57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
1
Chapter 6 Differential Analysis of Fluid Flow Fluid Element Kinematics Fluid element motion consists of translation, linear deformation, rotation, and angular deformation.
Types of motion and deformation for a fluid element. Linear Motion and Deformation:
Translation of a fluid element
Linear deformation of a fluid element 1
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
2
Change in δ∀ :
∂u δ x (δ yδ z ) δ t ∂x
δ∀ =
the rate at which the volume δ∀ is changing per unit volume due to the gradient ∂u/∂x is
∀) ( ∂u ∂x ) t ∂u 1 d (dd = lim = d t →0 dd t ∀ dt ∂x If velocity gradients ∂v/∂y and ∂w/∂z are also present, then using a similar analysis it follows that, in the general case,
1 d (d∀ ) ∂u ∂v ∂w = + + = ∇⋅V d∀ dt ∂x ∂y ∂z This rate of change of the volume per unit volume is called the volumetric dilatation rate. Angular Motion and Deformation For simplicity we will consider motion in the x–y plane, but the results can be readily extended to the more general case.
2
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
3
Angular motion and deformation of a fluid element The angular velocity of line OA, ωOA, is
δα δ t →0 δ t
ωOA = lim For small angles
tan δa = ≈ δa
x ) δ xδ t ( ∂v ∂=
δx
∂v δt ∂x
so that
( ∂v ∂x ) δ t ∂v ωOA lim = = δ t →0 δ t ∂x
Note that if ∂v/∂x is positive, ωOA will be counterclockwise. Similarly, the angular velocity of the line OB is
δβ ∂u = ωOB lim = δ t →0 δ t ∂y In this instance if ∂u/∂y is positive, ωOB will be clockwise.
3
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
4
The rotation, ωz, of the element about the z axis is defined as the average of the angular velocities ωOA and ωOB of the two mutually perpendicular lines OA and OB. Thus, if counterclockwise rotation is considered to be positive, it follows that 1 ∂v ∂u = ωz − 2 ∂x ∂y Rotation of the field element about the other two coordinate axes can be obtained in a similar manner: 1 ∂w ∂v wx = − 2 ∂y ∂z
wy =
1 ∂u ∂w − 2 ∂z ∂x
The three components, ωx,ωy, and ωz can be combined to give the rotation vector, ω, in the form: 1 1 ω= ω x i + ω y j + ω z k= curlV= ∇×V 2 2 since i j k
1 1 ∂ ∇×V = 2 2 ∂x u =
∂ ∂y v
∂ ∂z w
1 ∂w ∂v 1 ∂u ∂w 1 ∂v ∂u − i + − j + − k 2 ∂y ∂z 2 ∂z ∂x 2 ∂x ∂y 4
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
5
The vorticity, ζ, is defined as a vector that is twice the rotation vector; that is, ς = 2ω = ∇ × ς The use of the vorticity to describe the rotational characteristics of the fluid simply eliminates the (1/2) factor associated with the rotation vector. If ∇ × V = 0 , the flow is called irrotational. In addition to the rotation associated with the derivatives ∂u/∂y and ∂v/∂x, these derivatives can cause the fluid element to undergo an angular deformation, which results in a change in shape of the element. The change in the original right angle formed by the lines OA and OB is termed the shearing strain, δγ, δγ = δα + δβ The rate of change of δγ is called the rate of shearing strain or the rate of angular deformation: (𝜕𝜕𝜕𝜕 ⁄𝜕𝜕𝜕𝜕)𝛿𝛿𝛿𝛿 + (𝜕𝜕𝜕𝜕⁄𝜕𝜕𝜕𝜕)𝛿𝛿𝛿𝛿 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿 = lim =� �= + 𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿→0 𝛿𝛿𝛿𝛿 𝛿𝛿𝛿𝛿 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝛾𝛾̇𝑥𝑥𝑥𝑥 = lim
Similarly,
𝛾𝛾̇𝑥𝑥𝑥𝑥 = 𝛾𝛾̇𝑦𝑦𝑦𝑦 =
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 + 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 + 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
The rate of angular deformation is related to a corresponding shearing stress which causes the fluid element to change in shape. 5
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
6
The Continuity Equation in Differential Form The governing equations can be expressed in both integral and differential form. Integral form is useful for large-scale control volume analysis, whereas the differential form is useful for relatively small-scale point analysis. Application of RTT to a fixed elemental control volume yields the differential form of the governing equations. For example for conservation of mass
∂ρ dV CV ∂t
∑ ρV ⋅ A = − ∫ CS
net outflow of mass across CS
=
rate of decrease of mass within CV
6
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
7
Consider a cubical element oriented so that its sides are to the (x,y,z) axes ∂ (ρu )dx dydz ρu +
inlet mass flux ρudydz
∂x outlet mass flux
Taylor series expansion retaining only first order term
We assume that the element is infinitesimally small such that we can assume that the flow is approximately one dimensional through each face. The mass flux terms occur on all six faces, three inlets, and three outlets. Consider the mass flux on the x faces
∂ x flux = ρu + ρu dx dydz outflux − ρudydz influx ( ) ∂x =
∂ (ρu )dxdydz ∂x V
Similarly for the y and z faces ∂ y flux = (ρv)dxdydz ∂y ∂ z flux = (ρw )dxdydz ∂z
7
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
8
The total net mass outflux must balance the rate of decrease of mass within the CV which is ∂ρ − dxdydz ∂t Combining the above expressions yields the desired result ∂ρ ∂ ∂ ∂ ( u ) ( v ) ( w ) ρ + ρ + ρ + ∂t ∂x dxdydz = 0 y z ∂ ∂ dV
∂ρ ∂ ∂ ∂ + (ρu ) + (ρv) + (ρw ) = 0 ∂t ∂x ∂y ∂z
per unit V differential form of continuity equations
∂ρ + ∇ ⋅ (ρV) = 0 ∂t ρ∇ ⋅ V + V ⋅ ∇ρ Dρ + ρ∇ ⋅ V = 0 Dt
D ∂ = + V ⋅∇ Dt ∂t
Nonlinear 1st order PDE; ( unless ρ = constant, then linear) Relates V to satisfy kinematic condition of mass conservation Simplifications: 1. Steady flow: ∇ ⋅ (ρV) = 0 2. ρ = constant: ∇ ⋅ V = 0
8
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
i.e.,
Chapter 6
9
∂u ∂v ∂w + + =0 ∂x ∂y ∂z
3D
∂u ∂v + =0 ∂x ∂y
2D
The continuity equation in Cylindrical Polar Coordinates
The velocity at some arbitrary point P can be expressed as V = vr e r + vθ eθ + vz e z The continuity equation: vr ) 1 ∂ ( vθ ) ∂ ( vz ) ∂r 1 ∂ ( r rrr 0 + + + = ∂t r ∂r ∂z r ∂θ For steady, compressible flow vr ) 1 ∂ ( vθ ) ∂ ( vz ) 1 ∂ ( r rrr + + = 0 ∂r ∂z r r ∂θ For incompressible fluids (for steady or unsteady flow) 1 ∂ ( rvr ) 1 ∂vθ ∂vz 0 + + = r ∂r r ∂θ ∂z
9
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
10
The Stream Function Steady, incompressible, plane, two-dimensional flow represents one of the simplest types of flow of practical importance. By plane, two-dimensional flow we mean that there are only two velocity components, such as u and v, when the flow is considered to be in the x–y plane. For this flow the continuity equation reduces to ∂u ∂v + =0 ∂x ∂y
We still have two variables, u and v, to deal with, but they must be related in a special way as indicated. This equation suggests that if we define a function ψ(x, y), called the stream function, which relates the velocities as ∂y ∂y u= , v= − ∂y ∂x then the continuity equation is identically satisfied: ∂ ∂y ∂ ∂y ∂ 2y ∂ 2y − = 0 + − = ∂x ∂y ∂y ∂x ∂x∂y ∂x∂y
Velocity and velocity components along a streamline 10
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
11
Another particular advantage of using the stream function is related to the fact that lines along which ψ is constant are streamlines.The change in the value of ψ as we move from one point (x, y) to a nearby point (x + dx, y + dy) along a line of constant ψ is given by the relationship: ∂y ∂y dy = dx + dy = −vdx + udy = 0 ∂x ∂y and, therefore, along a line of constant ψ dy v = dx u
The flow between two streamlines The actual numerical value associated with a particular streamline is not of particular significance, but the change in the value of ψ is related to the volume rate of flow. Let dq represent the volume rate of flow (per unit width perpendicular to the x–y plane) passing between the two streamlines. ∂y ∂y dq = udy − vdx = dx + dy = dy ∂x ∂y Thus, the volume rate of flow, q, between two streamlines such as ψ1 and ψ2, can be determined by integrating to yield:
11
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
= q
ψ2
∫ψ
Chapter 6
12
= 2− 1 dψψψ
1
In cylindrical coordinates the continuity equation for incompressible, plane, two-dimensional flow reduces to 1 ∂ ( rvr ) 1 ∂vθ 0 + = r ∂r r ∂θ
and the velocity components, vr and vθ, can be related to the stream function, ψ(r, θ), through the equations ∂ 1 ∂ψψ , vθ = − vr = ∂r r ∂θ
Navier-Stokes Equations Differential form of momentum equation can be derived by applying control volume form to elemental control volume The differential equation of linear momentum: elemental fluid volume approach
12
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
∑𝐹𝐹 =
Chapter 6
13
𝜕𝜕 � 𝜌𝜌𝑉𝑉𝑑𝑑V + � 𝑉𝑉𝜌𝜌𝑉𝑉 ⋅ 𝑛𝑛�𝑑𝑑𝐴𝐴 𝜕𝜕𝜕𝜕 ���� �� �� ��������� 𝐶𝐶𝐶𝐶 � 𝐶𝐶𝐶𝐶 (1)
𝜕𝜕
(2)
𝜕𝜕𝑉𝑉
𝜕𝜕𝜕𝜕
(1) = 𝜕𝜕𝜕𝜕 �𝜌𝜌𝑉𝑉�𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = � 𝜕𝜕𝜕𝜕 𝑉𝑉 + 𝜌𝜌 𝜕𝜕𝜕𝜕 � 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝜕𝜕 𝜕𝜕 𝜕𝜕 (2) = ��� �𝜌𝜌𝜌𝜌𝑉𝑉� + �𝜌𝜌𝜌𝜌𝑉𝑉� + �𝜌𝜌𝜌𝜌𝑉𝑉� �� �� ��� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕��� 𝜕𝜕𝜕𝜕��� 𝜕𝜕𝜕𝜕 ��� �� �� 𝑥𝑥−face
𝜕𝜕𝑉𝑉
=�𝜌𝜌𝜌𝜌 𝜕𝜕𝜕𝜕 + 𝑉𝑉
𝑧𝑧−face
𝑦𝑦−face
𝜕𝜕𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
+ 𝜌𝜌𝜌𝜌
𝜕𝜕𝑉𝑉 𝜕𝜕𝜕𝜕
+ 𝑉𝑉
𝜕𝜕𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
+ 𝜌𝜌𝜌𝜌
𝜕𝜕𝑉𝑉 𝜕𝜕𝜕𝜕
+ 𝑉𝑉
𝜕𝜕𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
combining and making use of the continuity equation yields 𝜕𝜕𝜕𝜕
𝜕𝜕𝑉𝑉
∑𝐹𝐹 = �𝑉𝑉 ���+ ∇ ⋅ �𝜌𝜌𝑉𝑉� ���� ���� + 𝜌𝜌 � 𝜕𝜕𝜕𝜕
∴ ∑𝐹𝐹 = 𝜌𝜌
=0
𝐷𝐷𝑉𝑉 𝐷𝐷𝐷𝐷
𝜕𝜕𝜕𝜕
+ 𝑉𝑉 ⋅ ∇𝑉𝑉�� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 or
where ∑𝐹𝐹 = ∑𝐹𝐹body + ∑𝐹𝐹surface ∑f = ∑f𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + ∑f𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
∑f = 𝜌𝜌
𝐷𝐷𝑉𝑉 𝐷𝐷𝐷𝐷
𝑉𝑉 ⋅ ∇ = 𝑢𝑢 𝐷𝐷
𝐷𝐷𝐷𝐷
=
𝜕𝜕
𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
+ 𝑣𝑣
𝜕𝜕
𝜕𝜕𝜕𝜕
+ 𝑉𝑉 ⋅ ∇
+ 𝑤𝑤
𝜕𝜕
𝜕𝜕𝜕𝜕
13
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
14
Body forces are due to external fields such as gravity or magnetics. Here we only consider a gravitational field; that is, ∑ F body = d F gρav = ρgdxdydz and g = −gkˆ for g↓ z↑
i.e., f body = −ρgkˆ Surface forces are due to the stresses that act on the sides of the control surfaces symmetric (σij = σji) σij = - pδij + τij 2nd order tensor normal pressure
viscous stress =
-p+τxx τyx τzx
τxy -p+τyy τzy
δij = 1 δij = 0
i=j i≠j
τxz τyz -p+τzz
As shown before for p alone it is not the stresses themselves that cause a net force but their gradients. ∂ ∂ ∂ dFx,surf = (σ xx ) + (σ xy ) + (σ xz ) dxdydz ∂y ∂z ∂x ∂p ∂ ∂ ∂ = − + (τ xx ) + (τ xy ) + (τ xz ) dxdydz ∂y ∂z ∂x ∂x 14
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
15
This can be put in a more compact form by defining vector stress on x-face τ x = τ xx ˆi + τ xy ˆj + τ xz kˆ and noting that ∂p dFx,surf = − + ∇ ⋅ τ x dxdydz ∂x ∂p per unit volume fx,surf = − + ∇ ⋅ τ x ∂x
similarly for y and z ∂p fy,surf = − + ∇ ⋅ τ y ∂y
τ y = τ yx ˆi + τ yy ˆj + τ yz kˆ
∂p + ∇ ⋅ τz ∂z
τ z = τ zx ˆi + τ zy ˆj + τ zz kˆ
fz,surf = −
finally if we define τ ij = τ x ˆi + τ y ˆj + τ z kˆ
then
f surf = −∇p + ∇ ⋅ τ ij = ∇ ⋅ s ij
σ ij = − pδ ij + τ ij
15
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
16
Putting together the above results ∑ f = f body + f surf = r
DV Dt
f body = −ρgkˆ f surface = −∇p + ∇ ⋅ τ ij
= a
DV ∂ V = + V ⋅∇ V Dt ∂t
ρ a = − ρ gkˆ − ∇p + ∇ ⋅τ ij
inertia force
body force surface due to force due gravity to p
surface force due to viscous shear and normal stresses
16
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
17
For Newtonian fluid the shear stress is proportional to the rate of strain, which for incompressible flow can be written 𝜕𝜕𝑢𝑢𝑖𝑖
𝜏𝜏𝑖𝑖𝑖𝑖 = 2𝜇𝜇𝜀𝜀𝑖𝑖𝑖𝑖 = 𝜇𝜇 �
𝜕𝜕𝑥𝑥𝑗𝑗
+
𝜕𝜕𝑢𝑢𝑗𝑗 𝜕𝜕𝑥𝑥𝑖𝑖
�
where, 𝜇𝜇 = coefficient of viscosity 𝜀𝜀𝑖𝑖𝑖𝑖 = rate of strain tensor 𝜕𝜕𝜕𝜕
⎡ 𝜕𝜕𝜕𝜕 ⎢ 1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = ⎢ 2 �𝜕𝜕𝜕𝜕 + 𝜕𝜕𝜕𝜕� ⎢1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 ⎣2 � 𝜕𝜕𝜕𝜕 + 𝜕𝜕𝜕𝜕 �
1 𝜕𝜕𝜕𝜕
�
+
�
+
𝜕𝜕𝜕𝜕
2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
�
𝜕𝜕𝜕𝜕 1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
�
𝜌𝜌𝑎𝑎 = −𝜌𝜌𝜌𝜌𝑘𝑘� − ∇𝑝𝑝 + ∇ ⋅ �𝜏𝜏𝑖𝑖𝑖𝑖 �
1 𝜕𝜕𝜕𝜕
�
𝜕𝜕𝜕𝜕
�⎤ ⎥ � + �⎥ 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 ⎥ 𝜕𝜕𝜕𝜕 ⎦ 𝜕𝜕𝜕𝜕
2 𝜕𝜕𝜕𝜕 1 𝜕𝜕𝜕𝜕
+
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
Ex) 1-D flow 𝑑𝑑𝑑𝑑 𝜏𝜏 = 𝜇𝜇 𝑑𝑑𝑑𝑑
where,
∇ ⋅ �𝜏𝜏𝑖𝑖𝑖𝑖 � = 𝜇𝜇
2
𝜕𝜕𝑢𝑢𝑗𝑗 𝜕𝜕𝑢𝑢𝑖𝑖 𝜕𝜕 𝑢𝑢 𝜕𝜕 𝜕𝜕𝑢𝑢𝑗𝑗 ⎞ + � = 𝜇𝜇 ⎛ 2𝑖𝑖 + � 𝜕𝜕𝑥𝑥𝑗𝑗 𝜕𝜕𝑥𝑥𝑗𝑗 𝜕𝜕𝑥𝑥 𝜕𝜕𝑥𝑥𝑖𝑖 𝜕𝜕𝑥𝑥 𝜕𝜕𝑥𝑥 𝑖𝑖 �𝑗𝑗 �𝑗𝑗 =0 ⎠ ⎝ ∇2𝑉𝑉 𝜕𝜕
𝜌𝜌𝑎𝑎 = −𝜌𝜌𝜌𝜌𝑘𝑘� − ∇𝑝𝑝 + 𝜇𝜇∇2 𝑉𝑉
𝜌𝜌𝑎𝑎 = −∇(𝑝𝑝 + 𝛾𝛾𝛾𝛾) + 𝜇𝜇∇2 𝑉𝑉 ∇ ⋅ 𝑉𝑉 = 0
Navier-Stokes Equation Continuity Equation
17
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
18
Four equations in four unknowns: V and p Difficult to solve since 2nd order nonlinear PDE 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑢𝑢
𝜕𝜕2 𝑢𝑢
𝜕𝜕2 𝑢𝑢
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑣𝑣
𝜕𝜕2 𝑣𝑣
𝜕𝜕2 𝑣𝑣
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
x: 𝜌𝜌 � 𝜕𝜕𝜕𝜕 + 𝑢𝑢 𝜕𝜕𝜕𝜕 + 𝑣𝑣 𝜕𝜕𝜕𝜕 + 𝑤𝑤 𝜕𝜕𝜕𝜕 � = − 𝜕𝜕𝜕𝜕 + 𝜇𝜇 �𝜕𝜕𝑥𝑥2 + 𝜕𝜕𝑦𝑦2 + 𝜕𝜕𝑧𝑧 2 � y: 𝜌𝜌 � 𝜕𝜕𝜕𝜕 + 𝑢𝑢 𝜕𝜕𝜕𝜕 + 𝑣𝑣 𝜕𝜕𝜕𝜕 + 𝑤𝑤 𝜕𝜕𝜕𝜕� = − 𝜕𝜕𝜕𝜕 + 𝜇𝜇 �𝜕𝜕𝑥𝑥2 + 𝜕𝜕𝑦𝑦2 + 𝜕𝜕𝑧𝑧 2 � 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑤𝑤
𝜕𝜕2 𝑤𝑤
𝜕𝜕2 𝑤𝑤
z: 𝜌𝜌 � 𝜕𝜕𝜕𝜕 + 𝑢𝑢 𝜕𝜕𝜕𝜕 + 𝑣𝑣 𝜕𝜕𝜕𝜕 + 𝑤𝑤 𝜕𝜕𝜕𝜕 � = − 𝜕𝜕𝜕𝜕 − 𝜌𝜌𝜌𝜌 + 𝜇𝜇 � 𝜕𝜕𝑥𝑥2 + 𝜕𝜕𝑦𝑦2 + 𝜕𝜕𝑧𝑧 2 � ∂u ∂v ∂w + + =0 ∂x ∂y ∂z
Navier-Stokes equations can also be written in other coordinate systems such as cylindrical, spherical, etc. There are about 80 exact solutions for simple geometries. For practical geometries, the equations are reduced to algebraic form using finite differences and solved using computers.
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57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
19
Ex) Exact solution for laminar incompressible steady flow in a circular pipe
Use cylindrical coordinates with assumptions 𝜕𝜕
𝜕𝜕𝜕𝜕 𝜕𝜕
𝜕𝜕𝜕𝜕
= 0 : Steady flow
= 0 : Fully-developed flow
𝑣𝑣𝑟𝑟 = 0 : Flow is laminar and parallel to the wall 𝑣𝑣𝜃𝜃 =
𝜕𝜕
𝜕𝜕𝜕𝜕
= 0 : Flow is axisymmetric with no swirl
Continuity equation: 1 𝜕𝜕(𝑟𝑟𝑣𝑣𝑟𝑟 ) 𝑟𝑟
𝜕𝜕𝜕𝜕
+
1 𝜕𝜕𝑣𝑣𝜃𝜃 𝑟𝑟 𝜕𝜕𝜕𝜕
+
𝜕𝜕𝑣𝑣𝑧𝑧 𝜕𝜕𝜕𝜕
=0
Thus, (𝑣𝑣𝑟𝑟 , 𝑣𝑣𝜃𝜃 , 𝑣𝑣𝑧𝑧 ) satisfies the continuity equation 19
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
20
Momentum equation: 𝜌𝜌 �
𝜕𝜕𝑣𝑣𝑟𝑟
𝜌𝜌 �
𝜕𝜕𝑣𝑣𝜃𝜃
𝜌𝜌 �
𝜕𝜕𝑣𝑣𝑧𝑧
or
𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑟𝑟
𝜕𝜕𝑣𝑣𝑟𝑟
+ 𝑣𝑣𝑟𝑟
𝜕𝜕𝑣𝑣𝜃𝜃
=−
𝜕𝜕𝜕𝜕
=− 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑟𝑟
=−
𝜕𝜕𝜕𝜕
+
𝑣𝑣𝜃𝜃 𝜕𝜕𝑣𝑣𝑟𝑟
+
𝑣𝑣𝜃𝜃 𝜕𝜕𝑣𝑣𝜃𝜃
+
𝑣𝑣𝜃𝜃 𝜕𝜕𝑣𝑣𝑧𝑧
𝑟𝑟 𝜕𝜕𝜕𝜕
+ 𝜌𝜌𝑔𝑔𝑟𝑟 + 𝜇𝜇 � 𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝑣𝑣𝑧𝑧
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
+
+ 𝜇𝜇 �
+ 𝑣𝑣𝑧𝑧
�𝑟𝑟
𝑣𝑣𝑟𝑟 𝑣𝑣𝜃𝜃 𝑟𝑟
1 𝜕𝜕
�𝑟𝑟 𝑟𝑟 𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑧𝑧
+ 𝜌𝜌𝑔𝑔𝑧𝑧 + 𝜇𝜇 �
0 = −𝜌𝜌𝜌𝜌 cos 𝜃𝜃 − 𝜕𝜕𝜕𝜕
𝑟𝑟 1 𝜕𝜕
+ 𝜌𝜌𝑔𝑔𝜃𝜃 + 𝜇𝜇 �
0 = −𝜌𝜌𝜌𝜌 sin 𝜃𝜃 − 0=−
−
𝑣𝑣𝜃𝜃2
1 𝜕𝜕
1 𝜕𝜕2 𝑣𝑣𝑟𝑟
𝑣𝑣𝑟𝑟
� − 𝑟𝑟 2 + 𝑟𝑟 2
𝜕𝜕𝑣𝑣𝜃𝜃
𝜕𝜕𝜕𝜕
�
𝜕𝜕𝑣𝑣𝜃𝜃
�
𝜕𝜕𝜕𝜕 𝑣𝑣𝜃𝜃
𝜕𝜕𝑣𝑣𝑧𝑧
1 𝜕𝜕2 𝑣𝑣𝑧𝑧
� + 𝑟𝑟 2 𝜕𝜕𝜕𝜕
−
1 𝜕𝜕2 𝑣𝑣𝜃𝜃
� − 𝑟𝑟 2 + 𝑟𝑟 2 𝜕𝜕𝜕𝜕 �
𝜕𝜕𝜃𝜃 2
𝜕𝜕𝜃𝜃 2
𝜕𝜕𝜃𝜃 2
+
2 𝜕𝜕𝑣𝑣𝜃𝜃 𝑟𝑟 2 𝜕𝜕𝜕𝜕
+
𝜕𝜕2 𝑣𝑣𝑧𝑧 𝜕𝜕𝑧𝑧 2
+
2 𝜕𝜕𝑣𝑣𝑟𝑟
𝑟𝑟 2 𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑣𝑣𝑟𝑟
+
𝜕𝜕𝑧𝑧 2
�
𝜕𝜕2 𝑣𝑣𝜃𝜃 𝜕𝜕𝑧𝑧 2
�
�
(1)
𝑟𝑟 𝜕𝜕𝜕𝜕 1 𝜕𝜕 𝜕𝜕𝑣𝑣𝑧𝑧
where, 𝑔𝑔𝑟𝑟 = −𝑔𝑔 sin 𝜃𝜃 𝑔𝑔𝜃𝜃 = −𝑔𝑔 cos 𝜃𝜃
𝜕𝜕𝜕𝜕
+ 𝑣𝑣𝑧𝑧
�𝑟𝑟 𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕 1 𝜕𝜕𝜕𝜕
�𝑟𝑟
𝜕𝜕𝜕𝜕
𝜕𝜕𝑣𝑣𝑧𝑧
𝜕𝜕𝜕𝜕
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝑣𝑣𝑟𝑟
𝜕𝜕𝑣𝑣𝑟𝑟
𝜕𝜕𝜕𝜕
(2) ��
(3)
Equations (1) and (2) can be integrated to give 𝑝𝑝 = −𝜌𝜌𝜌𝜌(𝑟𝑟 sin 𝜃𝜃 ) + 𝑓𝑓1 (𝑧𝑧) = −𝜌𝜌𝜌𝜌𝜌𝜌 + 𝑓𝑓1 (𝑧𝑧)
⇒ pressure 𝑝𝑝 is hydrostatic and 𝜕𝜕𝜕𝜕⁄𝜕𝜕𝜕𝜕 is not a function of 𝑟𝑟 or 𝜃𝜃
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57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
21
Equation (3) can be written in the from 𝜕𝜕𝑣𝑣𝑧𝑧 1 𝜕𝜕𝜕𝜕 1 𝜕𝜕 �𝑟𝑟 �= 𝜇𝜇 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
and integrated (using the fact that 𝜕𝜕𝜕𝜕⁄𝜕𝜕𝜕𝜕 = constant) to give 𝑟𝑟
𝜕𝜕𝑣𝑣𝑧𝑧 1 𝜕𝜕𝜕𝜕 2 = � � 𝑟𝑟 + 𝐶𝐶1 𝜕𝜕𝜕𝜕 2𝜇𝜇 𝜕𝜕𝜕𝜕
Integrating again we obtain
B.C.
1 𝜕𝜕𝜕𝜕 2 𝑣𝑣𝑧𝑧 = � � 𝑟𝑟 + 𝐶𝐶1 ln 𝑟𝑟 + 𝐶𝐶2 4𝜇𝜇 𝜕𝜕𝜕𝜕 𝑣𝑣𝑧𝑧 (𝑟𝑟 = 0) ≠ ∞ ⇒ 𝐶𝐶1 = 0 𝑣𝑣𝑧𝑧 (𝑟𝑟 = 𝑅𝑅) = 0 ⇒ 𝐶𝐶2 = − ∴ 𝑣𝑣𝑧𝑧 =
1
𝜕𝜕𝜕𝜕
� � 𝑅𝑅2
4𝜇𝜇 𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕 � � (𝑟𝑟 2 − 𝑅𝑅2 ) 4𝜇𝜇 𝜕𝜕𝜕𝜕
⇒ at any cross section the velocity distribution is parabolic
21
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
1) Flow rate 𝑄𝑄:
𝑅𝑅
Chapter 6
22
𝑅𝑅
𝜋𝜋𝑅𝑅4 𝜕𝜕𝜕𝜕 � � 𝑄𝑄 = � 𝑣𝑣𝑧𝑧 𝑑𝑑𝑑𝑑 = 2𝜋𝜋 � 𝑣𝑣𝑧𝑧 𝑟𝑟𝑟𝑟𝑟𝑟 = − 𝜕𝜕𝜕𝜕 8𝜇𝜇 0 0
where, 𝑑𝑑𝑑𝑑 = (2𝜋𝜋𝜋𝜋)𝑑𝑑𝑑𝑑
If the pressure drops Δ𝑝𝑝 over a length ℓ:
2) Mean velocity 𝑉𝑉:
𝜋𝜋𝑅𝑅 4 Δ𝑝𝑝 𝑄𝑄 = 8𝜇𝜇ℓ
Δ𝑝𝑝 ℓ
=−
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝑄𝑄 1 𝜋𝜋𝑅𝑅4 Δ𝑝𝑝 𝑅𝑅 2 Δ𝑝𝑝 𝑉𝑉 = = � 2 � � �= 𝐴𝐴 𝜋𝜋𝑅𝑅 8𝜇𝜇ℓ 8𝜇𝜇ℓ
3) Maximum velocity 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 : 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚
⇒
𝑅𝑅2 𝜕𝜕𝜕𝜕 𝑅𝑅 2 Δ𝑝𝑝 = 𝑣𝑣𝑧𝑧 (𝑟𝑟 = 0) = − � � = = 2𝑉𝑉 4𝜇𝜇 𝜕𝜕𝜕𝜕 4𝜇𝜇ℓ 𝑣𝑣𝑧𝑧
𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚
𝑟𝑟 2 =1−� � 𝑅𝑅 22
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
23
4) Wall shear stress (𝜏𝜏𝑟𝑟𝑟𝑟 )𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 :
where
𝜕𝜕𝑣𝑣𝑟𝑟 𝜕𝜕𝑣𝑣𝑧𝑧 𝜕𝜕𝑣𝑣𝑧𝑧 𝜏𝜏𝑟𝑟𝑟𝑟 = 𝜇𝜇 � + � = 𝜇𝜇 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝑣𝑣𝑧𝑧 2𝑟𝑟 4𝑉𝑉𝑉𝑉 = 𝑣𝑣� �− � = − 𝑚𝑚𝑚𝑚𝑚𝑚 𝜕𝜕𝜕𝜕 𝑅𝑅 2 𝑅𝑅 2 =2𝑉𝑉
Thus, at the wall (i.e., 𝑟𝑟 = 𝑅𝑅),
(𝜏𝜏𝑟𝑟𝑟𝑟 )𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = −
and with 𝑄𝑄 = 𝜋𝜋𝑅𝑅 2 𝑉𝑉,
|(𝜏𝜏𝑟𝑟𝑟𝑟 )𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 | =
4𝜇𝜇𝜇𝜇 𝑅𝑅
4𝜇𝜇𝜇𝜇 𝜋𝜋𝑅𝑅 3
Note: Only valid for laminar flows. In general, the flow remains laminar for Reynolds numbers, Re = 𝜌𝜌𝜌𝜌(2𝑅𝑅)⁄𝜇𝜇, below 2100. Turbulent flow in tubes is considered in Chapter 8.
23
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
24
Differential Analysis of Fluid Flow We now discuss a couple of exact solutions to the NavierStokes equations. Although all known exact solutions (about 80) are for highly simplified geometries and flow conditions, they are very valuable as an aid to our understanding of the character of the NS equations and their solutions. Actually the examples to be discussed are for internal flow (Chapter 8) and open channel flow (Chapter 10), but they serve to underscore and display viscous flow. Finally, the derivations to follow utilize differential analysis. See the text for derivations using CV analysis. Couette Flow
boundary conditions First, consider flow due to the relative motion of two parallel plates Continuity
∂u =0 ∂x
Momentum
d 2u 0=µ 2 dy
u = u(y) v=o ∂p ∂p = =0 ∂x ∂y
or by CV continuity and momentum equations: 24
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
25
ρu1∆y = ρu 2 ∆y u1 = u2 ∑ Fx =∑ uρV ⋅ d A = ρQ(u 2 − u1 ) = 0 dp dτ = p∆y − p + ∆x ∆y − τ∆x + τ + dy ∆x = 0 dx dy dτ =0 dy d du i.e. µ = 0 dy dy d 2u µ 2 =0 dy from momentum equation du µ =C dy C u = y+D µ u(0) = 0 ⇒ D = 0 U u(t) = U ⇒ C = µ t U u= y t du µU τ=µ = = constant dy τ
25
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
26
Generalization for inclined flow with a constant pressure gradient
Continutity
∂u =0 ∂x
Momentum
d 2u ∂ 0 = − (p + γz ) + µ 2 ∂x dy
i.e.,
d 2u dh µ 2 =γ dx dy
u = u(y) v=o ∂p =0 ∂y
h = p/γ +z = constant plates horizontal plates vertical
dz =0 dx
dz =-1 dx
which can be integrated twice to yield du dh = γ y+A dy dx dh y 2 µu = γ + Ay + B dx 2
µ
26
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
27
now apply boundary conditions to determine A and B u(y = 0) = 0 ⇒ B = 0 u(y = t) = U µU dh t 2 dh t µU = γ + At ⇒ A = −γ dx 2 t dx 2 dh t γ dh y 2 1 µU u ( y) = + −γ dx 2 µ dx 2 µ t γ dh U ty − y 2 + y =− 2µ dx t
(
)
This equation can be put in non-dimensional form: u γt 2 dh y y y =− 1 − + U 2µU dx t t t define: P = non-dimensional pressure gradient p γt 2 dh h = +z =− γ 2µU dx γz 2 1 dp dz Y = y/t + =− 2µU γ dx dx u ⇒ = P ⋅ Y(1 − Y ) + Y U parabolic velocity profile
27
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
28
u Py Py 2 y = − 2 + U t t t t
q = ∫ udy 0 t
∫ U[ ]dy q 0 u= = t t
tu t P P y = ∫ y − 2 y 2 + dy U 0 t t t
=
Pt Pt t − + 2 3 2
u P 1 t 2 dh U = + ⇒u= − γ + U 6 2 12µ dx 2 For laminar flow
ut < 1000 ν
Recrit ∼ 1000 28
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
29
The maximum velocity occurs at the value of y for which: du 1 P 2P d u =0 y 0 − + = = dy t t t2 dy U
⇒y=
t (P + 1) = t + t @ umax 2P 2 2P
∴ u max = u (y max ) =
note: if U = 0:
u u max
=
for U = 0, y = t/2
UP U U + + 2 4P 4
P P 2 = 6 4 3
The shape of the velocity profile u(y) depends on P: dh 1. If P > 0, i.e., < 0 the pressure decreases in the dx direction of flow (favorable pressure gradient) and the velocity is positive over the entire width γ
dh d p dp = γ + z = − γ sin θ dx dx γ dx
a)
dp <0 dx
b)
dp < γ sin θ dx
29
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
30
1. If P < 0, i.e., dh dx > 0 the pressure increases in the direction of flow (adverse pressure gradient) and the velocity over a portion of the width can become negative (backflow) near the stationary wall. In this case the dragging action of the faster layers exerted on the fluid particles near the stationary wall is insufficient to overcome the influence of the adverse pressure gradient. dp − γ sin θ > 0 dx dp > γ sin θ dx 2. If P = 0, i.e.,
or
γ sin θ <
dp dx
dh = 0 the velocity profile is linear dx
U y t dp Note: we derived a) = 0 and θ = 0 dx this special case dp b) = γ sin θ dx u For U = 0 the form = PY (1 − Y ) + Y is not appropriate U u = UPY(1-Y)+UY γt 2 dh =− Y(1 − Y ) + UY 2µ dx γt 2 dh Now let U = 0: u=− Y(1 − Y ) 2µ dx u=
30
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
31
3. Shear stress distribution Non-dimensional velocity distribution u u * = =P ⋅ Y (1 − Y ) + Y U
u is the non-dimensional velocity, U γ t 2 dh P≡− is the non-dimensional pressure 2 µU dx
where
u* ≡
Y≡
gradient
y is the non-dimensional coordinate. t
Shear stress τ =µ
du dy
In order to see the effect of pressure gradient on shear stress using the non-dimensional velocity distribution, we define the non-dimensional shear stress: τ* =
Then t = *
= =
=
where
A≡
2µ >0 ρUt
τ
1 ρU 2 2
Ud ( u U ) 2 µ du * 1 µ = 1 td ( y t ) ρUt dY 2 ρU 2 2µ ( −2 PY + P + 1) ρUt 2µ ( −2 PY + P + 1) ρUt A ( −2 PY + P + 1)
is a positive constant.
So the shear stress always varies linearly with Y across any section. 31
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
32
At the lower wall (Y = 0 ) : * τ= A (1 + P ) lw
At the upper wall (Y = 1) : * τ= A (1 − P ) uw
For favorable pressure gradient, the lower wall shear stress is always positive: 1. For small favorable pressure gradient ( 0 < P < 1) : * τ lw* > 0 and τ uw >0 2. For large favorable pressure gradient ( P > 1) : * τ lw* > 0 and τ uw < 0
τ
τ ( 0 < P < 1)
( P > 1)
For adverse pressure gradient, the upper wall shear stress is always positive: 1. For small adverse pressure gradient ( −1 < P < 0 ) : * >0 τ lw* > 0 and τ uw 2. For large adverse pressure gradient ( P < −1) : * τ lw* < 0 and τ uw >0
32
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
33
τ
τ
( −1 < P < 0 )
( P < −1)
For U = 0 , i.e., channel flow, the above non-dimensional form of velocity profile is not appropriate. Let’s use dimensional form: γ t 2 dh γ dh u= − Y (1 − Y ) = − y (t − y ) 2 µ dx 2 µ dx
Thus the fluid always flows in the direction of decreasing piezometric pressure or piezometric head because dh γ > 0, y > 0 and t − y > 0 . So if is negative, 2µ dx dh tive; if dx is positive, u is negative.
Shear stress:
du γ dh 1 t= µ = − t − y dy
Since
1 t − y > 0 , 2
u is posi-
2 dx
2
the sign of shear stress τ is always oppodh
site to the sign of piezometric pressure gradient dx , and the magnitude of τ is always maximum at both walls and zero at centerline of the channel.
33
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
34 dh < 0, τ > 0 dx dh >0, τ < 0 dx
For favorable pressure gradient, For adverse pressure gradient,
τ
dh <0 dx
τ dh >0 dx
Flow down an inclined plane
uniform flow ⇒ velocity and depth do not change in x-direction Continuity
du =0 dx
34
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
Chapter 6
35
∂ d 2u x-momentum 0 = − (p + γz ) + µ 2 ∂x dy ∂ y-momentum 0 = − (p + γz ) ⇒ hydrostatic pressure variation ∂y dp ⇒ =0 dx d2u µ 2 = − γ sin θ dy γ du = − sin θy + c µ dy γ y2 u = − sin θ + Cy + D µ 2 du γ γ = 0 = − sin θd + c ⇒ c = + sin θd dy y=d µ µ
u(0) = 0 ⇒ D = 0 γ y2 γ u = − sin θ + sin θ dy µ 2 µ =
γ sin θ y(2d − y ) 2µ
35
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
u(y) =
Chapter 6
36
g sin θ y(2d − y ) 2n d
2 y3 γ q = ∫ udy = sin qdy − 2µ 3 0 0 d
=
V avγ
discharge per unit width
1γ 3 d sin θ 3µ
θ 1γ 2 γd 2 d siν θ = = = siν θ d 3µ 3ν
in terms of the slope So = tan θ ∼ sin θ gd 2So V= 3ν Exp. show Recrit ∼ 500, i.e., for Re > 500 the flow will become turbulent ∂p = − γ cos θ ∂y
Re crit =
Vd ∼ 500 ν
p = − γ cos θ y + C
p(d ) = p o = − γ cos θ d + C
36
57:020 Mechanics of Fluids and Transport Processes Professor Fred Stern Fall 2014
i.e.,
Chapter 6
37
p = γ cos θ (d − y ) + p o
* p(d) > po * if θ = 0 if θ = π/2
p = γ(d − y) + po entire weight of fluid imposed p = po no pressure change through the fluid
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