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# Chapter 11 Gd (continue)

## 11.7 Isothermal Flow

In this section a model dealing with gas that flows through a long tube is described. This model has a applicability to situations which occur in a relatively long distance and where heat transfer is relatively rapid so that the temperature can be treated, for engineering purposes, as a constant. For example, this model is applicable when a natural gas flows over several hundreds of meters. Such situations are common in large cities in U.S.A. where natural gas is used for heating. It is more predominant (more applicable) in situations where the gas is pumped over a length of kilometers.

Fig. 11.17 Control volume for isothermal flow.

The high speed of the gas is obtained or explained by the combination of heat transfer and the friction to the flow. For a long pipe, the pressure difference reduces the density of the gas. For instance, in a perfect gas, the density is inverse of the pressure (it has to be kept in mind that the gas undergoes an isothermal process.). To maintain conservation of mass, the velocity increases inversely to the pressure. At critical point the velocity reaches the speed of sound at the exit and hence the flow will be choked.

### 11.7.1 The Control Volume Analysis/Governing equations

Figure 11.17 describes the flow of gas from the left to the right. The heat transfer up stream (or down stream) is assumed to be negligible. Hence, the energy equation can be written as the following: \begin{align} \dfrac{d\, Q }{ \dot{m} } = c_p dT + d \dfrac{U^2 }{ 2} = c_p dT_{0} \label{isothermal:eq:CV} \end{align} The momentum equation is written as the following \begin{align} -A\, dP - \tau_{w}\, dA_{\text{wetted area}} = \dot{m}\, dU \label{isothermal:eq:momentum} \end{align} where $A$ is the cross section area (it doesn't have to be a perfect circle; a close enough shape is sufficient.). The shear stress is the force per area that acts on the fluid by the tube wall. The $A_{wetted\;\;area}$ is the area that shear stress acts on. The second law of thermodynamics reads \begin{align} {s_2 - s_1 \over C_p} = \ln {T_2 \over T_1 } - {k -1 \over k} \ln {P_2 \over P_1} \label{isothermal:eq:2law} \end{align} The mass conservation is reduced to \begin{align} \dot {m} = \text{constant} = \rho\, U\, A \label{isothermal:eq:mass} \end{align} Again it is assumed that the gas is a perfect gas and therefore, equation of state is expressed as the following: \begin{align} P = \rho\, R\, T \label{isothermal:eq:state} \end{align}

### 11.7.2 Dimensionless Representation

In this section the equations are transformed into the dimensionless form and presented as such. First it must be recalled that the temperature is constant and therefore, equation of state reads \begin{align} {dP \over P} = {d \rho \over \rho} \label{isothermal:eq:stateDa} \end{align} It is convenient to define a hydraulic diameter \begin{align} D_{H} = {4 \times \hbox{Cross Section Area} \over \hbox{ wetted perimeter }} \label{isothermal:eq:HDdef} \end{align} The Fanning friction factor is introduced, this factor is a dimensionless friction factor sometimes referred to as the friction coefficient as \begin{align} f = { \tau_w \over \dfrac{1}{2} \rho U^{2} } \label{isothermal:eq:f} \end{align} Substituting equation \eqref{isothermal:eq:f} into momentum equation qref{isothermal:eq:momentum} yields \begin{align} - dP - \dfrac{4\, dx }{ D_H} f \left( \dfrac{1}{2}\, \rho\, U^{2} \right) = { \overbrace{\rho\, U}^{\dot {m} \over A } dU } \label{isothermal:eq:momentumD1} \end{align} Rearranging equation \eqref{isothermal:eq:momentumD1} and using the identify for perfect gas $M^{2} = \rho U^{2} / k P$ yields: \begin{align} - \dfrac{dP }{ P} - \dfrac{4\,f\, dx }{ D_H} \left( \dfrac{ k\,P\, M^{2} }{ 2} \right) = \dfrac{k\,P\, M^{2}\, dU }{ U} \label{isothermal:eq:momentumD} \end{align} The pressure, $P$ as a function of the Mach number has to substitute along with velocity, $U$ as \begin{align} U^{2} = {k\,R\,T} M^{2} \label{isothermal:eq:MtoUfun} \end{align} Differentiation of equation \eqref{isothermal:eq:MtoUfun} yields \begin{align} d(U^{2}) = k\,R\,\left(M^{2}\; dT + T \; d (M^{2}) \; \right) \label{isothermal:eq:anotherMtoU} \end{align} \begin{align} \dfrac{d(M^2) }{ M^{2}} = \dfrac{d(U^2) }{ U^{2}} - \dfrac{ dT }{ T} \label{isothermal:eq:MtoUfunD} \end{align} It can be noticed that $dT = 0$ for isothermal process and therefore \begin{align} {d(M^2) \over M^{2}} = {d(U^2) \over U^{2}} = {2U \; dU \over U^{2}} = {2 dU \over U} \label{isothermal:eq:yetMtoUfunD} \end{align} The dimensionalization of the mass conservation equation yields \begin{align} {d \rho \over \rho} + {dU \over U } = {d \rho \over \rho} + {2\, U\, dU \over 2 U^{2} } = {d \rho \over \rho} + {d (U^{2}) \over 2\; U^{2}} = 0 \label{isothermal:eq:massD} \end{align} Differentiation of the isotropic (stagnation) relationship of the pressure \eqref{gd:iso:eq:pressureDless} yields \begin{align} \dfrac{dP_0 }{ P_0} = \dfrac{dP }{ P } + \left( \dfrac{ \dfrac { k\, M^{2} }{2}} { 1 + \dfrac{ k-1 }{ 2 } M^{2} } \right) \; {d M^{2} \over M^{2} } \label{isothermal:eq:PzeroD} \end{align} Differentiation of equation \eqref{gd:iso:eq:temperatureDless} yields: \begin{align} dT_{0} = dT \left( 1 + {k - 1 \over 2} M^{2} \right) + T \; {k - 1 \over 2} \; d M^{2} \label{isothermal:eq:dT0T} \end{align} Notice that $dT_{0} \neq 0$ in an isothermal flow. There is no change in the actual temperature of the flow but the stagnation temperature increases or decreases depending on the Mach number (supersonic flow of subsonic flow). Substituting $T$ for equation \eqref{isothermal:eq:dT0T} yields: \begin{align} dT_{0} = \dfrac{ T_{0} \; \dfrac{k - 1 }{ 2} \; d\,M^{2} }{ \left( 1 + \dfrac{k - 1 }{ 2} M^{2} \right) } \; {M^{2} \over M^{2} } \label{isothermal:eq:TOTtmp} \end{align} Rearranging equation \eqref{isothermal:eq:TOTtmp} yields \begin{align} { dT_{0} \over T_{0} } = \dfrac{ {(k - 1 )} \; M^{2} }{ 2 \left( 1 + \dfrac{k - 1 }{ 2} \right) } \; {dM^{2} \over M^{2} } \label{isothermal:eq:logT0T} \end{align} By utilizing the momentum equation it is possible to obtain a relation between the pressure and density. Recalling that an isothermal flow ($dT=0$) and combining it with perfect gas model yields \begin{align} {dP \over P } = { d\rho \over \rho} \label{isothermal:eq:stateD} \end{align} From the continuity equation (see equation \eqref{isothermal:eq:yetMtoUfunD}) leads \begin{align} {d M^{2} \over M^{2} } = { 2 dU \over U } \label{isothermal:eq:massAnother} \end{align} The four equations momentum, continuity (mass), energy, state are described above. There are 4 unknowns ($M, T, P, \rho$) and with these four equations the solution is attainable. One can notice that there are two possible solutions (because of the square power). These different solutions are supersonic and subsonic solution. The distance friction, $t\dfrac{4\,f\,L}{D}$, is selected as the choice for the independent variable. Thus, the equations need to be obtained as a function of $\dfrac{4f\,L}{D}$. The density is eliminated from equation \eqref{isothermal:eq:massD} when combined with equation \eqref{isothermal:eq:stateD} to become \begin{align} \dfrac{dP }{ P } = - \dfrac{dU }{ U} \label{isothermal:eq:stateMassD} \end{align} After substituting the velocity \eqref{isothermal:eq:stateMassD} into equation \eqref{isothermal:eq:momentumD}, one can obtain \begin{align} - \dfrac{dP }{ P} - \dfrac{4\,f\, dx }{ D_H} \left( \dfrac{ k\,P\, M^{2} }{ 2} \right) = {k\,P\, M^{2} }\, \dfrac{dP }{ P } \label{isothermal:eq:mm} \end{align} Equation \eqref{isothermal:eq:mm} can be rearranged into \begin{align} \dfrac{dP }{ P } = \dfrac{d\rho }{ \rho} = - \dfrac{dU }{ U} = - \dfrac{1 }{ 2} \dfrac{dM^{2} }{ M^{2} } = - \dfrac{ k \,M^{2} }{ 2\left( 1 - k\,M^{2} \right) } 4\,f \,\dfrac{dx }{ D} \label{isothermal:eq:pressureGov} \end{align} Similarly or by other paths, the stagnation pressure can be expressed as a function of $\dfrac{4\,f\,L}{D}$ \begin{align} \dfrac{dP_{0} }{ P_{0} } = {{k \,M^ {2} \left( 1 - \dfrac{k + 1 }{ 2}\,M^{2} \right) } \over {2 \left( k\,M^{2} - 1 \right) \left( 1 + \dfrac{k -1 }{ 2} M^{2} \right) }} 4f {dx \over D} \label{isothermal:eq:stangnationPressureGov} \end{align} \begin{align} {dT_{0} \over T_{0} } = {{k \left( 1 - k \right) M^{2} } \over {2 \left(1 - k\,M^{2} \right) \left( 1 + \dfrac{k -1 }{ 2} M^{2} \right) }}\, 4\,f {dx \over D} \label{isothermal:eq:stangnationTempGov} \end{align} The variables in equation \eqref{isothermal:eq:pressureGov} can be separated to obtain integrable form as follows \begin{align} \int_{0}^{L} { 4\,f\, dx \over D} = \int_{M^{2}}^{1/k} { 1 - k\,M{2} \over k\,M{2}} dM^{2} \label{isothermal:eq:integralMach} \end{align} It can be noticed that at the entrance $(x = 0)$ for which $M = M_{x=0}$ (the initial velocity in the tube isn't zero). The term $\dfrac{4f\,L}{D}$ is positive for any $x$, thus, the term on the other side has to be positive as well. To obtain this restriction $1= k\,M^{2}$. Thus, the value $M= {1\over \sqrt{k}}$ is the limiting case from a mathematical point of view. When Mach number larger than $M > {1\over \sqrt{k}}$ it makes the right hand side of the integrate negative. The physical meaning of this value is similar to $M=1$ choked flow which was discussed in a variable area flow in Section 11.4. Further it can be noticed from equation \eqref{isothermal:eq:stangnationTempGov} that when $M \rightarrow {1\over \sqrt{k}}$ the value of right hand side approaches infinity ($\infty$). Since the stagnation temperature ($T_{0}$) has a finite value which means that $dT_{0} \rightarrow \infty$. Heat transfer has a limited value therefore the model of the flow must be changed. A more appropriate model is an adiabatic flow model yet this model can serve as a bounding boundary (or limit). Integration of equation \eqref{isothermal:eq:integralMach} requires information about the relationship between the length, $x$, and friction factor $f$. The friction is a function of the Reynolds number along the tube. Knowing the Reynolds number variations is important. The Reynolds number is defined as \begin{align} Re = \dfrac{D\, U\, \rho}{\mu} \label{isothermal:eq:Re} \end{align} The quantity $U\,\rho$ is constant along the tube (mass conservation) under constant area. Thus, only viscosity is varied along the tube. However under the assumption of ideal gas, viscosity is only a function of the temperature. The temperature in isothermal process (the definition) is constant and thus the viscosity is constant. In real gas, the pressure effects are very minimal as described in Basic of fluid mechanics'' by this author. Thus, the friction factor can be integrated to yield

Friction Mach Isothermal Flow

\begin{align} \label{isothermal:eq:workingEq} {\left.\dfrac{4\,f\,L}{D}\right|_{max}} = { 1- k\, M^{2} \over k\, M^{2} } + \ln \left(k\,M^{2} \right) \end{align}
The definition for perfect gas yields $M^{2} = { U^{2} / k\,R\,T}$ and noticing that $T=constant$ is used to describe the relation of the properties at $M = 1 / \sqrt{k}$. By denoting the superscript symbol $*$ for the choking condition, one can obtain that \begin{align} { M^{2} \over U^{2} } = {{ 1/k} \over {U^{*}}^ {2}} \label{isothermal:eq:MachDef} \end{align} Rearranging equation \eqref{isothermal:eq:MachDef} is transformed into \begin{align} {U \over U^{*}} = \sqrt{k} M \label{isothermal:eq:Ubar} \end{align} Utilizing the continuity equation provides \begin{align} \rho U = \rho^{*} U^{*} \quad \Longrightarrow \quad \dfrac{\rho }{ \rho^{\star}} = \dfrac{ 1 }{ \sqrt{k} \, M} \end{align} Reusing the perfect–gas relationship

Pressure Ratio

\begin{align} \label{isothermal:eq:Pbar} \dfrac{P }{ P^{*}} = {\rho \over \rho^{*}} = \dfrac{ 1 }{ \sqrt{k} \, M} \end{align}
Utilizing the relation for stagnated isotropic pressure one can obtain \begin{align} {P_{0} \over P_{0}^{*}} = {P \over P^{*}} \left[ {1 + { k -1 \over 2 } M ^ {2} \over { 1 + {k -1 \over 2k} } } \right] ^ { k \over k -1 } \end{align} Substituting for ${P \over P^{*}}$ equation \eqref{isothermal:eq:Pbar} and rearranging yields

Stagnation Pressure Ratio

\begin{align} \label{isothermal:eq:P0bar} \dfrac{P_{0} }{ P_{0}^{*}} = {1 \over \sqrt{k}} \left( {2\,k \over 3\,k- 1} \right)^ {\dfrac{k }{ k -1} } \left( 1 + {k -1 \over 2} M ^{2}\right)^{\dfrac{k }{ k-1} } \dfrac{ 1 }{ M} \end{align}
And the stagnation temperature at the critical point can be expressed as

Stagnation Pressure Ratio

\begin{align} \label{isothermal:eq:T0bar} {T_{0} \over T_{0}^{*}} = { T \over T^{*}} \dfrac{ 1 + \dfrac{k -1 }{ 2} M ^{2} }{ 1 + \dfrac{k -1 }{ 2\,k} } = \dfrac{2\,k }{ 3\,k -1 } \left( 1 + {k -1 \over 2} \right) M ^{2} \end{align}
These equations \eqref{isothermal:eq:Ubar}-qref{isothermal:eq:T0bar} are presented on in Figure \eqref{isothermal:fig:basic}.

Fig. 11.18 Description of the pressure, temperature relationships as a function of the Mach number for isothermal flow.

### 11.7.3 The Entrance Limitation of Supersonic Branch

This setion deals with situations where the conditions at the tube exit have not arrived at the critical condition. It is very useful to obtain the relationships between the entrance and the exit conditions for this case. Denote {m 1} and {m 2} as the conditions at the inlet and exit respectably. From equation \eqref{isothermal:eq:pressureGov} \begin{align} \dfrac{4\,f\,L}{D} = \left. \dfrac{4\,f\,L}{D}\right|_{{max}_1} - \left.\dfrac{4\,f\,L}{D}\right|_{{max}_{2}} = { 1 - k\,{M_{1}}^{2} \over k\,{M_{1}}^{2}} - { 1 - k\,{M_{2}}^{2} \over k\,{M_{2}}^{2}} + \ln \left( {M_{1} \over M_{2}} \right)^{2} \label{isothermal:eq:workingFLD} \end{align} For the case that $M_1 > > M_2$ and $M_1 \rightarrow 1$ equation qref{isothermal:eq:workingFLD} is reduced into the following approximation \begin{align} \dfrac{4\,f\,L}{D} = 2 \ln \left( M_{1}\right) -1 - \overbrace{\dfrac{ 1 - k\,{M_{2}}^{2} }{ k {M_{2}}^{2}}}^{\sim 0} \label{isothermal:eq:workingFLDApprox} \end{align} Solving for $M_1$ results in \begin{align} M_1 \sim \text{e}^{\dfrac{1 }{ 2}\,\left(\dfrac{4\,f\,L}{D} +1\right)} \label{isothermal:eq:workingFLDAppSol} \end{align} This relationship shows the maximum limit that Mach number can approach when the heat transfer is extraordinarily fast. In reality, even small $\dfrac{4\,f\,L}{D} > 2$ results in a Mach number which is larger than 4.5. This velocity requires a large entrance length to achieve good heat transfer. With this conflicting mechanism obviously the flow is closer to the Fanno flow model. Yet this model provides the directions of the heat transfer effects on the flow.

# Example 11.14

Calculate the exit Mach number for pipe with $\dfrac{4\,f\,L}{D} = 3$ under the assumption of the isothermal flow and supersonic flow. Estimate the heat transfer needed to achieve this flow.

### 11.7.4 Supersonic Branch

Apparently, this analysis/model is over simplified for the supersonic branch and does not produce reasonable results since it neglects to take into account the heat transfer effects. A dimensionless analysis demonstrates that all the common materials that the author is familiar which creates a large error in the fundamental assumption of the model and the model breaks. Nevertheless, this model can provide a better understanding to the trends and deviations from Fanno flow model. In the supersonic flow, the hydraulic entry length is very large as will be shown below. However, the feeding diverging nozzle somewhat reduces the required entry length (as opposed to converging feeding). The thermal entry length is in the order of the hydrodynamic entry length (look at the Prandtl number , (0.7-1.0), value for the common gases.). Most of the heat transfer is hampered in the sublayer thus the core assumption of isothermal flow (not enough heat transfer so the temperature isn't constant) breaks down. The flow speed at the entrance is very large, over hundred of meters per second. For example, a gas flows in a tube with $\dfrac{4\,f\,L}{D}= 10$ the required entry Mach number is over 200. Almost all the perfect gas model substances dealt with in this book, the speed of sound is a function of temperature. For this illustration, for most gas cases the speed of sound is about $300 [m/sec]$. For example, even with low temperature like $200K$ the speed of sound of air is $283 [m/sec]$. So, even for relatively small tubes with $\dfrac{4\,f\,D}{D}= 10$ the inlet speed is over 56 [km/sec]. This requires that the entrance length to be larger than the actual length of the tub for air. \begin{align} L_{entrance} = 0.06\, \dfrac{ U \,D }{ \nu} \label{fanno:eq:entrceL} \end{align} The typical values of the the kinetic viscosity, $\nu$, are 0.0000185 kg/m-sec at 300K and 0.0000130034 kg/m-sec at 200K. Combine this information with our case of $\dfrac{4\,f\,L}{D} =10$ \begin{align*} {L_{entrance} \over D} = 250746268.7 \end{align*} On the other hand a typical value of friction coefficient $f = 0.005$ results in \begin{align*} {L_{max} \over D} = { 10 \over 4 \times 0.005} = 500 \end{align*} The fact that the actual tube length is only less than 1% of the entry length means that the assumption is that the isothermal flow also breaks (as in a large response time). If Mach number is changing from 10 to 1 the kinetic energy change is about $\mathbf{T_0 \over {T_0}^{*}} = 18.37$ which means that the maximum amount of energy is insufficient. Now with limitation, this topic will be covered in the next version because it provide some insight and boundary to the Fanno Flow model.

### 11.7.5 Figures and Tables

 $M$ 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.81 0.81879 0.82758 0.83637 0.84515 $\dfrac{4\,f\,L}{D}$ 785.97 439.33 279.06 192.12 139.79 105.89 82.704 66.1599 13.9747 7.9925 4.865 3.0677 1.9682 1.2668 0.80732 0.50207 0.29895 0.16552 0.08085 0.03095 0.00626 0.00371 0.00205 0.000896 0.00022 0 $\dfrac{P}{P^{\star}}$ 28.1718 21.1289 16.9031 14.0859 12.0736 10.5644 9.3906 8.4515 4.2258 3.3806 2.8172 2.4147 2.1129 1.8781 1.6903 1.5366 1.4086 1.3002 1.2074 1.1269 1.056 1.043 1.032 1.021 1.011 1 $\dfrac{P_{0}}{{P_{0}}^{\star}}$ 17.6651 13.2553 10.6109 8.8493 7.592 6.65 5.9181 5.3334 2.723 2.2126 1.8791 1.647 1.4784 1.3524 1.2565 1.1827 1.1259 1.0823 1.0495 1.0255 1.009 1.007 1.005 1.003 1.001 1 $\dfrac{\rho}{\rho^{\star}}$ 28.1718 21.1289 16.9031 14.0859 12.0736 10.5644 9.3906 8.4515 4.2258 3.3806 2.8172 2.4147 2.1129 1.8781 1.6903 1.5366 1.4086 1.3002 1.2074 1.1269 1.056 1.043 1.032 1.021 1.011 1 $\dfrac{T_{0}}{{T_{0}}^{\star}}$ 0.87516 0.87528 0.87544 0.87563 0.87586 0.87612 0.87642 0.87675 0.882 0.88594 0.89075 0.89644 0.903 0.91044 0.91875 0.92794 0.938 0.94894 0.96075 0.97344 0.987 0.98982 0.99232 0.99485 0.99741 1

### 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

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, $\dfrac{4\,f\,L}{D}$ $$\dfrac{4\,f\,L}{D} = \dfrac{4 \times 0.005 5000 }{ 0.25} = 400$$ Utilizing Table \eqref{isothermal:tab:basic} or the Potto–GDC provides

 Isothermal Flow Input: $\dfrac{4\,f\,L}{D}$ k = 1.31 $M$ $\dfrac{4\,f\,L}{D}$ $\dfrac{P}{P^{\star}}$ $\dfrac{P_0}{{P_0}^{\star}}$ $\dfrac{\rho}{\rho^{\star} }$ $\dfrac{T_0}{{T_0}^{\star}}$ 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 \begin{align*} \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 \begin{align} \label{isothermal:eq:incompressibleP-Loss} P_{1} - P_{2} = \dfrac{4\,f\,L}{D}\, { U ^{2} \over 2} \end{align} 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 \eqref{isothermal:eq:incompressibleP-Loss} one can obtained that \begin{align} M_{2} = M_{1} \,\dfrac{P_{1} }{ P_{2}} \label{isothermal:eq:m2} \end{align} 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 \eqref{isothermal:eq:m2} 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 \eqref{isothermal:tab:basic} or by utilizing the Potto–GDC one can obtain

 Isothermal Flow Input: $M$ k = 1.31 $M$ $\dfrac{4\,f\,L}{D}$ $\dfrac{P}{P^{\star}}$ $\dfrac{P_0}{{P_0}^{\star}}$ $\dfrac{\rho}{\rho^{\star} }$ $\dfrac{T_0}{{T_0}^{\star}}$ 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 $M$ $\dfrac{4\,f\,L}{D}$ $\dfrac{P}{P^{\star}}$ $\dfrac{P_0}{{P_0}^{\star}}$ $\dfrac{\rho}{\rho^{\star} }$ $\dfrac{T_0}{{T_0}^{\star}}$ 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

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 $M$ $\dfrac{4\,f\,L}{D}$ $\dfrac{P}{P^{\star}}$ $\dfrac{P_0}{{P_0}^{\star}}$ $\dfrac{\rho}{\rho^{\star} }$ $\dfrac{T_0}{{T_0}^{\star}}$ 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 $17[in]$ 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 $M$ $\dfrac{4\,f\,L}{D}$ $\dfrac{P}{P^{\star}}$ $\dfrac{P_0}{{P_0}^{\star}}$ $\dfrac{\rho}{\rho^{\star} }$ $\dfrac{T_0}{{T_0}^{\star}}$ 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

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. {\begin{align} \nonumber {\left.\dfrac{4\,f\,L}{D}\right|_{max}}= {4 \times 0.01 \times 4000 \over 0.4} = 400 \end{align}} 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 $M$ $\dfrac{4\,f\,L}{D}$ $\dfrac{P}{P^{\star}}$ $\dfrac{P_0}{{P_0}^{\star}}$ $\dfrac{\rho}{\rho^{\star} }$ $\dfrac{T_0}{{T_0}^{\star}}$ 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.

1. Guess reasonable value of $M_1$ and calculate $\dfrac{4\,f\,L}{D}$
2. Calculate the value of $\left.{\dfrac{4\,f\,L}{D}}\right|_{2}$ by subtracting $\left.{\dfrac{4\,f\,L}{D}}\right|_{1} -\dfrac{4\,f\,L}{D}$
3. Obtain $M_2$ from the Table ? or by using the Potto–GDC.
4. Calculate the pressure, $P_2$ bear 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 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

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