Intermediate fluid mechanics by Su.

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Example text

As we did there, we will look for a similarity solution, of the form √ Γ = Γ0 f (η), where η = r/ νt. (7) Plugging (7) into (4), we get f ∂η ∂r ∂η =ν f ∂t 2 +f ∂ 2 η 1 ∂η − f , ∂r 2 r ∂r (8) using the notation df /dη = f . For our definition of η, we have ∂η r = − 1/2 3/2 , ∂t 2ν t ∂η 1 = √ , and ∂r νt 2 ∂ η = 0. ∂r 2 Inserted in (8): −f Multiplying through by r ν √ r 2ν 1/2 t3/2 =ν f 1 1 1 − f √ . νt r νt νt on both sides: −f r2 r = f √ − f , giving 2νt νt 2 η −f = f η − f , or 2 η 1 −f − =f .

Xj (4) But observe that we can write ∂ (uj ui ) = ∇ · (ui u) ∂xj (think about this), and then we have for (4) ρ V ∂ (uj ui ) = ρ ∂xj V ∇ · (ui u) dV = ρ S (ui u) · n ˆ dS = ρ S ui uj nj dS (5) where we have applied the implied summation to the u · n ˆ dot product. For the right hand side of (2), our integral is − V ∂p dV = − ∂xi S p ni dS = − S pˆ n dS (using an identity in the text derived from the divergence theorem). The result of integrating (2) is thus − S pˆ n dS = ρ S u(u · n ˆ) dS which is exactly what you’d find from a control volume analysis.

Dz dz There is one theorem that will be particularly useful to us. It states that if a function w = f (z) = g + ih is differentiable at some point z0 , then ∂g ∂h = ∂x ∂y and ∂g ∂h =− , ∂y ∂x dw ∂g ∂h = +i . dz ∂x ∂x and (1) This can be verified easily for simple functions, say w = z 2 . 2 The complex potential Why is this interesting? Remember that we earlier said that for irrotational flow, we could write u = ∇φ where φ was called the velocity potential. Thus, in the two-dimensional case, u= ∂φ ∂φ ,v = ∂x ∂y (for 2-D, irrotational flow).

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