÷
QUESTION
-
2 Ca ) :
J = C- Gxy ) it C-3×2+3511
The velocity components for this
flow field
are :
U = -
Gxy
us . in:::?¥¥I
!
: .
u -
-
0¥ = -
6×9
→
Velocity potential , Io =
f -
a
bxydx
constant
÷¥:÷¥÷
= -
35g + f
←
am
Question 2 Ca ) :
-
fcy) ]
=
§ f- 3×3 + = v
3×2
It
dy
= -
t
§ fry)
: .
N =
-34435 = +
Fy fog)
f-
'
fog ) 3
y
=
f- Cy) ±
f 35 dy
fcy) ±
yrs
'
: .
Io = -
35g t
y Ans .
*
Questioning
superposition of a GOURI and
A UNIFORM flow .
Is given :
O
knew = Ur Sino
tzmno
Note that :
source
m
-
-
q → source strength
uniform
is Draw the resulting flow of superposition
U ay
-#
→
I
→ x
'
I
TF
streamlines
" "
Flow around half Rankine body
-
Question 2 Cb ) :
di ) As
given the stream function 4 .
Knew = Ur sin O t
Igo
To determine velocity potential . Io ;
we must determine velocity components .
fH÷÷÷÷÷÷÷.÷÷
Ur = JIO
.
.
Ur new,
=
toffee Ifa fur sina.info ]
-
-
✓
Ur ,
new = U
Lif
cos O t
Oftener
r
=
-
Question 2 Cb ) :
M
i .
d Ionw U cos O t -
=
g- 2M r
Ionw =
f @ as a +
Ifip
dr
Ur cost
Ionew +
IF In Ans
=
r .
*
-
iii ) Determine distance b .
Refer to the
Figure
2lb) ; stagnation
located at the point b)
point is co -
. ,
D- =
tf C 180 )
At the stagnation point ; ur and
Uo are all zero .
Ur , new
= U cos O t
Mz = O
Question
-
2lb )
m
iiil Up new
= U cos Ot -
= 0
,
217N
Substitute 0=4 ;
U cos CM ) t I = 0
217 r
-
U = - I
217T
: . r = Me metre .
2174
where r is the distance from
origin to any points (in this case
-
stagnation point ) co b) .
-
r
t¥y
-
-
Inu
-
-
I
:b =
Iftu * Ans -
2 Cb)
Question
-
:
Cv ) Determine the stream function
passes the stag point
.
.
.
'
.
4
new
= Ur sin O +
2M¥ a
ht stagnation point ,
r =
stfu ; -0=17
Oz
4kg )G Fact)
ICFTU
- +
MYS Ans
Ys tag Mz
.
