REE 307 Fluid Mechanics II Lecture 5

November 8, 2017

Dr./ Ahmed Mohamed Nagib Elmekawy

Zewail City for Science and Technology DIFFERENTIAL ANALYSIS OF FLUID FLOW

2 The fundamental differential equations of fluid motion are derived in this chapter, and we show how to solve them analytically for some simple flows. More complicated flows, such as the air flow induced by 2 a tornado shown here, cannot be solved exactly.

3 Objectives

• Understand how the differential equation of conservation of mass and the differential linear momentum equation are derived and applied

• Obtain analytical solutions of the equations of motion for simple flow fields

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4 LAGRANGIAN AND EULERIAN DESCRIPTIONS Kinematics: The study of motion. Fluid kinematics: The study of how fluids flow and how to describe fluidmotion. There are two distinct ways to describe motion: Lagrangian and Eulerian Lagrangian description: To follow the path of individual objects. This method requires us to track the position and velocity of each individual fluid parcel (fluid particle) and take to be a parcel of fixedidentity.

With a small number of objects, such In the Lagrangian description, one as billiard balls on a pool table, must keep track of the position and individual objects can be tracked. velocity of individual particles. 4

5 • A more common method is Eulerian description of fluid motion. • In the Eulerian description of fluid flow, a finite volume called a flow domain or control volume is defined, through which fluid flows in and out. • Instead of tracking individual fluid particles, we define field variables, functions of space and time, within the control volume. • The field variable at a particular location at a particular time is the value of the variable for whichever fluid particle happens to occupy that location at that time. • For example, the pressure field is a scalar field variable. We define the velocity field as a vector field variable.

Collectively, these (and other) field variables define the flow field. The velocity field can be expanded in Cartesian coordinates as

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6 • In the Eulerian description we don’t really care what happens to individual fluid particles; rather we are concerned with the pressure, velocity, acceleration, etc., of whichever fluid particle happens to be at the location of interest at the time of interest. • While there are many occasions in which the Lagrangian description is useful, the Eulerian description In the Eulerian description, one is often more convenient for fluid defines field variables, such as mechanics applications. the pressure field and the velocity field, at any location • Experimental measurements are and instant in time. generally more suited to the Eulerian description.

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7 CONSERVATION OF MASS—THE CONTINUITY EQUATION

The net rate of change of mass withinthe control volume is equal to the rate at which mass flows into the control volume minus the rate at which mass flows out of the control volume.

To derive a differential conservation equation, we imagine shrinking a control volume to infinitesimal size.

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8 Derivation Using an Infinitesimal Control Volume At locations away from the center of the box, we use a Taylor series expansion about the center of the box.

A small box-shaped control volume centered at point P is used for derivation of the differential equation for conservation of mass in Cartesian coordinates; the blue dots indicate the center of each face.

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9 1 0

10 1 1 Conservation of Mass: Alternative forms • Use product rule on divergence term

    V  u i  v j  w k          i  j  k x y z Continuity Equation in Cylindrical Coordinates

Velocity components and unit vectors in cylindrical coordinates: (a) two- dimensional flow in the xy- or r휃-plane, (b) three-dimensional flow. 12

13 Conservation of Mass: Cylindrical coordinates 1 5

The divergence operation in Cartesian and cylindrical coordinates. 1 6

Special Cases of the Continuity Equation

Special Case 1: Steady Compressible Flow 1 7

Special Case 2: Incompressible Flow 1 8 1 9 2 0 2 1 THE DIFFERENTIAL LINEAR MOMENTUM EQUATION

Derivation Using Newton’s Second Law

If the differential fluid element is a material element, it moves with the flow and Newton’s second law applies directly. Acceleration Field The equations of motion for fluid flow (such as Newton’s second law) are written for a fluid particle, which we also call a material particle. If we were to follow a particular fluid particle as it moves around in the flow, we would be employing the Lagrangian description, and the equations of motion would be directly applicable. For example, we would define the Newton’s second law applied to a fluid particle’s location in space in terms particle; the acceleration vector (gray arrow) of a material position vector is in the same direction as the force vector (black arrow), but the velocity vector (red (xparticle(t), yparticle(t), zparticle(t)). arrow) may act in a different direction.

22 Local Advective (convective) acceleration acceleration The components of the acceleration vector in cartesian coordinates:

Flow of water through the nozzle of a garden hose illustrates that fluid particles may accelerate, even in a steady flow. In this example, the exit speed of the water is much higher than the water speed in the hose, implying that fluid particles have accelerated even though the flow is steady.

24 Acceleration Components The components of the acceleration are:

VVVV    Vector equation: a  u  v  w particle t  x  y  z

x- component u u u u a   u  v  w x t x y z v v v v Y-component a   u  v  w y t x y z w w w w a   u  v  w z-component z t x y z

Question: Give examples of steady flows with acceleration

Incompressible Steady ideal flow in a variable-area duct Conservation of Momentum Types of forces: 1. Surface forces: include all forces acting on the boundaries of a medium though direct contact such as pressure, friction,…etc. 2. Body forces are developed without physical contact and distributed over the volume of the fluid such as gravitational and electromagnetic. • The force F acting on A may be resolved into two components, one normal and the other tangential to the area. Body Forces

Positive components of the stress tensor in Cartesian coordinates on the positive (right, top, and front) faces of an infinitesimal rectangular control volume. The blue dots indicate the center of each face. Positive components on the negative (left, bottom, and back) faces are in the opposite direction of those shown here.

27 28 Surface Forces

휏푖푗 called the viscous stress tensor

For fluids at rest, the only stress on a fluid element is the hydrostatic pressure, which always acts inward and normal to any surface. 3 0

Sketch illustrating the surface forces acting in the x- direction due to the appropriate stress tensor component on each face of the differential control volume; the blue dots indicate the center of each face. 3 1

If the differential fluid element is a material element, it moves with the flow and Newton’s second law applies directly. 3 2

Newtonian versus Non-Newtonian Fluids

Rheology: The study of the deformation of flowing fluids. Newtonian fluids: Fluids for which the shear stress is linearly proportional to the shear strain rate. Newtonian fluids: Fluids for which the shear stress is not linearly related to the shear strain rate. Viscoelastic: A fluid that returns (either fully or partially) to its original shape after the applied stress is released. Rheological behavior of fluids—shear stress as a function of shear strain rate. Some non-Newtonian fluids are called shear thinning fluids or pseudoplastic fluids, because the In some fluids a finite stress called the more the fluid is sheared, the less yield stress is required before the viscous it becomes. fluid begins to flow at all; such fluids Plastic fluids are those in whichthe are called Bingham plastic fluids. shear thinning effect is extreme. 3 3

Derivation of the Navier–Stokes Equation for Incompressible, Isothermal Flow

The incompressible flow approximation implies constant density, and the isothermal approximation implies constant viscosity. The Laplacian operator, shown here in both Cartesian and cylindrical coordinates, appears in the viscous term of the incompressible Navier–Stokes equation.

34 The Navier–Stokes equation is an unsteady, nonlinear, second order, partial differential equation. Equation 9–60 has four unknowns (three velocity components and pressure), yet it represents only three equations (three components since it is a vector equation). Obviously we need another equation to make the problem solvable. The fourth The Navier–Stokes equation is the equation is the incompressible continuity cornerstone of fluid mechanics. equation (Eq. 9–16). 28

35 Navier-Stokes Equations u u u u p  u 2   (  u  v  w )  g   [(2  .V )] t x y z x x x x 3 x-  v u  w u  [(  )] [(  )] (a) momentum y x y z x z v v v v p  v u (  u  v  w )  g y   [(  )] t x y z y x x y y-  v 2    v w momentum  [(2  .V )] [(  )] (b) y y 3 z z y w w w w p  w u (  u  v  w )  g z   [(  )] t x y z z x x z z-  v w  w 2   momentum  [(  )] [(2  .V )] (c) y z y z z 3 Navier-Stokes Equations • For incompressible fluids, constant µ: • Continuity equation: .V = 0

 u 2    v u  w u [(2  .V )] [(  )] [(  )]  x x 3 y x y z x z  u  v u  w u { [(2 )] [(  )] [(  )]}  x x y x y z x z  2u  2u  2u  2u  2v  2w (   )  (   )  x2 y 2 z 2 x2 xy xz  2u  2u  2u  u v w (   )   (   )  x2 y 2 z 2 x x y z  2u  2u  2u (   )  2u x2 y 2 z 2 Navier-Stokes Equations • For incompressible flow with constant dynamic viscosity:

Du p  2u  2u  2u • x- momentum   g   (   ) (a) Dt x x x2 y 2 z 2

2 2 2 • Dv p  v  v  v Y- momentum   g   (   ) (b) Dt y y x2 y 2 z 2

• z-momentum Dw p  2w  2w  2w   g   (   ) (c) Dt z z x2 y 2 z 2 • In vector form, the three equations are given by:

 DV     g  p  2V Incompressible NSE Dt written in vector form Continuity and Navier–Stokes Equations in Cartesian Coordinates Navier-Stokes Equation

• The Navier-Stokes equations for incompressible flow in vector form:

Incompressible NSE written in vector form

• This results in a closed system of equations! • 4 equations (continuity and 3 momentum equations) • 4 unknowns (u, v, w, p) • In addition to vector form, incompressible N-S equation can be written in several other forms including: • Cartesian coordinates • Cylindrical coordinates • Tensor notation Euler Equations • For inviscid flow (µ = 0) the momentum equations are given by:

u u u u p • x- momentum (  u  v  w )  g  (a) t x y z x x

v v v v p • Y- momentum (  u  v  w )  g  (b) t x y z y y

w w w w p • z-momentum (  u  v  w )  g  (c) t x y z z z • In vector form, the three equations are given by:  DV    g  p Euler equations Dt written in vector form Continuity and Navier–Stokes Equations in Cylindrical Coordinates Differential Analysis of Fluid Flow Problems

• Now that we have a set of governing partial differential equations, there are 2 problems we can solve 1. Calculate pressure (P) for a known velocity field 2. Calculate velocity (U, V, W) and pressure (P) for known geometry, boundary conditions (BC), and initial conditions (IC) • There are about 80 known exact solutions to the NSE • Solutions can be classified by type or geometry, for example: 1. Couette shear flows 2. Steady duct/pipe flows (Poisseulle flow) DIFFERENTIAL ANALYSIS OF FLUID FLOW PROBLEMS

•Calculating both the velocity and pressure fields for a flow of known geometry and known boundary conditions

A general three-dimensional incompressible flow field with constant properties requires four equations to solve for four unknowns.

44 Exact Solutions of the NSE Procedure for solving continuity and NSE 1. Set up the problem and geometry, identifying all relevant dimensions and parameters 2. List all appropriate assumptions, approximations, simplifications, and boundary conditions 3. Simplify the differential equations as much as possible 4. Integrate the equations 5. Apply BCs to solve for constants of integration 6. Verify results • Boundary conditions are critical to exact, approximate, and computational solutions. ▪ BC’s used in analytical solutions are • No-slip boundary condition • Interface boundary condition Exact Solutions of the Continuity Boundary Conditions and Navier–Stokes Equations

A piston moving at speed VP in a cylinder. A thin film of oil is sheared between the piston and the cylinder; a magnified viewof Procedure for solving the the oil film is shown. The no-slip boundary incompressible continuity and condition requires that the velocity of fluid Navier–Stokes equations. adjacent to a wall equal that of the wall.32

46 At an interface between two fluids, the velocity of the two fluids must be equal. In addition, the shear stress parallel to the interface must be the same in both fluids.

Along a horizontal free surface of water and air, the water and air velocities must be equal and the shear stresses must match.

However, since air << water, a good approximation is that the shear stress at the water surface is negligibly small.

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47 Other boundary conditions arise depending on the problem setup. For example, we often need to define inlet boundary conditions at a boundary of a flow domain where fluid enters the domain. Likewise, we define outlet boundary Boundary conditions along a plane of conditions at an outflow. symmetry are defined so asto ensure Symmetry boundary conditions are that the flow field on one side of the useful along an axis or plane of symmetry plane is a mirror image of symmetry. that on the other side, as shown here For unsteady flow problems we also for a horizontal symmetry plane. need to define initial conditions (at the starting time, usually t = 0). 34

48 Summary • Introduction • Conservation of mass-The continuity equation Derivation Using an Infinitesimal Control Volume Continuity Equation in Cylindrical Coordinates Special Cases of the Continuity Equation • The Navier-Stokes equation Derivation of the Navier–Stokes Equation for Incompressible, Isothermal Flow Continuity and Navier–Stokes Equations in Cartesian Coordinates Continuity and Navier–Stokes Equations in Cylindrical Coordinates • Differential analysis of fluid flow problems Exact Solutions of the Continuity and Navier– Stokes Equations

49 Example exact solution Fully Developed Couette Flow

• For the given geometry and BC’s, calculate the velocity and pressure fields, and estimate the shear force per unit area acting on the bottom plate

• Step 1: Geometry, dimensions, and properties Fully Developed Couette Flow

• Step 2: Assumptions and BC’s • Assumptions 1. Plates are infinite in x and z 2. Flow is steady, /t = 0 3. Parallel flow, v = 0 4. Incompressible, Newtonian, laminar, constant properties 5. No pressure gradient 6. 1D, w = 0, /z = 0 7. Gravity acts in the -y direction, • Boundary conditions  ˆ 1. Bottom plate (y=0) – no slip condition: u=0, v=0, w=0g  g j , g y  g 2. Top plate (y=h) : no slip condition: u=V, v=0, w=0 Fully Developed Couette Flow Note: these numbers refer 3 to the assumptions on the 6 previous slide • Step 3: Simplify u v w Continuity    0 x y z u This means the flow is “fully developed”  0 or not changing in the direction of flow x X-momentum 2 Cont. 3 6 5 7 Cont. 6 u u u u p  2u  2u  2u (  u  v  w )    g  (   ) t x y z x x x2 y 2 z 2

 2u  0 y 2 Fully Developed Couette Flow

y•-momentumStep 3: Simplify, cont. 2,3 3 3 3,6 3 3 3 v v v v p  2v  2v  2v (  u  v  w )    g  (   ) t x y z y y x2 y 2 z 2 p  g y p = p(y) y z-momentum 2,6 6 6 6 7 6 6 6 w w w w p  2w  2w  2w (  u  v  w )    g  (   ) t x y z z z x2 y 2 z 2 p p dp  0 p = p(y)   g y z y dy Fully Developed Couette Flow

• Step 4: Integrate X-momentum integrate integrate y-momentum dp integrate  g p(y)  gy  C dy 3 Fully Developed Couette Flow

• Step 5: Apply BC’s

• Y = 0, u = 0 = C1(0) + C2  C2 = 0 • Y = h, u =V =C1h  C1 = V/h • This gives

• For pressure, no explicit BC, therefore C3 can remain an arbitrary constant (recall only P appears in NSE).

• Let p = p0 at y = 0 (C3 renamed p0)

1. Hydrostatic pressure p(y)  p0  gy 2. Pressure acts independently of flow Fully Developed Couette Flow

• Step 6: Verify solution by back-substituting into differential equations • Given the solution (u,v,w)=(Vy/h, 0, 0)

• Continuity is satisfied 0 + 0 + 0 = 0 • X-momentum is satisfied u u u u p 2u 2u 2u (  u  v  w )    g  (   ) t x y z x x x2 y2 z2 Fully Developed Couette Flow

• Finally, calculate shear force on bottom plate v u V    (  )   u=V/h xy yx x y h Shear force per unit area acting on the wall

Note that w is equal and opposite to the shear stress acting on the fluid yx (Newton’s third law). Parallel Plates (Poiseuille Flow) • Given: A steady, fully developed, laminar flow of a Newtonian fluid in a rectangular channel of two parallel plates where the width of the channel is much larger than the height, h, between the plates.

• Find: The velocity profile and shear stress due to the flow.

Assumptions: • Entrance Effects Neglected • No-Slip Condition • No vorticity/turbulence Additional and Highlighted Important Assumptions

• The width is very large compared to the height of the plate. • No entrance or exit effects. • Fully developed flow. • Steady flow • THEREFORE… • Velocity can only be dependent on vertical location in the flow (u) • v = w = 0 • The pressure drop is constant and in the x-direction only. p p  Constant  , where L is a length in x. x L Boundary Conditions

• No Slip Condition Applies • Therefore, at y = -a and y = +a, u=v=w = 0 • The bounding walls in the z direction are often ignored. If we don’t ignore them we also need: • z = -W/2 and z = +W/2, u=v=w = 0, where W is the width of the channel.

y  a Fixed plate y x 2a Fluid flow direction

y  a Fixed plate Navier-Stokes Equations In Vector Form for incompressible flow:  v     v •v  p  2 v  g  t  Which we expand to component form: x - component :  u u u u  p  2u  2u  2u    u  v  w         g    2 2 2  x  t x y z  x x y z  y - component :  v v v v  p  2v  2v  2v   u  v  w         g    2 2 2  y  t x y z  y x y z  z - component :  w w w w  p  2w  2w  2w   u  v  w         g    2 2 2  z  t x y z  z  x y z  Reducing Navier-Stokes x - component :  u u u u  p 2u 2u 2u    u  v  w         g    2 2 2  x  t x y z  x x y z  y - component :  v v v v  p 2v 2v 2v   u  v  w         g    2 2 2  y  t x y z  y x y z  z - component :  w w w w  p 2w 2w 2w   u  v  w         g    2 2 2  z  t x y z  z  x y z  N-S equation therefore reduces to p  2u x - component:     g  0 x y 2 x p y - component:   g  0 y y p z - component:   g  0 z z

Ignoring gravitational effects, we get p 2u x - component:     0 x y 2 p y - component:   0 p is not a function of y y p is a function p of x only z - component:  0 p is not a function of z z 

Rewriting (4), we get 2u 1 p  (5) y 2  x

p is a function of x only and is a constant and therefore RHS is a function of x only

u is a function of y only and therefore LHS is a function of y only

Therefore (5) gives, function of (y) = function of (x) = constant p p It means   = constant L x That is, pressure gradient in the x-direction is a constant.

LHS = left hand side of the equation RHS = right hand side of the equation Rewriting (5), we get  2u 1 p  p   (6) y 2  x L p p where   is the constant pressure gradient in the L x x-direction Since u is only a function of y, the partial derivative becomes an ordinary derivative.

Therefore, (5) becomes integrate d 2u p du p   (7)   y  C dy 2 L dy L 1 Integrating again, we get p y 2 u    C y  C L 2 1 2 (8) y  a Fixed plate where C1 and C2 are constants to be determined using the boundary y conditions given below: x 2a no-slip u  0 at y  a Fluid flow direction boundary u  0 at y  a } condition y  a Fixed plate Substituting the boundary conditions in (8), we get p a2 C1  0 and C2  L 2

p Therefore, (8) reduces to u  a2  y 2  2L Parabolic velocity profile Steady, incompressible flow of Newtonian fluid in an infinite channel with one plate moving at uniform velocity - fully developed plane Couette-Poiseuille flow Moving plate p y 2 (8) at velocity U u    C1 y  C2 y  a L 2 y where C1 and C2 are constants to be x 2a determined using the boundary conditions given below: Fluid flow direction u  U at y  a no-slip y  a Fixed plate } boundary u  0 at y  a condition Substituting the boundary conditions in (8), we get U U p a2 C  and C   1 2a 2 2 L 2 Therefore, (8) reduces to Parabolic velocity profile is super imposed on a p U u  a2  y 2  a  y linear velocity profile 2L 2a Steady, incompressible flow of Newtonian fluid in an infinite channel with one plate moving at uniform velocity - fully developed plane Couette-Poiseuille flow

Starting from plane Couette-Poiseuille flow y  a Moving plate p U u  a2  y 2  a  y y at velocity U 2L 2a x 2a If zero pressure gradient is maintained in Fluid flow direction the flow direction, then ∆p = 0. y  a Fixed plate Therefore, we get U u  a  y Linear velocity profile 2a

Points to remember: - Pressure gradient gives parabolic profile - Moving wall gives linear profile Let us get back to the velocity profile of fully developed plane Couette-Poiseuille flow p U u  a2  y 2  a  y 2L 2a Let us non-dimensionalize it as follows: a2p  y 2  U  y  u  1   1 Rearranging gives  2    2L  a  2  a 

u a2p  y 2  1  y   1   1 Dividing by U gives  2    U 2LU  a  2  a  u y Introducing non-dimensional variables u  and y  , we get U a

a2p 1 u  1 y 2  1 y 2LU 2 Plot the non-dimensionalized velocity profile of the fully developed plane Couette-Poiseuille flow a2p 1 u  1 y 2  1 y 2LU 2 1 p y y  2LU / a2 a 1

0.25 0 -0.5 0 0.5 1 1.5 0

-0.25 u u  U -1 -1 R. Shanthini 15 March 2012 Plot the non-dimensionalized velocity profile of the fully developed plane Couette-Poiseuille flow a2p 1 u  1 y 2  1 y 2LU 2 1 y y  a p 2LU / a2

BACK FLOW 0 0 -0.25 0.75

-0.5 u u  U R. Shanthini -1 15 March 2012 Plot the non-dimensionalized velocity profile of the fully developed plane Couette-Poiseuille flow a2p 1 u  1 y 2  1 y 2LU 2

1 p y y  2LU / a2 a 0

-0.25 0 -1 0 1 -0.5

u -1.5 u  U -1 R. Shanthini 15 March 2012 Determine the volumetric flow rate of the fully developed plane Couette-Poiseuille flow

Let us determine the volumetric flow rate y  a Moving plate through one unit width fluid film along z- at velocity U direction, using the velocity profile y x 2a p U Fluid flow direction u  a2  y 2  a  y 2L 2a y  a Fixed plate

a a a 3 2  p 2 2 U   p  2 y  U  y  Q  udy  a  y  a  y dy   a y    ay     2L 2a  2L  3  2a  2  a a       a

3 2 3 2  p  3 a  U  2 a   p  3 a  U  2 a  Q   a    a      a     a   2L  3  2a  2  2L  3  2a  2  2a3 p Q  Ua 3 L Plot the non-dimensionalized volumetric flow rate of the fully developed plane Couette-Poiseuille flow

Non-dimensionalizing Q, we get Q 2 a2p 1 Ua 3 LU 4 Q Ua 3

2 Why the flow rate becomes zero? 1

0 -4 -2 0 2 a 2p 4 -1 U

-2 Steady, incompressible flow of Newtonian fluid in a pipe - fully developed pipe Poisuille flow

Fixed pipe r

z Fluid flow direction 2a 2a Laminar Flow in pipes

Steady Laminar pipe flow. Assumptions: vr  v  0 vz  vz r p dp  i.e. pressure is a linear of z. z dz The above “assumptions” can be obtained from the single assumption of “fully developed” flow. In fully developed pipe flow, all velocity components are assumed to be unchanging along the axial direction, and axially symmetric i.e.: Now look at the z-momentum equation. v v v v v ( z  v z   z  v z ) t r r r  z z dp  1   v  1  2v  2v      r z  z  z     2 2 2  dz  r r  r  r  dz  Results from Mass/Momentum The removal of the indicated terms yields: 1 dp 1   v    r z   0  dz r r  r  Or rearranging, and substituting for the pressure gradient: 1 d  dv  1 dp r z   r dr  dr   dz

In a typical situation, we would have control over dp/dz. That is, we can induce a pressure gradient by altering the pressure at one end of the pipe. We will therefore take it as the input to the system (similar to what an electrical engineer might do in testing a linear system). d  dv  r dp r z   dr  dr   dz Laminar flow in pipes

Integrating once gives:

2  dvz  r dp r    C1  dr  2 dz

But at r = 0 velocity = max or dvz/dr = 0 C1 = 0

2  dv  r dp Divide by r  dv  r dp r z    z    dr  2 dz  dr  2 dz

Integrating one more time gives: r 2 dp v   C z 4 dz 2 a2 dp a2 dp at pipe wall r = a velocity = 0 0   C C   4 dz 2 2 4 dz Laminar flow in pipes

Substituting in the velocity equation gives: r 2 dp a2 dp dp a2  r 2 v     ( ) z 4 dz 4 dz dz 4 Volume flow rate is obtained from: ra Q  v dA  z Q  2rvz dr crosssection r0 dp  ( ) 4 ra a  dp  Q  dz ra2  r 2 dr Q    2 r0 8  dz 

D4  dp  Q    128  dz  Laminar flow in pipes

Average velocity, vav= Q/area D4  dp  D2 D2  dp  vz,ave    /    128  dz  4 32  dz  Maximum velocity at r = 0: dp a2 dp D2 v   ( )   ( ) z,max dz 4 dz 16

vavergae = ½ vmax Shear stress: dv dp  2a dp D v    z  ( )( )   ( )  8( z,ave ) dr ra dz 4 dz 4 D Laminar flow in pipes

Drag Coefficient:

 w 16vave 16 C f   2  1 2 v D R v ave ave e 2

Coefficient of friction: f = 4 Cf 64 f  Re