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next up previous index
Next: The Table for Fanno Up: Fanno Flow Previous: The Approximation of the   Index

More Examples of Fanno Flow


\begin{examl}
To demonstrate the utility in Figure \eqref{fanno:fig:M1-FLD}
cons...
...agnation conditions at the entrance are
$300 K$ and $3[bar]$ air.
\end{examl}
Solution

First calculate the dimensionless resistance, $ \frac{4fL}{D}$ .

$\displaystyle \frac{4fL}{D} = {4 \times 0.05 \times 4 \over 0.02 } = 40 $

From Figure (9.21) for $ P_2 / P_1 = 0.1$ $ M_1 \approx 0.13$ etc.

or accurately utilizing the program as in the following table.

$ \rule[-0.1in]{0.pt}{0.3 in} \mathbf{M_1} $ $ \mathbf{M_2} $ $ \mathbf{4fL \over D} $ $ \mathbf{\left.{ 4fL \over D }\right\vert _{1}} $ $ \mathbf{\left.{ 4fL \over D }\right\vert _{2}} $ $ \mathbf{P_2 \over P_1} $
0.12728 1.0000 40.0000 40.0000 0.0 0.11637
0.12420 0.40790 40.0000 42.1697 2.1697 0.30000
0.11392 0.22697 40.0000 50.7569 10.7569 0.50000
0.07975 0.09965 40.0000 107.42 67.4206 0.80000
<>


Only for the pressure ratio of 0.1 the flow is choked.

$ \rule[-0.1in]{0.pt}{0.3 in}\mathbf{M} $ $ \mathbf{T \over T_0} $ $ \mathbf{\rho \over \rho_0} $ $ \mathbf{A \over A^{\star}} $ $ \mathbf{P \over P_0} $ $ \mathbf{A\times P \over A^{*} \times P_0}$
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 for the same the pressure. Hence, the mass rate is a function of the Mach number. The Mach number is indeed function of the pressure ratio but and therefore mass flow rate is function pressure ratio only through Mach number.

The mass flow rate is

$\displaystyle \dot{m} = P A M \sqrt{k \over R T} = 300000 \times {\pi \times 0....
...0.127 \times \sqrt{ 1.4 \over 287 300} \approx 0.48 \left( kg \over sec \right)$    

and for the rest

$\displaystyle \dot{m} \left( \mathbf{P_2 \over P_1} = 0.3 \right) \sim 0.48 \times {0.1242 \over 0.1273}=0.468 \left(kg\over sec\right)$    
$\displaystyle \dot{m} \left( \mathbf{P_2 \over P_1} = 0.5 \right) \sim 0.48 \times {0.1139 \over 0.1273}=0.43 \left(kg\over sec\right)$    
$\displaystyle \dot{m} \left( \mathbf{P_2 \over P_1} = 0.8 \right) \sim 0.48 \times {0.07975 \over 0.1273}=0.30 \left(kg\over sec\right)$    



next up previous index
Next: The Table for Fanno Up: Fanno Flow Previous: The Approximation of the   Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21