Conservation of Linear Momentum for a Differential Control Volume


 Richard Pitts
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1 Conservation of Linear Momentum for a Differential Control Volume When we applied the rateform of the conservation of mass equation to a differential control volume (open sstem in Cartesian coordinates, we obtained the continuit equation: ( ρv ( ρv ( ρv ρ z t z = + + We now want to do the same thing for conservation of linear momentum. As with the development of the continuit equation, we will first appl the rateform of conservation of linear momentum to a finitesize open sstem (a control volume and then consider what happens in the limit as the size of the control volume approaches zero. (1 Conservation of Linear Momentum for a TwoDimensional Flow For simplicit, we will consider a twodimensional, unstead flow subject to a bod force b as shown in Figure 1. The bod force, the densit, and the velocit all are field variables and ma var with position (, and (t. In words, the time rate of change of the linear momentum within the sstem equals the net transport rate of linear momentum into the sstem b eternal forces and b mass flow. To appl this principle, we select a small finitesize volume in the flow field, z, where z is the depth of the small volume into the paper, i.e. in the direction of the zais. Because of the multiple momentum transfers in a moving fluid, we will onl focus on onl one component of linear momentum at a time. We will begin with the component of linear momentum, referred to hereafter as the momentum. To identif the transfers of momentum, we enlarge the small volume and show all transfers of momentum on the diagram of the control volume (see Figure 2. Linear momentum can be transported into the sstem b surface forces acting on all four sides, b bod forces within the sstem, and b mass flows across all four sides of the control volume. Densit field: ρ = ρ(, t, Velocit field: V = V ˆi+ V ˆj where V = V (,, t and V = V (,, t and î, ĵ = unit vectors for, aes Bod force field: b= b ˆi+ b ˆj Figure 1 A differential control volume for a twodimensional flow Navier Stokes_01d.doc Prepared b D. E. Richards Page 1 of 9
2 mv + F, + F, mv F bod, F, + mv + F, mv Ke Force transport Mass transport Figure 2 Transport rates of the component of linear momentum for the differential control volume Writing the rateform of the conservation of linear momentum equation for the momentum gives: dp = Fnet, + F surface net, + ( mv ( bod + mv net dt net Rate of change of momentum inside the sstem of momentum into the sstem b surface forces of momentum into the sstem b bod forces of momentum of momentum into the sstem into the sstem b mass flow b mass flow in the direction in the direction The lefthand side of Eq. (2 represents the rate of change (storage or accumulation. The terms on the righthand side correspond to the transport rates of momentum illustrated in Figure 2. (2 Rate of change of momentum inside the control volume The momentum inside the control volume can be written as follows using the meanvalue theorem from calculus: P ss, = VρdV = ρv z (3 z where the tilde notation ρ V indicates an average or mean value. The meanvalue theorem sas that the integral of a continuous function ρv over a volume equals the product of the volume and the mean value of the continuous function within the volume; thus, a simple product can replace the integral. To find the rateofchange of the momentum inside the control volume, the lefthand side of Eq. (2, we evaluate the derivative Navier Stokes_01d.doc Prepared b D. E. Richards Page 2 of 9
3 ( dp d ρv = ( ρv z = ( z (4 dt dt t Because we are holding,, and z constant during our differentiation the result is a partial derivative. of momentum b eternal forces The net transport rate of momentum b bod forces is related to b, the component of the bod force acting in the positive direction, and can be written as net, bod z of momentum into the sstem b bod forces ρ ( F = b ρdv = b z The gravitation field of the earth produces the most common bod force; however, other bod forces can also be important. For eample, bod forces produced b electrostatic fields can be used to enhance the flow of hot air over baked goods and speed cooking time. A bod of fluid in a uniforml accelerating reference frame, for instance a glass of pop sitting on the floor of our car as it accelerates, can also be studied b treating the acceleration as a bod force. The net transport rate of momentum b surface forces is related to eternal forces acting on each surface of the control volume. For a twodimensional flow, the surface force for an surface can be decomposed into two components, a normal force and a shear force. Each force, shear or normal, is defined as the integral of the appropriate surface stress over the surface area. The component of the surface force acting on the surface at +, is wrtten as F, = σda= σ z (6 + + z where the minus sign occurs because the normal stress σ b convention points out from the surface. The component of the surface force acting on the surface at, is written as F, = σda= σ z (7 z The component of the surface force acting on the surface at +, is written as F, = σ da= σ z (8 + + z where the shear stress σ points in the positive direction on a surface whose normal vector points in the positive direction. (5 Navier Stokes_01d.doc Prepared b D. E. Richards Page 3 of 9
4 The component of the surface force acting on the surface at, is written as F, = σ da= σ z (9 z where again the minus sign comes from the sign convention on the shear stress σ. Combining these forces to find the component of the net surface force acting on the control volume gives { } + { + } F = F F + F F net, surface,,,, of momentum into the sstem b surface forces { } = { σ z σ z } σ z σ z = σ ( σ z + σ + σ ( z + σ σ ( = z + ( z σ σ = + ( z (10 of momentum b mass flow Finall, we will eamine the net transport rate of momentum b mass flow. Mass flow occurs on all four surfaces of the control volume. To begin, we will determine the momentum tranport rate of momentum into the sstem with mass flow at and + : ( mv = ( mv ( mv net + of momentum into the sstem b mass flow in the direction = ( ρv ( z V ρv z V in out = ( z ( ρvv ( ρvv + ( ρvv = ( z + (11 Now we can determine the transport rate of momentum into the sstem with mass flow at and + : Navier Stokes_01d.doc Prepared b D. E. Richards Page 4 of 9
5 : ( mv = ( mv ( mv of momentum into the sstem b mass flow in the direction net + ( ρ ( ρ = V z V V z V in out = ( z ( ρv ( V ρvv + ( ρvv = ( z + (12 Putting it all together (the first time Now that we have developed epressions for each momentum transport or storage rate, we can substitute these values back into Eq. (2 repeated here for reference: dp dt Rate of change of momentum inside the sstem = F + net, surface net, bod of momentum into the sstem b surface forces F of momentum into the sstem b bod forces ( mv ( mv + + Making the substitutions for each term we have the following epression: net net of momentum of momentum into the sstem into the sstem b mass flow b mass flow in the direction in the direction ( ρv ( σ z σ = + ( z t rate of change of momentum inside the sstem net transport rate of momentum into the sstem b surface forces (component of surface forces ( ρvv ( z net transport rate of momentum into the sstem b mass flow in the direction + ( ρb ( z + net tranport rate of momentum into the sstem b bod forces (component of bod forces ( ρvv + ( z net transport rate of momentum into the sstem b mass flow in the direction (13 Dividing through b the volume z and taking the limit as 0, 0, and z 0, the average terms indicated b the tilde notation ( approach the value at the point (,,t. Navier Stokes_01d.doc Prepared b D. E. Richards Page 5 of 9
6 This gives the conservation of momentum for a differential control volume (twodimensional flow: ( ρv σ σ ( ρvv ( ρvv = + + ρb + t (14 A similar epression can be developed for the conservation of momentum for a differential control volume (twodimensional flow: ( ρv σ σ ( ρvv ( ρvv = + + ρ g + t (15 Revisiting the surface forces Before we proceed, we need to reeamine the surface forces. Surface forces include both pressure forces or viscous forces. The pressure force is a normal force produced b the local pressure P acting over a surface. The local pressure is an intensive propert of the substance and is a normal stress whose value is independent of orientation. Viscous forces are produced b viscous stresses τ ij that depend on fluid viscosit and velocit gradients in the flow. Viscous stresses strongl depend on orientation and for an surface can be decomposed into both normal stresses and shear stresses. For a twodimensional flow the surface stress σ ij can be separated into a pressure term P and a viscous stress term τ ij as shown below: σ σ σ ij = σ σ P + τ τ P 0 τ τ = τ P τ = 0 P + τ τ + Pressure Stress Viscous Stress Now the surface stress terms can be rewritten to clearl separate the pressure and viscous stresses. For eample in Eq. (14, the surface stress terms can be rewritten as follows: σ σ P + = ( P + τ + = + + Pressure Viscous (16 (17 Navier Stokes_01d.doc Prepared b D. E. Richards Page 6 of 9
7 Putting all together (again Substituting the appropriate value for each stress back into Equations (14 and (15 and rearranging terms gives the following epressions conservation of linear momentum for a differential control volume in twodimensional flow: component component ( ρv P ( ρvv ( ρvv = ρb + t ( ρv P ( ρvv ( ρvv = ρb + t (18 Note that there is now a clear distinction between the pressure terms and the viscous stress terms. To evaluate the viscous stresses we need to have a constitutive model for the fluid that describes how the shear stresses are related to the velocit gradients in the flow and to the fluid propert known as viscosit. When a flow is inviscid, it has no viscosit and the viscous stress terms disappear. This greatl simplifies the mathematics and can give useful results under some conditions. NavierStokes Equations for Incompressible, TwoDimensional Flow Man important flows are essentiall incompressible and this leads to significant simplifications. Additionall, we will restrict ourselves to Newtonian fluids. With these assumptions, we reproduce the NavierStokes equations for incompressible, twodimensional flow. To develop these equations, we first assume that the flow is incompressible and consider the consequences. The continuit equation reduces, Eq. (1 becomes 0 = + (19 In Eq. (18, the conservation of linear momentum equation, incompressible flow brings the densit outside of the derivatives as shown below: Navier Stokes_01d.doc Prepared b D. E. Richards Page 7 of 9
8 component ρ ρb ρ ( VV ( VV P ρ = ρb ρ + t component ( VV ( VV P = t Further simplifications occur b epanding the terms in brackets and appling the incompressible continuit equation. For the term in brackets in the component equation in Eq. (20, we have the following simplification ( ( VV VV + = V + V + V + V = V + + V + V = 0 Wh? = V + V (20 (21 Substituting this result and its counterpart for the component back into Eq. (20 gives differential equation for conservation of linear in an incompressible, twodimensional flow: component P ρ = ρb ρ V + V t component P ρ = ρb ρ V + V t Please note the long list of qualifiers for these two equations. All of them are important and if used under an other conditions, Equation (22 will be in invalid. A Newtonian fluid is a fluid in which viscous stresses are proportional to the rate of angular deformation within the fluid. The constant of proportionalit is µ the dnamic viscosit. The viscous stresses for an incompressible, twodimensional flow of a Newtonian fluid become τ = 2 µ, τ = 2 µ, and τ = τ = µ + (23 Using these results, the viscous stress terms in Eq. (22 are replaced b terms containing the fluid viscosit and velocit gradients. Note viscous stresses depend on both the fluid and the flow. (22 Navier Stokes_01d.doc Prepared b D. E. Richards Page 8 of 9
9 The net step is to substitute the viscous stress terms in Eq. (23 back into Eq. (22 and use the incompressible continuit equation, Eq. (19, to simplif the epressions. Once done we recover the NavierStokes equations. The component of the NavierStokes Equation for a twodimensional, incompressible flow is shown below: 2 2 ρ = P + µ V + V 2 2 t Rate of change of momentum of momentum b pressure forces of momentum b viscous forces ρ V V b ρ V V + + of momentum b bod forces of momentum b mass flow This arrangement shows a clear connection to our original conservation of linear momentum equation; however, the more traditional arrangement found in most tetbooks moves the mass transport term to the other side of the equal sign. When arranged in this form we have the result we have been seeking (24 NavierStokes Equations for a TwoDimensional, Incompressible Flow component 2 2 P V V ρ + ρ V + V = + ρ b + µ t component 2 2 P V V ρ + ρ V + V = + ρ b + µ t (25 Following a similar process for threedimensions, we can derive the full set of equations. These are stated below without eplanation or development: NavierStokes Equations for an Incompressible Flow component: component: zcomponent: ρ + ρ V V V P V V V + V + V z = + ρ b + µ t z z V V V V V V V ρ + ρ V P V Vz ρ b µ + + = t z z z z z z z z z ρ + ρ V V V P V V V + V + V z = + ρ bz + µ t z z z Navier Stokes_01d.doc Prepared b D. E. Richards Page 9 of 9
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