completely or partially immersed. 𝐵 = 𝑤𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙 = 𝑚𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙𝑔 𝐵 = ρ𝑓𝑙𝑢𝑖𝑑𝑉𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙𝑔 • An object will float if it has a lower density than the fluid: ρ𝑓𝑙𝑢𝑖𝑑 ≥ ρ𝑜𝑏𝑗𝑒𝑐𝑡 𝐵 𝑉𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙𝑔 ≥ 𝑤 𝑉𝑜𝑏𝑗𝑒𝑐𝑡𝑔 • The buoyant force B and the weight of the object must cancel each other for the object to float. 𝑉𝑓𝑙𝑢𝑖𝑑𝑑𝑖𝑠𝑝𝑙 ≤ 𝑉𝑜𝑏𝑗𝑒𝑐𝑡 • No need to submerge the entire volume of the object just to float.
  • 12. Fluid Flow Steady flow - the flow pattern is not changing with time. Streamlines/ flow lines Streamlines in a steady flow do not intersect each other Nonviscous fluid - no internal friction (no viscosity) Non viscous fluid → fluid particles passing through a given cross-section will have the same speed. viscous fluid → fluid particles near the walls of the tube will have lower speed than the one at the center. The densities of the fluid on these regions are the same. Incompressible fluid - the density of the fluid remains constant on any sections of the flow
  • 13. Continuity Equation The mass of the fluid that enters a tube section should be equal to the mass of the fluid that leaves the tube section. ρ2𝐴2𝑣2 = ρ1𝐴1𝑣1 𝐴2𝑣2 = 𝐴1𝑣1 (Incompressible fluid) 𝑄 = 𝐴𝑣 The volume flow rate (Q) is constant for an incompressible fluid
  • 14. Bernoulli's Equation 𝑃 + 1 2 ρ𝑣2 + ρ𝑔𝑦 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 For an incompressible, non viscous fluid moving with a steady flow: 𝑃2 + 1 2 ρ𝑣2 2 + ρ𝑔𝑦2 = 𝑃1 + 1 2 ρ𝑣1 2 + ρ𝑔𝑦1 For two different points: Bernoulli's equation is derived using the conservation of mechanical energy. The fluid's speeds at points 1 and 2 are still related via: 𝐴2𝑣2 = 𝐴1𝑣1
  • 15. • Uniform pipe: 𝑃2 + ρ𝑔𝑦2 = 𝑃1 + ρ𝑔𝑦1 • The pressure at point with lower elevation is larger than those at higher elevation. • Result is similar to fluid at rest. Bernoulli's Equation