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Next: Supersonic Branch Up: Fanno Flow Previous: The working equations   Index

Examples of Fanno Flow

Figure 9.3: Schematic of Example (9.1)
\begin{figure}\centerline{\includegraphics {cont/fanno/example1}} \end{figure}

\begin{examl} Air flows from a reservoir and enters a uniform pipe with a diamet... ...ture and total pressure in the reservoir under the Fanno flow model. \end{examl} 9.4
Solution

For isentropic, the flow to the pipe inlet, the temperature and the total pressure at the pipe inlet are the same as the those in the reservoir. Thus, finding the total pressure and temperature at the pipe inlet is the solution. With the Mach number and temperature known at the exit, the total temperature at the entrance can be obtained by knowing the $ \fld$ . For given Mach number ($ M=0.9$ ) the following is obtained.

Fanno Flow Input: M k = 1.4
M fld P/P* P0/P0* ρ/ρ* U/U* T/T*
0.9 0.0145124 1.12913 1.00886 1.09338 0.914598 1.0327

So, the total temperature at the exit is

$\displaystyle \left.T^{*} \right\vert _2 = \left. { T^{*} \over T} \right\vert _{2} T_2 = { 300 \over 1.0327} = 290.5[K]$    

To ``move'' to the other side of the tube the $ \fld$ is added as

$\displaystyle \left. \frac{4fL}{D} \right\vert _{1} = \frac{4fL}{D} + \left. \f... ...\right\vert _{2} = { 4 \times 0.004 \times 10 \over 0.05} + 0.01451 \simeq 3.21$    

The rest of the parameters can be obtained with the new $ \fld$ either from the table by interpolations or utilizing attached program.

Fanno Flow Input: fld k = 1.4
M fld P/P* P0/P0* ρ/ρ* U/U* T/T*
0.358856 3.21 3.01404 1.74047 2.57639 0.388141 1.16987

Note that the subsonic branch is chosen. the stagnation ratios has to be added for $ M = 0.35886$

Fanno Flow Input: M k = 1.4
M fld P/P* P0/P0* ρ/ρ* U/U* T/T*
0.35886 3.20989 3.014 1.74046 2.57636 0.388145 1.16987

The total pressure $ P_{01}$ can be found from the combination of the ratios as follows:

$\displaystyle P_{01} =$ $\displaystyle \overbrace{ \overbrace{P_{2} \left.{P^{*}\over P}\right\vert _{2}... ...P \over P^{*} }\right\vert _{1} } ^{P_1} \left.{P_{0} \over P }\right\vert _{1}$    
$\displaystyle =$ $\displaystyle 1 \times { 1\over 1.12913} \times 3.014 \times { 1 \over 0.915} = 2.91 [Bar]$    

$\displaystyle T_{01} =$ $\displaystyle \overbrace{ \overbrace{T_{2} \left.{T^{*}\over T}\right\vert _{2}... ...T \over T^{*} }\right\vert _{1} } ^{T_1} \left.{T_{0} \over T }\right\vert _{1}$    
$\displaystyle =$ $\displaystyle 300 \times { 1 \over 1.0327} \times 1.17 \times {1 \over 0.975} \simeq 348 K = 75 \celsius$    


Figure 9.4: The schematic of Example (9.2)
Image example2
Another academic question/example:
\begin{examl} A system is composed of a convergent-divergent nozzle followed by ... ...}\par Take k = 1.4, R = 287 $\left[J/kg K\right]$ and $f = 0.005$. \end{examl}

Solution

(a)

Assuming that the pressure vessel very much larger than the pipe therefore, the velocity in the vessel can be assumed small enough so it can be neglected. Thus, the stagnation conditions can be approximated as the condition in the tank. It further assumed that the flow through the nozzle can be approximated as isentropic. Hence, $ T_{01} = 400 K$ and $ P_{01}= 29.65 [Par]$

The mass flow rate through the system is constant and for simplicity reason point 1 is chosen in which,

$\displaystyle \dot{m} = \rho A M c $

The density and speed of sound are unknowns and needed to be computed. With the isentropic relationship the Mach number at point one is known the following can be found either from the Table or Potto-GDC

Isentropic Flow Input: M k = 1.4
M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*
3 0.357143 0.0762263 4.23457 0.0272237 0.115281 0.653256
The temperature is
$\displaystyle T_{1} = {T_1 \over T_{01}} T_{01} = 0.357 \times 400 = 142.8 K$    

With the temperature the speed of sound can be calculated as

$\displaystyle c_{1} = \sqrt{kRT} = \sqrt{1.4 \times 287 \times 142.8} \simeq 239.54 [m/sec] $

The pressure at point 1 can be calculated as

$\displaystyle P_{1} = {P_1 \over P_{01}} P_{01} = 0.027 \times 30 \simeq 0.81 [Bar] $

The density as a function of other properties at point 1 is

$\displaystyle \rho_1 = \left. P \over R T \right\vert _1 = {8.1\times 10^4\over 287\times 142.8} \simeq 1.97\left[kg \over m^3 \right] $

The mass flow rate can be evaluated from equation (9.2)

$\displaystyle \dot{m} = 1.97 \times {\pi \times {0.025}^{2} \over 4} \times 3 \times 239.54 = 0.69 \left[{kg \over sec}\right] $

(b)
First, a check whether the flow is shockless by comparing the flow resistance and the maximum possible resistance. From the table or by using Potto-GDC, to obtain the flowing

$ \rule[-0.1in]{0.pt}{0.3 in}\mathbf{M} $ $ \mathbf{4fL \over D} $ $ \mathbf{P \over P^{*}} $ $ \mathbf{P_0 \over {P_0}^{*}} $ $ \mathbf{\rho \over \rho^{*}} $ $ \mathbf{U \over {U}^{*}} $ $ \mathbf{T \over T^{*}} $
3.0000 0.52216 0.21822 4.2346 0.50918 1.9640 0.42857


and the conditions of the tube are

$\displaystyle \fld = { 4 \times 0.005 \times 1.0 \over 0.025 } = 0.8 $

Since $ 0.8 > 0.52216$ the flow is chocked and with a shock wave.

The exit pressure determines the location of the shock, if a shock exists, by comparing ``possible'' $ P_{exit}$ to $ P_B$ . Two possibilities needed to be checked; one, the shock at the entrance of the tube, and two, shock at the exit and comparing the pressure ratios. First, the possibility that the shock wave occurs immediately at the entrance for which the ratio for $ M_x$ are (shock wave table)

$ \rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x} $ $ \mathbf{M_y} $ $ \mathbf{T_y \over T_x} $ $ \mathbf{\rho_y \over \rho_x} $ $ \mathbf{P_y \over P_x} $ $ \mathbf{{P_0}_y \over {P_0}_x } $
3.0000 0.47519 2.6790 3.8571 10.3333 0.32834

After shock wave the flow is subsonic with ``$ M_1$ ''$ =0.47519$ . (fanno flow table)

$ \rule[-0.1in]{0.pt}{0.3 in}\mathbf{M} $ $ \mathbf{4fL \over D} $ $ \mathbf{P \over P^{*}} $ $ \mathbf{P_0 \over {P_0}^{*}} $ $ \mathbf{\rho \over \rho^{*}} $ $ \mathbf{U \over {U}^{*}} $ $ \mathbf{T \over T^{*}} $
0.47519 1.2919 2.2549 1.3904 1.9640 0.50917 1.1481

The stagnation values for $ M=0.47519$ are

$ \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}$ $ \mathbf{F \over F^{*}} $
0.47519 0.95679 0.89545 1.3904 0.85676 1.1912 0.65326

The ratio of exit pressure to the chamber total pressure is

$\displaystyle {P_2 \over P_0 } = \overbrace{\left( P_2 \over P^{*} \right)}^{1... ..._y \over {P_0}_x \right) \overbrace{\left( {P_0}_x \over {P_0} \right) }^{1} $

$\displaystyle = 1 \times {1 \over 2.2549} \times 0.8568 \times 0.32834 \times 1 = 0.12476 $

The actual pressure ratio $ 1/29.65 = 0.0338 $ is smaller than the case in which shock occurs at the entrance. Thus, the shock is somewhere downstream. One possible way to find the exit temperature, $ T_2$ is by finding the location of the shock. To find the location of the shock ratio of the pressure ratio, $ P_2 \over P_1$ is needed. With the location of shock, ``claiming'' up stream from the exit through shock to the entrance. For example, calculating the parameters for shock location with known $ \frac{4fL}{D}$ in the ``y'' side. Then either utilizing shock table or the program to obtained the upstream Mach number.

The procedure of the calculations:

1)
Calculated the entrance Mach number assuming the shock occurs at the exit:
a)
set $ M_2^{'} = 1$ assume the flow in the entire tube is supersonic:
b)
calculated $ M_1^{'}$
Note this Mach number is the high Value.
2)
Calculated the entrance Mach assuming shock at the entrance.

a)
set $ M_2 = 1$
b)
add $ \fld$ and calculated $ M_1$ ' for subsonic branch
c)
calculated $ M_x$ for $ M_1$ '
Note this Mach number is the low Value.
3)
according your root finding algorithm9.5calculated or guess the shock location and then compute as above the new $ M_1$ .
a)
set $ M_2 = 1$
b)
for the new $ \fld$ and compute the new $ M_y$ ' as on the subsonic branch
c)
calculated $ M_x$ ' for the $ M_y$ '
d)
Add the leftover of $ \fld$ and calculated the $ M_1$
4)
guess new location for the shock according to your finding root procedure and according the result repeat previous stage until the solution is obtained.

$ \rule[-0.1in]{0.pt}{0.3 in} \mathbf{M_1} $ $ \mathbf{M_2} $ $ \mathbf{\left.{ 4fL \over D}\right\vert _{up}} $ $ \mathbf{\left.{ 4fL \over D }\right\vert _{down}} $ $ \mathbf{M_x} $ $ \mathbf{M_y} $
3.0000 1.0000 0.22019 0.57981 1.9899 0.57910


(c)

The way that numerical procedure of solving this problem is by finding $ \left.{ 4fL \over D }\right\vert _{up}$ that will produce $ M_1=3$ . In the process $ M_x$ and $ M_y$ must be calculated (see the chapter on the program with its algorithms.).


next up previous index
Next: Supersonic Branch Up: Fanno Flow Previous: The working equations   Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21