Download - Lecture Exam Practice
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K&C page 77, exercise 3
y,v
C (r=1)
x,u
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K&C page 77, exercise 3
x,u
y,v u= ay = a r sinCirculation:
= C u d s= 02
a sin2
d
=[ a 12
x cos x sin x ]0
2
= a
d s r [ i sin j cos ]d
ds, d
C (r=1)
Stokes theorem: surface integral of vorticity = line integral of velocitySometimes one of the two is far easier to calculate!
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K&C page 77, exercise 3
x,u
y,v
Circulation: = C dA= 12 a = a
u= ay = a r sinCirculation:
Vorticity:
z= a
= C u d s= 02
a sin2
d
=[ a 12
x cos x sin x ]0
2
= a
d s r [ i sin j cos ]d
z= v x
u y
ds, d
C (r=1)
Stokes theorem: surface integral of vorticity = line integral of velocitySometimes one of the two is far easier to calculate!
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K&C page 77, exercise 4
u i
x j= 1
2
u i
x j
u j
xi
1
2
u i
x j
u j
xir ij = ije ij
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K&C page 77, exercise 4
e11 = u1
x1= u
x= 0
e 22= u 2 x2
= v y
= 0
Linear strain rates:
ui x j
= 12
ui x j
u j xi
12
ui x j
u j
xi
r ij = ije ij
Shear strain rates:
e12 = 12 u1 x2
u 2 x1
= 12
U b
Vorticity:
z= r 21= 12 u2 x1
u1 x2
= U b
e 21=12
U b
z= r 12=U b
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K&C page 77, exercise 4
u i x j
= 12
u i x j
u j xi
12
ui x j
u j xi
r ij = ije ij
strain along principal axis (e12
); followed by rotation ( z)
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K&C page 77, exercise 4
u= Uy /b= y= U
2
2bC
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K&C page 77, exercise 6
Streamlines:dxu
= dyv
1 t x
dx= 2 t 2y
dy
1 t log x =12
2 t log 2y log f t NB: instant. streamlines, e.g. t = const.; ln f(t) for convenience
log x1 t =log 2y 1 t /2 log f t We combine last 2 terms in one log, so it becomes a multiplication
x1 t = f t 2y 1 t /2 This is our answer for the streamlines: x expressed as a function y (or vice versa)
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K&C page 77, exercise 6
Pathlines: x t = 0t dx
dt dt = 0
t u dt
dxdt
= x1 t
dx x
= dt 1 t
log x= log 1 t log f x0
x= f x0 1 t
x= x0 1 t
t = 0 x= x0,so f x0 = x0
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K&C page 77, exercise 6
Same for y: y t = 0t dy
dt dt = 0
t v dt
dydt
= 2y2 t
dy y
= 2dt2 t
log y= 2log 2 t 2log g y0
y= g y02 2 t 2
y= y02 /4 2 t 2
t = 0 y= y0, so g y02= 1 /4 y0
2
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K&C page 77, exercise 6
y= y02 /4 2 t 2
x= x0 1 t
The pathline for a particle that is at x0,y
0at t=0 is given by:
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K&C page 135, exercise 6
const. acceleration a
Steady, so no viscous deformation;only hydrostatic forces; Navier-Stokesequation reduce to:
0= 1 p x
a
g x
x
= tan
0= 1 p z
gcompareclassichydrostatics:
tan = ag
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K&C page 335, exercise 1
xy
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K&C page 335, exercise 1
D u D t
= 1 P g 2 u
0= g sin 2 u
y2
Fully developed, steady, no driving pressure
0= g sin y u
yC 1
0= g2
sin y2 u C 1 y C 2
u= g y2
2sin C 1 y C 2
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K&C page 335, exercise 1
D u D t
= 1 P g 2 u
0= g sin 2 u
y2
Fully developed, steady, no driving pressure
0= g sin y u
yC 1
0= g2
sin y2 u C 1 y C 2
u= g y2
2sin C 1 y C 2
No slip: u = 0 at y = h
0= g h2
2sin h2 C 2
Stress-free at surface:
du /dy = 0 at y= 0
C 1= 0
C 2=g h 2
2sin
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K&C page 335, exercise 1
u=g y2
2 sin C 1 y C 2 C 1= 0 C 2= g h2
2 sin
u= g y2
2sin g h
2
2sin
u= g
2sin h2 y2 QED
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K&C page 335, exercise 1
u= g2
sin h2 y2
Q= 0h
udy = 0h g
2sin h2 y2 dy =[ g sin
2h2 y y
3
3]0
h
= g sin h3
3
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K&C page 335, exercise 1
u= g2
sin h2 y2
W = u y y= h
W =h g sin = gh sin NB : / =
u y = g 2ysin2 = g y sin
NB: shear stress is related to column of water that drives the flow
y = h
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K&C page 335, exercise 1
u= g2
sin h2 y2
W = u y y= h
W =h g sin = gh sin NB : / =
u y = g 2ysin2 = g y sin
NB: shear stress is related to column of water that drives the flow
Question: would this also hold for rivers?
y = h
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Outer flow is irrotational,So no net viscous forces:Inviscid Bernoulli
Show ( x u =0)or recognize inviscid vortex
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Pressure in outer flow:(Bernoulli, compare withfar away u=0, p = p 0)
p F(z) =p 0 + g(h-z) u 2
The stress is continuous atsurface, so pressure fromboth sides should be equal(no pressure drop):
p F = p c
p 0+ g(h-z)=p 0+ cg(h-z) u2
Solving for u 2 gives answer
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Compare the two expressions:
u =2 r
u2= 2g ' h z
2 r
2
= 2g ' h z
outside
surface
R z 2=2
8 2 g ' 1
h z
Surface r=R, solve:
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Velocity scale: V =h
Froude influence of hydrostatic pressureversus flow-induced pressure (e.g. Y>>1, hardlyany influence of gravity)
u2
V 2= u
2
2 /h2= 2 g ' h z2 /h2
u ' = 2Y
1 z , with Y =2
g ' h 3
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SEE KUNDU:Stokes' second problem
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Surface tension creates a pressuredifference; Pressure differenceused to fill plenum cannot be higherthan the pressure due to surfacetension; so: calculate surfacetension, compare with Poiseuille
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