fluid_mechanics

11.6 Analogy between Compressible and Open-Channel Flows 633
Ma x 2.82. A considerable amount of available energy is lost
across the shock.
For a normal shock at x 0.3 m, we note from the table of
Example 11.8 that Ma x 2.14 and
From Fig. D.4 for Ma x 2.14 we obtain p yp x 5.2, Ma y 0.56,
and
From Fig. D.1 for Ma y 0.56 we get
For x 0.3 m, the ratio of duct exit area to local area 1A 2A y 2
is, using the area equation from Example 11.8,
A 2 0.1 10.522
(4)
A y 0.1 10.32 1.842 2
Using Eqs. 3 and 4 we get
p x
p 0, x
0.10
p 0,y
p 0, x
0.66
A y
A* 1.24
A 2
A* a A y
A* b aA 2
A y
b 11.24211.8422 2.28
(1)
(2)
(3)
Note that for the isentropic flow upstream of the shock, A*
0.10 m 2 1the actual throat area2, while for the isentropic flow downstream
of the shock, A* A 22.28 0.35 m 2 2.280.15 m 2 . With
A 2A* 2.28 we use Fig. D.1 and find Ma 2 0.26 and
p 2
p 0,y
0.95
Combining Eqs. 2 and 5 we obtain
p 2
p 0, x
a p 2
p 0,y
b a p 0,y
p 0, x
b 10.95210.662 0.63
(5)
(Ans)
When the back pressure, p 2 , is set equal to 0.63 times the inlet
stagnation pressure, p 0,x , the normal shock will be positioned at
x 0.3 m. The corresponding T–s diagrams are shown in
Figs. E11.19a and E11.19b.
COMMENT Note that p 2p 0,x 0.63 is less than the value of
this ratio for subsonic isentropic flow through the converging–
diverging duct, p 2p 0 0.98 1from Example 11.82 and is larger
than p IIIp 0,x 0.36, for duct flow with a normal shock at the exit
1see Fig. 11.132. Also the stagnation pressure ratio with the shock
at x 0.3 m, p 0,yp 0, x 0.66, is much greater than the stagnation
pressure ratio, 0.38, when the shock occurs at the exit
1x 0.5 m2 of the duct.
T, K
340
300
260
220
180
p 0, x = 101 kPa (abs)
0, x
T y = 275 K
Normal shock
p 0, y = 38 kPa (abs)
p y =
36 kPa (abs)
0, y
= p III
y, III
T 0, x =
T 0,y = 288 K
T, K
340
300
260
220
180
p 0, x =
101 kPa (abs)
0, x
Normal shock
0, y
2
y
T 2 = 284 K
T y = 271 K
p x = 10 kPa (abs)
p y = 52 kPa (abs)
p 0, y = 67 kPa (abs)
p 2 = 64 kPa (abs)
T 0, x = T 0,y =
288 K
140
p x = 4 kPa (abs)
140
x
T x = 150 K
100
x
T x = 112 K
100
Shock within nozzle (x = 0.3 m)
Shock at nozzle exit plane (x = 0.5 m)
0 80 160 240 320 400 480 0 80 160 240 320 400 480
s – s x , ______ J
s – s
(kg•K)
x , ______ J
(kg•K)
(a)
(b)
F I G U R E E11.19
11.6 Analogy between Compressible and Open-Channel Flows
During a first course in fluid mechanics, students rarely study both open-channel flows 1Chapter
102 and compressible flows. This is unfortunate because these two kinds of flows are strikingly
similar in several ways. Furthermore, the analogy between open-channel and compressible
flows is useful because important two-dimensional compressible flow phenomena can be simply