Basics of Fluid Mechanics, 2014a

11.7. FANNO FLOW 465
0.9
0.8
0.7
0.6
0.5
0.4
0.3
M 1
M 2
P 1
P 2
0 1 2 3 4 5 6 7
i, number of iteration
Fig. -11.35. Conversion of solution for given 4 fL
D
=0.5 and pressure ratio equal 0.8.
Example runs is presented in the Figure 11.35 for 4 fL
D
=0.5 and pressure ratio
equal to 0.8. The blue line in Figure 11.34 intersection with the horizontal line of
P 2 /P 1 =0.8 yield the solution of M ∼ 0.5. The whole solution obtained in 7 iterations
with accuracy of 10 −12 .
In Potto–GDC there is another older iterative method used to solve constructed
on the properties of several physical quantities must be in a certain range. The first fact
is that the pressure ratio P 2 /P 1 is always between 0 and 1 (see Figure 11.33). In the
figure, a theoretical extra tube is added in such a length that cause the flow to choke
(if it really was there). This length is always positive (at minimum is zero).
The procedure for the calculations is as the following:
′
′
1) Calculate the entrance Mach number, M 1 assuming the 4 fL
D
= 4 fL
D
∣
max
(chocked flow);
′
Calculate the minimum pressure ratio (P 2 /P 1 ) min
for M 1 (look at table (11.6))
2) Check if the flow is choked:
There are two possibilities to check it.
a) Check if the given 4 fL
4 fL
D
is smaller than
D
obtained from the given P 1/P 2 ,or
b) check if the (P 2 /P 1 ) min
is larger than (P 2 /P 1 ),
continue if the criteria is satisfied. Or if not satisfied abort this procedure and
continue to calculation for choked flow.
3) Calculate the M 2 based on the (P ∗ /P 2 )=(P 1 /P 2 ),
4) calculate Δ 4 fL
D based on M 2,