CONSERVATION OF MASS AND BALANCE OF LINEAR MOMENTUM

CONSERVATION OF MASS AND BALANCE OF LINEAR MOMENTUM Summary of integral theorems Material time derivative Reynolds transport theorem Principle of conservation of mass Principle of balance of linear momentum J. N. Reddy Mass-Momenta - 1

INTEGRAL THEOREMS Divergence Theorem Gradient Theorem Curl Theorem General Theorem nf ˆ d F d n ˆ * F ds * F d nf ˆ d F d n ˆ F d F d J. N. Reddy Mass-Momenta - 2

Gradient Theorem INTEGRAL THEOREMS nˆ F d Fd The gradient of a function F represents the rate of change of F with respect to the coordinate directions. the partial derivative with respect to, for eample, gives the rate of change of F in the direction. Level surfaces F normal to the surface J. N. Reddy Mass-Momenta - 3

F Gradient of a Scalar Function a first- order tensor,that is, a eˆ eˆ eˆ y y F F F F eˆ eˆ eˆ y y vector in rectangular Cartesian system 1 eˆ eˆ eˆ r r r F 1 F F F eˆ eˆ eˆ r r r in cylindrical coordinate system J. N. Reddy Mass-Momenta - 4

Divergence Theorem INTEGRAL THEOREMS nf ˆ d F d The divergence represents the volume density of the outward flu of a vector field F from an infinitesimal volume d around a given point. It is a local measure of its "outgoingness." d Flu d J. N. Reddy Mass-Momenta - 5

F Divergence of First-Order Tensors a eroth tensor,that is, a scalar eˆ eˆ eˆ y y F F y F F y 1 eˆ eˆ eˆ r r r 1 ( rfr ) F F F r r r in rectangular Cartesian system in cylindrical coordinate system J. N. Reddy Mass-Momenta - 6

S Divergence of Second-order Tensors a first- order tensor,that is, a vector S S y F S y Syy F y S eˆ eˆ y y S S y F eˆ y 1 ( rsrr ) S r S r S r S eˆ r r 1 ( rsr ) S S r S r r 1 ( rs ) S S r eˆ in cylindrical coordinate system eˆ r r J. N. Reddy r r Mass-Momenta - 7 in rectangular Cartesian system r r y

INTEGRAL THEOREMS Curl Theorem (Stoke s Theorem) nˆ F d Fd The curl of a vector F describes the infinitesimal rotation of F. A physical interpretation is as follows. Suppose the vector field describes the velocity field F = v of a fluid flow, say, in a large tank of liquid, and a small spherical ball is located within the fluid (the center of the ball being fied at a certain point but free to rotate about an ais perpendicular to the plane of the flow). If the ball has a rough surface, the fluid flowing past the ball will make it rotate. The rotation ais (oriented according to the right hand rule) points in the direction of the curl of the field at the center of the ball, and the angular speed of the rotation is half the magnitude of the curl at this point. J. N. Reddy Mass-Momenta - 8

Curl of a Vector Function in (,y,) system A a first- order tensor,that is, a vector A y A ˆ ˆ ˆ ˆ ˆ ˆ A A ee eey ee A ê A ê A y y y ê y eˆ eˆ eˆ eˆ eˆ eˆ y y y y ê A ê y A y eˆ eˆ eˆ eˆ eˆ eˆ A y A A y A A A y A eˆ eˆ eˆ y y ê y y J. N. Reddy Mass-Momenta - 9

Material Time Derivative Material time derivative is the time derivative of a function in spatial description with the material coordinate X held constant. We denote it with D/Dt (also called substantive derivative). Material (or Lagrange) description ( X, t): D Dt t t Xconst Spatial (or Eulerian) description D i (, t): Dt t t t D Dt v t Xconst const Xconst Local change Translational change J. N. Reddy Mass-Momenta - 1 i

Problem statement: Suppose that a motion is described by the one-dimensional mapping, 1 tx for t >. Determine (a) the velocities and accelerations in the spatial and material descriptions, and (b) the time derivative of a function 2 ( X,) t Xt in the spatial and material descriptions. Solution: The velocity v D / Dt can be epressed in the material and spatial coordinates as The acceleration EXAMPLE 5-1 D vxt (,) X, vt (,) X( t,) Dt t 1t a Dv / Dt in the two descriptions is Dv( X,) t v Dv(,) t v v a, a v Dt t Dt t J. N. Reddy Mass-Momenta - 11

EXAMPLE 5-1 (continued) The material time derivative of description is simply D( Xt,) ( Xt,) Dt t The material time derivative of in the spatial description is ( Xt,) 2Xt in the material 2 2 ( t,) X( tt,) t /( 1 t) 2 2 D 2t t t 2t v 2 Dt t 1t ( 1t) 1t 1 t 1t which is the same as that calculated before, ecept that it is epressed in terms of the current coordinate,. J. N. Reddy Mass-Momenta - 12

EXAMPLE 5-2 Problem statement: Consider the motion of a body described by the mapping Determine the density as a function position and time t. Solution: J. N. Reddy X, X, X 1 1 2 2 3 3 1 tx1 First, compute the velocity components D i v or v i Dt t t Thus, we have Xfied X v v v 2 1 1 2 1, 2 2, ( 1 tx1) 3 Xfied Mass-Momenta - 13

Net, compute EXAMPLE 5-2 (continued) D Dt 1 v 2 1 2. 1 tx1 Integrating this equation with respect to t, we obtain 1 1 1 D 2 Dt ln 2ln ln c 1 2 1 J. N. Reddy Mass-Momenta - 14 X X X 1tX 1tX 1 1 where c is a constant of integration. If at time t=, then wehave ln c ln c,and the density becomes 2 1tX ( 1 t )

REYNOLDS TRANSPORT THEOREM Let each element of mass in the volume () t with closed boundary moves with the velocity v (,) t. Let (,) t be any function. We are interested in the material time derivative of the integral D Dt (,) t d Since () t is changing with time, we cannot take the differentiation through the integral. However, if the integral were over the volume in the reference configuration (which is fied), it is possible to interchange the integration and differentiation because D/Dt is differential with respect to time keeping X constant. The transformation X (,) t and djd lets us to do eactly that. J. N. Reddy Mass-Momenta - 15

REYNOLDS TRANSPORT THEOREM We have D D D DJ (,) td [ X (,),t] t J d J d Dt ( t) Dt Dt Dt D D v Jd v d Dt ( t) Dt v v d v d ( t) t ( t) t Thus D Dt d nˆ ( t) t ( t) v d (,) td d nˆ t ( t) ( t) ( t) v d J. N. Reddy Mass-Momenta - 16

REYNOLDS TRANSPORT THEOREM D Dt (,) td d nˆ t ( t) ( t) ( t) v d Thus the time rate of change of the integral of a function (,) t over a moving volume is equal to the integral of the local time rate of (,) t plus the net outflow of (,) t over the surface of the moving volume. Here (,) t can be a scalar or a tensor of any order. D Dt (,) td ( t) ( t) ( t) t D Dt J. N. Reddy Mass-Momenta - 17 v d v d

The Principle of Conservation of Mass in Spatial Description If a continuous medium of density ρ fills the volume at time t, the rate of increase of the total mass inside the volume is ρ d t () t The rate of mass outflow through the surface element is ρvnd where v ˆ ( nˆ n vn is the outward normal). Hence, the rate of inflow through the entire surface is ρv d ρvnˆ d ρv d n d J. N. Reddy Mass-Momenta - 18

The Principle of Conservation of Mass (continued) If no mass is created or destroyed inside the volume () t, this must be equal to rate of mass inflow through the surface. Equating the rate of mass inflow to the rate of increase of mass, we obtain ρ t ρ d ρ d ρ d t v v In we shrink the volume to a point, we obtain ρ Dρ t Dt ρv or v J. N. Reddy Mass-Momenta - 19

Conservation of Mass (continued) This is the invariant form (i.e., valid in any coordinate system and dimension) of the statement of the principle of conservation of mass, also known as the continuity equation (this author does not prefer this name). ρ t Steady-state flows : = ( ρv) Dρ Dt Incompressible fluids : = v J. N. Reddy Mass-Momenta - 2

Conservation of Mass (continued) ρ + ( ρv) = t Cartesian component form ( ) ˆ ˆ ˆ v i j k v ˆ ˆ ˆ ivyjvk y ( v) ( vy) ( v) y v v y v t y J. N. Reddy Mass-Momenta - 21

The Principle of Conservation of Mass in one dimension Infinitesimal volume, V A ρv ρv + + If no mass is created or destroyed inside V, the time rate of change of mass must equal to the rate of inflow of mass through the surface. J. N. Reddy 22

The Principle of Conservation of Mass in one dimension (continued) A A tt t v A v A t A A v A v A tt t t v t v t J. N. Reddy 23

Conservation of Mass (continued) ρ t + ( ρv) = Cylindrical component form 1 eˆ eˆ eˆ r r r 1 ( rvr ) v v v r r r 1vr 1 r v ( v ) t r r r J. N. Reddy Mass-Momenta - 24

Alternative form of REYNOLDS TRANSPORT THEOREM Replace the function (,) t with (,) t (,) t in the Reynolds transport theorem and obtain D D (,) t(,) td v d Dt ( t) ( t) Dt D D D v d d ( t) Dt Dt ( t) Dt where the equation resulting from conservation of mass is used in arriving at the last step. We have D Dt D (,) t(,) td d Dt J. N. Reddy Mass-Momenta - 25

CONSERVATION OF MASS: Material Description Let be an arbitrary material volume occupied by the body in the reference configuration, and be the volume occupied by the body in the current configuration. Conservation of mass states that if the mass is neither created nor destroyed during the motion from to, the mass of the material volume is conserved: ρ d ρd Since the volumes in the two configurations are related by djd, we obtain ρ J. N. Reddy Mass-Momenta - 26 Jρ

AN EXAMPLE Problem statement: Consider a water hose with conical-shaped nole at its end, as shown in the figure. (a) Determine the pumping capacity required in order The velocity of the water (assuming incompressible for the present case) eiting the nole be 25 m/s. (b) If the hose is connected to a rotating sprinkler through its base, determine the average speed of the water Leaving the sprinkler nole. J. N. Reddy Mass-Momenta - 27

Solution: AN EXAMPLE (a) The principle of conservation of mass for steady onedimensional flow requires 1Av 1 1 2Av 2 2 If the eit of the nole is taken as the section 2, we can calculate the flow at section 1 as (for an incompressible fluid, 1 2) 3 2 ( 21 ) 3 Q1 Av 1 1 Av 2 2 25. 25 m / s. 4 (b) The average speed of the water leaving the sprinkler nole can be calculated using the principle of conservation of mass for steady one-dimensional flow. We obtain 2Q1. 5 Q12 Av 2 2 v2 32 2 3 2 d ( 12. 51 ) m/s. J. N. Reddy Mass-Momenta - 28

BALANCE OF LINEAR MOMENTUM The time rate of change of total linear momentum of a given continuum equals the vector sum of all eternal forces acting on the continuum. This also known as Newton s Second Law. t d D t d ρfd ρvd Dt f d d d Dv nˆ d ρfd ρ d Dt Dv ρf ρ d Dt Newton's First Law. Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an eternal force. J. N. Reddy Mass-Momenta - 29

Conservation of Linear Momentum (continued) Vector form of the equation of motion σρf Dv ρ Dt Cartesian Component Form eˆ ˆ ˆ ˆ k ji j i ρfi i ee e k Dv ( ˆ iei) ρ Dt σji Dv σ i ji Dv eˆ ˆ ˆ i ρfiei ρ ei ρfi ρ Dt Dt j j i J. N. Reddy Mass-Momenta - 3

Conservation of Linear Momentum (continued) Cartesian Component Form (epanded form) σ ji j ρf i Dvi ρ Dt σ σ σ Dv ρf ρ Dt 11 21 31 1 1 1 2 3 σ σ σ Dv ρf ρ Dt 12 22 32 2 2 1 2 3 σ σ σ Dv ρf ρ Dt 13 23 33 3 3 1 2 3 J. N. Reddy Mass-Momenta - 31

EQUATIONS OF MOTION in Rectangular Cartesian System Alternative approach to the derivation: Sum all the forces on the infinitesimal block of dimensions d1, d2, and d3 d 1 d 3 1 3 ρ f 3 ρ f 1 d 2 ρ f 2 2 σ σ σ 33 σ 33 + d σ 23 σ13 3 σ + d + d 3 3 3 σ 32 σ 32 + d2 13 3 σ 12 σ σ 22 31 31 d1 1 σ 32 ρf1, ρf2, ρ f3= body force components (per unit mass) Origin is at the center of the parallelopiped σ σ 23 σ + d σ 33 11 11 1 1 σ 21 σ + 31 σ 11 σ 13 σ σ σ 2 σ + d 22 22 2 2 σ + d 12 12 2 2 σ + d 21 21 1 1 23 3 J. N. Reddy Mass-Momenta - 32

Conservation of Linear Momentum Special Cases Fluid Mechanics Solid Mechanics, for steady state v ρf ρ vv t v ρf ρ vv, t ρf v ρ t, for solid bodies Viscous stress tensor PI Hydrostatic pressure J. N. Reddy Mass-Momenta - 33

@¾ 11 + @¾ 21 + @¾ 31 @ 1 @ 2 @ 3 @¾ 12 + @¾ 22 + @¾ 32 @ 1 @ 2 @ 3 @¾ 13 + @¾ 23 + @¾ 33 @ 1 @ 2 @ 3 @¾ @ @¾ y @ @¾ @ + @¾ y @y + @¾ yy @y + @¾ y @y EQUATIONS OF MOTION Fluid Mechanics + @¾ @ + @¾ y @ + @¾ @ µ @v1 + ½f 1 = ½ @t + v @v 1 @v 1 @v 1 1 + v 2 + v 3 @ 1 @ 2 @ µ 3 @v2 + ½f 2 = ½ @t + v @v 2 @v 2 @v 2 1 + v 2 + v 3 @ 1 @ 2 @ µ 3 @v3 + ½f 3 = ½ @t + v @v 3 @v 3 @v 3 1 + v 2 + v 3 @ 1 @ 2 @ 3 µ + ½f @v = ½ @t + v @v @ + v @v y @y + v @v @ µ + ½f @vy y = ½ @t + v @v y @ + v @v y y @y + v @v y @ µ + ½f @v = ½ @t + v @v @ + v @v y @y + v @v @ J. N. Reddy Mass-Momenta - 34

@¾ rr @r EQUATIONS OF MOTION in cylindrical coordinates + 1 r = ½ @¾ rµ + @¾ r @ Ã @µ @vr @t + v r + 1 r (¾ rr ¾ µµ ) + ½f r @v r @r + v µ r @v r @µ + v @v r @ v2 µ r! @¾ rµ @r @¾ r @r @¾ µµ + 1 r @µ + @¾ µ @ + 2¾ rµ + ½f µ µ r @vµ = ½ @t + v @v µ r @r + v µv r + v µ r r @¾ µ + 1 r @µ + @¾ @ + ¾ r µ r @v = ½ @t + v @v r @r + v µ r + ½f @v @µ + v @v @ @v µ @µ + v @v µ @ J. N. Reddy Mass-Momenta - 35

EQUATIONS OF MOTION in cylindrical coordinates Alternative approach to the derivation: Sum all the forces on the infinitesimal block of dimensions dr, rd, and d ê ê θ θ σθ σ θ σθ + d σ θ + dθ ê r θ σ r σ + d dθ σθθ σθθ + dθ σ θ rr σ rθ r σr θ + dθ σ θ θr σ σ rθ σ r r σ σ r + dr r r σ σθr θθ σ σθr + dr θ σ r dr σ θ σ rr σ rr + dr r d σ r σ r + d J. N. Reddy Mass-Momenta - 36

EQUATIONS OF MOTION in material description f ( ) d fx ( ) d, ˆ T σn d PNd P d, D 2 u u vd d d Dt t t t 2 σdapda, dv J dv (or dj d ). 2 T u P f 2 2 u S T T F f 2 J. N. Reddy Mass-Momenta - 37 t t

EXAMPLE: COUETTE FLOW Flow of a viscous fluid between parallel plates (Couette Flow) y Assume: Flow steady state: () t incompressible: no body forces: v 2 3.v =, v f f f v v v 1 v 1 2 3 1 1 d σ yy σ σ y y σ σ σ v v σ σ y f( ) 1 2 f( y) J. N. Reddy Linear momentum 25

For viscous fluids, the total stresses are given by The viscous stresses are proportional to the gradient of the velocity field, and they are independent of and. @¾ @ @¾ y @ @¾ @ + @¾ y @y + @¾ yy @y + @¾ y @y COUETTE FLOW (continued) P, P, P yy yy,, y y y y + @¾ @ + @¾ y @ + @¾ @ ij µ + ½f @v = ½ @t + v @v @ + v @v y @y + v @v @ µ + ½f @vy y = ½ @t + v @v y @ + v @v y y @y + v @v y @ µ + ½f @v = ½ @t + v @v @ + v @v y @y + v @v @ J. N. Reddy Linear momentum 39

4 J. N. Reddy J.N. Reddy-ENGR214- L7(Feb. 8, 2) Thus the linear momentum equations become This implies that the pressure P is only function of. Thus, we have one equation Thus we have COUETTE FLOW (continued) P y P P,, y y P y v P v y y y 2 2 y 2 2 y

J. N. Reddy AN EXAMPLE Problem statement: Given the following state of stress in a kinematically infinitesimal deformation ( ), ij ji 2 11 21, 12 74 1 2 3, 13 11 32, 2 2 22 31 22 53, 23, 33 51 32 33 determine the body force components for which the stress field describes a state of equilibrium. Solution: The body force components are f [( 4 ) ( 4 ) ], 11 21 31 1 1 1 1 2 3 f [( 4 ) ( 4 ) ], f 12 22 32 2 2 2 1 2 3 3 13 23 33 [ 13] 4. 1 2 3 Mass-Momenta - 41

1. 2. EXERCISES Determine if the following velocity fields for an incompressible flow satisfies the continuity equation: v(, ), v (, ) where r r r 1 2 2 2 2 1 1 2 2 2 1 2 2 1 2 The velocity distribution between two parallel plates separated by distance b is y y y v( y) v c 1, vy, v, yb, b b b where y is measured from and normal to the bottom plate, is taken along the plates, v is the velocity component parallel to the plates, v is the velocity of the top plate in the direction, and c is a constant. Determine if the velocity field satisfies the continuity equation and find the volume rate of flow and the average velocity. J. N. Reddy Mass-Momenta - 42

3. EXERCISES If the stress field in a body has the following components in a rectangular Cartesian coordinate system 2 2 2 1 2 ( b 2) 1 2 2 1 2 2 [ ] a ( b 2) 1 ( 2 3b ) 2 3 2 2b 3 where a and b constants, determine the body force components necessary for the body to be in equilibrium. 4. For a cantilevered beam bent by a point load at the free end, the bending moment M about the y-ais is given by M = P. The aial stress is given by M P, I I where I is the moment of inertia of the cross section about the y- ais. Starting with this equation, use the two-dimensional equilibrium equations to determine stresses and and as functions of and. J. N. Reddy Mass-Momenta - 43