Problem 13.28
[Difficulty: 3]
Given:
Data on flow in a passage
Find:
Exit temperature and mass flow rate of air assuming isentropic flow
Solution:
The given or available data is:
R
=
53.33
ft-lbf/lbm-
°
R
k
=
1.4
T
1
=
450
°R
p
1
=
45
psia
p
01
=
51
psia
A
1
=
4
ft
2
A
2
=
3
ft
2
Equations and Computations:
From the static and stagnation pressures we can calculate
M
1
:
M
1
=
0.427
From the
M
1
and
T
1
we can get
T
01
:
T
01
=
466.38
°R
From
M
1
, and Eq. 13.7d
(using built-in function
IsenA
(
M
,
k
))
A
*
1
=
2.649
ft
2
For isentropic flow (
p
02
=
p
01
,
T
02
=
T
01
,
A
*
2
=
A
*
1
)
p
02
=
51
psia
T
02
=
466.38
°R
A
*
2
=
2.649
ft
2
A
2
/A
*
2
=
1.1325

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