fluid_mechanics

632 Chapter 11 ■ Compressible Flow
SOLUTION
We assume that the flow along the stagnation pathline is isentropic
except across the shock. Also, the shock is treated as a normal
shock. Thus, in terms of the data we have
p 0,y
a p 0,y
b a p 0, x
b
(1)
p x p 0, x p x
where p 0,y is the stagnation pressure measured by the probe, and
p x is the static pressure measured by the wall tap. The stagnation
pressure upstream of the shock, p 0, x , is not measured.
Combining Eqs. 1, 11.156, and 11.59 we obtain
p 0,y
p x
531k 1224Ma 2 x 6 k 1k12
532k1k 124Ma 2 x 31k 121k 1246 1 1k12
which is called the Rayleigh Pitot-tube formula. Values of p 0,yp x
from Eq. 2 are considered important enough to be included in Fig.
D.4 for k 1.4. Thus, for k 1.4 and
p 0,y
p x
we use Fig. D.4 1or Eq. 22 to ascertain that
60 psia
12 psia 5
Ma x 1.9
(2)
(Ans)
To determine the flow velocity we need to know the static temperature
upstream of the shock, since Eqs. 11.36 and 11.46 can be
used to yield
V x Ma x c x Ma x 1RT x k
(3)
The stagnation temperature downstream of the shock was measured
and found to be
Since the stagnation temperature remains constant across a normal
shock 1see Eq. 11.1362,
For the isentropic flow upstream of the shock, Eq. 11.56 or
Fig. D.1 can be used. For Ma x 1.9,
or
With Eq. 3 we obtain
T 0,y 1000 °R
T 0, x T 0,y 1000 °R
T x
T 0, x
0.59
T x 10.59211000 °R2 590 °R
V x 1.87 2353.3 1ft # lb21lbm # °R241590 °R211.42
392 1ft # lblbm2 12 3132.2 lbm # fts 2 2lb4 1 2
2220 fts
(Ans)
COMMENT Application of the incompressible flow Pitot
tube results 1see Section 3.52 would give highly inaccurate
results because of the large pressure and density changes
involved.
E XAMPLE 11.19
Normal Shock in a Converging–Diverging Duct
normal shock at the exit of the duct. What value of
GIVEN Consider the converging–diverging duct of Example
1x 0.5 m2
pressure, p IIIp 0, x 1see Fig. 11.132, that will result in a standing
11.8.
the ratio of back pressure to inlet stagnation pressure would be
FIND Determine the ratio of back pressure to inlet stagnation
required to position the shock at x 0.3 m? Show related
temperature–entropy diagrams for these flows.
SOLUTION
For supersonic, isentropic flow through the nozzle to just upstream
of the standing normal shock at the duct exit, we have
from the table of Example 11.8 at x 0.5 m
and
Ma x 2.8
p x
p 0,x
0.04
From Fig. D.4 for Ma x 2.8 we obtain
p y
p x
9.0
Thus,
p y
p 0, x
a p y
p x
b a p x
p 0, x
b 19.0210.042
0.36 p III
p 0, x
(Ans)
When the ratio of duct back pressure to inlet stagnation pressure,
p IIIp 0, x , is set equal to 0.36, the air will accelerate through the
converging–diverging duct to a Mach number of 2.8 at the duct
exit. The air will subsequently decelerate to a subsonic flow
across a normal shock at the duct exit. The stagnation pressure
ratio across the normal shock, p 0,yp 0,x , is 0.38 1Fig. D.4 for