fluid_mechanics

600 Chapter 11 ■ Compressible Flow
Finally from Eq. 1 we have
(Ans)
(b) For p re 40 kPa1abs2 53.3 kPa1abs2 p*, we have
p th p* 53.3 kPa1abs2 and Ma th 1. The converging duct is
choked. From Eq. 2 1see also Eq. 11.662
or
From Eq. 5 1see also Eq. 11.642,
or
m # 11.04 kgm 3 211 10 4 m 2 21193 ms2
0.0201 kgs
r th
1.23 kgm 3 e 1
1 311.4 1224112 2 f 111.412
r th 0.780 kgm 3
T th
288 K 1
1 311.4 1224112 2
T th 240 K
From Eq. 4,
V th 112 23286.9 J1kg # K241240 K211.42
310 1Jkg2 12 310 ms
since 1 Jkg 1 N # mkg 1 1kg # ms 2 2 # mkg 1ms2 2 . Finally
from Eq. 1
m # 10.780 kgm 3 211 10 4 m 2 21310 ms2
0.0242 kgs
(Ans)
From the values of throat temperature and throat pressure calculated
above for flow situations 1a2 and 1b2, we can construct the
temperature–entropy diagram shown in Fig. E11.5b.
COMMENT Note that the flow from standard atmosphere to
the receiver for receiver pressure, p re , greater than or equal to the
critical pressure, p*, is isentropic. When the receiver pressure is
less than the critical pressure as in situation 1b2 above, what is the
flow like downstream from the exit of the converging duct? Experience
suggests that this flow, when p re 6 p*, is threedimensional
and nonisentropic and involves a drop in pressure
from p th to p re , a drop in temperature, and an increase of entropy
as are indicated in Fig. E11.5c.
Isentropic flow Eqs. 11.56, 11.59, and 11.60 have been used to construct Fig. D.1 in Appendix
D for air 1k 1.42. Examples 11.6 and 11.7 illustrate how these graphs of TT 0 , pp 0 , and rr 0
as a function of Mach number, Ma, can be used to solve compressible flow problems.
E XAMPLE 11.6
Use of Compressible Flow Graphs in Solving Problems
GIVEN Consider the flow described in Example 11.5.
duct throat to solve for mass flowrate from
m # r th A th V th (1)
T th 10.9421288 K2 271 K
FIND Solve Example 11.5 using Fig. D.1 of Appendix D.
SOLUTION
We still need the density and velocity of the air at the converging Thus, from Eqs. 2 and 3
(a) Since the receiver pressure, p re 80 kPa1abs2, is greater
and
than the critical pressure, p* 53.3 kPa1abs2, the throat pressure,
r th 10.85211.23 kgm 3 2 1.04 kgm 3
p th , is equal to the receiver pressure. Thus
Furthermore, using Eqs. 11.36 and 11.46 we get
p th 80 kPa1abs2
p 0 101 kPa1abs2 0.792
From Fig. D.1, for pp 0 0.79, we get from the graph
Ma th 0.59
T th
0.94
T 0
r th
0.85
r 0
(2)
(3)
V th Ma th 2RT th k
10.592 23286.9 J1kg # K241269 K211.42
194 1Jkg2 12 194 ms
since 1 Jkg 1 N # mkg 1 1kg # ms 2 2 # mkg 1ms2 2 . Finally,
from Eq. 1
m # 11.04 kgm 3 211 10 4 m 2 21194 ms2
0.0202 kgs
(Ans)