Mecanica De Fluidos

CHAPTER 3

Introduction to Fluids in Motion
3.1
pathline
streamline

3.2

Pathlines:

streakline

Release several at an instant in time and take a time exposure of the
subsequent motions of the bulbs.

Sreakline:
Continue to release the devises at a given location and after the last
one is released, take a snapshot of the “line” of bulbs. Repeat this
for several differentrelease locations for additional streaklines.
3.3
streakline
pathline
t =0

hose
time t

boy

3.4
y

streakline at t = 3 hr
pathline
t = 2 hr

streamlines
t = 2 hr
x

34

3.5

dx
= 2t + 2
dt
x = t 2 + 2t + c1

a) u =

v=

dy
= 2t
dt
y = t 2 + c2

y

streamlines
t=5s
(27, 21)

= y +2 y

39.8o

∴parabola.

∴ x − 2 xy + y = 4 y
2

2

(35, 25)

xb) x = t 2 + 2t + c 1 . ∴ c 1 = −8 , and c 2 = −4.
= y + 4 + 2( ± y + 4 ) − 8
∴parabola.

∴ x 2 − 2xy + y 2 + 8 x − 12 y = 0.

3.6
3.7

v
ˆ
ˆˆ
V = ui + vj + wk



v
ˆ
ˆ
dr = dxi + dy ˆ + d z k 
j


ˆ
ˆ ˆ ˆjˆ
using i × j = k , ˆ × i = −k .

Lagrangian: Several college students would be hired to ride bikes around the
various roads, making notes of quantities ofinterest.
Eulerian:

3.8

v
v
(V × dr ) z = udy − vdx

Several college students would be positioned at each intersection
and quantities would be recorded as a function of time.

a) At t = 2 a nd (0 ,0,0 ) V = 2 2 = 2 m / s .
At t = 2 a nd (1, −2 ,0) V = 3 2 + 2 2 = 3.606 m / s.
b) At t = 2 a nd (0 ,0, 0) V = 0.
At t = 2 and ( 1,−2 ,0 ) V = ( −2) 2 + ( −8 ) 2 = 8.246 m / s.
c) At t =2 a nd (0 ,0,0 ) V = (−4 ) 2 = 4 m / s.
At t = 2 a nd (1, −2 ,0)

3.9

(D)

V = 2 2 + ( −4 ) 2 + ( −4) 2 = 6 m / s.

( −51.4 × 10−5 ˆ )
j

A simultaneous solution yields n x = 4 /5 and n y = 3/5. (They must
both have the same sign.

3.10

v$
a) cos α = V ⋅ i / V = (1 + 2)/
v
$
V ⋅ n = 0.

∴nx =

3 2 + 2 2 = 0.832.

∴ α = 33.69o

3
ny = − nx
3n x + 2n y = 0 
$
2(3 $ + 2 $ ) ⋅ ( n x i + n y $ ) = 0.
i
j
j
∴
92
n x2 + n y2 = 1 
2
n x + nx = 1
4
2
3
1
$
$
, ny = −
or n =
( 2i − 3 $ ).
j
13
13
13

35

v$
b) cos α = V ⋅ i / V = −2 / ( −2) 2 + ( −8) 2 = −0.2425.
∴ α = 104 o
v
−2 n x − 8 n y = 0
n x = −4n y
$
$
V ⋅ n = 0. (−2i − 8 $ ) ⋅ (n x $ + n y $ ) = 0.
j
i
j
 ∴ 16n 2 + n 2 = 1
2
2
n x + ny = 1 
y
y
1
4
$, nx = −
or n =
17
17

∴ ny =

v$
c) cos α = V ⋅ i / V = 5 /
v
$
V ⋅ n = 0.

5 2 + ( −8) 2 = 0.6202.

$
( 5$ − 8 $ ) ⋅ ( n x i + n y $ ) = 0.
i
j
j

∴ α = −51.67 o

8
nx = ny
5n x − 8n y = 0
5
 ∴ 64
n 2 + n y2 = 1 
2
x
ny + n 2 = 1
y
25

5
8
1
$
, nx =
or n =
( 8$ + 5 $ ).
i
j
89
89
89

∴ny =
3.11

1
$j
( −4i + $ ).
17

[(x + 2)i$ + xtj$]× (dxi$ + dyj$) = 0.

v
v
a) V × dr = 0.

∴ ( x + 2) dy − xtdx = 0 or t

xdx
= dy .
x+2

xdx
= dy . t [ x − 2ln x + 2 ] = y + C.
x +2 ∫
2(1 − 2ln 3) = −2 + C.
∴ C = 0.8028.
t [ x − 2ln x + 2 ] = y + 0.8028

Integrate: t ∫

v
v
b) V × dr = 0.

[ xyi$ − 2 y $j ] × (dxi$ + dyj$) = 0.
2

dy
2dx
=− .
x
y
Integrate: 2 lnx = −ln ( y / C). 2ln(1) = − ln( −2 / C).
∴ xydy +2 y 2 dx = 0 o r
∴ C = −2.
v
v
c) V × dr = 0.

lnx 2 = − ln( y / −2).

[(x

2

∴ x 2 y = −2.

]

$
$
$
$
+ 4)i − y 2 tj × ( dxi + dyj ) = 0.

( x 2 + 4) dy + y 2 tdx = 0 o r

dy
tdx
=− 2.
2
x +4
y

t
x
2
1
1
1

 tan −1 + C = .
 tan −1 + C = − .
y

2
2
2
2
2
x


∴ C = −0.9636. yt  tan −1 − 0.9636 = 2


2
Integrate:

363.12

v
v
v
v
∂V
∂V
∂V
v ∂V
ˆ
ˆ
ˆˆ
ˆ
a=
+u
+v
+w
= 2 xy (2 yiˆ) − y 2 (2 xi − 2 yj ) = −16i − 8i + 16 j .
∂t
∂x
∂y
∂z

(C)

∴ a = 82 + 162 = 17.89 m/s
3.13

v
v
v
v
v
DV
∂V
∂V
∂V ∂V
a)
=u
+v
+w
+
=0.
Dt
∂x
∂y
∂z
∂t
v
v
v
v
∂V
∂V
∂V ∂V
$
$
$
$
b) u
+v
+w
+
= 2x ( 2i ) + 2 y ( 2 $ ) = 4 xi + 4 yj = 8i − 4 $
j
j
∂x
∂y
∂z
∂t
v
v...