{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

complete_sol3 - At every point P(r,z at every instant t the...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
At every point P(r, θ ,z) at every instant t, the velocity components u r , u θ and u z must satisfy this eq. The two u r can be combined into one term to give an equivalent alternate form, 1 r ∂r ( ru r ) + 1 r ∂u θ ∂θ + ∂u z ∂z = 0 . Question 3: Problem 4.20 Given 2D velocity field ~ U = u ˆ i + v ˆ j where u = K (1 - e - αy ) , and v = V 0 at y = 0 (all x) Conservation of mass requires ∂u ∂x + ∂v ∂y = 0 since ∂u ∂x = 0, then we must have ∂v ∂y = 0. Thus, v = F ( x ) at most. At y = 0, we know that v = V 0 for all x, therefore F ( x ) = V 0 and v = V 0 here. The solution is u = K (1 - e - αy ) v = V 0 K must have units of velocity (m/s). ( αy ) must be dimensionless, thus α must have units of ( 1 m ). Question 4: Problem 4.25 Conservation of mass in cylindrical coordinates is 1 r ∂r ( rv r ) + 1 r ∂v θ ∂θ = 0 Given v r = Kcosθ 1 - b r 2 then rV r = Krcosθ - Kbcosθ r ∂r ( rv r ) = Kcosθ + Kbcosθ r 2 Given v θ = - Ksinθ 1 + b r 2 = - Ksinθ - Kbsinθ r 2 then ∂v θ ∂θ = - Kcosθ - Kb cosθ r 2 Sub into LHS of cons. of mass, Kcosθ r + Kbcosθ r 3 - Kcosθ r - Kbcosθ r 3 = 0 LHS = RHS, the continuity is satisfied.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}