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11.7.12: More Examples of Fanno Flow

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    1339
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    Example

    To demonstrate the utility in Figure ?? 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, \(\dfrac{4\,f\,L}{D}\).
    \[
        \dfrac{4\,f\,L}{D} = {4 \times 0.05 \times 4 \over 0.02 } = 40   \tag{10}
    \]
    From Figure ?? for \(P_2 / P_1 = 0.1\) \(M_1 \approx 0.13\) 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
    \(M_1\) \(M_2\) \(\dfrac{4\,f\,L}{D}\) \(\left.\dfrac{4\,f\,L}{D}\right|_{1}\) \(\left.\dfrac{4\,f\,L}{D}\right|_{2}\) \(\dfrac{P_2}{P_1}\)
    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, \(T\approx T_0\) and 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*}

    Contributors

    • Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.