Unformatted text preview: Ch. IV Differential Relations for a Fluid Particle
This chapter presents the development and application of the basic differential
equations of fluid motion. Simplifications in the general equations and common
boundary conditions are presented that allow exact solutions to be obtained. Two
of the most common simplifications are 1). steady flow and 2). incompressible
flow.
The Acceleration Field of a Fluid
A general expression of the flow field velocity vector is given by: ˆ
ˆ
V ( r , t ) = i u(x, y, z, t ) + ˆ v(x, y, z, t ) + k w(x , y, z, t )
j
One of two reference frames can be used to specify the flow field characteristics:
eulerian – the coordinates are fixed and we observe the flow field
characteristics as it passes by the fixed coordinates.
lagrangian  the coordinates move through the flow field following individual
particles in the flow.
Since the primary equation used in specifying the flow field velocity is based on
Newton’s second law, the acceleration vector is an important solution parameter.
In cartesian coordinates, this is expressed as a= ∂V
∂V ∂ V
dV ∂ V ∂ V
=
+ u
+v
+w
+ (V ⋅ ∇ )V
=
∂t ∂ x
∂y
∂ z ∂t
dt
total local convective The acceleration vector is expressed in terms of three types of derivatives:
Total acceleration = total derivative of velocity vector
= local derivative + convective derivative of velocity vector  v v IV  1  eText Main Menu  Textbook Table of Contents  Study Guide Likewise, the total derivative (also referred to as the substantial derivative ) of
other variables can be expressed in a similar form, e.g., dP ∂P ∂P
∂P
∂ P ∂ P
=
+ u
+v
+w
+ (V ⋅ ∇ )P
=
∂y
∂ z ∂t
dt ∂ t ∂ x
Example 4.1
Given the eulerian velocityvector field ˆ
ˆ
V = 3 t i + x z ˆ + t y2 k
j
find the acceleration of the particle.
For the given velocity vector, the individual components are
u = 3t v= xz 2 w = ty Evaluating the individual components, we obtain ∂V
2
= 3i +y k
∂t
∂V
= zj
∂x ∂V
= 2tyk
∂y ∂V
= xj
∂z Substituting, we obtain dV
2
2
= ( 3 i + y k) + (3 t) (z j) + (x z) (2 t y k) + (t y ) (x j)
dt
After collecting terms, we have dV
2
2
= 3 i + (3 t z + t x y ) j + ( 2 x y z t + y ) k ans.
dt  v v IV  2  eText Main Menu  Textbook Table of Contents  Study Guide The Differential Equation of Conservation of Mass
If we apply the basic concepts of conservation of mass to a differential control
volume, we obtain a differential form for the continuity equation in cartesian
coordinates ∂ρ ∂
∂
∂
+
(ρ u) + (ρ v) + (ρ w) = 0
∂t ∂ x
∂y
∂z
and in cylindrical coordinates ∂ρ 1 ∂
+
(r ρ v r ) + 1 ∂ (ρ vθ ) + ∂ (ρ v z ) = 0
∂t r ∂r
∂z
r ∂θ
Steady Compressible Flow
For steady flow, the term ∂
= 0 and all properties are function of position only.
∂t The previous equations simplify to
Cartesian: ∂
∂
∂
(ρ u) + (ρ v ) + (ρ w) = 0
∂x
∂y
∂z Cylindrical: 1∂
(r ρ vr ) + 1 ∂ (ρ vθ ) + ∂ (ρ v z ) = 0
∂z
r∂r
r ∂θ Incompressible Flow
For incompressible flow, density changes are negligible, ρ = const., and
In the two coordinate systems, we have ∂u ∂ v ∂ w
+
+
=0
∂x ∂y ∂z Cartesian:  v v IV  3  eText Main Menu  Textbook Table of Contents  Study Guide ∂ρ
=0
∂t 1∂
(r vr ) + 1 ∂ (vθ ) + ∂ (v z ) = 0
∂z
r∂r
r ∂θ Cylindrical:
Key Point: It is noted that the assumption of incompressible flow is not restricted to fluids
which cannot be compressed, e.g. liquids. Incompressible flow is valid for
(1) when the fluid is essentially incompressible (liquids) and (2) for compressible
fluids for which compressibility effects are not significant for the problem being
considered.
The second case is assumed to be met when the Mach number is less than 0.3:
Ma = V/c < 0.3 Gas flows can be considered incompressible The Differential Equation of Linear Momentum
If we apply Newton’s Second Law of Motion to a differential control volume we
obtain the three components of the differential equation of linear momentum. In
cartesian coordinates, the equations are expressed in the form: Inviscid Flow: Euler’s Equation
If we assume the flow is frictionless, all of the shear stress terms drop out. The
resulting equation is known as Euler’s equation and in vector form is given by: ρg  ∇P = ρ dV
dt  v v IV  4  eText Main Menu  Textbook Table of Contents  Study Guide dV
is the total or substantial derivative of the velocity discussed
dt
previously and ∇ P is the usual vector gradient of pressure. This form of Euler’s where equation can be integrated along a streamline to obtain the frictionless Bernoulli’s
equation ( Sec. 4.9).
The Differential Equation of Energy
The differential equation of energy is obtained by applying the first law of
thermodynamics to a differential control volume. The most complex element of
the development is the differential form of the control volume work due to both
normal and tangential viscous forces. When this is done, the resulting equation has
the form ρ du
+ P (∇⋅ V) = ∇⋅ (k ∇ T ) + Φ
dt where Φ is the viscous dissipation function. The term for the total derivative of
internal energy includes both the transient and convective terms seen previously.
Two common assumptions used to simplify the general equation are:
1. du ≈ Cv dT and 2. Cv, µ, k, ρ ≈ constants With these assumptions, the energy equation reduces to ρ Cv dT
= k ∇2 T + Φ
dt It is noted that the flowwork term was eliminated as a result of the assumption of
constant density, ρ, for which the continuity equation becomes ∇ ⋅ V = 0 ,thus
eliminating the term P (∇ ⋅ V ) .
We now have the three basic differential equations necessary to obtain complete
flow field solutions of fluid flow problems.  v v IV  5  eText Main Menu  Textbook Table of Contents  Study Guide Boundary Conditions for the Basic Equations
In vector form, the three basic governing equations are written as
Continuity: ∂ρ
+ ∇ ⋅ (ρ V) = 0
∂t Momentum: ρ dV
= ρ g − ∇ P + ∇ ⋅τi j
dt Energy: ρ du
+ P (∇⋅ V) = ∇⋅ (k ∇ T ) + Φ
dt We have three equations and five unknowns: ρ, V, P, u, and T ; and thus need two
additional equations. These would be the equations of state describing the
variation of density and internal energy as functions of P and T, i.e.,
ρ = ρ (P,T) and u = u (P,T)
Two common assumptions providing this information are either:
1. Ideal gas: ρ = P/RT and du = Cv dT 2. Incompressible fluid: ρ = constant and du = C dT
Time and Spatial Boundary Conditions
Time Boundary Conditions: If the flow is unsteady, the variation of each of the
variables (ρ, V, P, u, and T ) must be specified initially, t = 0, as functions of
spatial coordinates e.g. x,y,z.
Spatial Boundary Conditions: The most common spatial boundary conditions are
those specified at a fluid – surface boundary. This typically takes the form of
assuming equilibrium (e.g., no slip condition – no property jump) between the fluid
and the surface at the boundary.  v v IV  6  eText Main Menu  Textbook Table of Contents  Study Guide This takes the form:
Vfluid = Vwall Tfluid = Twall Note that for porous surfaces with mass injection, the wall velocity will be equal to
the injection velocity at the surface.
A second common spatial boundary condition is to specify the values of V, P, and
T at any flow inlet or exit.
Example 4.6
For steady incompressible laminar flow through a long tube, the velocity
distribution is given by r2 v z = U 1 − 2 R vr = 0 vθ = 0 where U is the maximum or centerline velocity and R is the tube radius. If the wall
temperature is constant at Tw and the temperature T = T(r) only, find T(r) for this
flow.
For the given conditions, the energy equation reduces to k d d T
dT d vz ρ Cv v r
=
r + µ dr r dr dr
dr 2 Substituting for vz and realizing the vr = 0, we obtain
2 4U 2 µ r 2
k d d T d vz r = − µ =−
4 dr r dr dr
R
Multiply by r/k and integrate to obtain  v v IV  7  eText Main Menu  Textbook Table of Contents  Study Guide dT
µU r
=−
+ C1
k R4
dr
23 Integrate a second time to obtain T=− µ U 2 r4
4 k R4 + C1 ln r + C2 Since the term, ln r, approaches infinity as r approaches 0, C1 = 0.
Applying the wall boundary condition, T = Tw at r = R, we obtain for C2 C2 = Tw + µ U2
4k The final solution then becomes µ U2 r4 T (r ) = Tw +
1 − 4 4k R
The Stream Function
The necessity to obtain solutions for multiple variables in multiple governing
equations presents an obvious mathematical challenge. However, the stream
function, Ψ , allows the continuity equation to be eliminated and the momentum
equation solved directly for the single variable, Ψ . The use of the stream function
works for cases when the continuity equation can be reduced to only two terms.
For example, for 2D, incompressible flow, continuity becomes ∂u ∂ v
+
=0
∂x ∂y  v v IV  8  eText Main Menu  Textbook Table of Contents  Study Guide Defining the velocity components to be u= ∂Ψ
∂y and v= − ∂Ψ
∂x which when substituted into the continuity equation yields ∂ ∂ Ψ ∂ ∂ Ψ +
−
=0
∂x∂y ∂y ∂x
and continuity is automatically satisfied.
Geometric interpretation of Ψ
It is easily shown that lines of constant Ψ are flow streamlines. Since flow does
not cross a streamline, for any two points in the flow we can write
2 2 1 1 Q1→2 = ∫ (V ⋅ n ) d A = ∫ d Ψ = Ψ2 − Ψ1
Thus the volume flow rate between two points in the flow is equal to the difference
in the stream function between the two points. Steady Plane Compressible Flow
In like manner, for steady, 2D, compressible flow, the continuity equation is ∂
∂
(ρ u) + (ρ v ) = 0
∂x
∂y
For this problem, the stream function can be defined such that ρu = ∂Ψ
∂y and ρv=− ∂Ψ
∂x  v v IV  9  eText Main Menu  Textbook Table of Contents  Study Guide As before, lines of constant stream function are streamlines for the flow, but the
change in stream function is now related to the local mass flow rate by
2 2 1 1 &
m1− 2 = ∫ ρ (V ⋅ n ) d A = ∫ d Ψ = Ψ 2 − Ψ1 Vorticity and Irrotationality
The concept of vorticity and irrotationality are very useful in analyzing many fluid
problems. The analysis starts with the concept of angular velocity in a flow field.
Consider three points, A, B, &
C, initially perpendicular at
time t, that then move and
deform to have the position and
orientation at t + dt.
The lines AB and BC have both
changed length and incurred
angular rotation dα and dβ
relative to their initial
positions. Fig. 4.10 Angular velocity and strain rate of two
fluid lines deforming in the xy plane
We define the angular velocity ωz about the z axis as the average rate of counterclockwise turning of the two lines expressed as 1 dα d β
− 2 dt
dt ωz =  v v IV  10  eText Main Menu  Textbook Table of Contents  Study Guide Applying the geometric properties of the deformation shown in Fig. 4.10
and taking the limit as ∆t → 0, we obtain 1 d v d u
− 2 d x d y ωz = In like manner, the angular velocities about the remaining two axes are 1 d w d v
− 2 d y d z 1 d u d w
− 2 d z d x ωx = ωy = From vector calculus, the angular velocity can be expressed as a vector with
the form ω = i ω x + j ω y + k ωz = 1/2 the curl of the velocity vector, e.g. i 1
2 ω = (curl V) = 1∂
2 ∂x
u j k v w ∂
∂
∂y ∂z The factor of 2 is eliminated by defining the vorticity, ξ , as follows:
ξ = 2 ω = curl V
Frictionless Irrotational Flows
When a flow is both frictionless and irrotational, the momentum equation reduces
to Euler’s equation given previously by ρg  ∇P = ρ dV
dt  v v IV  11  eText Main Menu  Textbook Table of Contents  Study Guide As shown in the text, this can be integrated along the path, ds, of a streamline
through the flow to obtain
2
∂V
dP 1 2
2
ds + ∫
+ (V2 − V1 ) + g(z 2 − z1 ) = 0
∫
2
1 ∂t
1ρ
2 For steady, incompressible flow this reduces to P ρ + 12
V + gz =
2 constant along a streamline  v v IV  12  eText Main Menu  Textbook Table of Contents  Study Guide ...
View
Full
Document
This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
 Spring '08
 Sakar

Click to edit the document details