SOE3211/2 Fluid Mechanics lecture 2

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Mavis Higgins
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1 SOE311/ Fluid Mechanics lecture example

2 example Fluid ows governed by conservation of mass, momentum. We can use this to solve ow problems. Draw box (control volume) around region of interest, then equate mass ux into, out of region. { Integral of NSE

3 example We can also write the conservation of momentum in a similar form. The momentum of a small piece of uid will be udv. So the rate of change is ZZZ d udv dt What is the ux of momentum? V

4 example We can also write the conservation of momentum in a similar form. The momentum of a small piece of uid will be udv. So the rate of change is ZZZ d udv dt What is the ux of momentum? In fact it is (u)u:da through a bit of area da. V

5 example We can also write the conservation of momentum in a similar form. The momentum of a small piece of uid will be udv. So the rate of change is ZZZ d udv dt What is the ux of momentum? In fact it is (u)u:da through a bit of area da. Thus we can write d dt ZZZ V V ZZ X udv + (u)u:da = Forces S

6 example The forces are 1 body forces, eg. gravity, and surface forces { pressure, viscous stress, etc.. can be written as a stress, and so d dt ZZZ V udv + Can we make use of this? ZZ S (u)u:da = ZZ S :da

7 The forces are 1 body forces, eg. gravity, and surface forces { pressure, viscous stress, etc.. can be written as a stress, and so ZZ S (u)u:da = ZZ S :da example Can we make use of this? If we assume that the ow is steady, i.e. ZZZ d udv = dt V and choose our control volume V intelligently, then we can use this to calculate the forces on a body.

8 U o B A L δ C D u x y x example Assume u x = U sin y across CD What happens if we consider momentum uxes?

9 U o B A L δ C D u x y x example Assume u x = U sin y across CD What happens if we consider momentum uxes? In the x direction : ux AB F AB = (U ) U ( W ) = U W

10 ux CD : For a small element dy the momentum ux is (u x )u x (W dy) example

11 example ux CD : For a small element dy the momentum ux is (u x )u x (W dy) so integrating this F CD = W = W Z Z U = U W Z = U W = U W y u x dy y sin dy 1 1 cos y sin y dy

12 ux BC : We can guess that the uid owing out of BC shares the undisturbed ow velocity in the x direction. Hence F BC = (mass ux) BC U = U W 1 example

13 ux BC : We can guess that the uid owing out of BC shares the undisturbed ow velocity in the x direction. Hence F BC = (mass ux) BC U = U W 1 Thus (Net momentum ux) x = F in F out example = F AB F CD F BC = U W 1 = U W 1 1 1

14 example The only surface left is AD. There is no uid owing across this surface, so this must represent the force exerted on the plate AD by the uid ow. NB. We have implicitly assumed there are no viscous stresses of importance on AB, BC, CD.

15 example example Tests of a model underwater projectile in a water tunnel show that the velocity prole in a certain cross-section of the wake may be approximated to the shape of a cone. At this section the centreline velocity is equal to half the free stream velocity and the width of the wake is equal to twice the missile diameter. Use the analysis to estimate the drag coecient of the torpedo. U oo d d

16 The wake velocity is u u1 = r d example We work in cylindrical polar coordinates, so over the ends da = rdr. Consider mass uxes rst : () front = () sides + () back

17 The wake velocity is u u1 = r d example We work in cylindrical polar coordinates, so over the ends da = rdr. Consider mass uxes rst : i.e. Z d () front = () sides + () back Z d U1rdr = () sides + urdr

18 The wake velocity is u u1 = r d example We work in cylindrical polar coordinates, so over the ends da = rdr. Consider mass uxes rst : i.e. Rearanging, Z d () front = () sides + () back Z d U1rdr = () sides + urdr () sides = Z d (U1 u)rdr

19 example balance over control volume : Force = F front F sides F back = Z d U 1 rdr U 1 () sides Z d u rdr

20 balance over control volume : Force = F front F sides F back = Z d Substituting for () sides F D = U 1 rdr U 1 () sides and rearanging, we get Z d u (U1 u) rdr Z d u rdr example

21 example balance over control volume : Force = F front F sides F back = Z d Substituting for () sides F D = U 1 rdr U 1 () sides and rearanging, we get Z d u (U1 Now substituting for u we have Z F D = U 1 d 1 = U 1 r r 4 u) rdr r d rdr 4d d Z d = U 1 u rdr d 4

22 example Finally, the drag coecient is C d = F D 1 U 1 A M where A M is the frontal area of the projectile = d =4. So in this case C d = 1 This is often known as the

23 Measure wake velocities Draw apprpriate control volume Apply forms of mass, momentum equations Ignore viscous stresses example

24 example Measure wake velocities Draw apprpriate control volume Apply forms of mass, momentum equations Ignore viscous stresses Mathematics equivalent to integrating momentum decit over area of wake. In cylindrical coordinates In cartesian coordinates F z = Z Z F x W = u z (U1 u x (U1 u z ) rdr u x ) dy

25 example So far we have dened a volume ux V = u:da and a mass ux = u:da and a momentum ux m = (u) u:da

26 example So far we have dened a volume ux and a mass ux and a momentum ux V = u:da = u:da m = (u) u:da Dene uxes for any quantity of interest : Kinetic energy ke = Angular mom. 1 u u:da am = (r u) u:da

SOE3213/4: CFD Lecture 3

SOE3213/4: CFD Lecture 3 CFD { SOE323/4: CFD Lecture 3 @u x @t @u y @t @u z @t r:u = 0 () + r:(uu x ) = + r:(uu y ) = + r:(uu z ) = @x @y @z + r 2 u x (2) + r 2 u y (3) + r 2 u z (4) Transport equation form, with source @x Two