Basics of Fluid Mechanics, 2014a

396 CHAPTER 11. COMPRESSIBLE FLOW ONE DIMENSIONAL
The definition of Fliegner’s number (Fn) is
√
R ṁc0
Fn ≡ √ (11.59)
RA∗ P 0
Utilizing Fliegner’s number definition and substituting it into equation (11.53)
results in
Fliegner’s Number
(
Fn = kM 1+ k − 1 ) −
k+1
M 2 2(k−1)
(11.60)
2
and the maximum point for Fn at M =1is
Fn = k
( k +1
2
) −
k+1
2(k−1)
(11.61)
Example 11.7:
Why Fn is zero at Mach equal to zero? Prove Fliegner number, Fn is maximum at
M =1.
Example 11.8:
The pitot tube measured the temperature of a flow which was found to be 300 ◦ C. The
static pressure was measured to be 2 [Bar]. The flow rate is 1 [kg/sec] and area of the
conduct is 0.001 [m 2 ]. Calculate the Mach number, the velocity of the stream, and
stagnation pressure. Assume perfect gas model with k=1.42.
Solution
This exactly the case discussed above in which the the ratio of mass flow rate to the area
is given along with the stagnation temperature and static pressure. Utilizing equation
(??) will provide the solution.
) 2 ) 2
= =2.676275 (11.VIII.a)
(ṁ
RT 0
P 2 A
(
287 × 373 1
200, 000 2 × 0.001
According to Table 11.1 the Mach number is about M =0.74 ··· (the exact number
does not appear here demonstrate the simplicity of the solution). The Velocity can be
obtained from the
U = Mc= M √ kRT
(11.VIII.b)
The only unknown the equation (11.VIII.b) is the temperature. However, the temperature
can be obtained from knowing the Mach number with the “regular” table. Utilizing
the regular table or Potto GDC one obtained.
M
T
T 0
ρ
ρ 0
A
A ⋆
P
P 0
A×P
A ∗ ×P 0
F
F ∗
0.74000 0.89686 0.77169 1.0677 0.69210 0.73898 0.54281