Subsonic Fanno Flow for a Given Entrance Mach Number and Pressure Ratio

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, $M_1$ , and given pressure ratio calculate the flow parameters, like the exit Mach number, $M_2$ . The procedure is based on the fact that the entrance star pressure ratio can be calculated using $M_1$ . Thus, using the pressure ratio to calculate the star exit pressure ratio provide the exit Mach number, $M_2$ . 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 Potto-GDC 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

$${P_2 \over P^{*} } = {{P_2 \over P_1 } {P_1 \over P^{*} } } = 0.4 \times 4.3546 = 1.74184$$

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 Potto-GDC 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 $\dot{m}/A$ is not provided and the other hand $\frac{4fL}{D}$ 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 $M_1=0.2$ as

$$D = {4 f L \over \frac{4fL}{D}} = { 4\times 0.005 \times 5 / 3.1887 } = 0.03136[m]$$

The same was done for all the other Mach number. Now the area can be calculated and therefor the $\dot{m}/A$ can be calculated. With this information the ``naughty professor'' is given and the entrance Mach number can be calculated. For example for $M_1=0.2$ one can obtain the following:

$$\dot{m} /A = 0.1 / (\pi\times0.03136^2/4) \sim 129.4666798$$

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 ($T$ and $P_0$ are given

From this exhibit it can be estimated that $M_1 = 0.18$ . 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 $D \sim 0.0251[m]$

The flow rate is $\dot{m} \sim 202.1[kg/sec\times m^2$

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.