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This situation pose a simple mathematical problem while the physical
situation occurs in cases where a specific flow rate is required with
a given pressure ratio (range) (this problem was considered by some
to be somewhat complicated).
The specific flow rate can be converted to entrance Mach number and
this simplifies the problem.
Thus, the problem is reduced to find for given entrance Mach,
, and given
pressure ratio calculate the flow parameters, like the exit Mach number,
.
The procedure is based on the fact that the entrance star pressure ratio
can be calculated using
.
Thus, using the pressure ratio to calculate the star exit pressure ratio
provide the exit Mach number,
.
An example of such issue is the following example that combines also
the ``Naughty professor'' problems.
Solution
The star pressure can be obtained from a table or PottoGDC
as
Fanno Flow 
Input: M 
k = 1.4 
M 
fld 
P/P* 
P0/P0* 
ρ/ρ* 
U/U* 
T/T* 
0.25 
8.48341 
4.35465 
2.40271 
3.67423 
0.272166 
1.18519 
And the star pressure ratio can be calculated at the exit as following
And the corresponding exit Mach number for this pressure ratio reads
Fanno Flow 
Input: Pbar 
k = 1.4 
M 
fld 
P/P* 
P0/P0* 
ρ/ρ* 
U/U* 
T/T* 
0.60694 
0.464084 
1.74184 
1.18013 
1.55848 
0.641652 
1.11766 
A bit show off the PottoGDC can carry these calculations in one click as
Fanno Flow 
Input: M1 P2/P1

k = 1.4 
M1 
M2 
fld 
P2/P1 
0.25 
0.606934 
8.0193 
0.4 

While the above example show the most simple form of this question, in
reality this question is more complicated.
One common problem is situation that the diameter is not given
but the flow rate and length and pressure (stagnation or static) with some
combination of the temperature. The following example deal with one of
such example.
Solution
The direct mathematical solution isn't possible and some kind of iteration
procedure or root finding for a representative function.
For the first part the ``naughty professor'' procedure cannot be used
because
is not provided
and the other hand
is not provided (missing Diameter).
One possible solution is to guess the entrance Mach and check whether
and the mass flow rate with the ``naughty professor'' procedure are
satisfied.
For Fanno flow at for several Mach numbers the following is obtained
Fanno Flow 
Input: M1 P2/P1 
k = 1.4 


M1 
M2 
fld 
P2/P1 
Diameter 
0.1 
0.111085 
13.3648 
0.9 
0.00748 
0.15 
0.16658 
5.82603 
0.9 
0.01716 
0.2 
0.222018 
3.18872 
0.9 
0.03136 
From the last table the diameter can be calculated for example for
as
The same was done for all the other Mach number.
Now the area can be calculated and therefor the
can be
calculated. With this information the ``naughty professor'' is given
and the entrance Mach number can be calculated.
For example for
one can obtain the following:
The same order as the above table it shown in ``naughty professor''
(isentropic table).
Isentropic Flow 
Input: M 
k = 1.4 
M 
T/T0 
ρ/ρ0 
A/A* 
P/P0 
PAR 
F/F* 
1.5781 
0.667521 
0.364051 
1.2329 
0.243012 
0.299609 
0.560088 
0.36221 
0.974432 
0.9373 
1.7268 
0.913335 
1.57714 
0.777844 
0.10979 
0.997595 
0.993998 
5.30922 
0.991608 
5.26466 
2.23063 
The first result are not reasonable and this process can continue until the
satisfactory solution is achieved.
Here an graphical approximation is shown.
Figure 9.20:
Diagram for finding solution
when the pressure ratio and entrance properties (
and
are
given

From this exhibit it can be estimated that
.
For this Mach number the following can be obtained
Fanno Flow 
Input: M1 P2/P1

k = 1.4 
M1 
M2 
fld 
P2/P1 
0.18 
0.19985 
3.98392 
0.9 
Thus, the diameter can be obtained as
The flow rate is
Isentropic Flow 
Input: M 
k = 1.4 
M 
T/T0 
ρ/ρ0 
A/A* 
P/P0 
PAR 
F/F* 
0.1719 
0.994125 
0.985377 
3.42655 
0.979587 
3.35661 
1.45645 
The exact solution is between 0.17 and 0.18 if better accuracy is needed.

Next: The Approximation of the
Up: The Practical Questions and
Previous: Subsonic Fanno Flow for
Index
Created by:Genick BarMeir, Ph.D.
On:
20071121