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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