AFM Class 10
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Transcript of AFM Class 10
Class-10 (Navier-Stokes Equations)
5/7/2012 1
Angular deformation
5/7/2012 2
β’ Angular deformation of the fluid particle is the sum of the two angular deformations
βπΆ + βπ·
β’ Also βπΆ =βπΌ
βπ and βπ· =
βπ
βπ , βπΌ ππ§π βπ are given by βπΌ =
ππ
ππβπβπ and βπ =
ππ
ππβπβπ
β’ Rate of angular deformation in x-y- plane: limβπβπ
βπΆ+βπ· βπ
= limβπβπ
βπΌ
βπ+
βπ
βπ
βπ
β’ Hence Rate of angular deformation in the x-y- plane: ππ
ππ+
ππ
ππ
β’ Rate of angular deformation in y-z- plane and x-z- plane: ππ
ππ+
ππ
ππ and
ππ
ππ+
ππ
ππ
5/7/2012 3
Newtonian and Non-Newtonian Fluids β’ For Newtonian fluids, the shear stress is linearly proportional to shear strain
Ex: Air and gases, water, kerosene, oil based liquids
β’ Non-Newtonian fluids, shear stress is not linearly proportional to shear strain.
Ex: Slurries, colloidal suspensions, polymer solutions, blood, paste and cake batter
β’ Shear thinning and shear thickening fluids:
β’ Some non-Newtonian fluids become less viscous when more sheared. These are
called shear thinning fluids (Eg. Paint)
β’ Some non-Newtonian fluids become more viscous when more sheared. These are
called shear thickening or dilatant fluids (Eg. Quick sand and water mixture)
β’ If shear thinning is extreme, it is called plastic fluid
β’ If fluid retains the original shape after shear force is
removed, it is called visco-elastic liquids
The angular velocity about the z axis is
Similarly,
β’ A fluid at rest will have the following stress tensor (there is no shear stress)
β’ Where P is the hydrostatic pressure , and is equal to the thermodynamic pressure which depends on the type of gas, temperature and density
β’ When the fluid is in motion, pressure is there and
shear stresses are acting due to viscosity
The new stress tensor is called the viscous stress tensor
Or the deviatoric stress tensor 5/8/2012 4
Stresses on a fluid β at rest and in motion
Shear Stress and Viscosity β 2 dimensional flow
β’ The force is applied in the y-plane and in the x- direction. Hence shear stress
πππ =πΉππ
πΉπ¨π=
π ππ
π π¨π , the deformation rate = lim
πΏπβπ
πΉπΆ πΉπ
=π πΆ
π π
πΉπ = πΉππΉπ also for small angles πΉπ = πΉππΉπΆ
SO, πΉπΆ πΉπ
= πΉπ πΉπ
, taking limits on both sides, π πΆ π π
= π π π π
β’ Hence the shear rate of the fluid is equal to π π π π
β’ For a Newtonian fluid the shear stress is linearly related to the shear rate
πππ β π πΆ π π
= π π π π
β’ And then πππ = ππ π π π
π is the viscosity of the fluid
For a 3-dimensional flow, the shear stress becomes complicated, and πππ = π(π π π π
+π π π π
) 5/8/2012
5
β’ In an incompressible fluid, density is constant, and then the thermodynamic pressure becomes the mechanical pressure P
β’ It has been proved that
β’ Also, for a Newtonian fluid, the shear (viscous) stress is linearly related to the shear strain, and then where is the stress tensor and the strain tensor
β’ Hence the viscous stress tensor becomes
5/8/2012 6
Stresses on a fluid β Incompressible fluid
β’ Hence the total stress tensor becomes,
β’ ππ₯π₯ = βπ + 2πππ’
ππ₯ ππ¦π¦ = βπ + 2π
ππ£
ππ¦ ππ§π§ = βπ + 2π
ππ€
ππ§
β’ But π = β1
3(ππ₯π₯ + ππ¦π¦ + ππ§π§)
β’ So ππ₯π₯ =1
3(βπ + 2π
ππ’
ππ₯ + βπ + 2π
ππ£
ππ¦ + βπ + 2π
ππ€
ππ§) + 2π
ππ’
ππ₯= βπ β
2
3πβ. π + 2π
ππ’
ππ₯
β’ ππ¦π¦ = βπ β2
3ππ». π + 2π
ππ£
ππ¦ and ππ§π§ = βπ β
2
3ππ». π + 2π
ππ€
ππ§
5/8/2012 7
Stresses on a fluid β Incompressible fluid
The differential momentum equation
5/8/2012 8
The differential momentum equation is
πππ +ππππ
ππ+
ππππ
ππ+
ππππ
ππ= π(
ππ
ππ+ π
ππ
ππ+v
ππ
ππ+
ππ
ππ)
πππ +ππππ
ππ+
ππππ
ππ+
ππππ
ππ= π(
ππ
ππ+ π
ππ
ππ+v
ππ
ππ+
ππ
ππ)
πππ +ππππ
ππ+
ππππ
ππ+
ππππ
ππ= π(
ππ
ππ+ π
ππ
ππ+v
ππ
ππ+
ππ
ππ)
β’ Where p is the Mechanical pressure (incompressible flow)
5/7/2012 9
The Navier-Stokes Equations β’ Substituting the value of the stresses in the differential momentum equations,
5/8/2012 10
β’ For an incompressible flow,
For incompressible flow, π. π = π
ππ«π
π«π= πππ β
ππ
ππ+ π(
π ππ
π ππ +π ππ
π ππ +π ππ
π ππ) + π(π ππ
π ππ +π ππ
π ππ π+
π ππ
π ππ π)
= πππ βππ
ππ+ π(
π ππ
π ππ +π ππ
π ππ +π ππ
π ππ ) + ππ
π π(
π π
π π+
π π
π π+
π π
π π)
= πππ βππ
ππ+ π(
π ππ
π ππ +π ππ
π ππ +π ππ
π ππ )
Similarly
ππ«π
π«π = πππ β
ππ
ππ+ π(
π ππ
π ππ +π ππ
π ππ +π ππ
π ππ)
ππ«π
π«π = πππ β
ππ
ππ+ π(
π ππ
π ππ +π ππ
π ππ +π ππ
π ππ )
The Navier-Stokes Equations