Boundary Layers

Question 1

What is a boundary layer?

Answer 1

Layer around ball with flow flowing around it where flow cannot be, of width 𝛿

Question 2

What is the equation for the boundary layer 𝛿?

Answer 2

𝛿/d = sqrt(μ/ρv0d) = 1/(Re^1/2), where d is the diameter of the ball and v0 is the velocity of the flow

Question 3

What is the pressure on the surface of the ball like at different points?

Answer 3

On left side at 270, P> at 315 > at 360

Question 4

What is dP/dx greater/less tan upstream/downstream?

Answer 4

dP/dx < 0 upstream, >0 downstream

Question 5

What does the upstream pressure gradient do?

Answer 5

Acts to accelerate the flow along surface in direction of streamlines. Called favourable pressure gradient

Question 6

What does the downstream pressure gradient do?

Answer 6

Opposes the flow: adverse pressure gradient.

Question 7

What is the N-S equation near the boundary of a slab?

Answer 7

(v.∇)v(x) = -1/ρ *dP/dx + μ/ρ *d^2v(x)/dy^2, at all y=0, v= 0, so d^2v(x)/dy^2 = 1/μ *dP/dx

Question 8

What do we assume in this N-S equation and what do we do then?

Answer 8

Assume that dP/dx = A, so can then integrate twice to get an equation, then use boundary conditions to find the constants

Question 9

How does v(x) vary with the constant A?

Answer 9

If A<0, v(x) is +ve, if A>0, v(x) is negative if A >= 2v0/𝛿^2

Question 10

What does the graph of v(x) against x look like for favourable pressure and for adverse pressure?

Answer 10

Favourable pressure: starts at 0 and curves upwards. Adverse pressure: starts at 0, goes negative then curls back round to positive and then curves upwards

Question 11

What is the separation point?

Answer 11

The point at which the flow may be halted or reversed due to adverse pressure. Can happen when A = 2v0/𝛿

Question 12

What is it called when there is a separation point?

Answer 12

Boundary separation.

Question 13

What does the flow diagram with boundary separation look like?

Answer 13

Normal flow but with 1 stagnation point at left side of ball, and then 2 lung shaped flows on right side of all, with separation points x(s) at top and bottom of ball

Question 14

When does turbulence develop?

Answer 14

If energy at wavenumber k = 2π/λ is E(k), then turbulence exhibits cascade (cascade to higher k. In high Re flow boundary separation leads to turbulence downstream of the separation.

Question 15

For potential flow, what is the pressure at θ = 0 and θ = π on the balls surface?

Answer 15

P(θ = 0) = P0 +1/2 ρv^2, P(θ = π) = P0 +1/2 ρv^2

Question 16

For real flow, what is the pressure at θ = 0 and θ = π on the balls surface?

Answer 16

P(θ = 0) = P0, P(θ = π) = P0 +1/2 ρv^2

Question 17

What does the pressure at θ = 0 and θ = π for real flow suggest?

Answer 17

That there is a net force the right: in cylinder frame is a net drag.

Question 18

Does laminar flow give drag?

Answer 18

No

Question 19

What does a region of adverse pressure downstream lead to?

Answer 19

A stagnation point and boundary separation

Question 20

What does the diagram of irrotational flow look like and what does this mean?

Answer 20

Simple flow above and below with 2 stagnation points. Surface pressure fully recovers downstream, so no drag.

Question 21

With boundary separation and turbulence, what does the diagram of flow look like?

Answer 21

Flow above and below, one stagnation point and then turbulence lines coming from top and bottom of ball and following flow.

Question 22

With boundary separation and turbulence, what is the pressure like at different points on the ball?

Answer 22

Pa > Pb, Pc -> drag, where a is on left of ball, b is on top and c is on right

Question 23

What is the equation for force on surface area S of the ball?

Answer 23

F = b*ρv^2/2 *S, where b is the drag coefficient

Question 24

What is the first way to reduce drag?

Answer 24

Narrower region of turbulence reduces drag. Do this by delaying the separation by streamlining the flow.

Question 25

What is the second way to reduce drag?

Answer 25

Make the boundary layer turbulent. Turbulent eddies have a net effect of recirculating flow into boundary layer. This delays boundary separation so x(s) moves further downstream (e.g. gold ball with dimples).

Question 26

What is the drag crisis?

Answer 26

Where the Reynolds number Re = ρvD/μ increases and there is a sudden drop in drag coefficient - as v increases suddenly the drag drops.

Question 27

What does the graph of drag coefficient against Re look like for a smooth sphere?

Answer 27

Straight then sudden drop in drag at certain Re

Question 28

What is the explanation for the drag crisis?

Answer 28

As Re increases the boundary layer thickness reduces. Shear stresses increases and boundary layer becomes turbulent. Separation points move further downstream and reduce drag.