Conservation of Linear Momentum

Transcription

1 Conservation of Linear Momentum Once we have determined the continuit equation in di erential form we proceed to derive the momentum equation in di erential form. We start b writing the integral form of the momentum equation: t VdV ~ + ~ V ( ~ V ˆn)dA = ~ F As in previous derivation we will do our analsis for a 2D situation and then we will etend our results to 3D. Then, let s start b defining our di erential element with sides, and. The first term of the momentum equation: t ~ VdV Represents the amount of momentum accumulated within the control volume, and for a di erential element, as the one we are dealing with: t ~ VdV = t (~ V ) Now, we work on the second term: ~ V ( ~ V ˆn)dA 1

2 Which represents the momentum crossing the boundaries of the control volume, as discussed in the previous chapter this equation is a vector equation and therefore it has three components (one in each direction). Let s do the analsis of this second term using a 2D di erential element and then we etend our results to 3D. apple v ~ V + (v~ V ) u( ~ V ) apple u ~ V + (u~ V ) v( ~ V ) So in the direction we will have: : u( ~ V )+ # u V ~ + (u~ V ) ] Simplifing we get: : (u~ V ) Similarl in the and directions we will get: : (v~ V ) and : (w~ V ) 2

3 Thus, the second term on the LHS of the momentum equation as applied to our di erential control volume is: ~ V ( ~ V ˆn)dA = (u V ~ # ) + (v~ V ) + (~ V ) Therefore, the LHS of the momentum equation applied to a di erential element can be written as: # t (~ V )+ (u~ V ) + (v~ V ) + (~ V ) Let s anale the epression within the square brackets to obtain a simplified version of it. B epanding this epression we can write: V t ~ + V ~ t + V ~ (u) + u~ V + V ~ (v) + v V ~ + V ~ (w) + w V ~ Now, we rearrange this epression to obtain: ~V t + V ~ (u) + V ~ (v) + V ~ (w) + V ~ t + u~ V + v V ~ ++w V ~ Factoring out ~ V from the first four terms: ~V t + (u) + (v) + (w) + V ~ t + u~ V + v V ~ ++w V ~! As we found when deriving the continuit equation: t + (u) + (v) + (w) =0 Thus, the LHS of the momentum equation can be presented as: V ~ t + u~ V + v V ~ ++w V ~! 3

4 or: V ~! t +(~ V r) V ~ or also: D~ V Dt # Once we have determined the LHS of the momentum equation, we continue with the RHS: ~ F In this part of our analsis we need to consider all the forces acting in our di erential control volume. As we know, we have bod forces and the surface forces. The onl bod force that we take into consideration is the weight of the fluid inside the control volume ( W). W W = ~g Where ~g is a vector representing the acceleration of gravit. Now, we take into consideration the surface forces. We need to consider normal and shear stresses and the forces theses stresses eert on the surfaces of the control volume. Let s start b considering the forces produced b normal stresses. Again, we do our analsis for a 2D di erential element and then we etend the results to a 3D element. Then, the forces due to the normal stresses acting on the element are: : +( + : +( + ) = ) = 4

5 + + Similarl, for the direction we have: : Therefore, the forces due to the normal stresses acting on 3D di erential element are: + + Net, we anale the forces created b the shear stresses acting on the boundaries of the control volume. We first sketch all forces due to shear stresses acting on the direction on our control volume. + + Thus, the summation of shear forces in the direction is: :

6 Which can be written as: : + In a similar wa, the forces due to shear stresses in the other two directions are obtained to be: : + And: : + If we now write together all the forces acting on the element, i.e. bod, normal and shear forces, we obtain: apple F ~ = ~g + î + ˆ + ˆk + + î + + ˆ + + ˆk Now that we have all the terms of the momentum equation, we proceed to write them into: t VdV ~ + ~ V ( ~ V ˆn)dA = ~ F To obtain: V ~ t + u~ V + v V ~ + w V ~ +! + = ~g + + ˆ î + ˆk Eliminating since it is common to ever term we obtain: V ~ t + u~ V + v V ~ + w V ~ +! = (g î + g ˆ + g ˆk)+ + + ˆ ˆk + î 6

7 This is a vector equation and can be written in terms of its components as: u : t + uu + v u + w u = g v : t + u v + v v + w v = g w : t + uw + v w + w w = g These are the momentum equations in di erential form. Later we will stud some particular cases of these equations. 7