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next up previous index
Next: Unchoked situations in Fanno Up: Isothermal Flow Previous: Figures and Tables   Index

Isothermal Flow Examples

There can be several kinds of questions aside from the proof questions8.6Generally, the ``engineering'' or practical questions can be divided into driving force (pressure difference), resistance (diameter, friction factor, friction coefficient, etc.), and mass flow rate questions. In this model no questions about shock (should) exist8.7.

The driving force questions deal with what should be the pressure difference to obtain certain flow rate. Here is an example.
\begin{examl}
A tube of 0.25 [m] diameter and 5000 [m] in length is attached to ...
...the maximum flow rate and then check if this request is
reasonable.
\end{examl}

Solution

If the flow was incompressible then for known density, $ \rho $ , the velocity can be calculated by utilizing $ \Delta P = \frac{4fL}{D} {U^2 \over 2g}$ . In incompressible flow, the density is a function of the entrance Mach number. The exit Mach number is not necessarily $ 1/\sqrt{k}$ i.e. the flow is not choked. First, check whether flow is choked (or even possible).

Calculating the resistance, $ \frac{4fL}{D}$

$\displaystyle \frac{4fL}{D} = {4 \times 0.005 5000 \over 0.25} = 400 $

Utilizing the table 8.1 or the program provides

$ \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{T_0 \over {T_0}^{*}} $
0.04331 400.00 20.1743 12.5921 0.0 0.89446
<>


The maximum flow rate (the limiting case) can be calculated by utilizing the above table. The velocity of the gas at the entrance $ U =c M =
0.04331\times \sqrt{1.31\times290\times300} \cong 14.62
\left[ m \over sec\right]$ . The density reads

$\displaystyle \rho = {P \over R T} = { 2,017,450 \over 290 \times 300} \cong 23.19
\left[ kg \over m^{3}\right]
$

The maximum flow rate then reads

$\displaystyle \dot{m} = \rho A U = { 23.19 \times
{\pi \times (0.25)^{2} \over 4} \times 14.62 }
\cong 16.9 \left[ kg \over sec\right]
$

The maximum flow rate is larger then the requested mass rate hence the flow is not choked. It is note worthy to mention that since the isothermal model breaks around the choking point, the flow rate is really some what different. It is more appropriate to assume isothermal model hence our model is appropriate.

To solve this problem the flow rate has to be calculated as

$\displaystyle \dot{m} = \rho A U = 2.0 \left[ kg \over sec\right]
$

$\displaystyle \dot{m} = {P_1 \over R T} A {k U \over k} =
{P_1 \over \sqrt{k R T} } A {k U \over \sqrt{k R T} } =
{P_1 \over c } A {k M_1 }
$

Now combining with equation (8.40) yields

$\displaystyle \dot{m} = {M_2 P_2 A k \over c }
$

$\displaystyle M_2 = { \dot{m} c \over P_2 A k} =
{ 2 \times 337.59 \over 100000 \times {\pi \times (0.25)^{2} \over
4} \times 1.31} = 0.103
$

From the table (8.1) or utilizing the program provides

$ \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{T_0 \over {T_0}^{*}} $
0.10300 66.6779 8.4826 5.3249 0.0 0.89567
<>


The entrance Mach number obtained by

$\displaystyle \left. \frac{4fL}{D} \right\vert _1 = 66.6779 + 400 \cong 466.68 $

$ \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{T_0 \over {T_0}^{*}} $
0.04014 466.68 21.7678 13.5844 0.0 0.89442
<>

The pressure should be

$\displaystyle P = 21.76780 \times 8.4826 = 2.566 [bar]
$

Note that table here above for this example are for $ k=1.31$



\begin{examl}
A flow of gas was considered for a distance of 0.5 [km] (500 [m])....
...
You can assume that the soundings temperature to be
$27^{\circ}C$.
\end{examl}

Solution

At first, the minimum diameter will be obtained when the flow is choked. Thus, the maximum $ M_1$ that can be obtained when the $ M_2$ is at its maximum and back pressure is at the atmospheric pressure.
$\displaystyle M_1 = M_2 { P_2 \over P_1} = \overbrace{ 1 \over \sqrt{k} }^{M_{max}} {1 \over 10} = 0.0845$    

Now, with the value of $ M_1$ either utilizing Table (8.1) or using the provided program yields
$ \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{T_0 \over {T_0}^{*}} $
0.08450 94.4310 10.0018 6.2991 0.0 0.87625
<>


With $ {\left.\frac{4fL}{D}\right\vert _{max}}= 94.431 $ the value of minimum diameter.

$\displaystyle D = {4 f L \over {\left.\frac{4fL}{D}\right\vert _{max}}} \simeq {4 \times 0.02 \times 500 \over 94.43} \simeq 0.42359 [m] = 16.68 [in]$    

However, the pipes are provided only in 0.5 increments and the next size is $ 17[in]$ or $ 0.4318 [m]$ . With this pipe size the calculations are to be repeated in reversed to produces: (Clearly the maximum mass is determined with)

$\displaystyle \dot{m} = \rho A U = \rho A M c = { P \over R T} A M \sqrt{kRT}
= {P A M \sqrt{k} \over \sqrt{RT}}
$

The usage of the above equation clearly applied to the whole pipe. The only point that must be emphasized is that all properties (like Mach number, pressure and etc) have to be taken at the same point. The new $ \frac{4fL}{D}$ is

$\displaystyle {\frac{4fL}{D}} = {4 \times 0.02 \times 500 \over 0.4318} \simeq 92.64
$

$ \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{T_0 \over {T_0}^{*}} $
0.08527 92.6400 9.9110 6.2424 0.0 0.87627
<>


To check whether the flow rate is satisfied the requirement

$\displaystyle \dot{m} = { { 10^{6} } \times {\pi\times 0.4318^2 \over 4} \times 0.0853 \times \sqrt{1.4} \over \sqrt{287\times 300} } \approx 50.3 [kg/sec]$    

Since $ 50.3 \geq 0.2 $ the mass flow rate requirements is satisfied.

It should be noted that $ P$ should be replaced by $ P_0$ in the calculations. The speed of sound at the entrance is

$\displaystyle c = \sqrt{kRT } = \sqrt{1.4 \times 287 \times 300} \cong
347.2 \left[ m \over sec \right]
$

and the density is

$\displaystyle \rho = {P \over R T} = {1,000,000 \over 287 \times 300} =
11.61 \left[ kg \over m^3 \right]
$

The velocity at the entrance should be

$\displaystyle U = M *c = 0.08528 \times 347.2 \cong
29.6 \left[ m \over sec \right]
$

The diameter should be

$\displaystyle D = \sqrt{ 4\dot{m} \over \pi U \rho} =
\sqrt{ 4 \times 0.2 \over \pi \times 29.6 \times 11.61} \cong 0.027
$

Nevertheless, for sake of the exercise the other parameters will be calculated. This situation is reversed question. The flow rate is given with the diameter of the pipe. It should be noted that the flow isn't choked.



\begin{examl}
A gas flows of from a station (a) with pressure of 20[bar]
through...
...Calculate the Mach number at the entrance to pipe and the flow rate.
\end{examl}

Solution

First, the information whether the flow is choked needed to be found. Therefore, at first it will be assumed that the whole length is the maximum length.

$\displaystyle {\left.\frac{4fL}{D}\right\vert _{max}}= {4 \times 0.01 \times 4000 \over 0.4} = 400
$

with $ \left.\frac{4fL}{D}\right\vert _{max}=400$ the following can be written

$ \rule[-0.1in]{0.pt}{0.3 in}\mathbf{M} $ $ \mathbf{\frac{4fL}{D}} $ $ \mathbf{T_0 \over {T_0}^{*T}} $ $ \mathbf{\rho \over \rho^{*T}} $ $ \mathbf{P \over P^{*T}} $ $ \mathbf{P_{0} \over {P_{0}}^{*T}} $
0.0419 400.72021 0.87531 20.19235 20.19235 12.66915
<>

From the table $ M_1\approx 0.0419$ ,and $ {P_{0} \over {P_{0}}^{*T} } \approx 12.67$

$\displaystyle {P_{0}}^{*T} \cong {28 \over 12.67} \simeq 2.21 [bar]
$

The pressure at point (b) utilizing the isentropic relationship ($ M=1$ ) pressure ratio is 0.52828.

$\displaystyle P_2 = { {P_{0}}^{*T} \over
\left( { P_2 \over { {P_{0}}^{*T} }} \right)
}
= { 2.21 \times 0.52828 } = 1.17 [bar]
$

As the pressure at point (b) is smaller the actual pressure $ P^{*} < P_2$ than the actual pressure one must conclude that the flow is not choked. The solution is iterative process.

1.
guess reasonably the value of $ M_1$ and calculate $ \frac{4fL}{D}$
2.
Calculate the value of $ \left.{\frac{4fL}{D}}\right\vert _{2}$ by subtracting $ \left.{\frac{4fL}{D}}\right\vert _{1} -\frac{4fL}{D} $
3.
Obtain $ M_2$ from the Table ? or using the Potto-GDC.
4.
Calculate the pressure, $ P_2$ in mind that this isn't the real pressure but based on the assumption
5.
Compare the results of guessed pressure $ P_2$ with the actual pressure. and chose new $ M_1$ accordingly.
Now the process has been done for you and is provided in the figure ([*]) or in table resulted from the provided program.
$ \rule[-0.1in]{0.pt}{0.3 in} \mathbf{M_1} $ $ \mathbf{M_2} $ $ \mathbf{\left.{4fL/D}\right\vert _{1} } $ $ \mathbf{4fL/D} $ $ \mathbf{P_2 \over P_1} $
0.0419 0.59338 400.32131 400.00000 0.10000
<>
The flow rate is

$\displaystyle \dot{m} = \rho A M c = {P \sqrt{k} \over \sqrt {R T}}
{\pi \time...
...00 \sqrt{1.4} \over \sqrt{300 \times 287}}
{\pi \times 0.2^{2}} \times 0.0419
$

$\displaystyle \simeq 42.46[kg/sec]
$


In this chapter, there are no examples on isothermal with supersonic flow.


next up previous index
Next: Unchoked situations in Fanno Up: Isothermal Flow Previous: Figures and Tables   Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21