Basics of Fluid Mechanics, 2014a

448 CHAPTER 11. COMPRESSIBLE FLOW ONE DIMENSIONAL
The temperature is
T 1 = T 1
T 01 =0.357 × 400 = 142.8K
T 01
Using the temperature, the speed of sound can be calculated as
c 1 = √ kRT = √ 1.4 × 287 × 142.8 ≃ 239.54[m/sec]
The pressure at point 1 can be calculated as
P 1 = P 1
P 01 =0.027 × 30 ≃ 0.81[Bar]
P 01
The density as a function of other properties at point 1 is
ρ 1 =
P [ ]
8.1 × 104 kg
RT ∣ =
1
287 × 142.8 ≃ 1.97 m 3
The mass flow rate can be evaluated from equation (11.154)
ṁ =1.97 × π × 0.0252
4
[ ] kg
× 3 × 239.54 = 0.69
sec
(b) First, check whether the flow is shockless by comparing the flow resistance and
the maximum possible resistance. From the Table 11.6 or by using the famous
Potto–GDC, is to obtain the following
M
4fL
D
P
P ∗ P 0
P 0
∗
ρ
ρ ∗
U
U ∗
T
T ∗
3.0000 0.52216 0.21822 4.2346 0.50918 1.9640 0.42857
and the conditions of the tube are
4fL
D
=
4 × 0.005 × 1.0
0.025
=0.8
Since 0.8 > 0.52216 the flow is choked and with a shock wave.
The exit pressure determines the location of the shock, if a shock exists, by
comparing “possible” P exit to P B . Two possibilities are needed to be checked;
one, the shock at the entrance of the tube, and two, shock at the exit and
comparing the pressure ratios. First, the possibility that the shock wave occurs
immediately at the entrance for which the ratio for M x are (shock wave Table
11.3)