fluid_mechanics

11.5 Nonisentropic Flow of an Ideal Gas 621
T
( )
Ma b =
√ 1_ k
b
Ma < 1
a (Ma a = 1)
Ma > 1
s
F I G U R E 11.22
Rayleigh line.
E XAMPLE 11.15
Frictionless, Constant Area Compressible Flow
with Heat Transfer (Rayleigh Flow)
GIVEN Air 1k 1.42 enters [section 112] a frictionless, constant
flow cross-sectional area duct with the following properties
(the same as in Example 11.11):
T 0 518.67 °R
T 1 514.55 °R
p 1 14.3 psia
FIND For Rayleigh flow, determine corresponding values of
fluid temperature and entropy change for various levels of downstream
pressure and plot the related Rayleigh line.
SOLUTION
To plot the Rayleigh line asked for, use Eq. 11.111
p 1rV22 RT
constant
p
and Eq. 11.76
s s 1 c p ln T R ln p (2)
T 1 p 1
to construct a table of values of temperature and entropy change
corresponding to different levels of pressure downstream in a
Rayleigh flow.
Use the value of ideal gas constant for air from Table 1.7
R 1716 1ft # lb21slug # °R2
or in EE system units
R 53.3 1ft # lb21lbm # °R2
and the value of specific heat at constant pressure for air from Example
11.11, namely,
c p 187 1ft # lb21lbm # °R2
Also, from Example 11.11, rV 16.7 lbm1ft 2 # s2. For the
given inlet [section 112] conditions, we get
RT 1
353.3 1ft # lb21lbm # °R241514.55 °R2
p 1 14.3 psia 1144 in. 2 ft 2 2
13.3 ft 3 lbm
Thus, from Eq. 1 we get
p 1rV22 RT
14.3 psia 316.7 lbm1ft 2 # s24 2 113.3 ft 3 lbm2
p
14.3 psia 3720 lbm1ft # s 2 2 constant
(1)
or, since
32.22 lbft 2 ,
p 1rV22 RT
14.3 psia 31372032.22lbft 2 411 ft 2 144 in. 2 2
p
15.10 psia constant
(3)
With the downstream pressure of p 13.5 psia, we can obtain
the downstream temperature by using Eq. 3 with the fact that
Hence, from Eq. 3,
or
1rV2 2 R
p
1 lbm1ft # s 2 2 31132.221lb # s 2 ft241ft # s 2 2 11
316.7 lbm 1ft 2 # s24 2 353.3 1ft # lb21lbm # °R24
1144 in. 2 ft 2 2 13.5 psia
7.65 3lbm1ft # s 2 24°R 31 lb132.2 lbm # fts 2 24
0.238 1lbft 2 2°R11 ft 2 144 in. 2 2
1.65 10 3 1lbin. 2 2°R
13.5 psia 31.65 10 3 1lbin. 2 2°R4 T 15.10 psia
T 969 °R
From Eq. 2 with the downstream pressure p 13.5 psia and temperature
T 969 °R we get
s s 1 3187 1ft # lb21lbm #
969 °R
°R24 ln a
514.55 °R b
13.5 psia
353.3 1ft # lb21lbm # °R24 ln a
14.3 psia b
s s 1 121 1ft # lb21lbm # °R2
By proceeding as outlined above, we can construct the table of
values shown below and graph the Rayleigh line of Fig. E11.15.