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# Chapter 11 Gd (continue)

## 11.7 Fanno Flow Fig. 11.19 Control volume of the gas flow in a constant cross section for Fanno Flow.

This adiabatic flow model with friction is named after Ginno Fanno a Jewish engineer. This model is the second pipe flow model described here. The main restriction for this model is that heat transfer is negligible and can This model is applicable to flow processes which are very fast compared to heat transfer mechanisms with small Eckert number. This model explains many industrial flow processes which includes emptying of pressured container through a relatively short tube, exhaust system of an internal combustion engine, compressed air systems, etc. As this model raised from the need to explain the steam flow in turbines.

### 11.9.1 Introduction

Consider a gas flowing through a conduit with a friction (see Figure 11.19). It is advantages to examine the simplest situation and yet without losing the core properties of the process. The mass (continuity equation) balance can be written as \begin{align} \begin{array}{c} \dot{m} = \rho\, A \,U = constant\ \hookrightarrow \rho_1 \, U_1 = \rho_2\, U_2 \end{array} \label{fanno:eq:mass} \end{align} The energy conservation (under the assumption that this model is adiabatic flow and the friction is not transformed into thermal energy) reads \begin{align} \begin{array}{rl} {T_{0}}_1 &= {T_{0}}_2 \ \hookrightarrow T_1 + \dfrac{ {U_1}^2 }{ 2\,c_p} &= T_2 + \dfrac{ {U_2}^2 }{ 2\,c_p} \end{array} \label{fanno:eq:energy} \end{align} Or in a derivative from \begin{align} C_p\, dT +d \left( U^2 \over 2 \right) = 0 \label{fanno:eq:energyDerivative} \end{align} Again for simplicity, the perfect gas model is assumed is slower in reality. However, experiments from many starting with 1938 work by has shown that the error is not significant. Nevertheless, the comparison with reality shows that heat transfer cause changes to the flow and they need/should to be expected. These changes include the choking point at lower Mach number.

### 11.7.5.1 Maximum Length for the Supersonic Flow

It has to be noted and recognized that as opposed to subsonic branch the supersonic branch has a limited length. It also must be recognized that there is a maximum length for which only supersonic flow can exist. The maximum length of the supersonic can be evaluated when $M = \infty$ as follows: \begin{multline*} \dfrac{4\, f\,L_{max} }{ D} = \dfrac{1 - M^2 }{ k\, M^2} + \dfrac{k+1 }{ 2\,k}\ln \dfrac{\dfrac{k+1 }{2}\,M^2} {2\, \left(1+ \dfrac{k-1 }{ 2}\,M^2 \right)} = \\ ld \left( M\rightarrow\infty \right) \sim \dfrac{- \infty }{ k \times \infty} + \dfrac{k + 1 }{ 2\,k} \ln \dfrac{ (k+1)\, \infty }{ (k-1)\, \infty} = \\ \dfrac{-1 }{ k} + \dfrac{k + 1 }{ 2\,k} \,\ln \dfrac{ (k+1) }{ (k-1) } = ld ( M\rightarrow\infty , k=1.4) = 0.8215 \end{multline*} \begin{align} \dfrac{4 \,f\,L_{max} }{ D} = ld ( M\rightarrow\infty , k=1.4) = 0.8215 \end{align} The maximum length of the supersonic flow is limited by the above number. From the above analysis, it can be observed that no matter how high the entrance Mach number will be the tube length is limited and depends only on specific heat ratio, $k$.

### 11.7.6 Working Conditions Fig. 11.23 The effects of increase of $\dfrac{4\,f\,L}{D}$ on the Fanno line.

It has to be recognized that there are two regimes that can occur in Fanno flow model one of subsonic flow and the other supersonic flow. Even the flow in the tube starts as a supersonic in parts of the tube can be transformed into the subsonic branch. A shock wave can occur and some portions of the tube will be in a subsonic flow pattern. The discussion has to differentiate between two ways of feeding the tube: converging nozzle or a converging-diverging nozzle. Three parameters, the dimensionless friction, $\dfrac{4\,f\,L}{D}$, the entrance Mach number, $M_1$, and the pressure ratio, $P_2/P_1$ are controlling the flow. Only a combination of these two parameters is truly independent. However, all the three parameters can be varied and some are discussed separately here.

### 11.7.6.1 Variations of The Tube Length ($ld$) Effects

In the analysis of this effect, it should be assumed that back pressure is constant and/or low as possible as needed to maintain a choked flow. First, the treatment of the two branches are separated.

#### Fanno Flow Subsonic branch Fig. 11.24 The effects of the increase of $\dfrac{4\,f\,L}{D}$ on the Fanno Line.

For converging nozzle feeding, increasing the tube length results in increasing the exit Mach number (normally denoted herein as $M_2$). Once the Mach number reaches maximum ($M = 1$), no further increase of the exit Mach number can be achieved with same pressure ratio mass flow rate. For increase in the pipe length results in mass flow rate decreases. It is worth noting that entrance Mach number is reduced (as some might explain it to reduce the flow rate). The entrance temperature increases as can be seen from Figure 11.24. The velocity therefore must decrease because the loss of the enthalpy (stagnation temperature) is used.'' The density decrease because $\rho = \dfrac{P }{ R\, T}$ and when pressure is remains almost constant the density decreases. Thus, the mass flow rate must decrease. These results are applicable to the converging nozzle. In the case of the converging–diverging feeding nozzle, increase of the dimensionless friction, $ld$, results in a similar flow pattern as in the converging nozzle. Once the flow becomes choked a different flow pattern emerges.

### 11.7.6.2 Fanno Flow Supersonic Branch Fig. 11.25 The Mach numbers at entrance and exit of tube and mass flow rate for Fanno Flow as a function of the $ld$.}

There are several transitional points that change the pattern of the flow. Point $\mathbf{a}$ is the choking point (for the supersonic branch) in which the exit Mach number reaches to one. Point $\mathbf{b}$ is the maximum possible flow for supersonic flow and is not dependent on the nozzle. The next point, referred here as the critical point $\mathbf{c}$, is the point in which no supersonic flow is possible in the tube i.e. the shock reaches to the nozzle. There is another point $\mathbf{d}$, in which no supersonic flow is possible in the entire nozzle–tube system. Between these transitional points the effect parameters such as mass flow rate, entrance and exit Mach number are discussed. At the starting point the flow is choked in the nozzle, to achieve supersonic flow. The following ranges that has to be discussed includes (see Figure 11.25): The 0-$\mathbf{a}$ range, the mass flow rate is constant because the flow is choked at the nozzle. The entrance Mach number, $M_1$ is constant because it is a function of the nozzle design only. The exit Mach number, $M_2$ decreases (remember this flow is on the supersonic branch) and starts ($\dfrac{4\,f\,L}{D}=0$) as $M_2= M_1$. At the end of the range $\mathbf{a}$, $M_2=1$. In the range of $\mathbf{a}-\mathbf{b}$ the flow is all supersonic. In the next range $\mathbf{a}-\mathbf{b}$ the flow is double choked and make the adjustment for the flow rate at different choking points by changing the shock location. The mass flow rate continues to be constant. The entrance Mach continues to be constant and exit Mach number is constant. The total maximum available for supersonic flow $\mathbf{b}-\mathbf{b'}$, $\left(\dfrac{4\,f\,L}{D}\right)_{max}$, is only a theoretical length in which the supersonic flow can occur if nozzle is provided with a larger Mach number (a change to the nozzle area ratio which also reduces the mass flow rate). In the range $\mathbf{b} -\mathbf{c}$, it is a more practical point. In semi supersonic flow $\mathbf{b} - \mathbf{c}$ (in which no supersonic is available in the tube but only in the nozzle) the flow is still double choked and the mass flow rate is constant. Notice that exit Mach number, $M_2$ is still one. However, the entrance Mach number, $M_1$, reduces with the increase of $\dfrac{4\,f\,L}{D}$. It is worth noticing that in the $\mathbf{a} - \mathbf{c}$ the mass flow rate nozzle entrance velocity and the exit velocity remains constant! In the last range $\bbb{c} - \infty$ the end is really the pressure limit or the break of the model and the isothermal model is more appropriate to describe the flow. In this range, the flow rate decreases since ($\dot{m} \propto M_1$). To summarize the above discussion, Figures 11.25 exhibits the development of $M_1$, $M_2$ mass flow rate as a function of $\dfrac{4\,f\,L}{D}$. Somewhat different then the subsonic branch the mass flow rate is constant even if the flow in the tube is completely subsonic. This situation is because of the double'' choked condition in the nozzle. The exit Mach $M_2$ is a continuous monotonic function that decreases with $\dfrac{4\,f\,L}{D}$. The entrance Mach $M_1$ is a non continuous function with a jump at the point when shock occurs at the entrance moves'' into the nozzle. Fig. 11.26 $M_1$ as a function $M_2$ for various $ld$.

Figure ef{fanno:fig:fannoM1M2} exhibits the $M_1$ as a function of $M_2$. The Figure was calculated by utilizing the data from Figure 11.20 by obtaining the $\left.\dfrac{4\,f\,L}{D}\right|_{max}$ for $M_2$ and subtracting the given $\dfrac{4\,f\,L}{D}$ and finding the corresponding $M_1$. Fig. 11.27 $M_1$ as a function $M_2$ for different $ld$ for

The Figure \eqref{fanno:fig:fannoM1M2Supper} exhibits the entrance Mach number as a function of the $M_2$. Obviously there can be two extreme possibilities for the subsonic exit branch. Subsonic velocity occurs for supersonic entrance velocity, one, when the shock wave occurs at the tube exit and two, at the tube entrance. In Figure ef{fanno:fig:fannoM1M2Supper} only for $\dfrac{4\,f\,L}{D}=0.1$ and $\dfrac{4\,f\,L}{D}=0.4$ two extremes are shown. For $\dfrac{4\,f\,L}{D}= 0.2$ shown with only shock at the exit only. Obviously, and as can be observed, the larger $\dfrac{4\,f\,L}{D}$ creates larger differences between exit Mach number for the different shock locations. The larger $\dfrac{4\,f\,L}{D}$ larger $M_1$ must occurs even for shock at the entrance. For a given $\dfrac{4\,f\,L}{D}$, below the maximum critical length, the supersonic entrance flow has three different regimes which depends on the back pressure. One, shockless flow, tow, shock at the entrance, and three, shock at the exit. Below, the maximum critical length is mathematically \begin{align*} \dfrac{4\,f\,L}{D} > - \dfrac{1 }{k} + \dfrac{1+k }{ 2\,k} \ln\left(\dfrac{ k+1 }{ k-1}\right) \end{align*} For cases of $\dfrac{4\,f\,L}{D}$ above the maximum critical length no supersonic flow can be over the whole tube and at some point a shock will occur and the flow becomes subsonic flow.

### The Pressure Ratio, $\left.{P_2 }\right/{ P_1}$, effects

In this section the studied parameter is the variation of the back pressure and thus, the pressure ratio $\left(\left.P_2 \right/ P_1\right)$ variations. For very low pressure ratio the flow can be assumed as incompressible with exit Mach number smaller than $<0.3$. As the pressure ratio increases (smaller back pressure, $P_2$), the exit and entrance Mach numbers increase. According to Fanno model the value of $\dfrac{4\,f\,L}{D}$ is constant (friction factor, $f$, is independent of the parameters such as, Mach number, Reynolds number et cetera) thus the flow remains on the same Fanno line. For cases where the supply come from a reservoir with a constant pressure, the entrance pressure decreases as well because of the increase in the entrance Mach number (velocity). Again a differentiation of the feeding is important to point out. If the feeding nozzle is converging than the flow will be only subsonic. If the nozzle is converging–diverging'' than in some part supersonic flow is possible. At first the converging nozzle is presented and later the converging-diverging nozzle is explained. Fig. 11.28 The pressure distribution as a function of $\dfrac{4\,f\,L}{D}$ for a short $\dfrac{4\,f\,L}{D}$.

### 11.7.7.1 Choking explanation for pressure variation/reduction

Decreasing the pressure ratio or in actuality the back pressure, results in increase of the entrance and the exit velocity until a maximum is reached for the exit velocity. The maximum velocity is when exit Mach number equals one. The Mach number, as it was shown in Section 11.4, can increases only if the area increase. In our model the tube area is postulated as a constant therefore the velocity cannot increase any further. However, for the flow to be continuous the pressure must decrease and for that the velocity must increase. Something must break since there are conflicting demands and it result in a jump'' in the flow. This jump is referred to as a choked flow. Any additional reduction in the back pressure will not change the situation in the tube. The only change will be at tube surroundings which are irrelevant to this discussion. If the feeding nozzle is a converging–diverging'' then it has to be differentiated between two cases; One case is where the $\dfrac{4\,f\,L}{D}$ is short or equal to the critical length. The critical length is the maximum $\left.\dfrac{4\,f\,L}{D}\right|_{max}$ that associate with entrance Mach number. Fig. 11.29 $\,$ The pressure distribution as a function of $\dfrac{4\,f\,L}{D}$ for a long $\dfrac{4\,f\,L}{D}$.

### Short $\left.{4\,f\,L}\right/{D}$

Figure 11.29 shows different pressure profiles for different back pressures. Before the flow reaches critical point a (in the Figure ) the flow is subsonic. Up to this stage the nozzle feeding the tube increases the mass flow rate (with decreasing back pressure). Pressure between point a and point b the shock is in the nozzle. In this range and further reduction of the pressure the mass flow rate is constant no matter how low the back pressure is reduced. Once the back pressure is less than point b the supersonic reaches to the tube. Note however that exit Mach number, $M_2 < 1$ and is not 1. A back pressure that is at the critical point c results in a shock wave that is at the exit. When the back pressure is below point c , the tube is clean'' of any The back pressure below point c} has some adjustment as it occurs with exceptions of point \textbf{d. Fig. 11.30 The effects of pressure variations on Mach number profile as a function of $\dfrac{4\,f\,L}{D}$ when the total resistance $\dfrac{4\,f\,L}{D} = 0.3$ for Fanno Flow.

### Long $\dfrac{4\,f\,L}{D}$

In the case of $\dfrac{4\,f\,L}{D} > \left.\dfrac{4\,f\,L}{D}\right|_{max}$ reduction of the back pressure results in the same process as explained in the short $\dfrac{4\,f\,L}{D}$ up to point c. However, point c in this case is different from point c at the case of short tube $\dfrac{4\,f\,L}{D} < \left.\dfrac{4\,f\,L}{D}\right|_{max}$. In this point the exit Mach number is equal to 1 and the flow is double shock. Further reduction of the back pressure at this stage will not move'' the shock wave downstream the nozzle. At point c or location of the shock wave, is a function entrance Mach number, $M_1$ and the extra'' $\dfrac{4\,f\,L}{D}$. The procedure is (will be) presented in later stage. Fig. 11.31 Pressure ratios as a function of $\dfrac{4\,f\,L}{D}$ when the total $\dfrac{4\,f\,L}{D} = 0.3$. Fig. 11.32 The extra tube length as a function of the shock location, $\dfrac{4\,f\,L}{D}$ supersonic branch.

#### The Maximum Location of the Shock

The main point in this discussion however, is to find the furthest shock location downstream. Figure 11.32 shows the possible $\Delta \left(4\,f\,L \over D \right)$ as a function of retreat of the location of the shock wave from the maximum location. When the entrance Mach number is infinity, $M_1= \infty$, if the shock location is at the maximum length, then shock at $M_x = 1$ results in $M_y=1$. The proposed procedure is based on Figure 11.32. beginNormalEnumerate change startEnumerate=1

1. \parbox[t]{0.92\textwidth}{ Calculate the extra $\dfrac{4\,f\,L}{D}$ and subtract the actual extra $\dfrac{4\,f\,L}{D}$ assuming shock at the left side (at the max length). }
2. \parbox[t]{0.92\textwidth}{ Calculate the extra $\dfrac{4\,f\,L}{D}$ and subtract the actual extra $\dfrac{4\,f\,L}{D}$ assuming shock at the right side (at the entrance). }
3. According to the positive or negative utilizes your root finding procedure. Fig. 11.33 The maximum entrance Mach number, $M_1$ to the tube as a function of $ld$ supersonic branch.

From numerical point of view, the Mach number equal infinity when left side assumes result in infinity length of possible extra (the whole flow in the tube is subsonic). To overcome this numerical problem, it is suggested to start the calculation from $\epsilon$ distance from the right hand side. Let denote \begin{align} \Delta \left( \dfrac{4\,f\,L}{D} \right ) = \bar{\dfrac{4\,f\,L}{D}}_{actual} - \left.\dfrac{4\,f\,L}{D}\right|_{sup} \label{fanno:eq:deltaFLD} \end{align} Note that $\left.\dfrac{4\,f\,L}{D}\right|_{sup}$ is smaller than $\left.\dfrac{4\,f\,L}{D}\right|_{max_{\infty}}$. The requirement that has to be satisfied is that denote $\left.{\dfrac{4\,f\,L}{D}}\right|_{retreat}$ as difference between the maximum possible of length in which the supersonic flow is achieved and the actual length in which the flow is supersonic see Figure 11.33. The retreating length is expressed as subsonic but \begin{align} \left. \dfrac{4\,f\,L}{D} \right|_{retreat} = \left.\dfrac{4\,f\,L}{D}\right|_{max_{\infty}} - \left.\dfrac{4\,f\,L}{D}\right|_{sup} \label{fanno:eq:FLDretreat} \end{align} Figure 11.33 shows the entrance Mach number, $M_1$ reduces after the maximum length is exceeded.

# Example 11.20

Calculate the shock location for entrance Mach number $M_1 = 8$ and for $\dfrac{4\,f\,L}{D} = 0.9$ assume that $k = 1.4$ ($M_{exit} = 1$).

# Solution

The solution is obtained by an iterative process. The maximum $\left.\dfrac{4\,f\,L}{D}\right|_{max}$ for $k=1.4$ is 0.821508116. Hence, $\dfrac{4\,f\,L}{D}$ exceed the maximum length $\dfrac{4\,f\,L}{D}$ for this entrance Mach number. The maximum for $M_1 =8$ is $\dfrac{4\,f\,L}{D} = 0.76820$, thus the extra tube is $\Delta \left( \dfrac{4\,f\,L}{D} \right) = 0.9 - 0.76820 = 0.1318$. The left side is when the shock occurs at $\dfrac{4\,f\,L}{D} = 0.76820$ (flow is choked and no additional $\dfrac{4\,f\,L}{D}$). Hence, the value of left side is $-0.1318$. The right side is when the shock is at the entrance at which the extra $\dfrac{4\,f\,L}{D}$ is calculated for $M_x$ and $M_y$ is

 ShockWave Input: $M_2$ $k = 1.4$M_xM_y\dfrac{T_y}{T_x}\dfrac{\rho_y}{\rho_x}\dfrac{P_y}{P_x}\dfrac{{P_0}_y}{{P_0}_x}$8.00 0.39289 13.3867 5.5652 74.50 0.00849 With$(M_1)'$ Fanno Flow Input:$\dfrac{4\,f\,L}{D}$k = 1.4$M\dfrac{4\,f\,L}{D}\dfrac{P}{P^{\star}}\dfrac{P_0}{{P_0}^{\star}}\dfrac{\rho}{\rho^{\star} }\dfrac{U}{U^{\star}}\dfrac{T_0}{{T_0}^{\star}}$0.39289 2.4417 2.7461 1.6136 2.3591 0.42390 1.1641 The extra$\Delta\left(\dfrac{4\,f\,L}{D}\right)$is$2.442 - 0.1318 = 2.3102 $Now the solution is somewhere between the negative of left side to the positive of the right side. In a summary of the actions is done by the following algorithm: beginNormalEnumerate change startEnumerate=1 1. check if the$\dfrac{4\,f\,L}{D}$exceeds the maximum$\left.{\dfrac{4\,f\,L}{D}}\right|_{max}$for the supersonic flow. Accordingly continue. 2. Guess$\left.{\dfrac{4\,f\,L}{D}}\right|_{up}= \dfrac{4\,f\,L}{D} - \left.\dfrac{4\,f\,L}{D}\right|_{max}$3. Calculate the Mach number corresponding to the current guess of$\left.{\dfrac{4\,f\,L}{D}}\right|_{up}$, 4. Calculate the associate Mach number,$M_x$with the Mach number,$M_y$calculated previously, 5. Calculate$\dfrac{4\,f\,L}{D}$for supersonic branch for the$M_x$6. Calculate the new and improved''$\left.{\dfrac{4\,f\,L}{D}}\right|_{up}$7. Compute the new$\left.{\dfrac{4\,f\,L}{D}}\right|_{down} = {\dfrac{4\,f\,L}{D}} - \left.{\dfrac{4\,f\,L}{D}}\right|_{up}$8. Check the new and improved$\left.\dfrac{4\,f\,L}{D}\right|_{down}$against the old one. If it is satisfactory stop or return to stage b. Shock location is:  Fanno Flow Input:$M_1$and$\dfrac{4\,f\,L}{D}$k = 1.4$M_1M_2\left.\dfrac{4\,f\,L}{D}\right|_{up}\left.\dfrac{4\,f\,L}{D}\right|_{down}M_xM_y$8.00 1.00 0.57068 0.32932 1.6706 0.64830  Fanno Flow Input: previous k = 1.4$i\left.\dfrac{4\,f\,L}{D}\right|_{up}\left.\dfrac{4\,f\,L}{D}\right|_{down}M_xM_y$0 0.67426 0.22574 1.3838 0.74664 1 0.62170 0.27830 1.5286 0.69119 2 0.59506 0.30494 1.6021 0.66779 3 0.58217 0.31783 1.6382 0.65728 4 0.57605 0.32395 1.6554 0.65246 5 0.57318 0.32682 1.6635 0.65023 6 0.57184 0.32816 1.6673 0.64920 7 0.57122 0.32878 1.6691 0.64872 8 0.57093 0.32907 1.6699 0.64850 9 0.57079 0.32921 1.6703 0.64839 10 0.57073 0.32927 1.6705 0.64834 11 0.57070 0.32930 1.6706 0.64832 12 0.57069 0.32931 1.6706 0.64831 13 0.57068 0.32932 1.6706 0.64831 14 0.57068 0.32932 1.670 0.64830 15 0.57068 0.32932 1.6706 0.64830 16 0.57068 0.32932 1.6706 0.64830 17 0.57068 0.32932 1.6706 0.64830 This procedure rapidly converted to the solution. ### 11.7.9 The Practical Questions and Examples of Subsonic branch The Fanno is applicable also when the flow isn't choke. In this case, several questions appear for the subsonic branch. This is the area shown in Figure 11.25 in beginning This kind of questions made of pair given information to find the conditions of the flow, as oppose to only one piece of information given in choked flow. There many combinations that can appear in this situation but there are several more physical and practical that will be discussed here. ### Subsonic Fanno Flow for Given$\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 \begin{align} \label{shock:eq:largeFLDP2P1} M_{1} = \sqrt{ \dfrac{1- \left[ \dfrac{P_{2}}{P_{0}}\right]^{2} } { k\,{\dfrac{4\,f\,L}{D}} } } \end{align} Equation \eqref{shock:eq:largeFLDP2P1} is suggested to be used up toM_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}$pipe dimensionless length. Figure ef{fanno:fig:MfldP1P2} exhibits the pressure ratio for fix$\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$the maximum Mach number is 0.50874 as shown in Figure ef{fanno:fig:MfldP1P2} by orange line. For a given entrance Mach number, the pressure ratio,$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$can be obtained and is drawn in Figure ef{fanno:fig:MfldP1P2}. Hence, when the pressure ratio,$P_2/P_1$is given along with given pipe,$\dfrac{4\,f\,L}{D}$the solution can be obtained by drawing a horizontal line. The intersection of the horizontal line with the right curve of the pressure ratio yields the entrance Mach number. This can be done by a computer program such Potto–GDC (version 0.5.2 and above). The summary of the procedure is as the following. beginNormalEnumerate change startEnumerate=1 1. If the pressure ratio is$P_2/P_1 < 0.02then using the perturbed solution the entrance Mach number is very small and calculate using the formula \begin{align} M = \sqrt{(1 - \dfrac{\dfrac{P_2}{P_1}}{ k\, \left(\dfrac{4\,f\,L}{D} \right)}} \label{fanno:eq:MforSmallP2P1} \end{align} 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$and pressure ratio equal 0.8. Example runs is presented in the Figure ef{fanno:fig:fldP2P1} for$\dfrac{4\,f\,L}{D}=0.5$and pressure ratio equal to 0.8. The blue line in Figure ef{fanno:fig:MfldP1P2} intersection with the horizontal line of$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: beginNormalEnumerate change startEnumerate=1 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 \begin{align} \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} \end{align} 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}$) and if not satisfied returned to stage \eqref{fanno:en:subChoked} until the solution is obtained. To demonstrate how this procedure is working consider a typical example of$\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}$and pressure ratio. Figure ef{gd:fig:subFLDP2P1} demonstrates that the conversion occur at about 7-8 iterations. With better first guess this conversion procedure converts much faster but at a certain range it is unstable. ### 11.7.11 Subsonic Fanno Flow for a Given$M_1$and Pressure Ratio This situation pose a simple mathematical problem while the physical situation occurs in cases where a specific flow rate is required with a given pressure ratio (range) (this problem was considered by some to be somewhat complicated). The specific flow rate can be converted to entrance Mach number and this simplifies the problem. Thus, the problem is reduced to find for given entrance Mach,$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$. An example of such issue is the following example that combines also the Naughty professor'' problems. # Example 11.21 Calculate the exit Mach number for$P_2/P_1 =0.4$and entrance Mach number$M_1 = 0.25$. # Solution The star pressure can be obtained from a table or Potto-GDC as  Fanno Flow Input:$M_1$k = 1.4$M_1\dfrac{4\,f\,L}{D}\dfrac{P}{P^{\star}}\dfrac{P_0}{{P_0}^{\star}}\dfrac{\rho}{\rho^{\star}}\dfrac{U}{U^{\star}}\dfrac{T}{T^{\star}}0.2500 8.4834 4.3546 2.4027 3.6742 0.27217 1.1852 And the star pressure ratio can be calculated at the exit as following \begin{align*} {P_2 \over P^{*} } = {{P_2 \over P_1 } {P_1 \over P^{*} } } = 0.4 \times 4.3546 = 1.74184 \end{align*} And the corresponding exit Mach number for this pressure ratio reads  Fanno Flow Input:\dfrac{P}{P^{\star}}$k = 1.4$M_1\dfrac{4\,f\,L}{D}\dfrac{P}{P^{\star}}\dfrac{P_0}{{P_0}^{\star}}\dfrac{\rho}{\rho^{\star}}\dfrac{U}{U^{\star}}\dfrac{T}{T^{\star}}$0.60694 0.46408 1.7418 1.1801 1.5585 0.64165 1.1177 A bit show off the Potto–GDC can carry these calculations in one click as  Fanno Flow Input:$M_1$and$\dfrac{P_2}{P_1}$k = 1.4$M_1M_2\dfrac{4\,f\,L}{D}\dfrac{P_2}{P_1}$0.250 0.60693 8.0193 0.400 ## Fig. 11.38 The entrance Mach number as a function of dimensionless As it can be seen for the Figure ef{gd:fig:fanno:fannoP2P1} the dominating parameter is$\dfrac{4\,f\,L}{D}$. The results are very similar for isothermal flow. The only difference is in small dimensionless friction,$\dfrac{4\,f\,L}{D}$. ### 11.7.12 More Examples of Fanno Flow # Example To demonstrate the utility in Figure ef{gd:fig:fanno:fannoP2P1} consider the following example. Find the mass flow rate for$f=0.05$,$L= 4[m]$,$D=0.02[m]$and pressure ratio$P_2 / P_1 = 0.1, 0.3, 0.5, 0.8$. The stagnation conditions at the entrance are$300 K$and$3[bar]$air. # Solution First calculate the dimensionless resistance,$\dfrac{4\,f\,L}{D}$. $$\dfrac{4\,f\,L}{D} = {4 \times 0.05 \times 4 \over 0.02 } = 40$$ From Figure ef{gd:fig:fanno:fannoP2P1} for$P_2 / P_1 = 0.1M_1 \approx 0.13$etc. or accurately by utilizing the program as in the following table.  Fanno Flow Input:$\dfrac{P_2}{P_1}$and$\dfrac{4\,f\,L}{D}$k = 1.4$M_1M_2\dfrac{4\,f\,L}{D}\left.\dfrac{4\,f\,L}{D}\right|_{1}\left.\dfrac{4\,f\,L}{D}\right|_{2}\dfrac{P_2}{P_1}$0.12728 0.99677 0.99195 4.5910 0.98874 4.5393 0.12420 0.99692 0.99233 4.7027 0.98928 4.6523 0.11392 0.99741 0.99354 5.1196 0.99097 5.0733 0.07975 0.99873 0.99683 7.2842 0.99556 7.2519 Therefore,$T\approx T_0and is the same for the pressure. Hence, the mass rate is a function of the Mach number. The Mach number is indeed a function of the pressure ratio but mass flow rate is a function of pressure ratio only through Mach number. The mass flow rate is \begin{align*} \dot{m} = P\, A\, M\, \sqrt{\dfrac{k }{ R\, T}} = 300000\, \times \dfrac{\pi \times 0.02^2 }{ 4 } \times 0.127 \times \sqrt{\dfrac{ 1.4 }{ 287\, 300}} \approx 0.48 \left(\dfrac{ kg }{ sec} \right) \end{align*} and for the rest \begin{align*} \dot{m} \left( \dfrac{\mathbf{P_2 }{ P_1}} = 0.3 \right) \sim 0.48 \times \dfrac{0.1242 }{ 0.1273}=0.468 \left(\dfrac{kg }{ sec}\right) \\ \dot{m}\, \left( \dfrac{\mathbf{P_2 }{ P_1}} = 0.5 \right) \sim 0.48 \times \dfrac{0.1139 }{ 0.1273}=0.43 \left(\dfrac{kg }{ sec}\right) \\ \dot{m} \, \left( \dfrac{ \mathbf{P_2 }{ P_1}} = 0.8 \right) \sim 0.48 \times \dfrac{0.07975 }{ 0.1273}=0.30 \left(\dfrac{kg }{ sec}\right) \end{align*} ## 11.8 The Table for Fanno Flow M$0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 2 3 4 5 6 7 8 9 10 20 25 30 35 40 45 50 55 60 65 70$\dfrac{4\,f\,L}{D}$787.08 440.35 280.02 193.03 140.66 106.72 83.4961 66.9216 14.5333 8.4834 5.2993 3.4525 2.3085 1.5664 1.0691 0.72805 0.49082 0.32459 0.20814 0.12728 0.07229 0.03633 0.01451 0.00328 0 0.305 0.52216 0.63306 0.6938 0.72988 0.7528 0.76819 0.77899 0.78683 0.81265 0.81582 0.81755 0.8186 0.81928 0.81975 0.82008 0.82033 0.82052 0.82066 0.82078$\dfrac{P}{P^{\star}}$36.5116 27.3817 21.9034 18.2508 15.6416 13.6843 12.1618 10.9435 5.4554 4.3546 3.6191 3.0922 2.6958 2.3865 2.1381 1.9341 1.7634 1.6183 1.4935 1.3848 1.2893 1.2047 1.1291 1.061 1 0.40825 0.21822 0.13363 0.089443 0.063758 0.047619 0.03686 0.029348 0.023905 0.00609 0.0039 0.00271 0.002 0.00153 0.00121 0.000979 0.000809 0.00068 0.000579 0.0005$\dfrac{P_0}{{P_0}^{\star}}$19.3005 14.4815 11.5914 9.6659 8.2915 7.2616 6.4613 5.8218 2.9635 2.4027 2.0351 1.778 1.5901 1.4487 1.3398 1.2549 1.1882 1.1356 1.0944 1.0624 1.0382 1.0207 1.0089 1.002 1 1.688 4.235 10.72 25 53.18 100 190 330 540 15000 46000 110000 250000 480000 860000 1.5e+06 2.3e+06 3.6e+06 5.4e+06 7.8e+06$\dfrac{\rho}{{\rho}^{\star}}$30.4318 22.8254 18.262 15.22 13.0474 11.4182 10.1512 9.1378 4.5826 3.6742 3.0702 2.64 2.3184 2.0693 1.8708 1.7092 1.5753 1.4626 1.3665 1.2838 1.2119 1.1489 1.0934 1.044 1 0.61237 0.50918 0.46771 0.44721 0.43568 0.42857 0.4239 0.42066 0.41833 0.41079 0.40988 0.40938 0.40908 0.40889 0.40875 0.40866 0.40859 0.40853 0.40849 0.40846$\dfrac{U}{{U}^{\star}}$0.03286 0.04381 0.05476 0.0657 0.07664 0.08758 0.09851 0.10944 0.21822 0.27217 0.32572 0.37879 0.43133 0.48326 0.53452 0.58506 0.63481 0.68374 0.73179 0.77894 0.82514 0.87037 0.9146 0.95781 1 1.633 1.964 2.138 2.236 2.295 2.333 2.359 2.377 2.39 2.434 2.44 2.443 2.445 2.446 2.446 2.447 2.447 2.448 2.448 2.448$\dfrac{T}{{T}^{\star}}\$ 1.1998 1.1996 1.1994 1.1991 1.1988 1.1985 1.1981 1.1976 1.1905 1.1852 1.1788 1.1713 1.1628 1.1533 1.1429 1.1315 1.1194 1.1065 1.0929 1.0787 1.0638 1.0485 1.0327 1.017 1 0.66667 0.42857 0.28571 0.2 0.14634 0.11111 0.086957 0.069767 0.057143 0.014815 0.00952 0.00663 0.00488 0.00374 0.00296 0.0024 0.00198 0.00166 0.00142 0.00122

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