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Review for Exam 1
10. 13. 2014
Hyunse Yoon, Ph.D. Assistant Research Scientist
IIHR-Hydroscience & Engineering e-mail: [email protected]
Chapter 1. Introduction - Fluid properties
1. Fluids and No-slip condition 2. Dimensions and Units 3. Weight and Mass 4. Properties involving mass or weight of fluid 5. Viscosity 6. Vapor pressure and Cavitation 7. Surface tension
2 Review for Exam 1 2014
1. Fluids and No-slip condition
β’ Fluid: Deforms continuously (i.e., flows) when subjected to a shearing stress β Solid: Resists to shearing stress by a static deflection
β’ No-slip condition: No relative motion between fluid and boundary at the contact β The fluid βsticksβ to the solid
boundaries 3
A fluid flowing over a stationary surface comes to a complete stop at the surface because of the no-slip condition (left) and the development of a velocity profile due to the no-slip condition as a fluid flows over a blunt nose (right) Review for Exam 1 2014
2. Dimensions and Units
Variable Dimension SI unit BG unit
Velocity π½π½ πΏπΏ ππβ m sβ ft sβ
Acceleration ππ πΏπΏ ππ2β m s2β ft s2β
Force ππ πππΏπΏ ππ2β N (Kg β m s2β ) lbf
Pressure ππ πΉπΉ πΏπΏ2β Pa (N m2β ) lbf ft2β
Density ππ ππ πΏπΏ3β Kg m3β slug ft3β
Internal energy ππ πΉπΉπΏπΏ ππβ J Kgβ (N β m kgβ ) BTU lbmβ
β’ Dimension: Quantitative expression of a physical variable (without numerical values) o Primary dimensions: Mass (ππ), length (πΏπΏ), time (ππ or π‘π‘), and
temperature (Ξ or ππ). Also referred to as basic dimensions o Secondary dimensions: All other dimensions can be derived from the
primary dimensions
β’ Unit: A way of attaching a number to quantitative dimensions o For example, length is dimension and meter or feet are units o SI unit system, BG unit system, EE unit system, and etc.
4 Review for Exam 1 2014
2. Dimensions and Units βContd. β’ Dimensional homogeneity: All equations must be dimensionally
homogeneous. Each additive term in an equation must have the same dimensions
β’ Consistent units: Each additive terms must have the same units
Ex): Bernoulli equation
ππ +12ππππ2 + ππππππ = constant
πππΏπΏππβ2 + β πππΏπΏ3 πΏπΏ2ππβ2
= ππππππβ2+ πππΏπΏβ3 πΏπΏππβ2 πΏπΏ
= ππππππβ2= πππΏπΏππβ2
o SI units: Nm2 + β kg
m3ms
2+ kg
m3ms2
m ; (N) = (kg)(m/s2)
o BG units: lbfft2
+ β slugsft2
fts
2+ slugs
ft2fts2
ft ; lbf = slugs ft/s2
5 Review for Exam 1 2014
3. Weight and Mass
β’ Weight ππ is a force dimension, i.e., ππ = ππ β g β In SI unit: ππ N = ππ kg β g, where g = 9.81 m/s2
β In BG unit: ππ lbf = ππ slug β g, where g = 32.2 ft/s2
β In EE unit: ππ lbf = ππ lbm /gcβ g, where g = 32.2 ft/s2
o 1 N = 1 kg Γ 1 m/s2
o 1 lbf = 1 slug Γ 1 ft/s2
o gc = 32.2 lbm/slug o 1 slug = 32.2 lbm
6
Unit System Mass Weight
SI 1 kg 9.81 N (= 1 kgf)
BG 1 slug 32.2 lbf
EE 1 lbm 1 lbf
Review for Exam 1 2014
4. Properties involving mass or weight of fluid
β’ Density
ππ = mass (ππ)volume ππ
(kg/m3 or slugs/ft3)
β’ Specific Weight
πΎπΎ = weight (ππππ)volume (ππ)
= ππππ (N/m3 or lbf/ft3)
β’ Specific Gravity
ππππ =πΎπΎ
πΎπΎπ€π€π€π€π€π€π€π€π€π€ =
πππππ€π€π€π€π€π€π€π€π€π€
7 Review for Exam 1 2014
5. Viscosity β’ Newtonian fluid
ππ = ππππππππππ
- ππ: Shear stress (N/m2 or lbf/ft2) - ππ: Dynamic viscosity (Nβ s/m2 or lbfβ s/ft2) - ππ = ππ ππβ : Kinematic viscosity (m2/s or ft2/s) - Shear force = ππ β π΄π΄
β’ Non-Newtonian fluid
ππ βππππππππ
ππ
8 Review for Exam 1 2014
6. Vapor pressure and cavitation β’ Vapor pressure: Below which a liquid evaporates, i.e., changes
to a gas
β’ Cavitation: If the pressure drop is due to fluid velocity β Boiling: if the pressure drop is due to temperature effect
9
β’ Cavitation number:
πΆπΆπ€π€ =ππ β πππ£π£12ππππβ
2
Note: πΆπΆπ€π€ < 0 implies cavitation
Cavitation formed on a marine propeller Review for Exam 1 2014
7. Surface tension
β’ Surface tension force: The force developed at the interface of two immiscible fluids (e.g., liquid-gas) due to the unbalanced molecular cohesive forces at the fluid surface.
10
πΉπΉππ = ππ β πΏπΏ
- πΉπΉππ = Line force with direction normal to the cut - πΏπΏ = Length of cut through the interface - ππ = Surface tension [N/m], the intensity of the
molecular attraction per unit length along πΏπΏ
Attractive forces acting on a liquid molecule at the surface and deep inside the liquid
Review for Exam 1 2014
7. Surface tension β Contd.
β’ Capillary Effect: The rise (or fall) of a liquid in a small-diameter tube inserted into a the liquid.
11
β’ Capillary rise:
β =2ππππππππ
cosππ
Note: ππ = contact angle
The forces acting on a liquid column that has risen in a tube due to the capillary effect
Review for Exam 1 2014
Chapter 2. Fluid Statics
1. Absolute pressure, gage pressure, and vacuum 2. Pressure variation with elevation 3. Pressure measurements (Manometry) 4. Hydrostatic forces on plane surfaces 5. Hydrostatic forces on curved surfaces 6. Buoyancy 7. Stability 8. Fluids in rigid-body motion
12 Review for Exam 1 2014
1. Absolute pressure, gage pressure, and vacuum
13
β’ Absolute pressure: The actual pressure measured relative to absolute vacuum β’ Gage pressure: Pressure measured relative to local atmospheric pressure β’ Vacuum pressure: Pressures below atmospheric pressure
Review for Exam 1 2014
2. Pressure variation with elevation
14
β’ Force balance in an incompressible static fluid (Newtonβs 2nd law per unit volume):
ππ ππβ=0
= βπππππποΏ½ β π»π»ππ + ππ π»π»2πποΏ½=0
β΄ π»π»ππ = βπΎπΎπποΏ½
or, ππππππππ
= 0; ππππππππ
= 0; ππππππππ
= βπΎπΎ
β’ Since πΎπΎ = constant (e.g., liquids),
ππ = βπΎπΎππ + ππ0 By taking ππ0 = 0 (gage) at ππ = 0,
β΄ ππ = βπΎπΎππ Thus, the pressure increases linearly with depth
Review for Exam 1 2014
3. Pressure measurements (1): Piezometer tube
15
β’ The simplest type of manometer
πππ΄π΄ = πΎπΎ1β1
β’ The fluid must be a liquid β’ Suitable only if πππ΄π΄ > πππ€π€π€π€ππ β’ πππ΄π΄ must be relatively small so that β1 is reasonable
Review for Exam 1 2014
3. Pressure measurements (2) U-Tube manometer
16
β’ Starting from one end, add pressure when move downward and subtract when move upward:
πππ΄π΄ + πΎπΎ1β1 β πΎπΎ2β2 = 0 Thus,
πππ΄π΄ = πΎπΎ2β2 β πΎπΎ1β1 or
πππ΄π΄ = πΎπΎ2 β2 βπΎπΎ1πΎπΎ2β1
β’ If πΎπΎ1 βͺ πΎπΎ2 (e.g., πΎπΎ1 is a gas and πΎπΎ2 a
liquid),
β΄ πππ΄π΄ β πΎπΎ2β2
Review for Exam 1 2014
3. Pressure measurements (3) Differential U-Tube manometer
17
β’ To measure the difference in pressure: πππ΄π΄ + πΎπΎ1β1 β πΎπΎ2β2 β πΎπΎ3β3 = πππ΅π΅
or πππ΄π΄ β πππ΅π΅ = πΎπΎ2β2 + πΎπΎ3β3 β πΎπΎ1β1
β’ For a pipe flow where πΎπΎ1 = πΎπΎ3 as shown in the boxed figure,
πππ΄π΄ β πππ΅π΅ = πΎπΎ2 1 βπΎπΎ1πΎπΎ2β2
if πΎπΎ1 βͺ πΎπΎ2, β΄ πππ΄π΄ β πππ΅π΅ β πΎπΎ2β2
Ex:
Review for Exam 1 2014
4. Hydrostatic forces on plane surfaces (1) Horizontal surfaces
18
β’ Pressure is uniform on horizontal surfaces (e.g., the tank bottom) as
ππ = πΎπΎβ β’ The magnitude of the resultant
force is simply
πΉπΉπ π = πππ΄π΄ = πΎπΎβπ΄π΄
Review for Exam 1 2014
4. Hydrostatic forces on plane surfaces (2) Inclined surfaces
19
β’ Average pressure on the surface
οΏ½Μ οΏ½π = πΎπΎβππ β’ Resultant pressure force
πΉπΉπ π = οΏ½Μ οΏ½ππ΄π΄ = πΎπΎβπππ΄π΄
β’ Pressure center
πππ π = ππππ +πΌπΌπ₯π₯πππππππ΄π΄
Review for Exam 1 2014
5. Hydrostatic forces on curved surfaces
20
β’ Horizontal force component: πΉπΉπ»π» = οΏ½Μ οΏ½ππππ€π€ππππ β π΄π΄πππ€π€ππππ β’ Vertical force component: πΉπΉππ = πΎπΎππ (weight of fluid above surface)
β’ Resultant force: πΉπΉπ π = πΉπΉπ»π»2 + πΉπΉππ2
Review for Exam 1 2014
6. Buoyancy (1) Immersed bodies
21
πΉπΉπ΅π΅ = πΉπΉππ2 β πΉπΉππ1 = πΎπΎππ
β’ Fluid weight equivalent to body volume ππ
β’ Line of action (or the center of buoyancy) is through the centroid of ππ
Review for Exam 1 2014
7. Stability (1) Immersed bodies
23
β’ If πΆπΆ is above ππ: Stable (righting moment when heeled) β’ If ππ is above πΆπΆ: Unstable (heeling moment when heeled)
Review for Exam 1 2014
7. Stability (2) Floating bodies
24
β’ ππππ > 0: Stable (ππ is above ππ) β’ ππππ < 0: Unstable (ππ is above ππ)
ππππ =πΌπΌ00ππβ πΆπΆππ
Review for Exam 1 2014
8. Fluids in rigid-body motion (1) Translation
25
β’ π»π»ππ = ππ ππ β ππ o ππ = πππ₯π₯οΏ½ΜοΏ½π + πππ§π§πποΏ½ (constant) o ππ = βπππποΏ½
β’ ππππ
ππππ= βππππ
β’ ππ = ππππππ
o ππ = πππ₯π₯2 + ππ + πππ§π§ 212
o ππ = tanβ1 π€π€π₯π₯ππ+π€π€π§π§
Ex:
Review for Exam 1 2014
8. Fluids in rigid-body motion (2) Rotation
26
β’ π»π»ππ = ππ ππ β ππ o ππ = βππΞ©2πποΏ½ππ (constant Ξ©) o ππ = βπππποΏ½
β’ ππππ
πππ€π€= ππππΞ©2 and ππππ
πππ§π§= βππππ
β’ ππ = ππ
2ππ2Ξ©2 β ππππππ + πΆπΆ
β’ ππ = ππ0βππππππ
+ Ξ©2
2ππππ2 Ex:
Review for Exam 1 2014
Chapter 3. Elementary Fluid Dynamics - The Bernoulli equation
1. Flow patterns 2. Streamline coordinates 3. Bernoulli equation 4. Application of Bernoulli equation
27 Review for Exam 1 2014
1. Flow patterns
28
β’ Streamline: A line that is everywhere tangent to the velocity vector at a given instant
β’ Pathline: The actual path traveled by a given fluid particle β’ Streakline: The locus of particles which have earlier passed through
a particular point β’ For steady flow, all three lines coincide
Streamline Pathline Streakline Review for Exam 1 2014
2. Streamline coordinates
29
β’ Velocity ππ = π£π£πππποΏ½ + π£π£πποΏ½
=0πποΏ½
Note: ππ = π£π£ππ = ππ
β’ Acceleration
ππ = πππππποΏ½ + πππππποΏ½
o ππππ = πππ£π£π π πππ€π€
+ π£π£πππππ£π£π π ππππ
o ππππ = πππ£π£πππππ€π€
+ π£π£π π 2
β
Review for Exam 1 2014
2. Streamline coordinates β Contd.
30
β’ Euler equation: Application of Newtonβs 2nd law (ππππ = βπΉπΉ) to inviscid (i.e., frictionless or ππ = 0) and incompressible* fluid motions
πππποΏ½β 0
= βπππππποΏ½ β π»π»ππ + πππ»π»2ππ=0
β’ The Euler equation in streamline coordinates
ππππ = βπ»π» ππ + πΎπΎππ
where,
π»π» =πππππππποΏ½ +
ππππππ
πποΏ½
or ππ πππ£π£π π
πππ€π€+ π£π£ππ
πππ£π£π π ππππ
= β ππππππ
ππ + πΎπΎππ
ππ πππ£π£πππππ€π€
+ π£π£π π 2
β= β ππ
ππππ(ππ + πΎπΎππ)
* Note: There is another version of Euler equation available for compressible fluid flows as well.
Review for Exam 1 2014
3. Bernoulli equation (1) Along streamline
β’ By integrating the Euler equation along ππ-direction (i.e., along a streamline) for a steady flow,
οΏ½πππππ£π£πππππ‘π‘οΏ½=0
β΅steady
+ π£π£πππππ£π£ππππππ
ππππ = βοΏ½ππππππ
ππ + πΎπΎππ ππππ
or
οΏ½ππππππ
ππ +12πππ£π£ππ2 + πΎπΎππ ππππ
2
1= 0
β΄ ππ2 +12ππππ22 + πΎπΎππ2 = ππ1 +
12ππππ12 + πΎπΎππ1 β΅ π£π£ππ = ππ
31 Review for Exam 1 2014
3. Bernoulli equation (2) Across streamline
β’ By integrating the Euler equation along ππ-direction (i.e., across streamlines) for a steady flow,
οΏ½πππππ£π£πππππ‘π‘οΏ½=0
β΅steady
+π£π£ππ2
βππππ = βοΏ½
ππππππ
ππ + πΎπΎππ ππππ
or
οΏ½ πππ£π£ππ2
βππππ +
ππππππ
ππ + πΎπΎππ ππππ2
1= 0
β΄ ππ2 + πποΏ½ππ2
βππππ
2
1+ πΎπΎππ2 = ππ1 + πΎπΎππ1 β΅ π£π£ππ = ππ
32 Review for Exam 1 2014
3. Bernoulli equation (3) Restrictions and alternative forms
β’ Restrictions 1) Inviscid flow (i.e., no friction) 2) Incompressible flow (i.e., ππ = constant) 3) Steady flow
β’ Static, stagnation dynamic, and Total pressure
ππβstatic
pressure
+12ππππ2
dynamicpressure
stagnation pressure
+ πΎπΎπποΏ½hydrostaticpressure
= πππποΏ½Total
pressure
β’ Head form
β΄πππΎπΎβ
pressurehead
+ππ2
2πποΏ½velocityhead
+ ππβelevationhead
= π»π»βtotalhead
33 Review for Exam 1 2014
4. Application of Bernoulli equation Example (1): Stagnation tube
34
ππ1 + ππππ12
2+ πΎπΎππ1 = ππ2 + ππ
ππ22
2+ πΎπΎππ2
Since ππ2 = 0 (stagnation point) and ππ1 = ππ2,
ππ1 + ππππ12
2= ππ2
Solve for ππ1:
ππ1 =2 ππ2 β ππ1
ππ
Also, ππ1 = πΎπΎππ and ππ2 = πΎπΎ ππ + β
β΄ ππ1 = 2ππβ
Review for Exam 1 2014
4. Application of Bernoulli equation Example (2): Pitot tube
35
ππ1 + ππππ12
2+ πΎπΎππ1 = ππ2 + ππ
ππ22
2+ πΎπΎππ2
where ππ1 = 0 (stagnation point),
ππ1 + πΎπΎππ1 = ππ2 + ππππ22
2+ πΎπΎππ2
Solve for ππ2:
ππ2 = 2ππππ1πΎπΎ
+ ππ1
=β1
βππ2πΎπΎ
+ ππ2
=β2
Thus,
β΄ ππ = ππ2 = 2ππ β β1 β β2from manometer
Review for Exam 1 2014
4. Application of Bernoulli equation Example (3): Free jets
36
Applying the B.E. between (1) and (2),
ππ1 + ππππ12
2+ πΎπΎππ1 = ππ2 + ππ
ππ22
2+ πΎπΎππ2
Since ππ1 = ππ2 = 0 and ππ1 β 0, and ππ1 β ππ2 = β,
πΎπΎβ = ππππ22
2
Solve for ππ2:
ππ2 = 2πΎπΎβππ
= 2ππβ
Review for Exam 1 2014
Note: Simplified Form of Continuity Equation
β’ Volume flow rate ππ = πππ΄π΄
β’ Mass flow rate
οΏ½ΜοΏ½π = ππππ = πππππ΄π΄ β’ Conservation of mass
ππ1ππ1π΄π΄1 = ππ2ππ2π΄π΄2 β’ Incompressible flow (i.e., ππ = const.)
ππ1π΄π΄1 = ππ2π΄π΄2
37 Review for Exam 1 2014
4. Application of Bernoulli equation Example (4): Venturimeter
38
Since ππ1 = ππ2,
ππ1 + ππππ12
2= ππ2 + ππ
ππ22
2
From continuity,
ππ1 =π΄π΄2π΄π΄1
ππ2 =π·π·2π·π·1
2
ππ2
Thus,
ππ1 +12ππ
π·π·2π·π·1
2
ππ2
2
= ππ2 + ππππ22
2
Solve for ππ2,
ππ2 =2 ππ1 β ππ2
ππ 1 β π·π·2 π·π·1β 4
Then, ππ = ππ2π΄π΄2
Review for Exam 1 2014
Chapter 4. Fluid Kinematics
1. Velocity and Description methods 2. Acceleration and Material derivatives 3. Euler equation 4. Flow classification 5. Control-volume approach and RTT
39 Review for Exam 1 2014
1. Velocity and Description methods
40
β’ Lagrangian description: Keep track of individual fluid particles ππππ π‘π‘ = ππππ(π‘π‘)οΏ½ΜοΏ½π + π£π£ππ(π‘π‘)ππΜ + π€π€ππ(π‘π‘)πποΏ½
β’ Eulerian description: Focus attention on a fixed point in space
ππ ππ, π‘π‘ = ππ ππ, π‘π‘ οΏ½ΜοΏ½π + π£π£ ππ, π‘π‘ ππΜ + π€π€ ππ, π‘π‘ πποΏ½
Review for Exam 1 2014
2. Acceleration and Material derivatives
β’ Lagrangian:
ππππ =πππππππππ‘π‘
= πππ₯π₯π€π€Μ + πππ¦π¦π₯π₯Μ + πππ§π§πποΏ½
πππ₯π₯ =πππππππππ‘π‘
, πππ¦π¦ =πππ£π£πππππ‘π‘
,πππ§π§ =πππ€π€πππππ‘π‘
β’ Eulerian:
ππ =π·π·πππ·π·π‘π‘
= πππ₯π₯π€π€Μ + πππ¦π¦π₯π₯Μ + πππ§π§πποΏ½
πππ₯π₯ =πππππππ‘π‘
+ ππππππππππ
+ π£π£ππππππππ
+ π€π€ππππππππ
πππ¦π¦ =πππ£π£πππ‘π‘
+ πππππ£π£ππππ
+ π£π£πππ£π£ππππ
+ π€π€πππ£π£ππππ
πππ§π§ =πππ€π€πππ‘π‘
+ πππππ€π€ππππ
+ π£π£πππ€π€ππππ
+ π€π€πππ€π€ππππ
41 Review for Exam 1 2014
2. Acceleration and material derivatives βContd.
42
Lagrangian Eulerian
ππππ π‘π‘ ππππ π‘π‘ + πΏπΏπ‘π‘
ππππ π‘π‘ = limπΏπΏπ€π€β0
ππππ π‘π‘ + πΏπΏπ‘π‘ β ππππ π‘π‘πΏπΏπ‘π‘
β΄ ππππ(π‘π‘) =πππππππππ‘π‘
ππ(ππ, π‘π‘) ππ(ππ + πππΏπΏπ‘π‘, π‘π‘ + πΏπΏπ‘π‘)
ππ ππ, π‘π‘ = limπΏπΏπ€π€β0
ππ(ππ + πππΏπΏπ‘π‘, π‘π‘ + πΏπΏπ‘π‘) β ππ ππ, π‘π‘πΏπΏπ‘π‘
= limπΏπΏπ€π€β0
ππ(ππ + πππΏπΏπ‘π‘, π‘π‘ + πΏπΏπ‘π‘) β ππ ππ, π‘π‘ + πΏπΏπ‘π‘ + ππ ππ, π‘π‘ + πΏπΏπ‘π‘ β ππ(ππ, π‘π‘)πΏπΏπ‘π‘
= limπΏπΏπ€π€β0
ππ ππ + πππΏπΏπ‘π‘, π‘π‘ + πΏπΏπ‘π‘ β ππ ππ, π‘π‘ + πΏπΏπ‘π‘ππ + πππΏπΏπ‘π‘ β ππ
β ππ + πππΏπΏπ‘π‘ β ππ
πΏπΏπ‘π‘
+ limπΏπΏπ€π€β0
ππ ππ, π‘π‘ + πΏπΏπ‘π‘ β ππ ππ, π‘π‘πΏπΏπ‘π‘
=ππππππππ
β ππ +πππππππ‘π‘
β΄ ππ ππ, π‘π‘ =πππππππ‘π‘
+ ππππππππππ
For 1D flow,
Review for Exam 1 2014
2. Acceleration and material derivatives βContd.
β’ Material derivative*: π·π·π·π·π‘π‘
=πππππ‘π‘
+ ππ β π»π»
where
π»π» =ππππππ
π€π€Μ +ππππππ
π₯π₯Μ +πππππππποΏ½
β’ Acceleration
ππ =π·π·πππ·π·π‘π‘
=πππππππ‘π‘οΏ½Localacc.
+ ππ β π»π» ππConvective
acc.
oπππππππ€π€
= Local or temporal acceleration. Velocity changes with respect to time at a given point
o ππ β π»π» ππ = Convective acceleration. Spatial gradients of velocity
43
*Note: Also referred as substantial derivative or total derivative
Review for Exam 1 2014
3. Euler equation
β’ In Cartesian coordinates
ππππ = ππππ β π»π»ππ
or
πππππππππ‘π‘
+ ππππππππππ
+ π£π£ππππππππ
+ π€π€ππππππππ
= πππππ₯π₯ βππππππππ
πππππ£π£πππ‘π‘
+ πππππ£π£ππππ
+ π£π£πππ£π£ππππ
+ π€π€πππ£π£ππππ
= πππππ¦π¦ βππππππππ
πππππ€π€πππ‘π‘
+ πππππ€π€ππππ
+ π£π£πππ€π€ππππ
+ π€π€πππ€π€ππππ
= πππππ§π§ βππππππππ
44 Review for Exam 1 2014
4. Flow classification
β’ One-, Two-, and Three-dimensional flow β’ Steady vs. Unsteady flow β’ Incompressible vs. Compressible flow β’ Viscous vs. Inciscid flow β’ Rotational vs. Irrotational flow β’ Laminar vs. Trubulent viscous flow β’ Internal vs. External flow β’ Separated vs. Unseparated flow
45 Review for Exam 1 2014