{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solnsheet4

# solnsheet4 - Fluid Dynamics 3 Solutions to Sheet 4 C D 1(i...

This preview shows pages 1–2. Sign up to view the full content.

Fluid Dynamics 3 - Solutions to Sheet 4 1. (i) For the given flow field, u θ = f ( r ) in cylindrical polars. Thus, applying the formula from chapter 0, ∇ · u = 1 r ∂u θ ∂θ = 0 . (ii) We have to solve the steady Euler equation ( ∇ · u ) u = −∇ p/ρ. According to chapter 0, the convective term on the left is ( ∇ · u ) u = u 2 θ r ˆ r = f 2 r ˆ r . Thus p can only depend on r , so that p = p ( r r , and the required equation follows. (iii) According to chapter 0, ∇ × u = ˆ z r ∂rf ∂r , which vanishes if f = A/r . From (ii), we have p ( r ) = ρA 2 r 3 , and so p ( r ) = ρA 2 2 r 2 + p atm . 2. (i) Conservation of mass is volume flux in equals volume flux out, so Ud = U 1 b 1 + U 2 b 2 (ii) Use the momentum integral theorem. Need to choose a control surface - this should be a boundary where we know stuff about the solu- tion. So follow example in lectures and take it to be around the flow cutting the inflow and outflows far away from the impingement. Let S be the contour ABCDEF in the figure: Deal with pressure term first: A B C D E F integraldisplay S p n d S = integraldisplay S ( p p atm ) n d S + integraldisplay S p atm n d S The last term is zero by the divergence theo- rem (see lectures). The first term is zero along all parts of S apart from the section DE . The other term is integraldisplay S ρ u ( u . n )d S which is zero on AF , DE , BC since u . n = 0 (there is no flow across these boundaries). So

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

solnsheet4 - Fluid Dynamics 3 Solutions to Sheet 4 C D 1(i...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online