Basics of Fluid Mechanics, 2014a

11.4. ISENTROPIC FLOW 401
Table -11.1. Fliegner’s number and other parameters as function of Mach number (continue)
M Fn ˆρ
(
P 0 A ∗
AP
) 2
RT 0
P 2
( ṁ
) 2 1
A Rρ 0P
( ṁ
) 2 1
A Rρ 2 0 T
( ṁ
) 2
A
0.95 2.188 1.843 3.181 10.88 6.112 4.037
0.96 2.233 1.881 3.259 11.60 6.436 4.217
0.97 2.278 1.920 3.338 12.37 6.777 4.404
0.98 2.324 1.961 3.419 13.19 7.136 4.600
0.99 2.371 2.003 3.500 14.06 7.515 4.804
1.00 2.419 2.046 3.583 14.98 7.913 5.016
Example 11.10:
A gas flows in the tube with mass flow rate of 0.1 [kg/sec] and tube cross section is
0.001[m 2 ]. The temperature at chamber supplying the pressure to tube is 27 ◦ C. At
some point the static pressure was measured to be 1.5[Bar]. Calculate for that point
the Mach number, the velocity, and the stagnation pressure. Assume that the process
is isentropic, k =1.3, R = 287[j/kgK].
Solution
The first thing that need to be done is to find the mass flow per area and it is
ṁ
A =0.1/0.001 = 100.0[kg/sec/m2 ]
It can be noticed that the total temperature is 300K and the static pressure is 1.5[Bar].
It is fortunate that Potto-GDC exist and it can be just plug into it and it provide that
M
T
T 0
ρ
ρ 0
A
A ⋆
P
P 0
A×P
A ∗ ×P 0
F
F ∗
0.17124 0.99562 0.98548 3.4757 0.98116 3.4102 1.5392
The velocity can be calculated as
U = Mc= √ kRTM =0.17 × √ 1.3 × 287 × 300× ∼56.87[m/sec]
The stagnation pressure is
P 0 =
P
P/P 0
=1.5/0.98116 = 1.5288[Bar]
End Solution
11.4.6 Isentropic Tables