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Fundamentals of Aerodynamics 6th Edition Anderson Solutions Manual
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CHAPTER2
2.1
If p = constant = Poo
However, the integral of the surface vector over a closed surface is zero, i.e.,
Hence, combining Eqs. (1) and (2), we have
2.2
t/f’,Per W11 I/
/LL.L/tL
fQ.
I
/
7
7
7
7
7
7
i.OU1er Wa I I
7’t!
7
f.t {‘X )
7
7
/
19
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Denote the pressure distributions on the upper and lower walls by pu(x) and p e (x) respectively.
The walls are close enough to the model such that Pu and p 1 are not necessarily equal to Pooยท
Assume that faces ai and bh are far enough upstream and downstream of the model such that
and v = 0
and ai and bh.
Take they-component ofEq. (2.66)
# (p v.dS) v – H(p dS)y
L =-
S
abhi
The first integral= 0 over all surfaces, either because V ยท ds = 0 or because v = 0. Hence
->
fJ (pdS)y
L’ = –
b
J Pu dx – J p dx]
= – [
โขbM
h
l;
โข
1
Minus sign because y-component is in downward
Direction.
Note: In the above, the integrals over ia and bh cancel because p = Poo on both faces. Hence
h
2.3
b
L’ =
J pf dx – J Pu dx
dy
v
cy/(x 2 +y 2 )
u
2
dx
=-=
dy
dx
y
x
2
ex I ( x + y )
=
y
x
-=-
The streamlines are straight lines emanating from the origin.
streamline pattern for a source, to be discussed in Chapter 3.)
2.4
dy
v
x
dx
u
y
-=-==
(This is the velocity field and
y dy = – x dx
20
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y2 = -x2 + const
x2 + y2 = const.
The streamlines are concentric with their centers at the origin. (This is the velocity field and
streamline pattern for a vortex, to be discussed in Chapter 3.)
2.5 From inspection, since there is no radial component of velocity, the streamlines must be
circular, with centers at the origin. To show this more precisely,
u = – v e sin = – er
rr = – cy
x
v = Ve cos 8 = er – = ex
r
dy v
x
= =
dx u
y
&2 + x2 = const.J
This is the equation of a circle with the center at the origin. (This velocity field corresponds to
solid body rotation.)
2.6
dy v
y
=
=
dx u
x
~.
)) ~. ~
dy
dx
=
y
x
~
fn y = X ยฃn X + C1
~ (/’
y = C2/x
x
f
The streamlines are hyperbolas.
21
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__,
1o
1 OV
cr
roe
8
In polar coordinates: V ยท V = — (r Vr) + – – –
r
Transformation:
x = r cos 8
y = r sine
Vr = u cos 8 + v sin 8
ve = – u sin e + v cos 8
22
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u=
v=
ex
er cosB
c cosB
(x2 + y2)
?
c
r
cy
(x2 + y2)
er sine
r1
=
c sine
r
. 2
c
v r = -c cos2e + -c sm
e=r
r
r
Ve= – ~ cose sine+ ~ cose sine= 0
r
r
. ->
18
1 8(0)
V V = — (c) + – – = 0
r
r !JB
a:
(b) From Eq. (2.23)
V x V = ez [O + 0 – O] = ~
The flowfield is irrotational.
2.8
u=
c sine
cy
er sine
=
=
?
(x2 +y2)
r
c
v=
-ex
er cose
c cose
– =
?
(x2+y2)
r
c
Yr= ~ cose sine – ~ cose sine= 0
r
r
ยท2
c
2
c
v e = – -c sm
e – – cos e = – r
(a)
vยท
r
v L~a: (O)+
=
r
r
18(-c/r) =O+O=~
r
8()
23
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(b)
V’x
V= e–+ [O'(-c/r) _~_!b'(O)]
z
V’ x
a
r2
r
oe
V=~except at the origin, where r = 0. The flowfield is singular at the origin.
2.9
Ve=cr
V’x
V= e__,, [o(c/r) +er_..!_ o(O)J
z
a
–7
r
roe
–7
ez (c + c – 0) = 2c ez
The vorticity is finite. The flow is not irrotational; it is rotational.
2.10
c
b
a..
24
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Mass flow between streamlines = Ll f;t
Lllf/ =pV Lln
Ll f;t = (-p Ve) M + p Yr (r8)
Let cd approach ab
(1)
Also, since f;t = f;t (r,9), from calculus
–
3,;,.
3,;,.
a
8e
d If/ = _ ‘ f ‘ dr + _ ‘ f ‘ de
(2)
Comparing Eqs. (1) and (2)
8f;t
-pVe=-
a
and
3
f;t
P rV r = 8B
or:
8f;t
pVe=- –
2.11
u = ex =
a
8
1// : ljf = cxy + f(x)
(1)
07
v = – cy = –
8
lf : lJf = cxy + f(y)
&
(2)
25
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Comparing Eqs. (1) and (2), f(x) and f(y) =constant
ljf = c x y + const. I
(3)
u = ex = Olf/ : ~ = cx2 + f(y)
&
(4)
v = – cy = Olf/ : ~ = – cy2 + f(x)
(5)
0’
Comparing Eqs. (4) and (5), f(y) = – cy2 and f(x) = cx 2
(6)
Differentiating Eq. (3) with respect to x, holding jf = const.
0 =ex dy + cy
dx
or,
dy)
=-y/x
(dx lf=const
(7)
Differentiating Eq. (6) with respect to x, holding~= const.
dy
0=2cx-2cy dx
or,
( dy)
= x/y
dx ยข=const
(8)
Comparing Eqs. (7) and (8), we see that
Hence, lines of constant jf are perpendicular to lines of constant~ยท
26
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2.12.
The geometry of the pipe is shown below.
t{ =/CJC> ..-w /:s-ec.,..(I
As the flow goes through the U-shape bend and is turned, it exerts a net force Ron the internal
surface of the pipe. From the symmetric geometry, R is in the horizontal direction, as shown,
acting to the right. The equal and opposite force, -R, exerted by the pipe on the flow is the
mechanism that reverses the flow velocity. The cross-sectional area of the pipe inlet is nd2/4
where dis the inside pipe diameter. Hence, A= nd2/4 = n(0.5)214 = 0.196m2 . The mass flow
entering the pipe is
โข
m = P1 A V1=(1.23)(0.196)(100)=24.11 kg/sec.
Applying the momentum equation, Eq. (2.64) to this geometry, we obtain a result similar to Eq.
(2.75), namely
R =-
# (p V ยท dS) V
(1)
Where the pressure term in Eq. (2.75) is zero because the pressure at the inlet and exit are the
same values. In Eq. (1), the product (p V ยท dS) is negative at the inlet (V and dS are in opposite
directions), and is positive at the exit (V and dS) are in the same direction). The magnitude of p
โข
V ยท dS is simply the mass flow, m. Finally, at the inlet V 1 is to the right, hence it is in the
positive x-direction. At the exit, V2 is to the left, hence it is in the negative x-direction. Thus,
V2 = – V1. With this, Eq. (1) is written as
.
.
~
R = – [- m V 1 + m V2] = m (V 1 – V2)
โข
โข
= m [V1 -(-V1)] = m (2V1)
R = (24.11)(2)(100) = &822
NI
27
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2.13 From Example 2.1, we have
u = Vโ
(2.35)
v = – Vโ h
(2.36)
and
Thus,
= u = Vโ
(2.35a)
Integrating (2.35a) with respect to x, we have
= Vโ x
= Vโ x
+ f(y)
+ f(y)
(2.35b)
From (2.36)
= v = – Vโ h
(2.36a)
Integrating (2.36a) with respect to y, we have
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= Vโ h
+ f(x)
=
+ f(x)
(2.36b)
Comparing (2.35b) and (2.36b), which represent the same function for , we see
in (2.36b) that f(x) = Vโ x. So the velocity potential for the compressible subsonic
flow over a wavy well is:
________________________________________________________________
2.14 The equation of a streamline can be found from Eq. (2.118)
=
For the flow over the wavy wall in Example 2.1,
=
As y โ โ, then
โ 0. Thus,
โ
=0
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Fundamentals of Aerodynamics 6th Edition Anderson Solutions Manual
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The slope is zero. Hence, the streamline at y โ โ is straight.
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