Fluid Dynamics Crete

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<p>SUMMARY OF VECTOR AND TENSOR NOTATION-Bird, Stewart and Lightfoot "Transport Phenomena" -Bird, Armstrong and Hassager "Dynamics of Polymeric Liquids" The Physical quantities encountered in the theory of transport phenomena can be categorised into: - Scalars (temperature, energy, volume, and time) - Vectors (velocity, momentum, acceleration, force) - Second-order tensors (shear stress or momentum flux tensor) While for scalars only one type of multiplication is possible, for vectors and tensors several kinds are possible which are: - single dot . - double dot : - cross x The following types of parenthesis will also be used to denote the results of various operations. ( ) = scalar (u . w), ( : ) [ ] = vector [ u x w], [ . u] { } = tensor { . } The multiplication signs can be interpreted as follows: Multiplication sign None x . Order of Result -1 -2</p> <p>: -4 ________________________________________ Scalars can be interpreted as 0th order tensors, and vectors as first order tensors. Examples: s uxw : order is order is order is 0+2=2 which is a 2nd order tensor 1+1-1=1 which is a vector 2+2-4=0 which is a scalar</p> <p>2 Definition of a Vector: A vector is defined as a quantity of a given magnitude and direction. |u| is the magnitude of the vector u Two vectors are equal when their magnitudes are equal and when they point in the same direction. Addition and Subtraction of Vectors: w u+w u Dot Product of two Vectors: (u . w) = |u| |w| cos() commutative (u . v) = (v . u) not associative (u . v)w u(v . w) distributive (u . [v + w]) = (u . v) + (u . w) Cross Product of two Vectors: [uxw] = |u| |w| sin() n where n is a vector (unit magnitude) normal to the plane containing u and w and pointing in the direction that a right-handed screw will move if we turn u toward w by the shortest route. u w w u Area=(u.w) u w u-w</p> <p>Area of this equals the length of [uxw]</p> <p>3 not commutative distributive [uxw] = -[wxu] [[u + v] x w] = [u x w] + [v x w]</p> <p>not associative [u x [v x w]] [[u x v] x w]</p> <p>4</p> <p>VECTOR OPERATIONS FROM AN ANALYTICAL VIEWPOINTDefine rectangular co-ordinates: 1, 2, 3 x, y, z respectively Many formulae can be expressed more compactly in terms of the kronecker delta ij and the alternating unit tensor ijk, which are defined as: ij = 1 if ij =0 if and ijk=1 ijk= -1 ijk =0 if if if ijk=123, 231, 312 ijk=321, 132, 213 any two indices are alike i=j ij</p> <p>We will use the following definitions, which can be easily proved:</p> <p> j k ijk hjk = 2 ihand</p> <p>k ijk mnk = im </p> <p>jn - in jm</p> <p>The determinant of a three-by-three matrix may be written as:</p> <p> 11 12 13 21 22 23 = i j k ijk 1i 2j 3k 31 32 33</p> <p>5</p> <p>DEFINITION OF A VECTOR AND ITS MAGNITUDE: THE UNIT VECTORSA vector u can be defined completely by giving the magnitudes of its projections u1, u2, and u3 on the co-ordinate axis 1, 2, and 3 respectively. Thus one may write3</p> <p>u = 1u1 + 2u 2 + 3u3 = i uii =1</p> <p>where 1, 2, and 3 are the unit vectors in the direction of the 1, 2 and 3 axes respectively. The following identities between the vectors can be proven readily:1u3 u2 u u1</p> <p>1.1 = 2 . 2 = 3 . 3 = 12</p> <p> 1. 2 = 2 . 3 = 3 . 1 = 0 1 1 = 2 2 = 3 3 = 0</p> <p>3</p> <p>[ 1 2 ] = 3</p> <p>[ 2 3 ] = 1</p> <p>[ 3 1] = 2</p> <p>[ 2 1] = - 3</p> <p>[ 3 2 ] = -1</p> <p>[ 1 3 ] = - 2</p> <p>All these relations can be summarized as:</p> <p>( i . j) = ij3</p> <p>[ i j ] = ijk kk =1</p> <p>6 Addition of vectors:</p> <p>u + w = i i u i + i i w i = i i ( u i + w i )Multiplication of a Vector by a Scalar:</p> <p>s u = s [i i ui] = i i (su i)Dot Product:( u . w ) = [i i ui] . [ j j w j] = i j ( i . j) u i w j =</p> <p>= i j ij ui w j = i ui wi</p> <p>Cross Product:[ u w ] = [( j j u j) (k k w k )]</p> <p>= j k [ j k ] u j w k = i j k ijk i u j w k</p> <p>1= u1 w1</p> <p>2u2 w2</p> <p>3u3 w3</p> <p>Proofs of Identities (Example):</p> <p>Prove the following identity u [v w ] = v ( u . w ) - w ( u . v )</p> <p>This identity will be proven for the i-component, so the summation i will be dropped out for the</p> <p>7</p> <p>sake of simplicity.(u x [v x w])i =</p> <p>j k ijk uj [ v x w ]k = j k ijk uj [l mklmvl wm] = j k l m ijk klm uj vl wm j k l m ijk lmk uj vl wm j l m (iljm - im jl) uj vl wm j l m iljm uj vl wm - j l m im jl uj vl wm vi j m jm uj wm - wi j l jl uj vl vi j uj wj - wi j uj vj vi ( u . w) - wi ( u . v)v (u . w ) - w (u . v )</p> <p>= = = = = = = =</p> <p>set l=i in the first term and m=i in the second term set m=j in the first term and l=j in the second term</p> <p>8</p> <p>VECTOR DIFFERENTIAL OPERATIONSDefine first the del operator, which is a vector +2 +3 = i i x1 x2 xi x3</p> <p> = 1</p> <p>The Gradient of a Scalar Field:</p> <p> s = 1 not commutative not associative distributive</p> <p>s s s s + 2 +3 = i i x1 x2 xi x3</p> <p>s sL (r)s (rs) (r+s) = r + s</p> <p>The Divergence of a Vector Field:</p> <p> ( . u ) = i i xi </p> <p> . j j u j = i j [ i . j] u j xi </p> <p>[</p> <p>]</p> <p>= i j ij</p> <p> ui u j = i xi xi</p> <p>not commutative not associative distributive</p> <p>( . u ) ( u . ) (. s ) u (s . u ) . ( u + w ) = ( . u ) + ( . w )</p> <p>The Curl of a Vector Field:</p> <p>9</p> <p> [ u ] = </p> <p> j j x [ k k u k ] x j </p> <p> = j k [ j k ] uk xj </p> <p>1= x1 u1</p> <p>2 x2 u2</p> <p>3 x3 u3</p> <p> u u = 1 3 - 2 x 2 x3</p> <p> + 2 u1 - u 3 x3 x1 </p> <p> + 3 u 2 - u1 x1 x 2 </p> <p>[ x u ] = curl (u) = rot (u) It is distributive but not commutative or associative.The Laplacian Operator:</p> <p>The Laplacian of a scalar is:</p> <p>( . s ) = i</p> <p>2 2 s 2 s 2 s + 2 s= + xi2 xi2 x2 x3 2</p> <p>The Laplacian of a vector is:</p> <p>2 u = ( . u ) - [ [ u ] ]The Substantial Derivative of a Scalar Field:</p> <p>If u is assumed to be the local fluid velocity then:D = + (u . ) Dt t</p> <p>The substantial derivative for a scalar is:</p> <p>10</p> <p>s s Ds = + i u i xi Dt t</p> <p>The substantial derivative for a vector is:Du u + (u . ) u = i i = Dt t ui + (u . ) ui t</p> <p>This expression is only to be used for rectangular co-ordinates. For all co-ordinates:(u.)u = 1 (u.u ) [u [ u ]] 2</p> <p>11</p> <p>SECOND - ORDER TENSORSA vector u is specified by giving its three components, namely u1, u2, and u3. Similarly, a secondorder tensor is specified by giving its nine components.</p> <p> 11 = 21 31</p> <p> 12 22 32</p> <p> 13 23 33</p> <p>The elements 11, 22, and 33 are called diagonal while all the others are the non-diagonal elements of the tensor. If 12=21, 31=13, and 32=23 then the tensor is symmetric. The transpose of is defined as:</p> <p> 11 * = 12 13If is symmetric then =*.</p> <p> 21 22 23</p> <p> 31 32 33</p> <p>Dyadic Product of Two Vectors:</p> <p>This is defined as follows: u1 w1 uw = u 2 w1 u3 w1 u1 w 2 u2 w 2 u3 w 2 u1 w 3 u 2 w3 u3 w 3</p> <p>12 Unit Tensor:</p> <p>1</p> <p>0 1 0</p> <p>0 0 1</p> <p> = 00</p> <p>The components of the unit tensor are ij (kronecker delta for i,j=1,3)</p> <p>Unit Dyads:</p> <p>These are just the dyadic products of unit vectors, mn in which m,n=1,2,3. 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0</p> <p> 1 1 = 00 0</p> <p>1 2 = 00 0</p> <p> 1 3 = 00</p> <p> 3 2 = 00</p> <p>Thus, a tensor can be represented as:</p> <p> = i j i j ijand the dyadic product of two vectors as:u w = i j i j u i w j</p> <p>13</p> <p>Also note the following identities: ( i j : k l ) = il jk scalar</p> <p>[ i j . k ] = i jk</p> <p>vector</p> <p>[ i . j k ] = ij k</p> <p>vector</p> <p> i j . k l = jk i l</p> <p>tensor</p> <p>Addition of Tensors:</p> <p> + = i j i j ij + i j i j ij= i j i j ( ij + ij )</p> <p>Multiplication of a Tensor by a Scalar:s = s i j i j ij = i j i j ( s ij )</p> <p>Double Dot Product of Two Tensors:</p> <p> : = (i j i j ij) : (k l k l kl)</p> <p>14</p> <p>= i j k l ( i j : k l) ij kl</p> <p>= i j k l il </p> <p>jk ij kl</p> <p>set l=i and k=j to simplify to:= i j ij ji</p> <p>which is a scalar 2 + 2 - 4 = 0</p> <p>Dot Product of Two Tensors:</p> <p> . = (i j i j ij) . (k l k l kl)= i j k l ( i j . k l) ij kl = i j k l ( jk i l) ij kl</p> <p>= i l i l ( j ij jl )</p> <p>Vector Product (or Dot Product) of a Tensor with a Vector:</p> <p>[ . u ] = [ ( i j i j ij ) . ( k k u k ) ]</p> <p>= i j k i jk ij u k</p> <p>= i i ( j ij u j )</p> <p>15 Differential Operations: [ . ] = i i j k j k jk x i </p> <p>[</p> <p>]</p> <p>= i j k [ i . j k ] = i j k ij k jk xi</p> <p> jk xi</p> <p> = k k i ik xi </p> <p>Some other identities which can readily be proven are: uk xi</p> <p>w . u = i k k w i</p> <p> : u = i j ij</p> <p> ui xj</p> <p>16</p> <p>INTEGRAL THEOREMS FOR VECTORS AND TENSORSGauss - Ostrogradskii Divergence Theorem :</p> <p>If V is a closed region in space surrounded by a surface S then ( . u ) dV = ( n . u ) dS = ( u . n ) dS = u n dSV S S S</p> <p>where n is the outwardly directed normal vector.</p> <p> s dV = n s dSV S</p> <p>where s is a scalar quantity. [ . ] dV = [ n . ] dSVS</p> <p>where is a tensor.The Stokes Curl Theorem:</p> <p>If S is a surface bounded by a close curve C, then: ( [ x u ] . n ) dS = (u.t) dCSC</p> <p>where t is the tangential vector in the direction of the integration and n is the unit vector normal to S in the direction that a right-handed screw would move if its head were twisted in the direction of integration along C.</p> <p>17 The Leibnitz Formula for Differentiating a Triple Integral:</p> <p>s d s dV = dV + s ( u s . n ) dS dt V V t S where us is the velocity of any surface element, and s is a scalar quantity which can be a function of position and time i.e., s=s(x,y,z,t). Keep in mind that V=V(t) and S=S(t). If the surface of the volume is moving with the local fluid velocity (us=u), then</p> <p>d Ds s dV = dV dt V Dt V where is the fluid density.</p> <p>18</p> <p>CURVILINEAR COORDINATESThus far, we have considered only rectangular co-ordinates x, y and z. However, many times in fluid mechanics it is more convenient to work with curvilinear co-ordinates. The two most common curvilinear co-ordinate systems are the cylindrical and the spherical. In this development, we are interested in knowing how to write various differentials, such as s, [xv], and (:v) in curvilinear co-ordinates. It turns out that two are the useful tools in doing this a. The expression for in the curvilinear co-ordinates. b. The spatial derivatives of the unit vectors in curvilinear co-ordinates.Cylindrical Coordinates</p> <p>z (x,y,z) or (r,,z) y xx = r cos y = rsin z= z</p> <p>There a point is located by giving the values of r, , and z instead of x, y, and z which is the case for the Cartesian coordinates. From simple geometry one may derive the following expressions between these two systems of coordinates. These are:</p> <p>r</p> <p>r = + x2 + y 2</p> <p> = arctan (y/x)z= z</p> <p>To convert derivatives with respect to x, y, and z into derivatives with respect to r, , and z, the "chain"rule of differentiation is used. Thus one may derive sin = ( cos ) + + (0) z x r r </p> <p>19</p> <p> cos = ( sin ) + + (0) z y r r = ( 0 ) + ( 0 ) +( 1 ) z r z</p> <p>With these relations, derivatives of any scalar functions with respect to x, y and z can be expressed in terms of derivatives with respect to r, and z. Now we turn out attention to the interrelationship between the unit vectors. We note those in the Cartesian coordinates as x, y, and z and those in the cylindrical coordinates as r, , and z. To see how these are related consider the Figure below where it can be seen that as the point P is moving in the (x,y) plane the directions of r, change. Elementary trigonometrical arguments lead to the following relations:</p> <p>y</p> <p>y r x</p> <p> r = (cos ) x + (sin ) y + (0) z = (- sin ) x + (cos ) y + (0) z z = ( 0 ) x + ( 0 ) y + (1) zx</p> <p>P(x,y) or P(r,)</p> <p>These can be solved for x, y, and z to result</p> <p> x = (cos ) r + ( - sin ) + (0) z y = ( sin ) r + (cos ) + (0) z z = ( 0 ) r + ( 0 ) + (1) zVectors and tensors can be decomposed into components in all systems of co-ordinates just as with</p> <p>20</p> <p>respect to rectangular co-ordinates discussed previously. For example:[ v w ] = r ( v w z - vz w ) + ( vz w r - vr w z ) + z ( vr w - v w r ) ( . ) = r r ( rr rr + r r r + rz zr ) + r ( . ) + r z ( . ) + ...</p> <p>Spherical Coordinates</p> <p>The spherical co-ordinates are related to rectangular by the following relations z P(x,y,z) or (r,,) r Figure: A Spherical system of co-ordinates</p> <p>y rsin</p> <p>x</p> <p>x = r sin cos y = rsin sin z = rcos</p> <p>r = + x2 + y 2 + z 2</p> <p> = arctan (</p> <p>2 x2 + y /z )</p> <p> = arctan(y/z)</p> <p>The derivative operators are as follows: cos cos sin = ( sin cos ) + + x r r r sin cos sin cos = ( sin sin ) + + y r r r sin </p> <p>21</p> <p> sin = ( cos ) + z r r</p> <p> +( 0 ) </p> <p>The relations between the unit vectors are:</p> <p> r = (sin cos ) x + (sin sin ) y + ( cos ) z = (cos cos ) x + (cos sin ) y + (- sin ) z = ( - sin ) x + ( cos ) y + ( 0 ) zThese can be solved for x, y, and z to result:</p> <p> x = (sin cos ) r + ( cos cos ) + ( - sin ) y = ( sin sin ) r + ( cos sin ) + ( cos ) z = ( cos ) r + ( - sin ) + ( 0 ) Some example operations in spherical co-ordinates are:( : ) = rr rr + r r + r r + r r + + ++ r r + + </p> <p>22</p> <p>ur( u .[ v x w ] ) =</p> <p>u v w</p> <p>u v w</p> <p>vr wr</p> <p>These examples tells us that the relations (not involving !) discussed earlier can be written in terms of spherical components.</p> <p>DIFFERENTIAL OPERATIONS IN CURVILINEAR COORDINATESThe operator will now be derived in cylindrical and spherical co-ordinates.Cylindrical:</p> <p>The following relations can be obtained by differentiating the relations between the unit vectors in the cylindrical co-ordinates with those in the Cartesian ones. r=0 r r = r=0 z = 0 r = - r = 0 z z=0 r z=0 z=0 z</p> <p>The definition of in Cartesian co-ordinates is: + y + z x y z</p> <p> = x</p> <p>Substituting x, y, and z in terms of r, , and z and simplifying we obtain for cylindrical coordinates, that is:</p> <p>23</p> <p> = r</p> <p> 1 + z + r r z</p> <p>Spherical Coordinates</p> <p>The following relations can be obtained by differentiating the relations between the unit vectors in the spherical coordinates with those in the Cartesian ones. r=0 r r = r = sin = 0 r = - r = cos = 0 r = 0 z = - r sin - cos </p> <p>The definition of in Cartesian co-ordinates is: + y + z x y z</p> <p>= x</p> <p>Substituting x, y, and z in terms of r, , and , and simplifying we obtain for spherical coordinates, that is: 1 1 + + r r r sin </p> <p> = r</p> <p>For more details see: 1. R.B. Bird, W.E. Stewart and E.N. Lightfoot, "Transport Phenomena," Wiley, New York, 1960. 2. R.B. Bird, R.C. Armstrong and O. Hassager, "Dynamics of Polymeric Liquids," Vol.1, Mechanics," Wiley, New York, 1977. "Fluid</p> <p>24</p> <p>25</p> <p>26</p> <p>27</p> <p>28</p> <p>29</p> <p>INTRODUCTION - FUNDAMENTAL CONCEPTSDefinition of a Fluid All forms of matter can be classified in terms of their physical appearance or phase into three classes: solids, liquids and gases. Liquids and gases are called fluids. A fluid is defined as "a substance that under the action of an infinitesimal force deforms permanently and continuously. a. t h F...</p>