11.7.11: Subsonic Fanno Flow for a Given $$M_1$$ and Pressure Ratio

This situation pose a simple mathematical problem while the physical situation occurs in cases where a specific flow rate is required with a given pressure ratio (range) (this problem was considered by some to be somewhat complicated). The specific flow rate can be converted to entrance Mach number and this simplifies the problem. Thus, the problem is reduced to find for given entrance Mach, $$M_1$$, and given pressure ratio calculate the flow parameters, like the exit Mach number, $$M_2$$. The procedure is based on the fact that the entrance star pressure ratio can be calculated using $$M_1$$. Thus, using the pressure ratio to calculate the star exit pressure ratio provide the exit Mach number, $$M_2$$. An example of such issue is the following example that combines also the "Naughty professor'' problems.

Example 11.21

Calculate the exit Mach number for $$P_2/P_1 =0.4$$ and entrance Mach number $$M_1 = 0.25$$.

Solution 11.21

The star pressure can be obtained from a table or Potto-GDC as

 Fanno Flow Input: $$M_1$$ k = 1.4 $$M_1$$ $$\dfrac{4\,f\,L}{D}$$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Core/Chemical_Engineering/Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.7:_Fanno_Flow/11.7.11:_Subsonic_Fanno_Flow_for_a_Given_\(M_1$$_and_Pressure_Ratio), /content/body/div[2]/div/table[1]/tbody/tr[2]/td[4]/span, line 1, column 4 \) $$\dfrac{\rho}{\rho^{\star}}$$ $$\dfrac{U}{U^{\star}}$$ $$\dfrac{T}{T^{\star}}$$ 0.2500 8.4834 4.3546 2.4027 3.6742 0.27217 1.1852

And the star pressure ratio can be calculated at the exit as following
\begin{align*}
{P_2 \over P^{*} } = {{P_2 \over P_1 }   {P_1 \over P^{*} } } =
0.4 \times  4.3546   = 1.74184
\end{align*}
And the corresponding exit Mach number for this pressure ratio reads

 Fanno Flow Input: $$\dfrac{P}{P^{\star}}$$ k = 1.4 $$M_1$$ $$\dfrac{4\,f\,L}{D}$$ $$\dfrac{P}{P^{\star}}$$ $$\dfrac{P_0}ParseError: EOF expected (click for details)Callstack: at (Core/Chemical_Engineering/Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.7:_Fanno_Flow/11.7.11:_Subsonic_Fanno_Flow_for_a_Given_\(M_1$$_and_Pressure_Ratio), /content/body/div[2]/div/table[2]/tbody/tr[2]/td[4]/span, line 1, column 4 \) $$\dfrac{\rho}{\rho^{\star}}$$ $$\dfrac{U}{U^{\star}}$$ $$\dfrac{T}{T^{\star}}$$ 0.60694 0.46408 1.7418 1.1801 1.5585 0.64165 1.1177

A bit show off the Potto–GDC can carry these calculations in one click as

 Fanno Flow Input: (\M_1\) and $$\dfrac{P_2}{P_1}$$ k = 1.4 $$M_1$$ $$M_2$$ $$\dfrac{4\,f\,L}{D}$$ $$\dfrac{P_2}{P_1}$$ 0.250 0.60693 8.0193 0.400

Fig. 11.38 The entrance Mach number as a function of dimensionless

As it can be seen for the Figure 11.38 the dominating parameter is $$\dfrac{4\,f\,L}{D}$$. The results are very similar for isothermal flow. The only difference is in small dimensionless friction, $$\dfrac{4\,f\,L}{D}$$.

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.