CE 023 1 Introduction To Fluid Mechanics
CE 023 1 Introduction To Fluid Mechanics
CE 023 1 Introduction To Fluid Mechanics
MECHANICS
Mechanics is the oldest physical science that deals
with both stationary and moving bodies under the
influence of forces.
1. Introduction 2 3
Mechanics
Body Fluid
Mechanics Mechanics
Statics Dynamics
Rigid Deformable
Statics Dynamics
1. Introduction 2 3
What is a Fluid?
Fluid Mechanics
Continuum
History
Inviscid Flow
• Viscous forces are very low and can be neglected.
• Occurs in areas far from solid surfaces.
1 2. Classifications of Fluid Flow 3
Incompressible Flow
• When the change in density is insignificant, the fluid is considered
as incompressible.
• Liquids are often considered as incompressible.
1 2. Classifications of Fluid Flow 3
Turbulent Flow
• Chaotic flow.
• Occurs when the fluid flow has a
high Reynold’s Number, Re.
1 2. Classifications of Fluid Flow 3
Unsteady Flow
• Fluid parameters such as velocity, density, pressure, and
acceleration at a point varies with time.
1 2. Classifications of Fluid Flow 3
Non-uniform Flow
• The velocity changes from point to point in a flow at any given
instant of time
1 2 3. Properties of Fluids
Density (ρ)
• Mass (m) per unit Volume (V)
𝒎
𝝆=
𝑽
• In liquids, velocity changes from point to point in a flow at
any given instant of time.
• In gases, density is strongly influenced by both pressure and
temperature.
1 2 3. Properties of Fluids
𝑾 𝒎𝒈
𝜸= = = 𝝆𝒈
𝑽 𝑽
1 2 3. Properties of Fluids
𝝆𝒍𝒊𝒒𝒖𝒊𝒅
𝑺𝑮 𝒍𝒊𝒒𝒖𝒊𝒅 =
𝝆𝒘𝒂𝒕𝒆𝒓 𝒂𝒕 𝟒°𝑪
𝝆𝒈𝒂𝒔
𝑺𝑮 𝒈𝒂𝒔 =
𝝆𝒂𝒊𝒓
1 2 3. Properties of Fluids
𝑽 𝟏
𝒗= =
𝒎 𝝆
1 2 3. Properties of Fluids
Sample Problem 1
Given:
The weight of a substance is 𝑊 = 64.75𝑘𝑁
64.75kN. If its volume is half of 𝑉 = 0.5𝑚3
a cubic meter, calculate the
density of the substance in Required: 𝜌
kg/m3.
*Solve the density
Solution: 𝜌=
𝑚
*Solve the mass of the substance 𝑉
𝑊 = 𝑚𝑔 6602.4268𝑘𝑔
64750N = 𝑚(9.807𝑚/𝑠 2 ) 𝜌=
0.5𝑚3
𝒎 = 𝟔𝟔𝟎𝟐. 𝟒𝟐𝟔𝟖𝒌𝒈
𝝆 = 𝟏𝟑𝟐𝟎𝟒. 𝟖𝟓𝟑𝟕𝒌𝒈/𝒎𝟑
1 2 3. Properties of Fluids
Sample Problem 2
Given:
The specific volume of a liquid 𝑣 = 1.25𝑐𝑚3 /𝑔
is 1.25cm3/g. Solve the specific 𝜌𝑤𝑎𝑡𝑒𝑟 𝑎𝑡 4°𝐶 = 1𝑔/𝑐𝑚3
gravity of the liquid with
respect to the density of water Required: 𝑆𝐺
at 4 degrees Celsius (1g/cm3).
*Solve SG
Solution: 𝑆𝐺 =
𝜌
*Solve the density of the liquid 𝜌𝑤𝑎𝑡𝑒𝑟 𝑎𝑡 4°𝐶
1
𝑣= 0.8
𝜌 𝑆𝐺 =
3 1 1
1.25𝑐𝑚 /𝑔 =
𝜌
𝝆 = 𝟎. 𝟖𝒈/𝒄𝒎𝟑 𝑺𝑮 = 𝟎. 𝟖
1 2 3. Properties of Fluids
Viscosity (μ)
• Relates shearing stress and fluid motion.
• Highly dependent on temperature.
• Only mildly dependent on pressure and the effect of pressure
is usually neglected
• Measured in Poise, P.
1𝑑𝑦𝑛 ∙ 𝑠
1𝑃 =
𝑐𝑚2
1𝑁 = 100000𝑑𝑦𝑛
1 2 3. Properties of Fluids
Viscosity (μ)
• It is also defined as
the ratio between
the shearing stress
(τ) and rate of
shearing strain or
velocity gradient
(du/dy).
𝝉
𝝁=
𝒅𝒖
𝒅𝒚
1 2 3. Properties of Fluids
Viscosity (μ)
Deformation of fluid placed Free body diagram of the
between two plates upper plate
1 2 3. Properties of Fluids
Viscosity (μ)
• Steeper slope means
that the fluid is more
viscous.
1 2 3. Properties of Fluids
Viscosity (μ)
• Generally, viscosity of
liquids decreases as
temperature increases.
• Gases behave
oppositely.
1 2 3. Properties of Fluids
Sample Problem 4
Two 100cm by 100cm parallel plates are 1cm apart. Varnish
(0.20P) is introduced between the plates. What is the force
required to draw the upper plate at 10m/s.
1 2 3. Properties of Fluids
Required: 𝑃 𝐹
𝑑𝑦𝑛∙𝑠 10000𝑐𝑚2
0.20 = 10𝑐𝑚/𝑠
𝑐𝑚2
1𝑐𝑚
𝐹 = 20000𝑑𝑦𝑛
𝑭 = 𝟎. 𝟐𝑵
1 2 3. Properties of Fluids
𝝁
𝒗=
𝝆
• Expressed in Stokes (St), ft2/s
• 1St = 1cm2/s
1 2 3. Properties of Fluids
𝒅𝒑
𝑬𝒗 = −
𝒅𝑽
𝑽
Where:
𝑑𝑝 is the differential change in pressure
𝑑𝑉
is the differential change in volume
𝑉
1 2 3. Properties of Fluids
𝑷 = 𝝆𝑹𝑻
Where:
𝑃 is the absolute pressure
𝑇 is the absolute temperature
𝜌 is the density of the gas
𝑅 is the ideal gas constant (N.m/kg.K, ft.lb/slug.R)
1 2 3. Properties of Fluids
Surface Tension, σ
• The intensity of the molecular attraction per unit length along
any line in the surface.
1 2 3. Properties of Fluids
Surface Tension, σ
• Occurs at the interface between a liquid and a gas, or
between two immiscible liquids
Sample Problem 5
Determine the excess pressure inside a 0.5in diameter soap
bubble floating in air. The surface tension of the soap solution is
0.0035lb/ft.
1 2 3. Properties of Fluids
Solution:
Sample Problem 5 *Sum forces along the horizontal
Given: +→ σ 𝐹𝑥 = 0
𝐷 = 0.5𝑖𝑛 𝜋𝐷2
𝜎(𝜋𝐷) − 𝑃( ) = 0
𝜎 = 0.0035𝑙𝑏/𝑓𝑡 = 0.0002917𝑙𝑏/𝑖𝑛 4
𝜋(0.5)2
0.0002917(𝜋)(0.5) − 𝑃( ) =0
4
𝑃 = 0.002333𝑙𝑏/𝑖𝑛2
𝑷 = 𝟎. 𝟎𝟎𝟐𝟑𝟑𝟑𝒑𝒔𝒊
Required: 𝑃
1 2 3. Properties of Fluids
𝟐𝝈𝒄𝒐𝒔𝜽
𝒉=
𝜸𝑹
1 2 3. Properties of Fluids
Sample Problem 6
A 0.6mm diameter glass tube is inserted into water at 20
degrees Celsius (σ = 0.073N/m). Determine the capillary rise of
water in the tube if the contact angle is 0 degrees. The specific
weight of water is 9810N/m3.
Given: Solution:
2𝜎𝑐𝑜𝑠𝜃
𝜎 = 0.073𝑁/𝑚 ℎ=
𝛾𝑅
𝑅 = 0.0003𝑚 2(0.073𝑁/𝑚)𝑐𝑜𝑠0
𝛾 = 9810𝑁/𝑚3 ℎ= 𝑁
9810( 3 )(0.0003𝑚)
𝑚
𝜃 = 0°
𝒉 = 𝟎. 𝟎𝟒𝟗𝟔𝒎
Required: ℎ
END