fluid_mechanics

606 Chapter 11 ■ Compressible Flow
0.4
0.3
r, m 0.2
0.1
Ma
1.0
Subsonic
Subsonic
0
–0.5 –0.4 –0.2 0 0.2 0.4 0.5
0
–0.5 –0.4 –0.2 0 0.2 0.4 0.5
x, m x, m
(a)
(b)
p 0 = 101 kPa (abs)
296
1.0
p a = p c = 100 kPa (abs)
292
0.9
0 T 0 = 288 K
0.8 Subsonic T/T 0 Subsonic p/p
288
0 a, c T a = T c = 285 K
0.7
284
___ T
T 0.6 0
280
0.5
p b = 86 kPa (abs)
___ p
276
T
p 0
0.4
b
b = 276 K
272
0.3
0.2
268
0.1
264
0.0
260
–0.5 –0.4 –0.2 0 0.2 0.4 0.5
_______ J
s,
x, m
(kg • K)
(c)
(d)
F I G U R E E11.10
T, K
A more precise solution for the flow of this example could
have been obtained with isentropic flow equations by following
the steps outlined below.
1. Use Eq. 11.59 to get pp 0 at x 0 knowing k and
Ma 0.48.
2. From Eq. 11.71, obtain value of AA* at x 0 knowing
k and Ma.
3. Determine A* knowing A and AA* at x 0.
4. Determine AA* at different axial locations, x.
5. Use Eq. 11.71 and AA* from step 4 above to get values
of Mach numbers at different axial locations.
6. Use Eqs. 11.56 and 11.59 and Ma from step 5 above to
obtain TT 0 and pp 0 at different axial locations, x.
COMMENT There are an infinite number of subsonic, isentropic
flow solutions for the converging–diverging duct considered
in this example 1one for any given Ma 1 at x 02.
Calculated,
From Fig. D.1
x (m) AA* Ma TT 0 pp 0 State
0.5 5.0 0.12 0.99 0.99 a
0.4 3.7 0.16 0.99 0.98
0.3 2.7 0.23 0.99 0.96
0.2 2.0 0.31 0.98 0.94
0.1 1.6 0.40 0.97 0.89
0 1.4 0.48 0.96 0.85 b
0.1
0.2
0.3
0.4
0.5
1.6 0.40 0.97 0.89
2.0 0.31 0.98 0.94
2.7 0.23 0.99 0.96
3.7 0.16 0.99 0.98
5.0 0.12 0.99 0.99 c
F l u i d s i n t h e N e w s
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