# 11.7.9: The Practical Questions and Examples of Subsonic branch

The Fanno is applicable also when the flow isn't choke. In this case, several questions appear for the subsonic branch. This is the area shown in Figure 11.25 in beginning This kind of questions made of pair given information to find the conditions of the flow, as oppose to only one piece of information given in choked flow. There many combinations that can appear in this situation but there are several more physical and practical that will be discussed here.

## Subsonic Fanno Flow for Given $$\dfrac{4\,f\,L}{D}$$ and Pressure Ratio

Fig. 11.34 Unchoked flow calculations showing the hypothetical "full'' tube when choked.

This pair of parameters is the most natural to examine because, in most cases, this information is the only provided information. For a given pipe $$\left(\dfrac{4\,f\,L}{D}\right)$$, neither the entrance Mach number nor the exit Mach number are given (sometimes the entrance Mach number is given see the next section). There is no known exact analytical solution. There are two possible approaches to solve this problem: one, by building a representative function and find a root (or roots) of this representative function. Two, the problem can be solved by an iterative procedure. The first approach require using root finding method and either method of spline method or the half method or the combination of the two. In the past, this book advocated the integrative method. Recently, this author investigate proposed an improved method. This method is based on the entrance Mach number as the base. The idea based on the idea that the pressure ratio can be drawn as a function of the entrance Mach number. One of difficulties lays in the determination the boundaries of the entrance Mach number. The maximum entrance Mach number is chocking Mach number. The lower possible Mach number is zero which creates very large $$\dfrac{4\,f\,L}{D}$$. The equations are solve for these large $$\dfrac{4\,f\,L}{D}$$ numbers by perturbation method and the analytical solution is

$\label{shock:eq:largeFLDP2P1} M_{1} = \sqrt{ \dfrac{1- \left[ \dfrac{P_{2}}{P_{0}}\right]^{2} } { k\,{\dfrac{4\,f\,L}{D}} } }$

Equation (7) is suggested to be used up to $$M_1< 0.02$$. To have small overlapping zone the lower boundary is $$M_1< 0.01$$.

Fig. 11.35 Pressure ratio obtained for a fix $$\dfrac{4\,f\,L}{D}$$ as a function of Mach number

The process is based on finding the pressure ratio for given $$\dfrac{4\,f\,L}{D}$$ pipe dimensionless length. Figure 11.35 exhibits the pressure ratio for fix $$\dfrac{4\,f\,L}{D}$$ as function of the entrance Mach number. As it can be observed, the entrance Mach number lays between zero and the maximum of the chocking conditions. For example for a fixed pipe, $$\dfrac{4\,f\,L}{D}=1$$ the maximum Mach number is 0.50874 as shown in Figure ?? by orange line. For a given entrance Mach number, the pressure ratio, $$P_1/P^{*} and \left.\dfrac{4\,f\,L}{D}\right|_1$$ can be calculated. The exit pipe length, $$\left.\dfrac{4\,f\,L}{D}\right|_2$$ is obtained by subtracting the fix length $$\dfrac{4\,f\,L}{D}$$ from $$\left.\dfrac{4\,f\,L}{D}\right|_1$$. With this value, the exit Mach number, $$M_2$$ and pressure ratio $$P_2/P^{*}$$ are calculated. Hence the pressure ratio, $$P_2/P_1$$ can be obtained and is drawn in Figure 11.35 Hence, when the pressure ratio, $$P_2/P_1$$ is given along with given pipe, $$\dfrac{4\,f\,L}{D}$$ the solution can be obtained by drawing a horizontal line. The intersection of the horizontal line with the right curve of the pressure ratio yields the entrance Mach number. This can be done by a computer program such Potto–GDC (version 0.5.2 and above). The summary of the procedure is as the following.
1. If the pressure ratio is $$P_2/P_1 < 0.02$$ then using the perturbed solution the entrance Mach number is very small and calculate using the formula

$M = \sqrt{(1 - \dfrac{\dfrac{P_2}{P_1}}{ k\, \left(\dfrac{4\,f\,L}{D} \right)}} \label{fanno:eq:MforSmallP2P1}$

If the pressure ratio smaller than continue with the following.
2. Calculate the $$\left.\dfrac{4\,f\,L}{D}\right|_1$$ for $$M_1=0.01$$
3. Subtract the given $$\dfrac{4\,f\,L}{D}$$ from $$\left.\dfrac{4\,f\,L}{D}\right|_1$$ and calculate the exit Mach number.
4. Calculate the pressure ratio.
5. Calculate the pressure ratio for choking condition (given $$\dfrac{4\,f\,L}{D}$$.
6. Use your favorite to method to calculate root finding (In potto–GDC Brent's method is used)

Fig. 11.36 Conversion of solution for given $$\dfrac{4\,f\,L}{D}=0.5$$ and pressure ratio equal 0.8.

Example runs is presented in the Figure 11.36 for $$\dfrac{4\,f\,L}{D}=0.5$$ and pressure ratio equal to 0.8. The blue line in Figure 11.35 intersection with the horizontal line of $$P_2/P_1=0.8$$ yield the solution of $$M\sim 0.5$$. 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 11.34). 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 $$\dfrac{4\,f\,L}{D} = {\left.\dfrac{4\,f\,L}{D}\right|_{max}}^{'}$$ (chocked flow); Calculate the minimum pressure ratio $$\left(P_2/P_1\right)_{min}$$ for $${M_1}^{'}$$ (look at table)
2. Check if the flow is choked: There are two possibilities to check it.
1. Check if the given $$\dfrac{4\,f\,L}{D}$$ is smaller than $$\dfrac{4\,f\,L}{D}$$ obtained from the given $$P_1/P_2$$, or
2. Check if the $$\left(P_2/P_1\right)_{min}$$ is larger than $$\left(P_2/P_1\right)$$, 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 $$\left(P^{*} / P_2\right) = \left(P_1 / P_2\right)$$,
4. Calculate $$\Delta \dfrac{4\,f\,L}{D}$$ based on $$M_2$$,
5. calculate the new $$\left(P_2 / P_1\right)$$, based on the new $$f\left(\left(\dfrac{4\,f\,L}{D}\right)_1, \left(\dfrac{4\,f\,L}{D}\right)_2\right)$$, (remember that $$\Delta\dfrac{4\,f\,L}{D} = \left(\dfrac{4\,f\,L}{D}\right)_2$$),
6. Calculate the corresponding $$M_1$$ and $$M_2$$,
7. Calculate the new and "improved'' the $$\Delta \dfrac{4\,f\,L}{D}$$ by

$\left(\Delta \dfrac{4\,f\,L}{D}\right)_{new} = \left(\Delta \dfrac{4\,f\,L}{D} \right)_{old} *{ \left(P_2 \over P_1 \right)_{given} \over \left(P_2 \over P_1 \right)_{old} } \label{fanno:eq:improvedFLD}$

Note, when the pressure ratios are matching also the $$\Delta \dfrac{4\,f\,L}{D}$$ will also match.
8. Calculate the "improved/new'' $$M_2$$ based on the improve $$\Delta \dfrac{4\,f\,L}{D}$$
9. Calculate the improved $$\dfrac{4\,f\,L}{D}$$ as $$\dfrac{4\,f\,L}{D} = \left(\dfrac{4\,f\,L}{D}\right)_{given} + \Delta \left(\dfrac{4\,f\,L}{D}\right)_{new}$$
10. Calculate the improved $$M_1$$ based on the improved $$\dfrac{4\,f\,L}{D}$$.
11. Compare the abs ($$\left(P_2/P_1\right)_{new} - \left(P_2/P_1\right)_{old}$$ )and if not satisfied returned to stage (??) until the solution is obtained.

To demonstrate how this procedure is working consider a typical example of $$\dfrac{4\,f\,L}{D}=1.7$$ and $$P_2/P_1 = 0.5$$. Using the above algorithm the results are exhibited in the following figure.

Fig. 11.37 The results of the algorithm showing the conversion rate for unchoked Fanno flow model with a given $$\dfrac{4\,f\,L}{D}$$ and pressure ratio.

Figure 11.37 demonstrates that the conversion occur at about 7-8 iterations. With better first guess this conversion procedure converts much faster but at a certain range it is unstable.