Chapter 9: Surface Resistance

57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 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 Chapter 6 Professor Fred Stern Fall 2006 2 Change ind": 骣u d" = 琪 dx( d y d z) d t 桫x the rate at which the volume d" is changing per unit volume due to the gradient ∂u/∂x is 1 d(d") 轾(抖 u x) d t u =lim 犏 = d" dtd t 0 臌 d t 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") 抖u v w = + + = 炎 V d" dt抖 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.

Angular motion and deformation of a fluid element

The angular velocity of line OA, ωOA, is

2 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 3 da wOA = lim d t 0 dt For small angles (抖v x)d x d t v tanda� da= d t d x x so that 轾(抖v x)d t v wOA =lim 犏 = d t 0 臌 dt x Note that if ∂v/∂x is positive, ωOA will be counterclockwise.

Similarly, the angular velocity of the line OB is db u wOB =lim = d t 0 dt y

In this instance if ∂u/∂y is positive, ωOB will be clockwise.

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 骣抖v u wz =琪 - 2 桫抖x y Rotation of the field element about the other two coordinate axes can be obtained in a similar manner: 1 骣抖w v wx =琪 - 2 桫抖y z

3 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 4 1 骣抖u w w y =琪 - 2 桫抖z x

The three components, ωx,ωy, and ωz can be combined to give the rotation vector, ω, in the form: 1 1 ω=w i + w j + w k =curl V = 汛 V x y z 2 2 since i j k 1 1 抖汛 V = 2 2 抖x y z u v w 1骣抖w v 1骣抖 u w 1 骣抖 v u =琪 -i +琪 - j + 琪 - k 2桫抖y z 2桫抖 z x 2 桫抖 x y The vorticity, ζ, is defined as a vector that is twice the rotation vector; that is, V =2ω = 汛 V 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 汛 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,

4 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 5 dg= da + db The rate of change of δγ is called the rate of shearing strain or the rate of angular deformation: dg 轾(抖v x)d t+ ( 抖 u y) d t 抖v u g˙ =lim = lim 犏 = + dt 0dt d t 0 臌 d t抖 x y The rate of angular deformation is related to a corresponding shearing stress which causes the fluid element to change in shape.

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

 V A    dV CS CV t net outflow of mass = rate of decrease across CS of mass within CV

5 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 6 Consider a cubical element oriented so that its sides are to the (x,y,z) axes    u  udx dydz  x  outlet mass flux inlet mass flux udydz 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

轾 xρuflux = ρu + dx( dydz) ρudydz - 臌犏 x outflux influx  = (u)dxdydz x V

Similarly for the y and z faces  y  (v)dxdydz flux y  z  (w)dxdydz flux z

6 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 7 The total net mass outflux must balance the rate of decrease of mass within the CV which is   dxdydz t

Combining the above expressions yields the desired result        (u)  (v)  (w)dxdydz  0  t x y z  dV

     (u)  (v)  (w)  0 per unit V t x y z differential form of continuity equations    (V)  0 t  V  V 

D D    V  0   V  Dt Dt t

Nonlinear 1st order PDE; ( unless  = constant, then linear) Relates V to satisfy kinematic condition of mass conservation

Simplifications: 1. Steady flow:  (V)  0

2.  = constant:  V  0

7 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 8 u v w i.e.,    0 3D x y z

u v   0 2D x y

The continuity equation in Cylindrical Polar Coordinates

The velocity at some arbitrary point P can be expressed as

V=vr e r + vq e q + v z e z The continuity equation: r 1抖(rr v) 1 ( r v) ( r v ) +r +q + z = 0 抖t r r r 抖q z For steady, compressible flow 1抖(rr v) 1 ( r v) ( r v ) r+q + z = 0 r抖 r rq z For incompressible fluids (for steady or unsteady flow) 1(rv ) 1 v v r +q +z = 0 r抖 r rq z

8 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 9 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 抖y y u=, v = - 抖y x then the continuity equation is identically satisfied: 抖骣y 抖骣 y 抖2 y 2 y 琪 +琪 - = - = 0 抖x桫 y 抖 y桫 x 抖 x y 抖 x y

Velocity and velocity components along a streamline

9 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 10 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: 抖y y dy = dx + dy = - vdx + udy = 0 抖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. 抖y y dq= udy - vdx = dx + dy = dy 抖x y Thus, the volume rate of flow, q, between two streamlines such as ψ1 and ψ2, can be determined by integrating to yield:

10 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 11

y 2 q= dy = y2 - y 1 y1

In cylindrical coordinates the continuity equation for incompressible, plane, two-dimensional flow reduces to 1(rv ) 1 v r +q = 0 r抖 r r q

and the velocity components, vr and vθ, can be related to the stream function, ψ(r, θ), through the equations 1 抖y y v=, v = - r r抖q q r

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

d F=蝌r Vd " + V r V dA 1-D flow approximation dt CV CS m˙ V- m ˙ V   = 邋( ii)out( i i ) in where m˙  AV  dydzu x-face mass flux d  = Vdxdydz dt

      =  uV vV wVdxdydz x y z  x-face y-face z-face combining and making use of the continuity equation yields DV DV V F   dxdydz = + V籽 V Dt Dt t where F  Fbody  Fsurface Body forces are due to external fields such as gravity or magnetics. Here we only consider a gravitational field; that is,

Fbody  dFgrav  gdxdydz and g  gkˆ for g z

12 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 13 ˆ i.e., f body  gk

Surface forces are due to the stresses that act on the sides of the control surfaces

symmetric (ij = ji) nd ij = - pij + ij 2 order tensor

ij = 1 i = j normal pressure viscous stress ij = 0 i  j

= -p+xx xy xz

yx -p+yy yz

zx zy -p+zz

As shown before for p alone it is not the stresses themselves that cause a net force but their gradients.

     dFx,surf =  xx  xy  xz dxdydz x y z 

 p     =   xx  xy  xz dxdydz  x x y z  This can be put in a more compact form by defining ˆ ˆ ˆ x  xx i  xy j xzk vector stress on x-face and noting that  p  dFx,surf =    x dxdydz  x 

13 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 14 p fx,surf =      per unit volume x x similarly for y and z p ˆ ˆ ˆ fy,surf =        i   j  k y y y yx yy yz

p ˆ ˆ ˆ fz,surf =        i   j  k z z z zx zy zz finally if we define ˆ ˆ ˆ ij  x i  y j zk then

f surf  p    ij   ij ij  pij  ij

Putting together the above results DV f  f  f   body surf Dt ˆ f body  gk

f surface  p   ij DV V a= = + V籽 V Dt t ˆ ra= - r gk -� p 炎 t ij inertia body force force surface surface force due to force due due to viscous gravity to p shear and normal stresses

For Newtonian fluid the shear stress is proportional to the rate of strain, which for incompressible flow can be written

ij  ij  = coefficient of viscosity

ij = rate of strain tensor

u  v u   w u  =       x  x y   x z   u v  v  w v         y x  y  y z   u w   v w  w        z x   z y  z

du    1-D flow dy rate of strain ˆ a  gk  p   ij 

 2  ij   V xi

a  gkˆ  p  2 V

15 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 16 a  p  z 2 V Navier-Stokes Equation  V  0 Continuity Equation

Four equations in four unknowns: V and p Difficult to solve since 2nd order nonlinear PDE

u u u u p  2u  2u  2u x:   u  v  w          2 2 2   t x y z  x x y z 

v v v v p  2v  2v  2v y:   u  v  w          2 2 2   t x y z y x y z 

w w w w  p  2w  2w  2w  z:   u  v  w          2 2 2   t x y z  z  x y z  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.

Exact solution for laminar flow in a pipe (neglect g for now)

16 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 17 use cylindrical coordinates: vx = u vr = v

u = u(r) only v = w = 0

 Continuity: rv  0  rv = constant = c r v = c/r v(r = 0) = 0  c = 0 i.e., v = 0

Momentum:

Du p  2u 1  2u 1 u  2u      2  2 2   2  Dt x x r  r r r 

骣抖u u 抖 u w u 抖 p轾 1 u2 u r琪 +u + v + = - + m犏 + 2 桫抖t z 抖 r r 秖抖 x臌 r r r

1   u  1 p r     r r  r   x

u  r  r 2  A r 2

 ur  r 2  Aln r  B 4

17 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 18 u(r = 0)    A = 0  2 2 u(r = ro) = 0  ur  r  r  4 o

1 p i.e. ur  r 2  r 2  parabolic velocity profile 4 x o

Differential Analysis of Fluid Flow

We now discuss a couple of exact solutions to the Navier- Stokes 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

u Continuity  0 u = u(y) x v = o p p d 2u   0 Momentum 0   x y dy2 or by CV continuity and momentum equations: u1y  u 2y u1 = u2

Fx uV dA  Qu 2  u1   0  dp   d   py  p  xy  x    dyx = 0  dx   dy 

d  0 dy d  du  i.e.    0 dy  dy  d 2u   0 dy2 from momentum equation du   C dy C u  y  D  u(0) = 0  D = 0

19 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 20 U u(t) = U  C =  t U u  y t du U      constant dy t Generalization for inclined flow with a constant pressure gradient

u Continutity  0 u = u(y) x v = o p  d2u  0 Momentum 0   p  z  y x dy2

d 2u dh i.e.,    h = p/ +z = constant dy2 dx dz plates horizontal  0 dx dz plates vertical =-1 dx

20 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 21 which can be integrated twice to yield

du dh    y  A dy dx

dh y2 u    Ay  B dx 2 now apply boundary conditions to determine A and B u(y = 0) = 0  B = 0 u(y = t) = U

dh t 2 U dh t U    At  A    dx 2 t dx 2

 dh y2 1 U dh t  u(y)      dx 2   t dx 2  dh U = ty  y2  y 2 dx t

This equation can be put in non-dimensional form: u t 2 dh  y  y y   1   U 2U dx  t  t t define: P = non-dimensional pressure gradient t 2 dh p =  h   z 2U dx  z2 1 dp dz  Y = y/t      2U  dx dx

21 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 22 u   P Y(1 Y)  Y U parabolic velocity profile

u Py Py2 y    U t t 2 t

t q   udy 0 t  U dy q u   0 t t tu t P P y  y  y2  dy   2  U 0  t t t  Pt Pt t =   2 3 2 u P 1 t 2  dh  U    u      U 6 2 12  dx  2

ut For laminar flow 1000 Recrit  1000 

The maximum velocity occurs at the value of y for which: du d  u  P 2P 1  0    0   y  dy dy  U  t t 2 t

23 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 24 t t t  y  P 1   @ umax for U = 0, y = t/2 2P 2 2P

UP U U  u  uy     max max 4 2 4P

u P P 2 note: if U = 0:   u max 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

dh d  p  dp      z   sin  dx dx    dx

dp a)  0 dx

dp b)  sin  dx

dh 1. If P < 0, i.e.,  0 the pressure increases in the dx 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

24 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 25 case the dragging action of the faster layers exerted on the fluid particles near the stationary wall is insufficient to over come the influence of the adverse pressure gradient

dp  sin   0 dx

dp dp  sin  or sin   dx dx dh 2. If P = 0, i.e.,  0 the velocity profile is linear dx U u  y t Note: we derived this dp a)  0 and  = 0 special case dx dp b)  sin  dx

u For U = 0 the form  PY1 Y Y is not appropriate U

u = UPY(1-Y)+UY

t 2 dh =  Y1 Y UY 2 dx t 2 dh Now let U = 0: u   Y1 Y 2 dx

25 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 26 3. Shear stress distribution Non-dimensional velocity distribution u u* = = P� Y(1 + Y) Y U u where u* is the non-dimensional velocity, U g t2 dh P � is the non-dimensional pressure gradient 2mU dx y Y is the non-dimensional coordinate. t Shear stress du t= m 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: t t * = 1 rU 2 2 Then 1Ud( u U ) 2m du* t* = m = 1 rU 2 td( y t) r Ut dY 2 2m =( -2PY + P + 1) rUt 2m =( -2PY + P + 1) rUt =A( -2 PY + P + 1) 2m A � 0 where rUt is a positive constant. So the shear stress always varies linearly with Y across any section.

At the lower wall (Y = 0) : * t lw =A(1 + P) At the upper wall (Y =1) : * t uw =A(1 - P)

For favorable pressure gradient, the lower wall shear stress is always positive: 1. For small favorable pressure gradient (0

0 and t uw > 0 2. For large favorable pressure gradient (P >1) : * * t lw > 0 and t uw < 0

For adverse pressure gradient, the upper wall shear stress is always positive: 1. For small adverse pressure gradient (-1

0 and t uw > 0 2. For large adverse pressure gradient (P < -1) : * * t lw < 0 and t uw > 0

For U = 0 , i.e., channel flow, the above non-dimensional form of velocity profile is not appropriate. Let’s use dimensional form: gt2 dh g dh u= - Y(1 - Y) = - y( t - y) 2mdx 2 m dx Thus the fluid always flows in the direction of decreasing piezometric pressure or piezometric head because g dh >0,y > 0 and t- y > 0 . So if is negative, u is 2m dx dh positive; if is positive, u is negative. dx

Shear stress: dug dh 骣 1 t= m = -琪t - y dy2 dx 桫 2

骣 1 Since 琪t- y > 0 , the sign of shear stress t is always 桫 2 dh opposite to the sign of piezometric pressure gradient , dx and the magnitude of t is always maximum at both walls and zero at centerline of the channel.

28 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 29 dh For favorable pressure gradient, < 0 , t > 0 dx dh For adverse pressure gradient, > 0 , t < 0 dx

dh dh < 0 > 0 dx dx

Flow down an inclined plane

uniform flow  velocity and depth do not change in x-direction

du Continuity  0 dx  d2u x-momentum 0   p  z  x dy2

29 57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006 30  y-momentum 0   p  zhydrostatic pressure variation y dp   0 dx

d 2u   sin  dy2

du    sin y  c dy 

 y2 u   sin   Cy  D  2

du    0   sin d  c  c   sin d dy yd  

u(0) = 0  D = 0

 y2  u   sin   sin dy  2 

 = sin y2d  y 2

gsin  u(y) = y2d  y 2

d d 3   2 y  discharge per q   udy  sin dy   2 3 unit width 0  0

1  = d3 sin  3 

2 q 1  2 gd Vavg   d sin   sin  d 3  3

in terms of the slope So = tan   sin 

gd2S V  o 3

Exp. show Recrit  500, i.e., for Re > 500 the flow will become turbulent

p Vd   cos Re   500 y crit 

p  cosy  C

pd  po  cosd  C i.e., p  cosd  y po

* p(d) > po

* if  = 0 p = (d  y) + po entire weight of fluid imposed

if  = /2 p = po no pressure change through the fluid