Question 5,
exam: 14_March_2013
A ) (2 points)
The following Cauchy’ s momentum equation is is written in the vector form.
r
r
DV
ρ
= ∇ ⋅ σ + ρg ( *)
Dt
Here, ρ is the density of the fluid,
r r
V = V (t , x, y , z ) = (u (t , x, y , z ), v (t , x, y , z ), w(t , x, y , z ) ) is the velocity vector field.
σ=
is the Cauchy stress tensor
r
and g = ( g x , g y , g z ) is the gravity acceleration vector,
The equation (*) can be expressed as 3 scalar equations ( x-component , y component and zcomponent).Write the x-component of the equation (*), free from the operators ∇ and div.
B (2 points) We consider a scalar field f ( r , θ , z ) given in cylindrical coordinates, where
x = r cos θ ,
y = r sin θ , z = z , and a new ( not standard) basis
r r r r
r r r r
r r r
B=( e1 , e2 , e3 ), where e1 = i − k e2 = 2 j , e3 = i + k .
Find the expression for the gradient, grad ( f ( r , θ , z )) , in cylindrical coordinates for the basis
r r r
B=( e1 , e2 , e3 ), that is:
r
r
r ∂f ∂f
∂f
,
and
.
∂r ∂θ
∂z
express grad ( f ( r , θ , z )) in terms of r ,θ , z , e1 , e2 , e3 ,
C ( 2 points)
We consider an incompressible, Newtonian steady ( components du not depend on time t, and
consequently, derivatives with respect to time t are 0) flow. The velocity field V=(ur, uθ, uz)
for this flow is given:
5
5
u r = 0 , uθ =
V= (0, ,0) , i. e.
and u z = 0 .
r
r
The gravity acceleration vector in our model is g =(0, 0, -g) (where g=9,81 m/sec2).
Use the Navier- Stokes equations ( where ( ρ and μ are constants) to find the pressure field
P (r,θ,z).
Cylindrical coordinates ( r , θ , z ) :
We consider an incompressible , isothermal Newtonian flow (density ρ =const, viscosity
r
μ =const), with a velocity field V = (u r , uθ , u z ).
Incompressible continuity equation:
1 ∂(ru r ) 1 ∂ (uθ ) ∂u z
+
+
=0
eq a)
∂z
r ∂r
r ∂θ
r-component:
⎛ ∂u r
∂u r uθ ∂u r uθ2
∂u ⎞
⎜
ρ⎜
+ ur
+
−
+ u z r ⎟⎟
∂r
∂z ⎠
r ∂θ
r
⎝ ∂t
eq b)
⎡ 1 ∂ ⎛ ∂u r ⎞ u r
∂P
1 ∂ 2ur
2 ∂uθ ∂ 2 u r ⎤
=−
+ ρg r + μ ⎢
− 2
+
⎜r
⎟− 2 + 2
⎥
2
∂r
r
∂
r
∂
r
∂
θ
r
r
∂
θ
r
∂z 2 ⎦
⎝
⎠
⎣
θ -component:
u ∂u
uu
∂u
∂u ⎞
⎛ ∂uθ
+ ur θ + θ θ + r θ + u z θ ⎟
r ∂θ
r
∂r
∂z ⎠
⎝ ∂t
ρ⎜
⎡ 1 ∂ ⎛ ∂uθ ⎞ uθ
1 ∂P
1 ∂ 2 uθ
2 ∂u r ∂ 2 uθ ⎤
+
+
=−
+ ρg θ + μ ⎢
⎜r
⎟− 2 + 2
⎥
r ∂θ
r ∂θ 2 r 2 ∂θ
∂z 2 ⎦
⎣ r ∂r ⎝ ∂r ⎠ r
z-component:
u ∂u
∂u
∂u ⎞
⎛ ∂u
ρ⎜ z + ur z + θ z + u z z ⎟
∂r
∂z ⎠
r ∂θ
⎝ ∂t
⎡ 1 ∂ ⎛ ∂u z ⎞ 1 ∂ 2 u z ∂ 2 u z ⎤
∂P
=−
+ ρg z + μ ⎢
+
⎜r
⎟+ 2
⎥
2
∂z
∂z 2 ⎦
⎣ r ∂r ⎝ ∂r ⎠ r ∂θ
eq d)
eq c)
SOLUTION
A ) (2 points)
The following Cauchy’ s momentum equation is a vector partial differential equation
r
r
DV
ρ
= ∇ ⋅ σ + ρg ( *)
Dt
where ρ is the density of the fluid,
r r
V = V (t , x, y , z ) = (u (t , x, y , z ), v (t , x, y , z ), w(t , x, y , z ) ) is the velocity vector field.
σ=
is the Cauchy stress tensor
r
and g = ( g x , g y , g z ) is the gravity acceleration vector,
is written in the vector form.
The equation (*) can be expressed as 3 scalar equations ( x-component , y component and zcomponent).Write the x-component of the equation (*), free from the operators ∇ and div.
A) Answer:
∙
= g +
+
+
or
∙(
+ +
+
) = g +
+
+
B (2 points) We consider a scalar field f ( r , θ , z ) given in cylindrical coordinates, where
x = r cos θ ,
y = r sin θ , z = z , and a new ( not standard) basis B=(
r r r r
r r r r
r r r
e1 , e2 , e3 ), where e1 = i − k e2 = 2 j , e3 = i + k . Find the expression for the gradient,
r r r
grad ( f ( r , θ , z )) , in cylindrical coordinates for the basis B=( e1 , e2 , e3 ), that is:
r
r
r ∂f ∂f
∂f
,
and
.
∂r ∂θ
∂z
express grad ( f ( r , θ , z )) in terms of r ,θ , z , e1 , e2 , e3 ,
Solution (B)
In x,y, z variables we have
∂f ∂f ∂f
∂f r ∂f r ∂f r
grad ( f ) = ( , , ) =
i+
j+ k
∂x ∂y ∂z
∂x
∂y
∂z
( eq1)
For cylindrical coordinates we have x = r cos θ ,
First we write the derivatives
y = r sin θ ,
z=z
∂f
∂f
, and
in r ,θ coordinates (the variable z is in both
∂y
∂x
coord systems) .
Solving the following system for
∂f
∂f
, and
,
∂y
∂x
∂f ∂f
∂f
=
⋅ cos θ +
⋅ sin θ
∂r ∂x
∂y
∂f
∂f ∂f
⋅ (r cos θ )
⋅ (−r sin θ ) +
=
∂y
∂θ ∂x
we get
∂f
∂f sin θ
= cos θ
−
∂x
∂r
r
∂f
∂f cos θ
= sin θ
+
∂y
∂r
r
∂f
∂θ
∂f
∂θ
(**)
We substitute the derivatives (**) in ( eq1) and get
∂f sin θ ∂f r
∂f cos θ
)i + (sin θ
−
+
r
∂r r r ∂θ
∂r
r r
Now we express i , j , k as a linear combinations of
grad ( f ) = (cos θ
r r r
From e1 = i − k
∂f r ∂f r
) j + k (* * *)
∂z
∂θ
r r r
e1 , e2 , e3 and substitute them into (***) .
r r r r
r
e2 = 2 j , e3 = i + k . we find
r r r
r er1 + er3
r r
i =
, j = e2 / 2, k = (e3 − e1 ) / 2
(eq b)
2
r r r
Then we put i , j , k from ( eq b) into (***) and get
r r
∂f sin θ ∂f e1 + e3
∂f cosθ ∂f r
∂f r r
−
)
+ (sin θ
+
)e2 / 2,+ (e3 − e1 ) / 2
grad ( f ) = (cosθ
∂r
2
∂r
∂z
r ∂θ
r ∂θ
r
After collecting components for e1
r
r
e2 , e3 we have
r
∂f sin θ ∂f ∂f e1
∂f cosθ ∂f r
)e 2 / 2
grad ( f ) = (cosθ
−
− ) + (sin θ
+
∂r
r ∂θ ∂z 2
∂r
r ∂θ
r
∂f sin θ ∂f ∂f e3
+ (cosθ
−
+ )
∂r
r ∂θ ∂z 2
C ( 2 points)
We consider an incompressible, Newtonian steady ( components du not depend on time t, and
consequently, derivatives with respect to time t are 0) flow. The velocity field V=(ur, uθ, uz)
for this flow is given:
5
5
u r = 0 , uθ =
V= (0, ,0) , i. e.
and u z = 0 .
r
r
The gravity acceleration vector in our model is g =(0, 0, –g) (where g=9,81 m/sec2).
Use the Navier- Stokes equations ( where ( ρ and μ are constants) to find the pressure field
P (r,θ,z).
Solution:
i) Since u r = 0 ,
uθ =
5
and u z = 0 . (according to c3), continuity equation in cylindrical
r
coordinates
1 ∂(ru r ) 1 ∂(uθ ) ∂u z
+
+
=0
∂z
r ∂r
r ∂θ
is identically fulfilled.
ii) The r-component of the Navier-Stokes equation reduces to:
uθ2
∂P
∂P
25
25 ∂P
−ρ
=−
⇒ −ρ 3 = −
⇒ρ 3 =
r
∂r
∂r
∂r
r
r
From the θ-component we have:
( eq r-c)
θ -component gives:
⎡ 1 ∂ ⎛ ∂uθ ⎞ uθ ⎤
⎡1 ∂ ⎛ − 5 ⎞ 5 ⎤
1 ∂P
1 ∂P
+ μ⎢
+ μ⎢
⎜r
⎟− 2 ⎥ ⇒0= −
⎜
⎟− 3⎥
r ∂θ
r ∂θ
⎣ r ∂r ⎝ r ⎠ r ⎦
⎣ r ∂r ⎝ ∂r ⎠ r ⎦
∂P
1 ∂P
1 ∂P
⎡5 5⎤
⇒0=
⇒0=−
+ μ⎢ 3 − 3 ⎥ ⇒ 0 = −
r ∂θ
∂θ
r ∂θ
r ⎦
⎣r
The z-component reduces to:
∂P
∂P
∂P
( eq z-c)
0=−
+ ρg z ⇒
= ρg z ⇒
= − ρg
∂z
∂z
∂z
In order to find the pressure P we solve ( eq r-c), ( eq θ-c) and ( eq z-c) :
0=−
25
∂P
=ρ 3 ,
∂r
r
∂P
= 0,
∂θ
From these equations we get
P = −ρ
25
− ρgz + C
2r 2
∂P
= − ρg ,
∂z