Next: Examples of Fanno Flow Up: Fanno Flow Previous: The Trends Index

The working equations

Integration of equation (9.25) yields

$\displaystyle {4 \over D} \int^{L_{max}}_{L} f dx = {1 \over k} {1 - M^2 \over M^2} + {k+1 \over 2k}\ln {{k+1 \over 2}M^2 \over 1+ {k-1 \over 2}M^2}$

(9.35)

A representative friction factor is defined as

$\displaystyle \bar{f} = { 1 \over L_{max}} \int ^{L_{max}} _{0} {f dx}$

(9.36)

By utilizing the mean average theorem equation (9.36) yields

$\displaystyle {4 \bar{f}L_{max}\over D} = {1 \over k} {1 - M^2 \over M^2} + {k+1 \over 2k}\ln {{k+1 \over 2}M^2 \over 1+ {k-1 \over 2}M^2}$

(9.37)

It is common to replace the $\bar{f}$ with

which is adopted in this book.

Equations (9.24), (9.27), (9.28), (9.29), (9.29), and (9.30) can be solved. For example, the pressure as written in equation (9.23) is represented by $\fld$ , and Mach number. Now equation (9.24) can eliminate term $\fld$ and describe the pressure on the Mach number. Dividing equation (9.24) in equation (9.26) yields

$\displaystyle {{dP \over P } \over {dM^{2} \over M^{2}} } = - {{ 1 + (k -1 M^{2}} \over 2 M^{2} \left( 1 + { k-1 \over 2} M^{2} \right)} dM^{2}$

(9.38)

The symbol ``*'' denotes the state when the flow is choked and Mach number is equal to 1. Thus,

when $P=P^{*}$ Equation (9.39) can be integrated to yield:

$\displaystyle {P \over P^{*}} = { 1 \over M} \sqrt{{k+1 \over 2} \over { 1 + {k - 1 \over 2} M^{2}} }$

(9.39)

In the same fashion the variables ratio can be obtained

$\displaystyle { T \over T^{*}} = {c^{2} \over {c^{*}}^{2} } = {{ k + 1 \over 2} \over { 1 + {k - 1 \over 2} M^{2}} }$

(9.40)

$\displaystyle {\rho \over \rho^{*}} = { 1 \over M} \sqrt{ { 1 + {k - 1 \over 2} M^{2}} \over {k+1 \over 2} }$

(9.41)

$\displaystyle { U \over U ^{*}} = \left( {\rho \over \rho^{*}} \right)^{-1} = M \sqrt{{k+1 \over 2} \over { 1 + {k - 1 \over 2} M^{2}} }$

(9.42)

The stagnation pressure decreases and can be expressed by

$\displaystyle {P_{0} \over {P_{0}}^{*}} = {\overbrace{ P_0 \over P }^{\left(1+ ... ...race{ {P_0}^{*} \over P^{*}} _{\left(2 \over k+1\right)^{k \over k -1}} P^{*} }$

(9.43)

Using the pressure ratio in equation (9.40) and substituting it into equation (9.44) yields

$\displaystyle {P_{0} \over {P_0}^{*}} = \left({ { 1 + {k - 1 \over 2} M^{2}} \o... ...er k -1} { 1 \over M} \sqrt{ { 1 + {k - 1 \over 2} M^{2}} \over {k+1 \over 2} }$

(9.44)

And further rearranging equation (9.45) provides

$\displaystyle {P_{0} \over {P_{0}}^{*}} = { 1 \over M} \left({ { 1 + {k - 1 \over 2} M^{2}} \over {k+1 \over 2} } \right)^{k +1 \over 2(k -1)}$

(9.45)

The integration of equation (9.34) yields

$\displaystyle {s - s^{*} \over c_p} = \ln M^{2} \sqrt{\left({{k+1}\over 2 M^{2} \left( 1 + {k -1 \over 2 }M^{2} \right) }\right)^{ k +1 \over k} }$

(9.46)

The results of these equations are plotted in Figure (9.2)

**Figure:** `Various parameters in Fanno flow as a function of Mach number`
$\begin{figure}\centerline{\includegraphics {calculations/fanno}} \end{figure}$

The Fanno flow is in many cases shockless and therefore a relationship between two points should be derived. In most times, the ``star'' values are imaginary values that represent the value at choking. The real ratio can be obtained by two star ratios as an example

$\displaystyle {T_2 \over T_1} ={ \left. T \over T^{*} \right\vert _{M_2} \over \left. T \over T^{*} \right\vert _{M_1} }$

(9.47)

A special interest is the equation for the dimensionless friction as following

$\displaystyle \int_{L_1}^{L_2} {4fL \over D} dx = \int_{L_1}^{L_{max}} {4fL \over D} dx - \int_{L_2}^{L_{max}} {4fL \over D} dx$

(9.48)

Hence,

$\displaystyle \left( {4f L_{max} \over D} \right)_{2} = \left( {4{f} L_{max} \over D} \right)_{1} - {4{f} L \over D}$

(9.49)

Next: Examples of Fanno Flow Up: Fanno Flow Previous: The Trends Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21

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