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11.4.5: Mass Flow Rate (Number)

One of the important engineering parameters is the mass flow rate which for ideal gas is

    \dot{m} = \rho\, U\, A = \dfrac{P }{ R\,T}\, U\, A    
    \label{gd:iso:eq:massFlowRate}   \tag{54}

This parameter is studied here, to examine the maximum flow rate and to see what is the effect of the compressibility on the flow rate. The area ratio as a function of the Mach number needed to be established, specifically and explicitly the relationship for the chocked flow. The area ratio is defined as the ratio of the cross section at any point to the throat area (the narrow area). It is convenient to rearrange the equation (52) to be expressed in terms of the stagnation properties as

    \dfrac{\dot{m} }{ A} = \dfrac{ P }{ P_{0} } \;
            \dfrac{P_{0}\, U }{ \sqrt{k\,R\,T}}
            \sqrt{\dfrac{k }{ R} } \sqrt{ \dfrac{T_{0} }{ T}}
            \; {1 \over \sqrt{T_{0}}}  
        =  { P_{0} \over \sqrt{T_0}} M \sqrt{k \over R }  

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        \left( {\sqrt{k} P_0 \over \sqrt{R T_{0}}} \right)
            \left( 1 + {k -1 \over 2} \right)^ {-\dfrac{ k+ 1 }{ 2\,(k -1)} }
    \label{gd:iso:eq:massFlowRateStar}  \tag{55}
Since the mass flow rate is constant in the duct, dividing equations (55) by equation (54) yields

Mass Flow Rate Ratio

    {A \over A^{*}} = \dfrac{ 1 }{ M}  
        \left( { 1 + {k -1 \over 2} M^{2} \over {k +1\over 2}} \right) ^ {-\dfrac{ k+ 1 }{ 2\,(k -1)} }  \tag{56}

Equation (56) relates the Mach number at any point to the cross section area ratio. The maximum flow rate can be expressed either by taking the derivative of equation (??) in with respect to \(M\) and equating to zero. Carrying this calculation results at \(M=1\). 

    \left(\dot{m} \over A^{*}\right)_{max}  { P_0  \over \sqrt{T_0}} =
    \sqrt{k \over R}
    \left(  {k +1 \over 2} \right)^ {-{ k+ 1 \over 2(k -1)}}
    \label{gd:iso:eq:fliegner}  \tag{57}
For specific heat ratio, \(k=1.4\)  
    \left(\dot{m} \over A^{*}\right)_{max} {P_0 \over \sqrt{T_0}}  
        \sim {0.68473 \over \sqrt{R}}
    \label{gd:iso:eq:fliegnerk}  \tag{58}
The maximum flow rate for air (\(R=287 \dfrac{j }{ kg\, K}\)) becomes,
    \dfrac{ \dot{m} \sqrt{T_0} }{ A^{\star} P_0 } = 0.040418  
    \label{gd:iso:eq:fliegnerAir}  \tag{59}
Equation (59) is known as Fliegner's Formula on the name of one of the first engineers who observed experimentally the choking phenomenon. It can be noticed that Fliegner's equation can lead to definition of the Fliegner's Number.
     \dfrac{ \dot{m} \sqrt{T_0}}{ A^{*} P_0 } =  
    {\dot{m} \overbrace{\sqrt{k\,R\, T_0}}^{c_0} \over\sqrt{k\,R}A^{*} P_0}  
    = \dfrac{1}{\sqrt{R}}\,\overbrace{ \dfrac{\dot{m} \, c_0 }{ A^{*} P_0 }}^{Fn}  
    \label{gd:iso:eq:fnA}  \tag{60}
The definition of Fliegner's number (Fn)  is
    \pmb{Fn} quiv  
    \dfrac{\sqrt{R}\,\dot{m} \,c_0 }{ \sqrt{R} \, A^{*} \, P_0 }
    \label{gd:iso:eq:fnDef}  \tag{61}
it into equation (55) results in

Fliegner's Nember

    \pmb{Fn} =  k\,M\,\left( 1+   {k -1 \over 2}M^2 \right)^  
        {-{ k+ 1 \over 2(k -1)}}  \tag{62}

and the maximum point for \(Fn\) at \(M=1\) is
    \pmb{Fn} =  k\left(  {k +1 \over 2} \right)^ {-{ k+ 1 \over 2(k -1)}}
    \label{gd:iso:eq:fmMax}  \tag{63}

Example 11.7

Why \(\pmb{Fn}\) is zero at Mach equal to zero? Prove Fliegner number, \(\pmb{Fn}\) is maximum at \(M=1\).


  • 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.