fluid_mechanics

11.4 Isentropic Flow of an Ideal Gas 599
SOLUTION
To determine the mass flowrate through the converging duct we
use Eq. 11.40. Thus,
or in terms of the given throat area, A th ,
We assume that the flow through the converging duct is isentropic
and that the air behaves as an ideal gas with constant c p and
c v . Then, from Eq. 11.60
The stagnation density, r 0 , for the standard atmosphere is
1.23 kgm 3 and the specific heat ratio is 1.4. To determine the
throat Mach number, Ma th , we can use Eq. 11.59,
p th
r th
m # rAV constant
m # r th A th V th
1
e
r 0 1 31k 1224Ma 2 th
1
e
p 0 1 31k 1224Ma 2 th
11k12
f
k1k12
f
(1)
(2)
(3)
Standard
atmosphere
Flow
T, K
300
290
280
270
260
250
240
230
Converging duct Receiver pipe
(a)
p 0 = 101 kPa (abs)
T 0 = 288 K
p th, a = 80 kPa (abs)
T th, a = 269 K
Situation (a)
p th, b = 53.3 kPa (abs) = p*
T th, b = 240 K
Situation (b)
The critical pressure, p*, is obtained from Eq. 11.62 as
If the receiver pressure, p re , is greater than or equal to p*, then
p th p re . If p re 6 p*, then p th p* and the flow is choked. With
p th , p 0 , and k known, Ma th can be obtained from Eq. 3, and r th can
be determined from Eq. 2.
The flow velocity at the throat can be obtained from Eqs.
11.36 and 11.46 as
The value of temperature at the throat, T th , can be calculated from
Eq. 11.56,
T th
1
(5)
T 0 1 31k 1224Ma 2 th
Since the flow through the converging duct is assumed to be isentropic,
the stagnation temperature is considered constant at the
standard atmosphere value of T 0 15 K 273 K 288 K.
Note that absolute pressures and temperatures are used.
(a) For p re 80 kPa1abs2 7 53.3 kPa1abs2 p*, we have
p th 80 kPa1abs2. Then from Eq. 3
or
From Eq. 2
or
p* 0.528p 0 0.528p atm
10.52823101 kPa1abs24 53.3 kPa1abs2
V th Ma th c th Ma th 2RT th k
80 kPa1abs2
1.411.412
101 kPa1abs2 e 1
f
1 311.4 1224Ma 2 th
Ma th 0.587
r th
1.23 kgm 3 e 1
1 311.4 122410.5872 2 f 111.412
r th 1.04 kgm 3
(4)
T, K
From Eq. 5
or
220
300
290
280
270
260
250
240
230
220
_______ J
s,
(kg • K)
(b)
F I G U R E E11.5
p 0 = 101 kPa (abs)
T 0 = 288 K
p* = 53.3 kPa (abs)
T* = 240 K
p re < p*
_______ J
s,
(kg • K)
(c)
T th
288 K 1
1 311.4 122410.5872 2
T th 269 K
T re < T*
Substituting Ma th 0.587 and T th 269 K into Eq. 4 we obtain
V th 0.587 23286.9 J1kg # K241269 K211.42
193 1Jkg2 1 2
Thus, since 1 Jkg 1 N # mkg 1 1kg # ms 2 2 # mkg
1ms2 2 , we obtain
V th 193 ms