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2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum Conservation of Linear Momentum Chapter 3 The linear momentum of a mass (m) with a velocity v vxi v y j vzk is defined to be mv . Important: linear momentum is a vector. Accumulationof linear momentum Linear momentum Linear momentum entering system leaving system within system during time period t during time period t during time period t P (mv) (mv) F out ext in t 1 2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum or, for a time period t: Psyst t t P syst t (mv)in t (mv)out t Fext t For a differential volume element xyz with density , the linear momentum P is given by ( xyz )v . 2 2001, W. E. Haisler 3 Chapter 3: Conservation of Linear Momentum Consider mass flow in the x direction with density and velocity vx flowing into the volume xyz through the surface yz during a time increment of t. Total mass entering and leaving through the “x face” in time t is: y y ( vx ) yzt x (x,y,z) z x z x ( vx ) yzt x+ x 2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum 4 For the “left x-face”, the total mass input is: ( vx ) yzt x and its momentum is: ( vxv ) yzt . x For the “right x-face”, the total mass output is: ( vx ) yzt xx and its momentum is: ( vxv ) yzt xx vxv momentum flux in x direction = momentum of mass flow in x direction (per unit area and per unit time) 2001, W. E. Haisler 5 Chapter 3: Conservation of Linear Momentum The momentum of the mass flow will tend to try to move the control volume (body) and therefore external surface tractions (forces/unit area) may be required to keep the body in equilibrium. For example, consider mass flowing through a pipe: pipe flow out flow in Think of the fluid within the pipe as the "system." For such a system, the pipe boundary must exert forces on the fluid boundary. These normal boundary forces on the fluid are what make the fluid turn as it flows through the pipe. The freebody of the fluid in the pipe is shown by: 2001, W. E. Haisler 6 Chapter 3: Conservation of Linear Momentum fluid boundary g fluid flow g fluid boundary forces on fluid boundary Note that the boundary forces are distributed along the pipe in some fashion that has yet to be determined. Because of viscosity effects, we will find that the pipe will also exert shear (or drag type) forces on the fluid which tend to slow the fluid near the pipe boundary. Traction forces will exist in both solid and fluid systems. Consider a frame problem from ENGR 211: 2001, W. E. Haisler 8 kN 7 Chapter 3: Conservation of Linear Momentum y 8 kN M x V 2m 2m 10 kN F 10 kN 5m Reactions at support F = 10 kN V = 8 kN M = -20 kNm 5m Freebody of structure t(n) n M F = = V Actual force distribution Idealized force and moment resultants At some point x Actual force distribution Traction Vector (force per unit area) At some point x in the beam, there is a distribution of normal (axial and shear forces) that can be represented by a "traction" vector. 2001, W. E. Haisler 8 Chapter 3: Conservation of Linear Momentum Now consider a general shaped body (system) with forces applied to its external surface as shown below: F3 F3 F1 F4 F4 F1 y F F1 -n x z A F1 A F1 F1 F2 n F1 Freebody 1 F2 F Freebody 2 With a cutting plane "A" which has an outward unit normal n on freebody 2, make two freebody diagrams. At the cut section we must place internal reactions (forces) as shown. In general, these forces will be forces distributed over the entire area of the cut section. For freebody 2, consider the force F applied to an area A that has unit normal n . Now define the traction t(n ) on this surface as 2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum t(n ) 9 F lim A0 A The notation t(n ) means the traction vector (vector force per unit area) acting on the surface with unit normal n . This vector is not necessarily normal to the surface. t(n ) is shown below: t ( n ) A n A n t (n ) t n lim F F lim t n A A 2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum 10 For freebody 1, we note that the force at a given point must be equal and opposite to the force on freebody 2 and hence the tractions must also be equal and opposite. The normal on FB 1also points in the opposite direction to FB 2. The traction on the surface with unit normal n is denoted as t(-n ) and we have that t(-n ) t(n ) Now lets consider a differential volume system in the fluid (or solid). Just as for the pipe example, the differential system (xyz) can also have tractions on its boundary. Define the tractions on the x-face as shown below: 2001, W. E. Haisler 11 Chapter 3: Conservation of Linear Momentum y t( i ) t( i ) y x t( i ) y x x t( i ) x (x,y,z) t( i ) z x x z components of t( i ) t(i ) t(i ) x i t(i ) y j t(i ) z k z The traction vector on the left x face is denoted by: t (i) x right face by t (i) xx . and on the Note: Traction is defined as positive when it acts in the positive coordinate direction (just like fluid velocity v ). 2001, W. E. Haisler 12 Chapter 3: Conservation of Linear Momentum The traction vector on the +i face may be written in terms of its components as: t(i) t(i) i t(i) j t(i) k . The x y z subscripts on the vector components mean the following: 1st subscript: face the force acts on (i means x) 2nd subscript: direction of the force. For example, t(i) is the traction on the i face acting in the y y direction. For the tractions acting on the -i and +i faces, the sum of the external forces (on x-faces only) is given by [ t(i) t(i) ]yz . x xx 2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum 13 We should also consider body forces that act on the volume. Assume that the body force per unit mass (usually due to gravity) is given g g xi g y j g zk . The total body force (vector) on the volume during a time increment t is ( gxyz)t 2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum 14 Now, add the momentum terms in the y and z directions: In the above, we have looked at momentum associated with mass flow in the x direction only. We can write similar momentum terms for mass flow in the y and z directions. Similarly, we can define the forces on the volume due to tractions on the y and z faces. Now, the change in momentum with respect to time: ( v v )xyz t t t The conservation of linear momentum (including linear momentum terms, tractions and boundary forces for each of the 3 coordinate directions) becomes 2001, W. E. Haisler ( v Chapter 3: Conservation of Linear Momentum 15 v ) xyz t t t =[( vxv ) (vxv ) ] yzt x xx + [(v yv ) (v yv ) ] xzt y yy + [(vz v) (vz v) ] xyt z zz + [(t(i) t(i) )xz]t )yz] t + [(t( j) t( j) y yy x xx + [(t(k ) t(k ) ) xy] t z zz + ( g xyz) t Divide by xyzt and take the limit of each term; x, y, z, t 0, to obtain 2001, W. E. Haisler 16 Chapter 3: Conservation of Linear Momentum ( v v ) ( v v ) ( v v ) y ( v ) x z x y z t t tj t i k g x y z Use the calculus product rule on the 1st and 2nd term in above equation. The second term (double underlined) becomes ( v v ) ( vxv ) ( vz v ) y x y z ( v ) ( vx ) ( vz ) y v v v v v v v y y z z x y z x x 2001, W. E. Haisler 17 Chapter 3: Conservation of Linear Momentum and the first term (single underline) becomes ( v ) v v t t t Group these two expansions as follows ( vx ) ( v y ) ( vz ) v v x y z t v v v vx v y vz v x y z t ( vx ) ( v y ) ( vz ) Conservation of mass is: x z z t so that the two triple-underlined terms (multiplied by v ) are conservation of mass and sum to zero. 2001, W. E. Haisler 18 Chapter 3: Conservation of Linear Momentum Thus, after incorporating conservation of mass (continuity) into the linear momentum equation, and rearranging the result, we obtain: t t t v v v ( i ) ( j ) v v x v y v z g (k ) x y z x y z t Note that the above is a vector equation. Note that since conservation of mass was incorporated into the above, this linear momentum equation automatically satisfies conservation of mass. Also, the term in brackets can be written in vector notation: [...](v )v so that t(i) t( j) t(k ) v (v )v g x y z t