7. Fluid mechanics
Introduction
In this chapter we will study the mechanics of fluids. A non-viscous fluid, i.e. fluid with no inner friction is called an ideal fluid. The stress tensor for an ideal fluid is given by T1= −p , which is the most simplest form of the stress tensor used in fluid mechanics.
A viscous material is a material for which the stress tensor depends on the rate of deformation D. If this relation is linear then we have linear viscous fluid or Newtonian fluid. The stress tensor for a linear viscous fluid is given by T = - p 1 + λ tr D 1 + 2μ D , where λ and μ are material constants. If we assume the fluid to be incompressible, i.e. tr D 1 = 0 and the stress tensor can be simplified to contain only one material coefficient μ .
Most of the fluids can not be modelled by these two simple types of stress state assumption and they are called non-Newtonian fluids. There are different types of non-Newtonian fluids. We have constitutive models for fluids of differential, rate and integral types. Examples non-Newtonian fluids are Reiner-Rivlin fluids, Rivlin-Ericksen fluids and Maxwell fluids.
In this chapter we will concentrate on ideal and Newtonian fluids and we will study some classical examples of hydrodynamics.
7.1 The general equations of motion for an ideal fluid
For an arbitrary fluid (liquid or gas) the stress state can be describe by the following constitutive relation, T1= −p (7.1) where p= p (r ,t) is the fluid pressure and T is the stress. A material for which the constitutive relation above can be used is called an ideal or perfect fluid.
When we assume the fluid to be modelled as an ideal, we usually take into consideration the complex form of the equations of motion, which a fluid with inner friction will lead to. So we have to choose between the possibility to find the solution with an ideal fluid assumption and not finding the solution at all.
All fluids are viscous in different degree. This is due to the inner friction. The inner friction leads to the so called adhesion at the boundaries and the fluid at the boundary has the same velocity as the boundary surface itself. Even if we assume the fluid to be an ideal, i.e. the stress tensor is given by (7.1), we still assume the adhesion at the boundary exists. Otherwise there is not possible to find a solution for an ideal fluid.
For many problems involving fluid flow with high velocity, the inner friction is significant only near the boundary surfaces. We can model the fluid as viscous in the region near the boundary surface and the rest of the fluid can be modelled as an ideal.
The conservation and balance equations for an ideal fluid are the continuity equation given by,
∂ρ +div(ρ rr) =+= ρ ρ div 0 (7.2) ∂t Cauchy’s first equation of motion, ρ rf=−+grad p ρ (7.3) In the following we will consider both compressible and incompressible fluids. For a incompressible material the motion is isochoric. According to the continuity equation that is equivalent to the condition,ρ=0 since the velocity field is free from divergence i.e. div r = 0 . The condition ρ=0 means that density has a time independent value for all times t. If in addition to that the material is homogeneous, then ρ has the same value in all material point and by that ρ is a material constant. We can now introduce the a function Π=Π(p) in the following way,
dΠ1 grad Π= grad p = grad p (7.4) dp ρ If we assume that there exists a potential field U=U( r,t) to the body force i.e. f = - grad U , Cauchy’s second equation of motion (7.7) can be rewritten as, r = - grad (Π+U) (7.5)
7.2 Bernoulli’s equations
In the classical hydrodynamics the equation of motion for the fluid are usually written in a form that allows us to obtain simple solution for some special cases. This form of the equation of motion is known as Bernoulli’s equations.
For an ideal fluid Cauchy’s first equation of motion (7.9) can be written as
1 2 ∂r grad( v +Π + U) + 2ωr ×+ =0 (7.6) 2t∂ ∂r 1 where the Euler’s acceleration formula r = Lr +∂r = +2ωr ×+ grad () v2 have been used. t ∂t2 Consider an ideal fluid with density ρ ,which flows stationary under the influence of a conservative specific body force fr= grad U( ) . Bernoulli’s equation yields that, rr⋅ B = B(r ) = +Π+ U (7.7) 2 is constant along every curve in the flow field with tangent perpendicular to ωr× . ω is the vorticity vector introduced in chapter 2. B is constant along the stream lines and vorticity lines (lines that ate parallel to the vorticity vector). If the flow is free from divergence, then B is constant at every point.
7.3 The general equations of motion for a linear viscous fluid (Newtonian fluid)
Fluids are in general slow-flowing i.e. viscous. A general class of fluids for which the constitutive relation is constituted as T = - p 1 + T ( D) T(0) = 0 (7.8)
are called Stokes fluids. The first term - p 1 in (7.6) describes the stress state of the fluid at rest, whereas the function T is a tensor function of the rate of deformation tensor D. If T = 0 then we have the special case of an ideal fluid. The tensor function T can be assumed to be a potential function of D, i.e., ∞ k T(DD) = ∑a k (7.9) k=0 where a k are real number, which should be interpreted as material constants. Cayley-Hamilton’s theorem yields that the stress tensor satisfies its own characteristic equation
321+α ++ α αα DDDD=−+θθθ123 (7.10) This implies that powers greater than 3 can be expressed in terms of lower powers.
2 T = - p 11D + b01 + b + b 2 D (7.11) where in the new constitutive relation the coefficients are functions of the invariants i.e.
bi= b iDDD( I,, II II ) of the rate of deformation tensor D,
1 2 I = tr DD , II=−= (tr D) tr D2 , III det (7.12) DD2 ( ) D Fluids that fulfils the constitutive relation (7.15) are called Reiner-Rivlin fluids or non-Newtonian fluids. Linearization of the constitutive relation (7.15) yields,
T = - p 1 + λtr D 1 + 2μ D (7.13) where λ and µ are constants and they are called the coefficients of viscosity. A fluid, for which the stress can be constituted by (7.13) is called a Newtonian fluid or non-viscous fluid. The continuity equation for a linear viscous fluid is ∂∂ρ/ t+ div ( ρ r) = ρ + ρ div r = 0 . Cauchy’s first equation of motion for a linear viscous fluid can be written in the form,
ρ r = grad p + grad (λ div rD ) + 2 div ( μ) + ρ f (7.14) If the constants λ and µ are independent of r, then
ρ r = -grad p + (λµ +) grad div r + μΔ r + ρ f (7.15) The constitutive relation (7.13) is valid for both compressible and incompressible material. For an incompressible material we have that div rD = tr = 0 and the viscous coefficient λ does not occur in the constitutive relation.
The equation of motion for a homogeneous incompressible linear viscous fluid in component format is called Navier-Stokes equation and it reads, ∂∂∂∂xxxx ∂ p ρ + x + y + z =−++ μΔ x ρ fx ∂∂∂∂t xyz ∂ x ∂∂∂∂yyyy ∂ p ρ + x + y + z =−++ μΔ y ρ fy (7.16) ∂∂∂∂t xyz ∂ y ∂∂∂∂zzzz ∂ p ρ + x + y + z =−++ μΔ z ρ fz ∂∂∂∂t xyz ∂ z
where µ and ρ are constants.
7.5 Special flow of incompressible linear viscous fluid
In this section w shall study some special cases of flow. We shall study flow over an infinite plate, flow between two plates, flow in a pipe with circular cross section and flow between two concentric pipes with circular cross section.
We start with rewriting Navier-Stokes equation on dimensionless form. Let l and v be positive numbers. Let l be the characteristic length and v the characteristic velocity e.g. mean value of the speed or maximum values of the speed. We introduce the dimensionless variables for the position vector r, dimensionless position vector r* according to, r r* = (7.17) l the dimensionless time t* as, v tt* = (7.18) l the dimensionless velocity v* as, 1 v*( r ** ,t )= vr ( ,t) (7.19) v and the dimensionless pressure potential Π* as, 1 Π* (rr ** ,t )= Π( ,t) (7.20) v2 With these definitions we have that, ∂∂l v* (r ** ,t )= v(r,t) (7.21) ∂∂tt*2v
* l grad v* (r ** ,t )= grad v(r,t) (7.22) v
∗∗ l gradΠ= grad Π (7.23) v2 If the specific body force is zero, f = 0, using the Navier-Stokes equation, the equation of motion can be written in dimensionless form. Navier-Stokes equation , f = 0
∂v ρρr= + vv grad ∂t ∂v ρr=−grad pμ +∆+ rf ρ , ⇒ ρ+vv grad =− ρμ grad Π+ ∆r l ∂t grad Π= grad p ρ In dimensionless form we obtain,
vv22∂v∗ v2 μ v + ∗∗∗∗ =−* Π+ * ∆ ** ⇒ ρρ∗ vvgrad grad 2 v ll∂t l l
∂ ∗∗∗∗∗∗ * *1 ** vv+grad v =− grad Π+ ∆v (7.24) ∂t∗ Re lvρ where Re = is a dimensionless quantity called Reynold’s number. μ The solution of the Navier-Stokes equation for a given Reynold’s number can be used to generate solutions to problems with different length and velocity scales but with same Reynold’s number. This mean that in laboratory test we can model the real flow in smaller scale by choosing the density of the fluid and the size of the model in such way that the Reynold’s number is kept constant.
I.Examples of incompressible linear stationary flows
For stationary isochoric and linear flow the velocity field is given by.
r = ( u,0,0) (7.25)
with u= u(r ) independent of time t. If the flow is isochoric flow , then
∂u div r =⇒=00 (7.26) ∂x
i.e. u= u(y,z) is independent of x and the velocity is constant along every stream line. This leads to acceleration must be zero,
rr=const ⇒= 0 (7.27)
Assume that the specific body force is conservative, f = -grad U(r) and introduce the generalized pressure as, pp= + ρU (7.28)
Insertion in Navier-Stokes equation (7.20) yields,
∂∂∂∂xxxx ∂ p ρ+ x + y + z =− +∆+μ x ρf ∂ x p ∂∂∂∂t xyz ∂ x 0 =−++μΔx ρfx ∂x ∂∂∂∂yyyy ∂ p ∂ ρ+ x + y + z =−+μΔy + ρf ⇒ p y 0 =−+ρfy ∂∂∂∂t xyz ∂ y ∂y ∂∂∂∂ ∂ ∂p zzzz p 0 =−+ρf ρ+ x + y + z =−+μΔz + ρfz z ∂∂∂∂tx yz ∂ z ∂z
∂U(r) ∂∂ p p where f =− ⇒− +ρf =− , i =x ,y,z and finally we obtain, ii∂∂ii ∂ i
∂p ∂∂22 uu =−+ + 0 μ 22 ∂x ∂∂ yz ∂p 0 = − (7.29) ∂y ∂p 0 = − ∂z
From the last two equation in (7.101) follow that the generalized pressure is independent of y and z, i.e. can be written as p= p(x) . According to the first equation in (7.101) the derivative with respect to x of the generalized pressure must be independent of x , which implies independent of x, y and z,
p= p0 − bx (7.30)
where p0 and b are constants, b is called the driving force. Insertion in x-component of Navier- Stokes equation (7.101) yields,
∂∂22uub Δu = + =−=const (7.31) ∂∂yz22μ
We can conclude that a incompressible linear viscous fluid, influenced by conservative specific body force, is flowing stationary and the generalized pressure pp= + ρU has a constant pressure drop per unit length equal to b. The velocity satisfies equation (7.103).
We shall now study two special solutions to the last equation,
-b 2 u= z ++ cz u(0) (7.32) 2μ
b 2 22 u= (a −− y z ) (7.33) 4μ The velocity field (7.104) represents a planar flow in the (x,z) plane and (7.105) represents flow in a pipe.
Case A: Flow over a infinite plate (x=0)
On upper boundary surface of the flow z=h, the atmospheric pressure pA is constant. The following boundary conditions are fulfilled,
u(0)= 0 (adhesion to the plate) (7.34)
The normal stress on the upper boundary surface is continuous i.e.,
Tzz (h)= - p(h)= -pA (7.35)
The shear stress on the upper boundary surface is continuous as well, i.e.,
∂u -b Txz (h) = μ = μ( h+c)= 0 (7.36) ∂z z=h μ
The velocity field is assume to have the same form as (7.104),
b 2 u= (2hz - z ) (7.37) 2μ
According to (7.100), (7.101) and the boundary conditions (7.106)-(7.108) we have,
pA0= p - bx - ρU(x,y,h) (7.38)
If the specific body force is constantf=g, then g0y = and (7.110) leads to,
pA0= p - bx - ρ(g x x + g z h) (7.39)
And we must have that,
pA0= p + ρg z h (7.40)
b = ρgx (7.41)
Accordingly we obtain the solution, (g can be interpreted as the specific force of gravity; if the inclination angle is α , then gx = g sinα and gz = g cosα ),
z gρsinα u= (2hz - z2 ) (7.114) 2μ u = u(z) p= p0 + ρg cosα (h-z) (7.115)
The volume of the fluid flow through u cross section with unit width and per unit time is,
h ρgh3 sin α ∫ u dz= (7.42) 0 3μ
Case B: Flow between two parallel plates
The distance between the plates is h. The bottom plate is fix, whereas the upper plate is moving with velocity (u0 ,0,0) . We have the following boundary conditions, u(0)= 0 (adhesion to the bottom plate) (7.43)
-b u= h2 + ch=u 2μ 0 (adhesion to the upper plate) (7.44)
The solution (7.104) leads to the following velocity field,
b u u= (h-z)z + 0 z (7.45) 2μh
The flow can be interpreted as the sum of two superimposed flows, one between two fix plates and the so called shear flow.
Exercise:
1. A linearly viscous incompressible fluid is transported from one container up to another (see figure). The flow is steady and the transportation line is inclined the angleα . (The fuid is homogeneous and its density is ρ , the viscosity coefficient is µ and the free surface of the fluid is d over the transportation line. Assume that the pressure drop is zero.) a) Determine by using Navier-Stoes equation, the velocity field x = uy( ) and the speed of the
transportation line u0 , if the volume flowing through a cross section per unit time is Q.
y
x = u(y) d g
x
u0
α
2. A linearly viscous, incompressible fluid with the density ρ and the viscosity coefficient µ is filling the upper half space y0≥ (see figure). In the plane y=0 is a plate oscillating in the x-
direction with the frequency ω and in a such a way that v0 sin(ω t) will represent the velocity in the x-direction at the time t. Determine, by using Naier-Stokes equation and the restrictions given by the boundary conditions, the constants A,B and c for the velocity field, which is given by
x ( y,t) =AeBy sin (ωt+Cy) (The body forces can be neglected and there is no pressure drop in the x.direction.)
y
v0 sin ωt x
3. To reducue the friction and to ease and to ease off the sliding between bodies a liquid can be added to the contact surface between the bodies (e.g. oil can be used). Consider a linearly viscous, incompessible fluid (density ρ and viscosity coefficient µ ) in a simple shear flow
between two parallell plates ( see fig); the upper plate has the speed v0 in the x-direction while the lower plate is fixed (the distance between the plates is d). The shear force per unit area (i.e.
the shear stress) on the upper plate (from the fluid) is denoted T1 . In order to further reduce the friction another layer of a fluid was added (with the thickness 0,1d); (see fig. 2.). This second fluid has approximately the same density bur the viscosity coefficient is 0,1µ (the distance between the plates is still d, and the upper plate has the speed
v0 ). Let now T2 denote the shear force per unit area on the upper plate (from the fluid) in this
case. Determine the ratio TT21/ . (The two fluids are assumed to be immiscible.)
y y
V0 V0
d d
0.1 d
x x
Figure 1. Figure 2.
Answers to Exercise
ραg sin u( y) =−y( 2d −+ y) u 2µ 0 1. Qρα g sin ud= + 2 0 d3µ
ρω − y ρω = 2µ ω − 2. x( y,t) v0 e sin t y 2µ
1 3. ττ= 211.9