Page 205 - 48Fundamentals of Compressible Fluid Mechanics

9.9. MORE EXAMPLES 167
9.9 More Examples

Example 9.4:
To demonstrate the utility of the figure 9.18 consider the flowing example. Find the
, b
and K air.
and pressure ratio
H
SOLUTION QS . The stagnation conditions at the entrance are UQ
Q
.
mass flow rate for H QS Q-9 , H V
H Q0S Q "
Q0S IQ0S
QS/9
First calculate the dimensionless resistance, WYXaZ [
QS Q-98V
W`XaZ [
H
etc. H V Q
QS Q "

or accurately utilizing the program as in the following table. 7 7
b
V
From Figure 9.18 for
H QS_I N
Q0S I
b
4
4

L
L
V Q^S QUQ
QUQ
Q^S IUI + )
VUQ^S Q
QUQ
Q
Q^S_I ")&"0
Q0S Q
IUS QUQ
Q
Q
L
L
"S I.+1")
V " S_I +-")
V Q^S QUQ
QUQ
Q0S V Q),"UQ
Q^S_I "=V "=Q
Q^S UQ
QUQ
Q
1" "
Q0S " "0+1"=)
9=Q^S*)091+1"
I(QS/)190+1"
Q^S 9
Q
QUQ
Q
Q^S_I
I
V Q^S QUQ
QUQ
IaQ) SV "
Q^S Q)0")09
+) SV "=Q1+
Q0S Q-"1"1+9
V Q^S QUQ
QUQ
Q^S UQ
QUQ
Q
Only for the pressure ratio of 0.1 the flow is choked. VS 9 1" 7
4
5
7
6
5
6 8
7 8
: 7 8
8
VS/90"0I(Q
Q^S "-"I."-9
Q0S "- 1 )V
Q^S_I ")&"0
Q0S "-"1+)-)

Q^S "-" "

VS +9 "
"
Q0S "- 1"
"-)
"1
Q0S "-"1+1"
VS*)Q
Q^S_I "=V "=Q
1" "
Q0S "-"
Q1"=)
9 SQ=)
9 S_I
I."1+
Q^S "-" 9V
Q0S "-") V I
Q^S_I
I
"
) S "-9^I."
Q^S Q)0")09
) S#"0
V
Q0S "-"1 )
Q0S "-"-9191+
Q^S "-"1+-
and for the same the pressure. Hence, the mass rate is
a function of the Mach number. The Mach number is indeed function of the pres-
Therefore, L
L
sure ratio but and therefore mass flow rate is function pressure ratio only through
Mach number.

The mass flow rate is 8Q0S I "-)$ ! "0 =) UQ
Q QS V- % )(G* F'&
F
QS Q "
L
V
!

IUS V
H N "
H
Q
QUQ
QUQ#