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MIT Department of Mechanical Engineering
2.25 Advanced Fluid Mechanics
Problem 10.16
This problem is from “Advanced Fluid Mechanics Problems by A.H. Shapiro and A.A. Sonin
Consider the twodimensional, steady, nonviscous flow of an incompressible fluid, with no body forces
present. The flow has vorticity.
a) Show that the vorticity remains constant on each streamline.
b) Show that the stream function is governed by the equation
� 3 3 � � 3 3 �
∂ψ ∂ ψ + ∂ ψ = ∂ψ ∂ ψ + ∂ ψ (10.16a)
3 2 3 2
∂y ∂x ∂x∂y ∂x ∂y ∂x ∂y
c
2.25 Advanced Fluid Mechanics 1 Copyright @ 2010, MIT
Vorticity Theorems A.H. Shapiro and A.A. Sonin 10.16
Solution:
a) First let us consider how many components the vorticity vector ω¯ has for this two dimensional flow. The
vorticity vector is defined
� !� � !� � !�
ω¯ = ∇×v¯ = ∂w − ∂v eˆ + ∂u − ∂w eˆ + ∂v − ∂u eˆ (10.16b)
∂y ∂z x ∂z ∂x y ∂x ∂y z
Since the zvelocity is zero, w = 0, and there are no gradients along the zdirection, ∂ = 0, we see that only
∂z
the component of vorticity along the zdirection can be nonzero, ω¯ = ω eˆ . Hence we only need to consider
z z
how the quantity ω changes along a streamline to prove that the vorticity remains constant on it.
z
The governing equations of motion for an incompressible fluid in a twodimensional flow are
∇·v¯= 0 (10.16c)
Dv¯ 1 2
=¯g − ∇p + ∇ v¯ (10.16d)
Dt ρ
Or alternatively
∂u ∂v
+ = 0 (10.16e)
∂x ∂y
!
� 2 2 �
∂u ∂u ∂u 1 ∂p ∂ u ∂ u
+ u + v = g − + + (10.16f)
x 2 2
∂t ∂x ∂y ρ∂x ∂x ∂y
� 2 2 !�
∂v ∂v ∂v 1 ∂p ∂ v ∂ v
+ u + v = g − + + (10.16g)
y 2 2
∂t ∂x ∂y ρ∂y ∂x ∂y
Although we have been told that the flow is steady and free from body forces, we seek to derive as general
a result as possible, so we include them in the following derivation. First let us take the curl of our two
dimensional momentum equation, Eq. (10.16d):
Dv¯ 1 2
∇× =¯g − ∇p + ∇ v¯ (10.16h)
Dt ρ
This operation is also the same as taking the cross derivatives of Eq. (10.16f) and (10.16g) and then sub
tracting them. More precisely
Dv¯ 1 2 � ∂ ∂ �! Dv¯ 1 2
∇× =¯g − ∇p + ∇ v¯ = eˆ − eˆ · =¯g − ∇p + ∇ v¯ eˆ (10.16i)
Dt ρ ∂x y ∂y x Dt ρ z
c
2.25 Advanced Fluid Mechanics 2 Copyright @ 2010, MIT
Vorticity Theorems A.H. Shapiro and A.A. Sonin 10.16
Taking the first term on the right hand side of Eq. (10.16i), we have
( � 2 2 !)�
∂ ∂v ∂v ∂v 1 ∂p ∂ v ∂ v
+ u + v = g − + + (10.16j)
y 2 2
∂x ∂t ∂x ∂y ρ∂y ∂x ∂y
which is equal to
∂2v ∂u∂v ∂2v ∂v∂v ∂2v ∂g 1 ∂2p � ∂3v ∂3v �!
+ + u + + v = y − + + (10.16k)
2 3 2
∂t∂x ∂x ∂x ∂x ∂x ∂y ∂x∂y ∂x ρ ∂x∂y ∂x ∂x∂y
Taking now the first term on the right hand side of Eq. (10.16i), we have
( � 2 2 !)�
∂ ∂u ∂u ∂u − 1 ∂p + ∂ u + ∂ u (10.16l)
+ u + v = g
x 2 2
∂y ∂t ∂x ∂y ρ∂x ∂x ∂y
which is equal to
!
2 2 2 2 � 3 3 �
∂ u ∂u∂u ∂ u ∂v∂u ∂ u ∂g 1 ∂ p ∂ u ∂ u
+ + u + + v = x − + + (10.16m)
2 2 3
∂t∂y ∂y ∂x ∂x∂y ∂y ∂y ∂y ∂y ρ∂x∂y ∂x ∂y ∂y
Combining Eq. (10.16k) and (10.16m) into Eq. (10.16i), we note that the crossderivatives of pressure cancel
and that if g¯ is spatially uniform any gradients in g are identically zero, and we obtain
∂2 2 � 2 2 �! � 2 2 !� � 3 3 3 3 �!
v ∂ u ∂u∂v ∂u∂u ∂v∂v ∂v∂u ∂ v ∂ u ∂ v ∂ u ∂ v ∂ u ∂ v ∂ u
− + − + − +u − +v − = − + −
2 2 3 2 2 3
∂t∂x ∂t∂y ∂x∂x ∂y∂x ∂x∂y ∂y∂y ∂x ∂x∂y ∂x∂y ∂y ∂x ∂x ∂y ∂x∂y ∂y
(10.16n)
This result may be suitably rearranged to obtain
! ! ! ! !
� � � � � � � � � �
∂ ∂v − ∂u + ∂v ∂u+ ∂v − ∂u ∂u+ ∂v + u ∂ ∂v − ∂u + v ∂ ∂v − ∂u =
∂t ∂x ∂y ∂x ∂x ∂y ∂y ∂x ∂y ∂x ∂x ∂y ∂y ∂x ∂y
� 2 � !� 2 � !!��
∂ ∂v − ∂u + ∂ ∂v − ∂u
2 2
∂x ∂x ∂y ∂y ∂x ∂y
Recalling our definitions from Eq. (10.16b) and (10.16e), we substitute these expressions into the above
equation to obtain
∂ω ∂ω ∂ω � ∂2ω ∂2ω !�
z + u z + v z = z + z (10.16o)
2 2
∂t ∂x ∂y ∂x ∂y
c
2.25 Advanced Fluid Mechanics 3 Copyright @ 2010, MIT
Vorticity Theorems A.H. Shapiro and A.A. Sonin 10.16
or alternatively, we can write Eq. (10.16o) as
Dω
z 2
Dt = ∇ ωz (10.16p)
This result reveals, that if a flow is nonviscous, (i.e. = 0), Dω /Dt = 0 and hence the vorticity of a
z
material element will not change as it moves along a streamline, so the vorticity remains constant on each
streamline.
b) Recall that the stream function ψ(x,y) is related to the velocity field by the equations
∂ψ ∂ψ
u = & v = − (10.16q)
∂y ∂x
If we substitute Eq. (10.16q) into our definition for ω , we have the relation
z
∂v ∂u ∂2ψ ∂2ψ
ω 2
z = − = − 2 − 2 = −∇ ψ (10.16r)
∂x ∂y ∂x ∂y
Substituting this result into Eq. (10.16o) with = 0 and ∂/∂t = 0 since we have steady flow, we obtain
! !
� 2 2 � � 2 2 �
∂ψ ∂ ∂ ψ ∂ ψ ∂ψ ∂ ∂ ψ ∂ ψ
− 2 + 2 + 2 + 2 = 0 (10.16s)
∂y ∂x ∂x ∂y ∂x∂y ∂x ∂y
which can be rewritten to obtain our final result
� 3ψ ∂3ψ �! ∂ψ �∂3ψ ∂3ψ !�
∂ψ ∂ + = + (10.16t)
3 2 3 2
∂y ∂x ∂x∂y ∂x ∂y ∂x ∂y
D
Problem Solution by TJO, Fall 2010
c
2.25 Advanced Fluid Mechanics 4 Copyright @ 2010, MIT
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