Basics of Fluid Mechanics, 2014a

168 CHAPTER 5. MASS CONSERVATION
Example 5.12:
Calculate the velocity, U x for a cross section of circular shape (cylinder).
Solution
The relationship for this geometry needed to be expressed.
The length of the line Y (x) is
√
(
Y (x) =2r 1 − 1 − x ) 2
(5.XII.a)
r
A
− x
Y(x) x
α
(r − x)
U e
y
r
A e
This relationship also can be expressed in the term of Fig. -5.10. Circular cross section
α as
for finding U x and various cross
sections.
Y (x) =2r sin α (5.XII.b)
Since this expression is simpler it will be adapted. When the relationship between radius
angle and x are
x = r(1 − sin α)
(5.XII.c)
The area A − x is expressed in term of α as
A − x =
(α − 1 )
2 , sin(2α) r 2
(5.XII.d)
Thus the velocity, U x is
A e
A
(α − 1 2 sin(2α) )
r 2 U e + U x 2 r sin αh=0
(5.XII.e)
U x = A (
e r α −
1
2 sin(2α)) U e
(5.XII.f)
A h sin α
Averaged velocity is defined as
U x = 1 ∫
UdS
(5.XII.g)
S S
Where here S represent some length. The same way it can be represented for angle
calculations. The value dS is r cos α. Integrating the velocity for the entire container
and dividing by the angle, α provides the averaged velocity.
which results in
Example 5.13:
U x = 1
2 r
∫ π
0
A e
A
U x =
r
h
(
α −
1
2 sin(2α)) U e rdα
tan α
(π − 1) A e r
4 A h U e
End Solution
(5.XII.h)
(5.XII.i)