This preview shows page 1. Sign up to view the full content.
Problem 6.14
[Difficulty: 3]
Given:
Velocity field
Find:
The acceleration at several points; evaluate pressure gradient
Solution:
The given data is
q2
m
3
s
m
⋅
=
K1
m
3
s
m
⋅
=
ρ
1000
kg
m
3
⋅
=
V
r
q
2
π
⋅
r
⋅
−
=
V
θ
K
2
π
⋅
r
⋅
=
The governing equations for this 2D flow are
The total acceleration for this steady flow is then
r
 component
a
r
V
r
r
V
r
∂
∂
⋅
V
θ
r
θ
V
r
∂
∂
⋅
+
V
θ
2
r
−
=
a
r
q
2
K
2
+
4
π
2
⋅
r
3
⋅
−
=
θ
 component
a
θ
V
r
r
V
θ
∂
∂
⋅
V
θ
r
θ
V
θ
∂
∂
⋅
+
V
r
V
θ
⋅
r
+
=
a
θ
0
=
Evaluating at point (1,0)
a
r
0.127
−
m
s
2
=
a
θ
0
=
Evaluating at point (1,
π
/2)
a
r
0.127
−
m
s
2
=
a
θ
0
=
Evaluating at point (2,0)
a
r
0.0158
−
m
s
2
=
a
θ
0
=
From Eq. 6.3, pressure gradient is
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics

Click to edit the document details