Esame di Fluidodinamica delle Macchine (I
modulo)
Cognome/Nome
Matricola
N=
1 Marzo 2013
1. Come illustrato in Fig. 1, un’onda d’urto si propaga all’interno di un condotto a sezione costante la cui estremità destra sfocia nell’ambiente esterno, in
cui regna pressione pa . Sapendo che la velocità (assoluta) del gas è ovun-
Figura 1: Riflessione di un’onda d’urto da parte di un confine libero
que subsonica, si calcoli la velocità u3 ed il numero di Mach (M3 ) con cui
il fluido abbandona il condotto, a seguito della riflessione dell’urto sul confine libero. Siano noti: u1 = 50 m/s, T1 = 300 K, p1 = pa = 101.3 kPa,
Ms = (Ws − u1 )/a1 = 1.15+(N/100).
u3 =
m/s;
1
M3 =
.
0.2
A
TE
LE
Y
0
B
LE
0
C
Y
-0.2
-0.02
-0.4
-0.2
0
X
-0.52
0.2 -0.5
-0.48
0.4
-0.46
X
Figura 2: Flusso attorno ad un profilo di Joukowsky simmetrico
Si consideri il flusso irrotazionale di un fluido incomprimibile attorno ad un profilo di Joukowsky simmetrico, Fig. 1. Il profilo ha corda l = 1 m; rapporto spessore/corda pari a t/l = 0.05 ed è investito da un flusso d’aria (ρ∞ = 1.2 kg/m3 )
con velocità all’infinito pari a U∞ = 75 m/s ed angolo di incidenza pari ad
α∞ = 5◦ . Si calcoli:
(a) il valore della circolazione Γ per cui il bordo di uscita è punto di ristagno
ed il coefficiente di portanza.
(b) La posizione, nel piano fisico del bordo di attacco e dei punti di massimo
spessore (quella riportata in figura è solo indicativa);
(c) La posizione (esatta o approssimata), nel piano fisico del punto di ristagno
di monte (punto C in Fig. 1).
(d) Il valore delle componenti della velocità nei punti di massimo spessore e
nel bordo d’attacco.
Suggerimento
Per il calcolo delle velocità può essere più semplice utilizzare la regola di derivazione delle funzioni composte, piuttosto che cercare di ottenere la velocità nel
piano fisico.
Γ=
xLE =
xA/B =
xC =
uA =
uB =
uLE =
m2 /s;
m;
m;
m;
m/s;
m/s;
m/s;
2
CL =
yLE =
yA/B = ±
yC =
vA =
vB =
vLE =
m;
m;
m;
m/s;
m/s;
m/s;
Solution
Let us write the complex potential in the transformed plane for the flow around
a circle of radius a with its centre in ζ0 = −m:
ζ
iΓ
a2 /c2 iα
ζ/c + ǫ −iα
−iα
F (ζ) = U∞ c
(1)
+ǫ e
+
e +
log
e
c
(ζ/c + ǫ)
2π
a/c
Eq. (1) has been obtained from [?, Eq. (4.23d)] making ζ non-dimensional using
the length scale c.
Differentiating (1), we get the velocity around the circle in the transformed
plane:
"
#
2
dF
(1
+
ǫ)
iΓ/c
iα
W (ζ) =
= U∞ e−iα −
(2)
+
2e
dζ
2π (ζ + ǫ)
(ζ + ǫ)
In Eq. (2) we have used ζ as a shorthand notation for ζ/c, so that from now on
ζ will be non-dimensional.
Kutta condition and lift coefficient
The Kutta condition requires that the velocity vanishes at the TE of the airfoil,
i.e. where ζ = 1. Replacing ζ = 1 in Eq. (2), we find:
U∞ e−iα − eiα +
iΓ/c
=0
2π (1 + ǫ)
(3)
The real part in Eq. (3) is identically zero, whereas the imaginary part becames
zero with the following choice of the circulation:
Γ = 4 π U∞ c (1 + ǫ) sin α
= 4 π U∞ a sin α = 21.3260 m2 /s2
(4a)
(4b)
The lift coefficient is:
CL = 2 π (1 + ǫ) sin α = 0.56869346
Once we replace in Eq.(2) the particular value (4) of the circulation that satisfies
the Kutta condition, the complex velocity becomes
#
"
2
2
i
sin
α
(1
+
ǫ)
(1
+
ǫ)
iα
(5)
W (ζ) = U∞ e−iα −
2e +
(ζ + ǫ)
(ζ + ǫ)
Forward stagnation point
In order to compute the location of the forward stagnation point, refer to Fig. 1,
which shows, beside the ζ = ξ + i η reference frame, a second reference frame:
ζ ′ = ξ ′ + i η ′ that has the origin in ζ0 = −ǫ and the ξ ′ axis aligned with the
freestream velocity. Without a vortex, the two stagnation points would be at
3
the intersection between the ξ ′ axis and the circle; however, the circulation (4)
has to move the rear stagnation point in ζT E = 1; the forward stagnation point
′
moves counter-clockwise up to the point C: ζC = a/c ei(α+π) = − (1 + ǫ) eiα .
Since the coordinates of point C in the two reference frames are related by the
following expression:
′
ζC = (ζC − ζ0 ) e−iα
we can get the coordinate of the forward stagnation point in the ζ reference
frame:
ζC = − (1 + ǫ) e2iα − ǫ
= [− (1 + ǫ) cos 2α − ǫ] − i sin 2α (1 + ǫ)
(6a)
(6b)
The coordinates of point C can also be easily obtained by means of geometrical
arguments from Fig. 1.
Once the coordinates (6) of point C are replaced within the complex velocity,
this becomes identically zero:
W (ζC ) /U∞ = e−iα − e−3iα − 2 i sin α e−i2α
(7a)
−iα
−2iα
−iα
= e
1−e
− 2 i sin α e
(7b)
= [1 − (cos 2α − i sin 2α)
− 2 i sin α (cos α − i sin α)] e−iα .
(7c)
Indeed, separating the real and imaginary part of the term in square brackets
of Eq. (7c), we get:
u (ζC ) /U∞ = 1 − cos 2α − 2 sin2 α ≡ 0
−v (ζC ) /U∞ = sin 2α − 2 sin α cos α ≡ 0
(8a)
(8b)
Observe, however, that Eq. (6) gives the coordinates of point C within the
transformed plane; its image in the physical plane is obtained by replacing
Eq. (6) in the Joukowski transformation:
zC
c
= ζC +
1
ζC
(9)
Replacing Eq. (6) into (9) we get:
xC = −0.49427056 m
yC = −6.17378578 10−3 m
(10a)
(10b)
Alternatively, we can get the coordinates of the forward stagnation point directly
in the physical plane using the approximate equations of the profile, derived
in [?]:
x/c = 2 cos ν
y/c = 2 ǫ (1 − cos ν) sin ν
4
(11a)
(11b)
η’
η
Α
2/3 π
rear stagnation point
ξ’
νc
α
LE
α
α
ζ =−m/c
TE
0
m/c
C
ξ
1
4/3 π
forward stagnation point
Β
Figura 3: Flusso attorno ad un cerchio con centro in ζ = −ǫ
Observe that in Eq. (11) the angle νC (shown in red in Fig. 1) has to be measured
from the positive ξ axis:
νC = tan−1
Replacing νC in (11) we get:
ℑ (ζC )
= 189.64424◦
ℜ (ζC )
xC = −0.49293348 m
yC = −6.40266389 10−3 m
(12a)
(12b)
Observe that the approximate (12) and true position (10) of the forward stagnation point are very close.
Leading edge
Within the transformed plane, the LE is given by the intersection of the negative
ξ axis with the circle, i.e.
ζLE = − (1 + 2ǫ)
(13)
Replacing (13) into the complex velocity (5), we get:
"
#
2
(1 + ǫ) iα
W (ζLE ) /U∞ = e−iα −
e − 2 i sin α
(1 + ǫ)2
= [(cos α − i sin α) − cos α − i sin α − 2 i sin α]
= −4 i sin α
5
(14a)
(14b)
(14c)
Eq. (14) gives the velocity at the LE of the circle; in order to get the velocity
at the LE of the profile, we have to compute:
−1
dz
dF
(15)
W (zLE ) =
dζ ζLE dζ ζLE
where:
dz
dζ
−1
ζLE
2
ζLE
(1 + 2ǫ)2
= 2
=
ζLE − 1
4ǫ (1 + ǫ)
1
= 1+
4ǫ (1 + ǫ)
Replacing (16) and (14c) into (15), we obtain:
W (ζLE ) /U∞ = −4 i sin α 1 +
1
4ǫ (1 + ǫ)
(16a)
(16b)
(17)
Eq. (17) shows that in the LE of the airfoil the fluid velocity has only a vertical
component; this is because, due to simmetry of the profile w.r.t. the x axis, its
normal in the LE is −ex ; therefore, the boundary condition at the LE becomes:
u · −ex = −u = 0.
Points of maximum thickness
Points where the profile has maximum thickness are found at νA/B = 2 π/3 and
4 π/3. These location are found from the approximate equations that define the
Joukowsky profile in the physical plane, i.e. Eq. (11). The angles νA/B (shown
in red in Fig. 1) are measured in the transformed plane. We are interested in
finding the image in the transformed plane of the points A and B where the
profile has maximum thickness so that the chain rule can be used to compute
the velocity in the physical plane:
−1
dz
dF
(18)
W zA/B =
dζ ζA/B dζ ζA/B
The points on the circle of the transformed plane that correspond to points A
and B of the physical plane are therefore at:
ζA/B = 1 + 1 − cos νA/B ǫ eiνA/B
(19a)
√ !
1
3
= [1 + 3ǫ/2] − ± i
(19b)
2
2
= −0.52886754 ± i 0.91602546
(19c)
In Eq. (19) we have used the approximate equation of the circle in the ζ plane:
ζ = [1 + (1 − cos ν) ǫ] eiν
6
since this is the equation that has been used to derive the equations of the
approximate profile (11).
Writing Eq. (15) within the points A and B, we get:
ζA2
dF
= 91.962418 − i 6.74382690 10−3 m/s (20a)
W (zA ) =
2
dζ ζA ζA − 1
dF
ζB2
W (zB ) =
= 68.520424 − i 6.92010950 10−3 m/s (20b)
2
dζ ζB ζB − 1
Observe two things. First, at the two points of maximum thickness, the normal
to the profile is aligned with the y axis, i.e. nA/B = ±ey ; therefore, due to the
boundary condition, the vertical velocity component should be identically zero.
In Eqs. (20) a small vertical velocity component is found, due to the fact that
the locations of the points A and B has been found by using an approximate
equation. Second, we see that, since the freestream flow is at incidence, even
though the profile is symmetric, the velocity within the point B that belongs
to the pressure side is lower than that measured in point A which is located
on the suction side. This observation is consistent with what we expect to see,
since lower velocity implies higher pressure and vice-versa.
2. Aria entra in un condotto a sezione costante (d = 2 cm) e isolato termicamente
dall’ambiente esterno, a M1 = 2.5 e pressione e temperatura (valori statici)
pari, rispettivamente, a p1 = 70 kPa e t1 = 40◦ C. In una sezione a valle, dove
il flusso ha raggiunto M2 = 2, si verifica un urto retto. Nella sezione di uscita
dal condotto il flusso di aria si muove a M4 = 0.8.
Assumendo il fattore di attrito pari a f = 0.005, si calcoli la distanza dell’urto
dalla sezione di ingresso, nonchè la lunghezza complessiva del condotto.
Si calcoli, inoltre, la pressione nella sezione di valle e la portata massica che
attraversa il condotto.
Si indichino con i pedici 1,2,3,4 le grandezze riferite, rispettivamente, alle sezioni:
di ingresso, a monte dell’urto, a valle dell’urto, in uscita dal condotto.
l12 =
p4 =
m;
kPa;
7
l14 =
ṁ =
m;
kg/s.