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11.7.6: Isothermal Flow Examples

There can be several kinds of questions aside from the proof questions. Generally, 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) exist. The driving force questions deal with what should be the pressure difference to obtain a certain flow rate. Here is an example.

Example 11.15

A tube of 0.25 [m] diameter and 5000 [m] in length is attached to a pump. What should be the pump pressure so that a flow rate of 2 \([kg/sec]\) will be achieved? Assume that friction factor \(f=0.005\) and the exit pressure is \(1 [bar]\). The specific heat for the gas, \(k=1.31\), surroundings temperature \(27^{\circ}C\), \(R = 290 \left[ \dfrac{J}{  K\, kg} \right]\). Hint: calculate the maximum flow rate and then check if this request is reasonable.  

Solution 11.15

If the flow was incompressible then for known density, $\rho$,  the velocity can be calculated by utilizing \( \Delta P = \dfrac{4\,f\,L}{D} {U^2 \over 2g}\). 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, \(\dfrac{4\,f\,L}{D}\)
\[
\dfrac{4\,f\,L}{D} = \dfrac{4 \times 0.005 5000 }{ 0.25} = 400 \tag{41}
\]
Utilizing Table (??) or the Potto–GDC provides

Isothermal Flow Input: \(\dfrac{4\,f\,L}{D}\) k = 1.31
\(M\) \(\dfrac{4\,f\,L}{D}\) \(\dfrac{P}{P^{\star}}\) \(\dfrac{P_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[2]/div/table[1]/tbody/tr[2]/td[4]/span, line 1, column 4
\)
\(\dfrac{\rho}{\rho^{\star} }\) \(\dfrac{T_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[2]/div/table[1]/tbody/tr[2]/td[6]/span, line 1, column 4
\)
.04331 400.00 20.1743 12.59921 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

\[
    \rho = {P \over R T} = { 2,017,450 \over 290 \times 300} \cong 23.19  
        \left[ kg \over m^{3}\right]
\end{align*}
The maximum flow rate then reads
\begin{align*}
    \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]
\end{align*}

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 an isothermal model hence our model is appropriate. For incompressible flow, the pressure loss is expressed as follows

\[
   \label{isothermal:eq:incompressibleP-Loss}
   P_{1} - P_{2}  = \dfrac{4\,f\,L}{D}\, { U ^{2}   \over 2} \tag{42}
\]
Now note that for incompressible flow $U_{1} = U_{2}= U$ and $\dfrac{4\,f\,L}{D}$
represent the ratio of the  traditional \(h_{12}\). To obtain a similar expression for isothermal flow, a relationship between \(M_{2}\) and \(M_{1}\) and pressures has to be derived. From equation (42) one can obtained that  
\[
   M_{2} = M_{1} \,\dfrac{P_{1} }{ P_{2}}    
   \label{isothermal:eq:m2} \tag{43}
\]
To solve this problem the flow rate has to be calculated as
\begin{align*}
    \dot{m} = \rho A U  = 2.0 \left[ kg \over sec\right]
\end{align*}
\begin{align*}
    \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 }  
\end{align*}
Now combining with equation (43) yields
\begin{align*}
    \dot{m} = \dfrac{M_2\, P_2\, A\, k }{ c }  
\end{align*}
\begin{align*}
    M_2 = \dfrac{  \dot{m} \,c }{ P_2\, A\, k} =  
        { 2 \times 337.59 \over 100000 \times {\pi \times (0.25)^{2} \over
            4} \times 1.31} = 0.103
\end{align*}
From Table (??) or by utilizing the Potto–GDC one can obtain

Isothermal Flow Input: \(M\) k = 1.31
\(M\) \(\dfrac{4\,f\,L}{D}\) \(\dfrac{P}{P^{\star}}\) \(\dfrac{P_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[2]/div/table[2]/tbody/tr[2]/td[4]/span, line 1, column 4
\)
\(\dfrac{\rho}{\rho^{\star} }\) \(\dfrac{T_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[2]/div/table[2]/tbody/tr[2]/td[6]/span, line 1, column 4
\)
0.10300 66.6779 8.4826 5.3249 0.0 0.89567

The entrance Mach number is obtained by  
\begin{align*}
    \left. \dfrac{4\,f\,L}{D} \right|_1 = 66.6779 + 400 \cong 466.68
\end{align*}
Hence,

Isothermal Flow Input: \(\dfrac{4\,f\,L}{D}\) k = 1.31
\(M\) \(\dfrac{4\,f\,L}{D}\) \(\dfrac{P}{P^{\star}}\) \(\dfrac{P_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[2]/div/table[3]/tbody/tr[2]/td[4]/span, line 1, column 4
\)
\(\dfrac{\rho}{\rho^{\star} }\) \(\dfrac{T_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[2]/div/table[3]/tbody/tr[2]/td[6]/span, line 1, column 4
\)
0.04014 466.68 21.7678 13.5844 0.0 0.89442

The pressure should be  
\begin{align*}
    P = 21.76780 \times 8.4826 = 2.566 [Bar]
\end{align*}
Note that tables in this example are for \(k=1.31\)

 

Example 11.16

A flow of gas was considered for a distance of 0.5 [km] (500 [m]). A flow rate of 0.2 [kg/sec] is required. Due to safety concerns, the maximum pressure allowed for the gas is only 10[bar]. Assume that the flow is isothermal and k=1.4, calculate the required diameter of tube. The friction coefficient for the tube can be assumed as 0.02 (A relative smooth tube of cast iron.). Note that tubes are provided in increments of 0.5 [in]. You can assume that the soundings temperature to be \(27^{\circ}C\).

Solution 11.16

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.  

\begin{align*}
    M_1 = M_2  { P_2 \over P_1}     =  
    \overbrace{ 1 \over \sqrt{k} }^{M_{max}} {1 \over 10} = 0.0845
\end{align*}

Now, with the value of \(M_1\) either by utilizing Table 11.4 or using the provided program yields

Isothermal Flow   Input: \(M\)     k = 1.31
\(M\) \(\dfrac{4\,f\,L}{D}\) \(\dfrac{P}{P^{\star}}\) \(\dfrac{P_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[4]/div/table[1]/tbody/tr[2]/td[4]/span, line 1, column 4
\)
\(\dfrac{\rho}{\rho^{\star} }\) \(\dfrac{T_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[4]/div/table[1]/tbody/tr[2]/td[6]/span, line 1, column 4
\)
0.08450 94.4310 10.0018 6.2991 0.0 0.87625

 With \({\left.\dfrac{4\,f\,L}{D}\right|_{max}}= 94.431\), the value of minimum diameter.  
\begin{align*}
    D = {4 f L \over {\left.\dfrac{4\,f\,L}{D}\right|_{max}}} \simeq  
        {4 \times 0.02 \times 500 \over 94.43}     \simeq
        0.42359 [m] = 16.68 [in]        
\end{align*}

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

\begin{align*}
    \dot{m} = \rho A U = \rho A M c = { P \over R\, T} \,A\, M \,\sqrt{k\,R\,T}   
        = {P\, A\, M \sqrt{k} \over \sqrt{R\,T}}  
\end{align*}
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 \(\dfrac{4\,f\,L}{D}\) is
\begin{align*}
    \dfrac{4\,f\,L}{D} =  \dfrac{4 \times   0.02 \times 500 }{ 0.4318} \simeq 92.64
\end{align*}

Isothermal Flow   Input: \(M\)     k = 1.31
\(M\) \(\dfrac{4\,f\,L}{D}\) \(\dfrac{P}{P^{\star}}\) \(\dfrac{P_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[4]/div/table[2]/tbody/tr[2]/td[4]/span, line 1, column 4
\)
\(\dfrac{\rho}{\rho^{\star} }\) \(\dfrac{T_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[4]/div/table[2]/tbody/tr[2]/td[6]/span, line 1, column 4
\)
0.08527 92.6400 9.9110 6.2424 1.0 0.87627

To check whether the flow rate satisfies the requirement

\begin{align*}
    \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]       
\end{align*}
Since \( 50.3 \geq 0.2 \)  the mass flow rate requirement 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
\begin{align*}
    c = \sqrt{k\,R\,T } = \sqrt{1.4 \times 287 \times 300} \cong  
        347.2 \left[ m \over sec \right]
\end{align*}
and the density is  
\begin{align*}
    \rho = \dfrac{P }{ R\, T} = \dfrac{1,000,000 }{ 287 \times 300} =  
        11.61 \left[ kg \over m^3 \right]  
\end{align*}
The velocity at the entrance should be  
\begin{align*}
    U = M *c = 0.08528 \times 347.2 \cong 29.6 \left[ m \over sec \right]  
\end{align*}
The diameter should be  
\begin{align*}
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
\end{align*}
Nevertheless, for the 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.

Example 11.17

A gas flows of from a station (a) with pressure of 20[bar] through a pipe with 0.4[m] diameter and 4000 [m] length to a different station (b). The pressure at the exit (station (b)) is 2[bar]. The gas and the sounding temperature can be assumed to be 300 K. Assume that the flow is isothermal, k=1.4, and the average friction f=0.01. Calculate the Mach number at the entrance to pipe and the flow rate.

Solution 11.17

First, the information whether the flow is choked needs to be found. Therefore, at first it will be assumed that the whole length is the maximum length.
\[
    \nonumber
    {\left.\dfrac{4\,f\,L}{D}\right|_{max}}= {4 \times 0.01 \times 4000 \over 0.4} = 400    \tag{44}
\]
with \(\left.\dfrac{4\,f\,L}{D}\right|_{max}=400\) the following can be written

Isothermal Flow Input: \(\dfrac{4\,f\,L}{D}\) k = 1.31
\(M\) \(\dfrac{4\,f\,L}{D}\) \(\dfrac{P}{P^{\star}}\) \(\dfrac{P_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[6]/div/table[1]/tbody/tr[2]/td[4]/span, line 1, column 4
\)
\(\dfrac{\rho}{\rho^{\star} }\) \(\dfrac{T_0}
ParseError: EOF expected (click for details)
Callstack:
    at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.6:_Isothermal_Flow_Examples), /content/body/div[6]/div/table[1]/tbody/tr[2]/td[6]/span, line 1, column 4
\)
0.04014 466.68 21.7678 13.5844 0.0 0.89442

From the table \(M_1\approx 0.0419\), and \(\dfrac{P_{0} }{ {P_{0}}^{*T} } \approx 12.67\)
\begin{align*}
     {P_{0}}^{*T}  \cong {28 \over 12.67} \simeq 2.21 [bar]
\end{align*}
The pressure at point (b) by utilizing the isentropic relationship (\(M=1\))  pressure ratio is \(0.52828\).
\begin{align*}
    P_2 = \dfrac{    {P_{0}}^{*T}  }{  
        \left(  \dfrac{ P_2 }{ { {P_{0}}^{*T} }} \right)
            }
        = { 2.21 \times 0.52828 } = 1.17 [bar]
\end{align*}

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

Guess reasonable value of \(M_1\) and calculate \(\dfrac{4\,f\,L}{D}\)

Calculate the value of \(\left.{\dfrac{4\,f\,L}{D}}\right|_{2}\) by subtracting \(\left.{\dfrac{4\,f\,L}{D}}\right|_{1} -\dfrac{4\,f\,L}{D}\)

Obtain \(M_2\) from the Table ? or by using the Potto–GDC.

Calculate the pressure, \(P_2\) bear in mind that this isn't the real pressure but based on the assumption.

Compare the results of guessed pressure \(P_2\) with the actual pressure and choose new Mach number \(M_1\) accordingly.

 

Isothermal Flow Input: \(P_2/P_1\) and \(\dfrac{4\,f\,L}{D}\) k = 1.4
\(M_1\) \(M_2\) \(\left.\dfrac{4\,f\,L}{D}\right|_{max}\) \(\dfrac{4\,f\,L}{D}\) \(\dfrac{P_2}{P_1}\)
0.0419 0.59338 400.32131 400.00 0.10

The flow rate is
\begin{align*}
    \dot{m} = \rho A M c = \dfrac{P \sqrt{k} }{ \sqrt {R T}}  
        \dfrac{\pi \times D^{2} }{ 4} \,M   
        = \dfrac{2000000\, \sqrt{1.4 }} { \sqrt{300 \times 287}}
        {\pi \times 0.2^{2}} \times 0.0419 \\
        \simeq 42.46[kg/sec]    
\end{align*}

 

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

Isothermal Flow Input: \(P_2/P_1\) and \(\dfrac{4\,f\,L}{D}\) k = 1.4
\(M_1\) \(M_2\) \(\left.\dfrac{4\,f\,L}{D}\right|_{max}\) \(\dfrac{4\,f\,L}{D}\) \(\dfrac{P_2}{P_1}\)
0.7272 0.84095 0.05005 0.05000 0.10
0.6934 0.83997 0.08978 0.08971 0.10
0.6684 0.84018 0.12949 0.12942 0.10
0.6483 0.83920 0.16922 0.16912 0.10
0.5914 0.83889 0.32807 0.32795 0.10
0.5807 0.83827 0.36780 0.36766 0.10
0.5708 0.83740 0.40754 0.40737 0.10

template('TranscludeAutoNum',{'PageNum':'11.7'});

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.