Basics of Fluid Mechanics
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Chapter 11 Compressible Flow One Dimensional

11.1 What is Compressible Flow?

This Chapters deals with an introduction to the flow of compressible substances (gases). The main difference between compressible flow and ``almost'' incompressible flow is not the fact that compressibility has to be considered. Rather, the difference is in two phenomena that do not exist in incompressible The first phenomenon is the very sharp discontinuity (jump) in the flow in properties. Choking is referred to as the situation where downstream conditions, which are beyond a critical value(s), doesn't affect the flow. The shock wave and choking are not intuitive for most people. However, one has to realize that $\textcolor{red}{\mathbf{intuition}}$ is really a condition where one uses his past experiences to predict other situations. Here one has to build his intuition tool for future use. Thus, not only engineers but other disciplines will be able use this ``intuition'' in design, understanding and even research.

11.2 Why Compressible Flow is Important?

Compressible flow appears in many natural and many technological processes. Compressible flow deals, including many different material such as natural gas, nitrogen and helium, etc not such only air. For instance, the flow of natural gas in a pipe system, a common method of heating in the U.S., should be considered a compressible flow. These processes include flow of gas in the exhaust system of an internal combustion engine. The above flows that were mentioned are called internal flows. Compressible flow also includes flow around bodies such as the wings of an airplane, and is categorized as external flow. These processes include situations not expected to have a compressible flow, such as manufacturing process such as the die casting, injection molding. The die casting process is a process in which liquid metal, mostly aluminum, is injected into a mold to obtain a near final shape. The air is displaced by the liquid metal in a very rapid manner, in a matter of milliseconds, therefore the compressibility has to be taken into account. Clearly, mechanical or aero engineers are not the only ones who have to deal with some aspects of compressible flow. Even manufacturing engineers have to deal with many situations where the compressibility or compressible flow understating is essential for adequate design. Another example, control engineers who are using pneumatic systems must consider compressible flow aspects of the substances used. The compressible flow unique phenomena also appear in zoology (bird fly), geological systems, biological system (human body) etc. These systems require consideration of the unique phenomena of compressible flow. In this Chapter, a greater emphasis is on the internal flow while the external flow is treated to some extend in the next Chapter. It is recognized that the basic fluid mechanics class has a limited time devoted to these topics. Additional information (such as historical background) can be found in ``Fundamentals of Compressible Flow'' by the same author on Potto Project web site.

11.3 Speed of Sound

Slow Moving Piston

Fig. 11.1 A very slow moving piston in a still gas.

Most of compressible flow occurs at relative high velocity as compere to the speed of sound. Hence, the speed of sound has to discussed initially. Outside the ideal gas, limited other situations will be discussed.

11.3.1 Introduction

People had recognized for several hundred years that sound is a variation of pressure. This velocity is referred to as the speed of sound and is discussed first.

Stationary Sound Wave

Fig. 11.2 Stationary sound wave and gas moves relative to the pulse.

To answer this question consider a piston moving from the left to the right at a relatively small velocity (see Figure 11.1). The information that the piston is moving passes thorough a single ``pressure pulse.'' It is assumed that if the velocity of the piston is infinitesimally small, the pulse will be infinitesimally small. Thus, the pressure and density can be assumed to be continuous. In the control volume it is convenient to look at a control volume which is attached to a pressure pulse (see Figure 11.2). Applying the mass balance yields \begin{align} \rho\, c = (\rho + d\rho)\,(c-dU) \label{gd:sd:eq:cvMass1} \end{align} or when the higher term $dU\,d\rho$ is neglected yields \begin{align} \rho d U = c \, d\rho \Longrightarrow dU = \dfrac{c d \rho }{ \rho} \label{gd:sd:eq:cvMass2} \end{align} From the energy equation (Bernoulli's equation), assuming isentropic flow and neglecting the gravity results \begin{align} \dfrac{ \left( c - dU\right)^2 - c^{2} }{ 2} + \dfrac{dP }{ \rho} = 0 \label{gd:sd:eq:cvEnergy1} \end{align} neglecting second term ($dU^2$) yield \begin{align} -c dU + \dfrac{dP }{ \rho} = 0 \label{gd:sd:eq:cvCombined} \end{align} Substituting the expression for $dU$ from equation \eqref{gd:sd:eq:cvMass2} into equation \eqref{gd:sd:eq:cvCombined} yields

Sound Speed

\begin{align} \label{gd:sd:eq:cvCombined2} c^{2} \left( { d\rho \over \rho } \right) = {dP \over \rho} \Longrightarrow c^2 = \dfrac{dP }{ d\rho} \end{align}
An expression is needed to represent the right hand side of equation \eqref{gd:sd:eq:cvCombined2}. For an ideal gas, $P$ is a function of two independent variables. Here, it is considered that $P= P(\rho, s)$ where $s$ is the entropy. The full differential of the pressure can be expressed as follows: \begin{align} dP = \left. \dfrac{\partial P}{\partial \rho} \right|_{s} d\rho + \left. {\dfrac{\partial P}{\partial s} } \right|_{\rho} ds \label{gd:sd:eq:insontropic1} \end{align} In the derivations for the speed of sound it was assumed that the flow is isentropic, therefore it can be written \begin{align} \dfrac{\partial P}{\partial \rho} = \left. \dfrac{\partial P}{\partial \rho} \right|_{s} \label{gd:sd:eq:insontropic2} \end{align} Note that the equation \eqref{gd:sd:eq:cvCombined2} can be obtained by utilizing the momentum equation instead of the energy equation.

Example 11.1

Demonstrate that equation \eqref{gd:sd:eq:cvCombined2} can be derived from the momentum equation.

Solution

The momentum equation written for the control volume shown in Figure 11.2 is \begin{align} \overbrace{(P + dP) - P}^{\sum F} = \overbrace{(\rho +d\rho)(c - dU)^{2} - \rho \, c^2}^{\int_{cs} U\,(\rho\, U\, dA)} \label{gd:sd:eq:cvMomentumEx} \end{align} Neglecting all the relative small terms results in \begin{align} dP = (\rho + d\rho) \left( c^{2} - \cancelto{\sim 0}{2\,c\,dU} + \cancelto{\sim 0}{dU^{2}} \right) - \rho c^{2} \label{gd:sd:eq:mem} \end{align} And finally it becomes \begin{align} dP = c ^{2} \, d \rho \label{gd:sd:eq:memSimple} \end{align} This yields the same equation as \eqref{gd:sd:eq:cvCombined2}.

11.3.2 Speed of Sound in Ideal and Perfect Gases

The speed of sound can be obtained easily for the equation of state for an ideal gas (also perfect gas as a sub set) because of a simple mathematical expression. The pressure for an ideal gas can be expressed as a simple function of density, $\rho$, and a function ``molecular structure'' or ratio of specific heats, $k$ namely \begin{align} P= constant\times \rho^{k} \label{gd:sd:eq:iserhoToP} \end{align} and hence \begin{multline} \label{gd:sd:eq:idealGas1} c = \sqrt{\dfrac{\partial P}{\partial \rho}} = k \times constant \times \rho^{k-1} =k \times \dfrac{\overbrace{ constant \times \rho^k}^{P} }{ \rho} = k \times \dfrac{P }{\rho } \end{multline} Remember that $P / \rho $ is defined for an ideal gas as $RT$, and equation \eqref{gd:sd:eq:idealGas1} can be written as

Ideal Gas Speed Sound

\begin{align} \label{gd:sd:eq:sound} c = \sqrt{ k\, R\, T} \end{align}

Example 11.2

Calculate the speed of sound in water vapor at $20 [bar]$ and $350^{\circ}C$, (a) utilizes the steam table, and (b) assuming ideal gas.

Solution

The solution can be estimated by using the data from steam table \begin{align} c \sim \sqrt{ \dfrac{\Delta P}{ \Delta \rho} }_{s=constant} \end{align} At $20[bar]$ and $350^{\circ}C$: s = 6.9563 $\left[ \dfrac{kJ }{ K\, kg}\right]$ $\rho $ = 6.61376 $\left[ \dfrac{kg }{ m^3} \right]$
At $18[bar]$ and $350^{\circ}C$: s = 7.0100 $\left[\dfrac{ kJ }{ K\, kg}\right]$ $\rho $ = 6.46956 $\left[ \dfrac{ kg }{ m^3}\right]$
At $18[bar]$ and $300^{\circ}C $: s = 6.8226 $\left[\dfrac{ kJ }{ K\, kg}\right]$ $\rho $ = 7.13216 $\left[ \dfrac{kg }{ m^3} \right] $
After interpretation of the temperature:
At $18[bar]$ and $335.7^{\circ}C $: s $\sim$ 6.9563 $\left[\dfrac{ kJ }{ K\, kg} \right]$ $\rho \sim$ 6.94199 $\left[ \dfrac{kg }{ m^3} \right]$
and substituting into the equation yields \begin{align} c = \sqrt{ 200000 \over 0.32823} = 780.5 \left[ m \over sec \right] \end{align} for ideal gas assumption (data taken from Van Wylen and Sontag, Classical Thermodynamics, table A 8.) \begin{align*} c = \sqrt{k\,R\,T} \sim \sqrt{ 1.327 \times 461 \times (350 + 273)} \sim 771.5 \left[ \dfrac{m }{ sec} \right] \end{align*} Note that a better approximation can be done with a steam table, and it $\cdots$

11.3.3 Speed of Sound in Almost Incompressible Liquid

Every liquid in reality has a small and important compressible aspect. The ratio of the change in the fractional volume to pressure or compression is referred to as the bulk modulus of the material. For example, the average bulk modulus for water is $2.2 \times 10^9$ $N/m^2$. At a depth of about 4,000 meters, the pressure is about $4 \times 10^7$ $N/m^2$. The fractional volume change is only about 1.8% even under this pressure nevertheless it is a change. The compressibility of the substance is the reciprocal of the bulk modulus. The amount of compression of almost all liquids is seen to be very small as given in the Book ``Fundamentals of Compressible Flow.'' The mathematical definition of bulk modulus as following \begin{align} B_T = \rho \,{\dfrac{\partial P}{\partial \rho}} \label{gd:sd:eq:bulkModulus} \end{align} In physical terms can be written as

Liquid/Solid Sound Speed

\begin{align} \label{gd:sd:eq:sondLiquid} c = \sqrt{\dfrac{elastic\;\; property }{ inertial\;\; property} } = \sqrt{\dfrac{B_T }{ \rho}} \end{align}
For example for water \begin{align} \nonumber c = \sqrt{\dfrac{2.2 \times 10^9 N /m^2 }{ 1000 kg /m^3}} = 1493 m/s \end{align} This value agrees well with the measured speed of sound in water, 1482 m/s at $20^{\circ}C$. A list with various typical velocities for different liquids can be found in ``Fundamentals of Compressible Flow'' by by this author. The interesting topic of sound in variable compressible liquid also discussed in the above book. It can be shown that velocity in solid and and slightly compressible liquid is expressed by In summary, the speed of sound in liquids is about 3 to 5 relative to the speed of sound in gases.

11.3.4 Speed of Sound in Solids

The situation with solids is considerably more complicated, with different speeds in different directions, in different kinds of geometries, and differences between transverse and longitudinal waves. Nevertheless, the speed of sound in solids is larger than in liquids and definitely larger than in gases. Young's Modulus for a representative value for the bulk modulus for steel is 160 $10^9$ N /$m^2$. A list of materials with their typical velocity can be found in the above book. Speed of sound in solid of steel, using a general tabulated value for the bulk modulus, gives a sound speed for structural steel of \begin{align} \nonumber c = \sqrt{ E \over \rho} = \sqrt{160 \times 10^{9} N/m^{2} \over 7860 Kg /m^3} = 4512 m/s \end{align} Compared to one tabulated value the example values for stainless steel lays between the speed for longitudinal and transverse waves.

11.3.5 The Dimensional Effect of the Speed of Sound

What is the significance of the speed of sound? This speed of sound determines what regime the flow will be. In Chapter 9 that Mach number was described as important parameter. It will be shown later in this Chapter that when Mach number is around 0.25-0.3 a significant change occur in the situation of flow. To demonstrate this point, consider a two dimensional situation where a particle is moving from the left to the right. A particle movement creates a pressure change which travels toward outside in equal speed relative to the particle. Figure 11.3 depicts an object with three different relative velocities. Figure 11.3(a) demonstrates that the whole surroundings is influenced by the object (depicted by red color). While Figure 11.3 (b) that there small zone a head object that is ``aware'' if the object arriving. In Figure 11.3 (c) the zone that aware of the object is practically zero.

Object travels at 0.005 of the speed of sound Object travels at 0.05 of the speed of sound
Object travels at 0.15 of the speed of sound

Figure 11.3 Moving object at three relative velocities. The gray point in the first circle is the initial point the object. The finial point is marked by red circled with gray filled. Notice that the circle line thickness is increase with the time i.e the more green wider circle line thickness. The transition from the blue fresher lines to the green older lines is properly marked.

In fact, when the object velocity is about or larger than the speed of sound then the object arrive to location where the fluid does not aware or informed about the object. The reason that in gas the compressibility plays significant role is because the ratio of the object or fluid velocity compared to speed of sound. In gases the speed of sound is smaller as compare to liquid and defendtly to solid. Hence, gases are media where compressebilty effect must be considered in realtionshp compressebilty. approaching to one.

11.4 Isentropic Flow

In this section a discussion on a steady state flow through a smooth and without an abrupt area change which include converging– diverging nozzle is presented. The isentropic flow models are important because of two main reasons: One, it provides the information about the trends and important parameters. Two, the correction factors can be introduced later to account for deviations from the ideal state.

Nozzle

Fig. 11.4 Flow of a compressible substance (gas) through a converging–diverging nozzle.

11.4.1 Stagnation State for Ideal Gas Model

It is assumed that the flow is quasi one–dimensional (that is the fluid flows mainly in one dimension). Figure \eqref{gd:iso:fig:nozzle} describes a gas flow through state is very useful in simplifying the solution and treatment of the flow. The stagnation state is a theoretical state in which the flow is brought into a complete motionless conditions in isentropic process without other forces (e.g. gravity force). Several properties that can be represented by this theoretical process which include temperature, pressure, and density et cetera and denoted by the subscript ``$0$.'' First, the stagnation temperature is calculated. The energy conservation can be written as \begin{align} h + \dfrac{U^2 }{ 2} = h_0 \label{gd:iso:eq:energy0} \end{align} Perfect gas is an ideal gas with a constant heat capacity, $C_p$. For perfect gas equation \eqref{gd:iso:eq:energy0} is simplified into \begin{align} C_{p}\, T + {U^2 \over 2} = C_{p} \, T_{0} \label{gd:iso:eq:temperature2} \end{align} $T_{0}$ is denoted as the stagnation temperature. Recalling from thermodynamic the relationship for perfect gas $R = C_{p} -C_{v}$ and denoting $k quiv C_{p} \div C_{v}$ then the thermodynamics relationship obtains the form \begin{align} C_{p} = \dfrac{k\, R }{ k -1 } \label{gd:iso:eq:CpCvRN} \end{align} and where $R$ is the specific constant. Dividing equation \eqref{gd:iso:eq:temperature2} by $(C_p T)$ yields \begin{align} 1 + { U ^{2} \over 2 \,C_{p}\, T} = { T_{0} \over T} \label{gd:iso:eq:tempSimple} \end{align} Now, substituting $c^2 = {k\,R\,T}$ or $T = c^2 /k\, R$ equation \eqref{gd:iso:eq:tempSimple} changes into \begin{align} 1 + {k\, R\, U ^{2} \over 2 \,C_{p}\, c^{2} } = { T_{0} \over T} \label{gd:iso:eq:T0T} \end{align} By utilizing the definition of $k$ by equation \eqref{thermo:eq:k} and inserting it into equation qref{gd:iso:eq:T0T} yields \begin{align} 1 + {{k -1} \over 2} \; { U ^{2} \over c^{2} } = { T_{0} \over T} \label{gd:iso:eq:tempSimple3} \end{align} It very useful to convert equation \eqref{gd:iso:eq:T0T} into a dimensionless form and denote Mach number as the ratio of velocity to speed of sound as

Mach Number Definition

\begin{align} \label{gd:iso:eq:Mach} M quiv \dfrac{U }{ c} \end{align}
into equation \eqref{gd:iso:eq:tempSimple3} reads

Isentropic Temperature relationship

\begin{align} \label{gd:iso:eq:temperatureDless} \dfrac{T_0 }{ T} = 1 + \dfrac{ k -1 }{ 2 } M^{2} \end{align}
Comparison Pressure Temperature Two Scale

Fig. 11.5 Perfect gas flows through a tube.

The usefulness of Mach number and equation \eqref{gd:iso:eq:temperatureDless} can be demonstrated by the following simple example. In this example a gas flows through a tube (see Figure 11.5) of any shape can be expressed as a function of only the stagnation temperature as opposed to the function of the temperatures and velocities. The definition of the stagnation state provides the advantage of compact writing. For example, writing the energy equation for the tube shown in Figure 11.5 can be reduced to \begin{align} \dot{Q} = C_{p} \, \left({T_{0}}_{B} - {T_{0}}_{A}\right) \,\dot{m} \label{gd:iso:eq:simpleEnergy} \end{align} The ratio of stagnation pressure to the static pressure can be expressed as the function of the temperature ratio because of the isentropic relationship as

Isentropic Pressure Definition

\begin{align} \label{gd:iso:eq:pressureDless} \dfrac{P_0 }{ P } = \left( { T_0 \over T} \right) ^ {k \over k -1} = \left( 1 + { k -1 \over 2 }\, M^{2} \right)^ {k \over k -1} \end{align}
In the same manner the relationship for the density ratio is

Isentropic Density

\begin{align} \label{gd:iso:eq:densityDless} \dfrac{\rho_0 }{ \rho } = \left( { T_0 \over T} \right) ^ {1 \over k -1} = \left( 1 + \dfrac{ k -1 }{ 2 } \, M^{2} \right)^{1 \over k -1} \end{align}
New useful definitions are introduced for the case when $M=1$ and denoted by superscript ``$*$.'' The special cases of ratio of the star values to stagnation values are dependent only on the heat ratio as the following:

Star Relationship

\begin{align} \label{thermo:eq:starRelationship} \begin{array}{rcccl} \dfrac{T^{*} }{ T_0} = &\dfrac{{c^{*}}^2 }{ {c_0}^2} \\ \dfrac {V_2} {V_1} = &\left(\dfrac{T_1}{T_2} \right)^{\dfrac{1}{k-1}} = &\left(\dfrac{\rho_1}{\rho_2} \right) &= &\left(\dfrac{P_1}{P_2} \right )^{\dfrac{1}{k}} \\ \dfrac{P^{*} }{ P_0} = & \left(\dfrac{2}{ k+1} \right)^{\dfrac{k}{ k-1}} \\ \dfrac{\rho^{*} }{ \rho_0} = & \left(\dfrac{2}{ k+1} \right)^{\dfrac {1 }{ k-1}} \end{array} \end{align}
Using all the definitions above relationship between the stagnation properties to star speed of sound are \begin{align} \label{gd:iso:eq:soundStagnation} c^{*} = \sqrt{k\,R\, \dfrac{2\,T_0}{ k +2}} \end{align}
Stagantion Properties

Fig. 11.6 The stagnation properties as a function of the Mach number, {m k=1.4}.

11.4.2 Isentropic Converging–Diverging Flow in Cross Section

Control Volume Converging-Diverging Nozzle

Fig. 11.7 Control volume inside a converging-diverging nozzle.

The important sub case in this chapter is the flow in a converging–diverging nozzle. The control volume is shown in Figure \eqref{gd:iso:fig:cvNozzle}. There are two models that assume variable area flow: First is isentropic and adiabatic model. Second is isentropic and isothermal model. Here only the first model will be described. Clearly, the stagnation temperature, $T_0$, is constant through the adiabatic flow because there isn't heat transfer. Therefore, the stagnation pressure is also constant through the flow because the flow isentropic. Conversely, in mathematical terms, equation \eqref{gd:iso:eq:temperatureDless} and equation qref{gd:iso:eq:pressureDless} are the same. If the right hand side is constant for one variable, it is constant for the other. In the same vein, the stagnation density is constant through the flow. Thus, knowing the Mach number or the temperature will provide all that is needed to find the other properties. The only properties that need to be connected are the cross section area and the Mach number. Examination of the relation between properties can then be carried out.

11.4.3 The Properties in the Adiabatic Nozzle

When there is no external work and heat transfer, the energy equation, reads \begin{align} dh + U\, dU = 0 \label{gd:iso:eq:energy} \end{align} Differentiation of continuity equation, $\rho\, A\, U = \dot{m} = constant$, and dividing by the continuity equation reads \begin{align} {d\rho \over \rho} + { dA \over A} + {dU \over U} = 0 \label{gd:iso:eq:mass} \end{align} The thermodynamic relationship between the properties can be expressed as \begin{align} T\,ds = dh - {dP \over \rho} \label{gd:iso:eq:thermo} \end{align} For isentropic process $ds quiv 0$ and combining equations (11.31) with \eqref{gd:iso:eq:thermo} yields \begin{align} {dP \over \rho} + U\, dU = 0 \label{gd:iso:eq:thermo2} \end{align} Differentiation of the equation state (perfect gas), $P = \rho R T$, and dividing the results by the equation of state ($\rho\, R\, T$) yields \begin{align} {dP \over P} = {d\rho \over \rho} + {dT \over T} \label{gd:iso:eq:stateDless} \end{align} Obtaining an expression for $dU/U$ from the mass balance equation \eqref{gd:iso:eq:mass} and using it in equation \eqref{gd:iso:eq:thermo2} reads \begin{align} \dfrac{dP }{ \rho} - U^{2} \overbrace{\left[ \dfrac{dA }{ A} + \dfrac{d\rho }{ \rho} \right]}^{\dfrac{dU }{ U} } = 0 \label{gd:iso:eq:combine1} \end{align} Rearranging equation \eqref{gd:iso:eq:combine1} so that the density, $\rho$, can be replaced by the static pressure, $dP/\rho$ yields \begin{align} \dfrac{dP }{ \rho} = U^{2}\, \left( {dA \over A} + {d\rho \over \rho}\, {dP \over dP} \right) = U^{2} \, \left( {dA \over A} + \overbrace{d\rho \over dP}^{\dfrac{ 1}{ c^2 }} {dP \over \rho} \right) \label{eq::varibleArea:combine2} \end{align} Recalling that $dP/d\rho = c^2$ and substitute the speed of sound into equation ( ef{eq::varibleArea:combine2}) to obtain \begin{align} {dP \over \rho } \left[ 1 - \left(U \over c\right)^2 \right] = U^2 {dA \over A} \label{eq::varibleArea:combine3} \end{align} Or in a dimensionless form \begin{align} {dP \over \rho } \left( 1 -M^{2} \right) = U^2 {dA \over A} \label{gd:iso:eq:areaChangeVelocity} \end{align} Equation \eqref{gd:iso:eq:areaChangeVelocity} is a differential equation for the pressure as a function of the cross section area. It is convenient to rearrange equation \eqref{gd:iso:eq:areaChangeVelocity} to obtain a variables separation form of \begin{align} dP = {\rho\, U^{2} \over A} \; {dA \over 1 -M^2} \label{gd:iso:eq:areaChangeMach} \end{align}

11.4.3.1 The pressure Mach number relationship

Before going further in the mathematical derivations it is worth looking at the physical meaning of equation \eqref{gd:iso:eq:areaChangeMach}. The term ${\rho\, U^{2} / A}$ is always positive (because all the three terms can be only positive). Now, it can be observed that $dP$ can be positive or negative depending on the $dA$ and Mach number. The meaning of the sign change for the pressure differential is that the pressure can increase or decrease. It can be observed that the critical Mach number is one. If the Mach number is larger than one than $dP$ has opposite sign of $dA$. If Mach number is smaller than one $dP$ and $dA$ have the same sign. For the subsonic branch $M<1$ the term $1 /(1 -M^2)$ is positive hence \begin{align*} dA > 0 \Longrightarrow dP> 0 \\ dA < 0 \Longrightarrow dP< 0 \end{align*} From these observations the trends are similar to those in incompressible fluid. An increase in area results in an increase of the static pressure (converting the dynamic pressure to a static pressure). Conversely, if the area decreases (as a function of $x$) the pressure decreases. Note that the pressure decrease is larger in compressible flow compared to incompressible flow. For the supersonic branch $M> 1$, the phenomenon is different. For $M > 1$ the term $1 / 1 - M^{2}$ is negative and change the character of the equation. \begin{align*} dA > 0 \Rightarrow dP< 0 \\ dA < 0 \Rightarrow dP> 0 \end{align*} This behavior is opposite to incompressible flow behavior. For the special case of $M=1$ (sonic flow) the value of the term $1 - M^{2}= 0$ thus mathematically $dP \rightarrow \infty$ or $dA = 0$. Since physically $dP$ can increase only in a finite amount it must that $dA = 0$. It must also be noted that when $M=1$ occurs only when $dA = 0$. However, the opposite, not necessarily means that when $dA = 0$ that $M=1$. In that case, it is possible that $dM=0$ thus the diverging side is in the subsonic branch and the flow isn't choked. The relationship between the velocity and the pressure can be observed from equation \eqref{gd:iso:eq:thermo2} by solving it for $dU$. \begin{align} dU = - \dfrac{ dP }{ P \,U} \label{gd:iso:eq:relationshipU-rho} \end{align} From equation \eqref{gd:iso:eq:relationshipU-rho} it is obvious that $dU$ has an opposite sign to $dP$ (since the term $PU$ is positive). Hence the pressure increases when the velocity decreases and vice versa. From the speed of sound, one can observe that the density, $\rho$, increases with pressure and vice versa (see equation \eqref{gd:iso:eq:relationshipP-rho}). \begin{align} d\rho = \dfrac{ 1 }{ c^2} dP \label{gd:iso:eq:relationshipP-rho} \end{align} It can be noted that in the derivations of the above equations (qref{gd:iso:eq:relationshipU-rho} - qref{gd:iso:eq:relationshipP-rho}), the equation of state was not used. Thus, the equations are applicable for any gas (perfect or imperfect gas). The second law (isentropic relationship) dictates that $ds = 0$ and from thermodynamics $ ds = 0 = C_p \,\dfrac{dT }{ T} - R \,\dfrac{dP }{ P}$ and for perfect gas \begin{align} \dfrac{dT }{ T} = \dfrac{k - 1 }{ k } \; \dfrac{dP }{ P} \label{gd:iso:eq:isentropicDiff} \end{align} Thus, the temperature varies in the same way that pressure does. The relationship between the Mach number and the temperature can be obtained by utilizing the fact that the process is assumed to be adiabatic $dT_0 = 0$. Differentiation of equation \eqref{gd:iso:eq:temperatureDless}, the relationship between the temperature and the stagnation temperature becomes \begin{align} dT_0 = 0 = dT \left( 1 + \dfrac{ k - 1 }{ 2 } M^2 \right) + T (k -1) M dM \label{gd:iso:eq:temp-M1} \end{align} and simplifying equation \eqref{gd:iso:eq:temp-M1} yields \begin{align} \dfrac{ dT }{ T } = - \dfrac{(k -1) \,M\, dM }{ 1 + \dfrac{ k - 1 }{ 2 }\, M^{2} } \label{gd:iso:eq:temp-M} \end{align}

11.4.3.2 Relationship Between the Mach Number and Cross Section Area

The equations used in the solution are energy \eqref{gd:iso:eq:temp-M}, second law qref{gd:iso:eq:isentropicDiff}, state \eqref{gd:iso:eq:stateDless}, mass \eqref{gd:iso:eq:mass}. Note, equation \eqref{gd:iso:eq:areaChangeVelocity} isn't the solution but demonstration of certain properties of the pressure profile. The relationship between temperature and the cross section area can be obtained by utilizing the relationship between the pressure and temperature \eqref{gd:iso:eq:isentropicDiff} and the relationship of pressure with cross section area \eqref{gd:iso:eq:areaChangeVelocity}. First stage equation \eqref{gd:iso:eq:temp-M} is combined with equation \eqref{gd:iso:eq:isentropicDiff} and becomes \begin{align} {(k - 1) \over k}\, { dP \over P } = - { (k -1) \, M \,dM \over 1 + \dfrac{ k - 1 }{ 2 } M^{2} } \label{gd:iso:eq:press-M} \end{align} Combining equation \eqref{gd:iso:eq:press-M} with equation qref{gd:iso:eq:areaChangeVelocity} yields \begin{align} {1 \over k} {\dfrac {\rho \,U^{2}} { A} \, \dfrac{dA }{ 1 - M^2} \over P } = - \dfrac{ M \, dM }{ 1 + \dfrac{ k - 1 }{ 2 } M^{2} } \label{gd:iso:eq:M-A-0} \end{align} The following identify, $ \rho \, U ^{2} = k\,M\, P$ can be proved as \begin{align} k\, M^2\, P = k \overbrace{ U^{2} \over c^2}^{M^2} \overbrace{\rho R T}^{P} = k { U ^{2} \over k\,R\,T} \overbrace{\rho\, R \, T}^{P} = \rho U ^{2} \label{gd:iso:eq:rU-kMP} \end{align} Using the identity in equation \eqref{gd:iso:eq:rU-kMP} changes equation \eqref{gd:iso:eq:M-A-0} into \begin{align} \dfrac{dA }{ A} = { M^2 -1 \over M \left( 1 + \dfrac{k-1 }{ 2} M^2 \right)} dM \label{gd:iso:eq:M-A} \end{align}

A vs. M Relationship

Fig. 11.8 The relationship between the cross section and the Mach number on the subsonic branch.

Equation \eqref{gd:iso:eq:M-A} is very important because it relates the geometry (area) with the relative velocity (Mach number). In equation \eqref{gd:iso:eq:M-A}, the factors $M \,\left( 1 + \dfrac{k-1 }{ 2} M^2 \right)$ and $A$ are positive regardless of the values of $M$ or $A$. Therefore, the only factor that affects relationship between the cross area and the Mach number is $M^2 -1$. For $M <1$ the Mach number is varied opposite to the cross section area. In the case of $M > 1$ the Mach number increases with the cross section area and vice versa. The special case is when $M=1$ which requires that $dA=0$. This condition imposes that internal flow has to pass a converting–diverging device to obtain supersonic velocity. This minimum area is referred to as ``throat.'' Again, the opposite conclusion that when $dA=0$ implies that $M=1$ is not correct because possibility of $dM=0$. In subsonic flow branch, from the mathematical point of view: on one hand, a decrease of the cross section increases the velocity and the Mach number, on the other hand, an increase of the cross section decreases the velocity and Mach number (see Figure \eqref{gd:iso:fig:MvA}).

11.4.4 Isentropic Flow Examples

Example 11.3

Air is allowed to flow from a reservoir with temperature of $21^{\circ}C$ and with pressure of 5[MPa] through a tube. It was measured that air mass flow rate is 1[kg/sec]. At some point on the tube static pressure was measured to be 3[MPa]. Assume that process is isentropic and neglect the velocity at the reservoir, calculate the Mach number, velocity, and the cross section area at that point where the static pressure was measured. Assume that the ratio of specific heat is $ k=C_p / C_v = 1.4$.

Solution

The stagnation conditions at the reservoir will be maintained throughout the tube because the process is isentropic. Hence the stagnation temperature can be written $T_{0} = constant$ and $P_{0} = constant$ and both of them are known (the condition at the reservoir). For the point where the static pressure is known, the Mach number can be calculated by utilizing the pressure ratio. With the known Mach number, the temperature, and velocity can be calculated. Finally, the cross section can be calculated with all these information. In the point where the static pressure known \begin{align*} \bar{P} = \dfrac{P }{ P_0} = \dfrac{3 [MPa] }{ 5 [MPa] } = 0.6 \end{align*} From Table \eqref{variableArea:tab:basicIsentropic} or from Figure qref{gd:iso:fig:basic} or utilizing the enclosed program, Potto-GDC, or simply using the equations shows that

$\bbb{M}$ $\bbb{\dfrac{T }{ T_0}} $ $\bbb{\dfrac{\rho }{ \rho_0}} $ $\bbb{\dfrac{A }{ A^{\star} }} $ $\bbb{\dfrac{P }{ P_0}} $ $\bbb{\dfrac{A \,P }{ A^{\star} \,P_0 }} $ $\bbb{\dfrac{F }{ F^{\star} }} $
0.88639 0.86420 0.69428 1.0115 0.60000 0.60693 0.53105
With these values the static temperature and the density can be calculated. \begin{align*} T = 0.86420338 \times (273+ 21) = 254.076 K \\ \rho = \dfrac{\rho }{ \rho_0} \overbrace{\dfrac{P_{0}}{ R\, T_0}}^{\rho_0}= 0.69428839 \times {5 \times 10^6 [Pa] \over 287.0 \left[\dfrac{J }{ kg\, K}\right] \times 294 [K] } \\ = 41.1416 \left[\dfrac{kg }{ m^3 }\right] \end{align*} The velocity at that point is \begin{align*} U = M \,\overbrace{\sqrt{k\,R\,T}}^{c} = 0.88638317 \times \sqrt { 1.4 \times 287 \times 254.076} \sim 283 [m /sec] \end{align*} The tube area can be obtained from the mass conservation as \begin{align*} A = \dfrac{\dot {m} }{ \rho\, U} = 8.26 \times 10^{-5} [m^{3}] \end{align*} For a circular tube the diameter is about 1[cm].

Example 11.4

The Mach number at point $\bbb{A}$ on tube is measured to be $M=2$1 and the static pressure is $2 [Bar]$. Downstream at point B the pressure was measured to be 1.5[Bar]. Calculate the Mach number at point B under the isentropic flow assumption. Also, estimate the temperature at point B. Assume that the specific heat ratio $k=1.4$ and assume a perfect gas model.

1

Well, this question is for academic purposes, there is no known way for the author to directly measure the Mach number. The best approximation is by using inserted cone for supersonic flow and measure the oblique shock. Here it is subsonic and this technique is not suitable.

Solution

With the known Mach number at point $\bbb{A}$ all the ratios of the static properties to total (stagnation) properties can be calculated. Therefore, the stagnation pressure at point $\bbb{A}$ is known and stagnation temperature can be calculated. At $M=2$ (supersonic flow) the ratios are

Isentropic Flow Input: M k = 1.4
$M$ $\dfrac{T}{T_0}$ $\dfrac{\rho}{\rho_0}$ $\dfrac{A}{A^{\star} }$ $\dfrac{P}{P_0}$ $\dfrac{A\, P }{ A^{\star} \, P_0} $ $\dfrac{F }{ F^{\star}} $
2.0 0.555556 0.230048 1.6875 0.127805 0.21567 0.593093

With this information the pressure at point B can be expressed as \begin{align*} \dfrac{P_{A} }{ P_{0}} = \overbrace{\dfrac{P_{B} }{ P_{0}} }^ {\text{from the table isentropic @ M = 2}} \times \dfrac{P_{A} }{ P_{B}} = 0.12780453 \times \dfrac{2.0 }{ 1.5} = 0.17040604 \end{align*} The corresponding Mach number for this pressure ratio is 1.8137788 and $T_{B} = 0.60315132$ ${P_{B} \over P_{0} }= 0.17040879$. The stagnation temperature can be ``bypassed'' to calculate the temperature at point $\mathbf{B}$ \begin{align*} T_{B} = T_{A}\times \overbrace{T_{0} \over T_{A} }^{M=2} \times \overbrace{T_{B} \over T_{0}} ^{M=1.81..} = 250 [K] \times {1 \over 0.55555556} \times {0.60315132} \simeq 271.42 [K] \end{align*}

Example 11.5

Gas flows through a converging–diverging duct. At point ``A'' the cross section area is $50$ [$cm^{2}$] and the Mach number was measured to be 0.4. At point B in the duct the cross section area is 40 [$cm^{2}$]. Find the Mach number at point B. Assume that the flow is isentropic and the gas specific heat ratio is 1.4.

Solution

To obtain the Mach number at point B by finding the ratio of the area to the critical area. This relationship can be obtained by \begin{align*} \dfrac{A_{B} }{ A{*} } = \dfrac{A _{B} }{ A_{A} } \times \dfrac{A_{A} }{ A^{\star}} = \dfrac{40 }{ 50} \times \overbrace{ 1.59014 }^{\text{from the 11.2}} = 1.272112 \end{align*} With the value of $\dfrac{A_{B} }{ A{\star} } $ from the Table or from Potto-GDC two solutions can be obtained. The two possible solutions: the first supersonic M = 1.6265306 and second subsonic M = 0.53884934. Both solution are possible and acceptable. The supersonic branch solution is possible only if there where a transition at throat where M=1.

Isentropic Flow Input: $\dfrac{A}{A^{\star}}$ k = 1.4
$M$ $\dfrac{T}{T_0}$ $\dfrac{\rho}{\rho_0}$ $\dfrac{A}{A^{\star} }$ $\dfrac{P}{P_0}$ $\dfrac{A\, P }{ A^{\star} \, P_0} $ $\dfrac{F }{ F^{\star}} $
0.538865 0.945112 0.868378 1.27211 0.820715 1.04404 0.611863
1.62655 0.653965 0.345848 1.27211 0.226172 0.287717 0.563918

Example 11.6

Engineer needs to redesign a syringe for medical applications. They complained that the syringe is ``hard to push.'' The engineer analyzes the flow and conclude that the flow is choke. Upon this fact, what engineer should do with the syringe; increase the pushing diameter or decrease the diameter? Explain.

Solution

This problem is a typical to compressible flow in the sense the solution is opposite the regular intuition. The diameter should be decreased. The pressure in the choke flow in the syringe is past the critical pressure ratio. Hence, the force is a function of the cross area of the syringe. So, to decrease the force one should decrease the area.

11.4.5 Mass Flow Rate (Number)

One of the important engineering parameters is the mass flow rate which for ideal gas is \begin{align} \dot{m} = \rho\, U\, A = \dfrac{P }{ R\,T}\, U\, A \label{gd:iso:eq:massFlowRate} \end{align} This parameter is studied here, to examine the maximum flow rate and to see what is the effect of the compressibility on the flow rate. The area ratio as a function of the Mach number needed to be established, specifically and explicitly the relationship for the chocked flow. The area ratio is defined as the ratio of the cross section at any point to the throat area (the narrow area). It is convenient to rearrange the equation \eqref{gd:iso:eq:massFlowRate} to be expressed in terms of the stagnation properties as \begin{align} \dfrac{\dot{m} }{ A} = \dfrac{ P }{ P_{0} } \; \dfrac{P_{0}\, U }{ \sqrt{k\,R\,T}} \sqrt{\dfrac{k }{ R} } \sqrt{ \dfrac{T_{0} }{ T}} \; {1 \over \sqrt{T_{0}}} = { P_{0} \over \sqrt{T_0}} M \sqrt{k \over R } \overbrace{{ P \over P_{0} } \sqrt{ {T_{0} \over T}} } ^{f(M,\,k)} \label{gd:iso:eq:massFlowRate1} \end{align} Expressing the temperature in terms of Mach number in equation qref{gd:iso:eq:massFlowRate1} results in \begin{align} \label{gd:iso:eq:massFlowRate2} {\dot{m} \over A} = \left(\dfrac{k\, M\, P_0 }{ \sqrt{k\,R\, T_{0}} }\right) \left( 1 + {k -1 \over 2} M^{2} \right)^ {-\dfrac{ k+ 1 }{ 2\,(k -1)} } \end{align} It can be noted that equation \eqref{gd:iso:eq:massFlowRate2} holds everywhere in the converging-diverging duct and this statement also true for the throat. The throat area can be denoted as by $A^{*}$. It can be noticed that at the throat when the flow is chocked or in other words $M=1$ and that the stagnation conditions (i.e. temperature, pressure) do not change. Hence equation ( ef{gd:iso:eq:massFlowRate2}) obtained the form \begin{align} {\dot{m} \over A^{*}} = \left( {\sqrt{k} P_0 \over \sqrt{R T_{0}}} \right) \left( 1 + {k -1 \over 2} \right)^ {-\dfrac{ k+ 1 }{ 2\,(k -1)} } \label{gd:iso:eq:massFlowRateStar} \end{align} Since the mass flow rate is constant in the duct, dividing equations \eqref{gd:iso:eq:massFlowRateStar} by equation \eqref{gd:iso:eq:massFlowRate2} yields

Mass Flow Rate Ratio

\begin{align} \label{gd:iso:eq:massFlowRateRatio} {A \over A^{*}} = \dfrac{ 1 }{ M} \left( { 1 + {k -1 \over 2} M^{2} \over {k +1\over 2}} \right) ^ {-\dfrac{ k+ 1 }{ 2\,(k -1)} } \end{align}
Equation \eqref{gd:iso:eq:massFlowRateRatio} relates the Mach number at any point to the cross section area ratio. The maximum flow rate can be expressed either by taking the derivative of equation qref{gd:iso:eq:massFlowRateStar} in with respect to M and equating to zero. Carrying this calculation results at $M=1$. \begin{align} \left(\dot{m} \over A^{*}\right)_{max} { P_0 \over \sqrt{T_0}} = \sqrt{k \over R} \left( {k +1 \over 2} \right)^ {-{ k+ 1 \over 2(k -1)}} \label{gd:iso:eq:fliegner} \end{align} For specific heat ratio, $k=1.4$ \begin{align} \left(\dot{m} \over A^{*}\right)_{max} {P_0 \over \sqrt{T_0}} \sim {0.68473 \over \sqrt{R}} \label{gd:iso:eq:fliegnerk} \end{align} The maximum flow rate for air ($R=287 \dfrac{j }{ kg\, K}$) becomes, \begin{align} \dfrac{ \dot{m} \sqrt{T_0} }{ A^{\star} P_0 } = 0.040418 \label{gd:iso:eq:fliegnerAir} \end{align} Equation \eqref{gd:iso:eq:fliegnerAir} is known as Fliegner's Formula on the name of one of the first engineers who observed experimentally the choking phenomenon. It can be noticed that Fliegner's equation can lead to definition of the Fliegner's Number. \begin{align} \dfrac{ \dot{m} \sqrt{T_0}}{ A^{*} P_0 } = {\dot{m} \overbrace{\sqrt{k\,R\, T_0}}^{c_0} \over\sqrt{k\,R}A^{*} P_0} = \dfrac{1}{\sqrt{R}}\,\overbrace{ \dfrac{\dot{m} \, c_0 }{ A^{*} P_0 }}^{Fn} \dfrac{1}{\sqrt{k}} \label{gd:iso:eq:fnA} \end{align} The definition of Fliegner's number (Fn) is \begin{align} \pmb{Fn} quiv \dfrac{\sqrt{R}\,\dot{m} \,c_0 }{ \sqrt{R} \, A^{*} \, P_0 } \label{gd:iso:eq:fnDef} \end{align} it into equation \eqref{gd:iso:eq:massFlowRateStar} results in

Fliegner's Number

\begin{align} \label{gd:iso:eq:fmG} \pmb{Fn} = k\,M\,\left( 1+ {k -1 \over 2}M^2 \right)^ {-{ k+ 1 \over 2(k -1)}} \end{align}
and the maximum point for $Fn$ at $M=1$ is \begin{align} \pmb{Fn} = k\left( {k +1 \over 2} \right)^ {-{ k+ 1 \over 2(k -1)}} \label{gd:iso:eq:fmMax} \end{align}

Example 11.7

Why $\pmb{Fn}$ is zero at Mach equal to zero? Prove Fliegner number, $\pmb{Fn}$ is maximum at $M=1$.

11.4.7 The Impulse Function

Significance of Impulse Function

Fig. 11.9 Schematic to explain the significances of the Impulse function.

One of the functions that is used in calculating the forces is the Impulse function. The Impulse function is denoted here as $F$, but in the literature some denote this function as $I$. To explain the motivation for using this definition consider the calculation of the net forces that acting on section shown in Figure \eqref{gd:iso:fig:impulse}. To calculate the net forces acting in the $x$–direction the momentum equation has to be applied \begin{align} F_{net}= \dot{m} (U_2 -U_1) + P_2 A_2 - P_1 A_1 \label{gd:iso:eq:monImp} \end{align} The net force is denoted here as $F_{net}$. The mass conservation also can be applied to our control volume \begin{align} \dot{m} = \rho_1A_1U_1 = \rho_2A_2U_2 \label{gd:iso:eq:massImp} \end{align} Combining equation \eqref{gd:iso:eq:monImp} with equation qref{gd:iso:eq:massImp} and by utilizing the identity in equation \eqref{gd:iso:eq:rU-kMP} results in \begin{align} F_{net}= kP_2A_2{M_2}^2 - kP_1A_1{M_1}^2 + P_2 A_2 - P_1 A_1 \label{gd:iso:eq:MomMassImp} \end{align} Rearranging equation \eqref{gd:iso:eq:MomMassImp} and dividing it by $P_0 A^{*}$ results in \begin{align} {F_{net} \over P_0 A^{*}} = \overbrace{P_2A_2 \over P_0 A^{*}}^{f(M_2)} \overbrace{\left( 1 + k{M_2}^2 \right)}^{f(M_2)} - \overbrace{P_1A_1 \over P_0 A^{*}}^{f(M_1)} \overbrace{\left( 1 + k{M_1}^2 \right)}^{f(M_1)} \label{gd:iso:eq:beforeDefa} \end{align} Examining equation \eqref{gd:iso:eq:beforeDefa} shows that the right hand side is only a function of Mach number and specific heat ratio, $k$. Hence, if the right hand side is only a function of the Mach number and $k$ than the left hand side must be function of only the same parameters, $M$ and $k$. Defining a function that depends only on the Mach number creates the convenience for calculating the net forces acting on any device. Thus, defining the Impulse function as \begin{align} F = PA\left( 1 + k{M_2}^2 \right) \label{gd:iso:eq:impulsDef} \end{align} In the Impulse function when $F$ ($M=1$) is denoted as $F^{*}$ \begin{align} F^{*} = P^{*}A^{*}\left( 1 + k \right) \label{gd:iso:eq:impulsDefStar} \end{align} The ratio of the Impulse function is defined as \begin{align} {F \over F^{*}} = {P_1A_1 \over P^{*}A^{*}} {\left( 1 + k{M_1}^2 \right) \over \left( 1 + k \right) } = {1 \over \underbrace{P^{*}\over P_{0} }_ {\left(2 \over k+1 \right)^{k \over k-1}}} \overbrace{{P_1A_1 \over P_0A^{*} } {\left( 1 + k{M_1}^2 \right) }}^{\hbox{see function qref{gd:iso:eq:beforeDefa}}} {1 \over \left( 1 + k \right) } \label{gd:iso:eq:ImpulseRatio} \end{align} This ratio is different only in a coefficient from the ratio defined in equation \eqref{gd:iso:eq:beforeDefa} which makes the ratio a function of $k$ and the Mach number. Hence, the net force is \begin{align} F_{net} = P_0 A^{*} (1+k) {\left( k+1 \over 2 \right)^{k \over k-1}} \left( {F_2 \over F^{*} } - { F_1 \over F^{*}}\right) \label{gd:iso:eq:NetForce} \end{align} To demonstrate the usefulness of the this function consider a simple situation of the flow through a converging nozzle.

Example 11.11

Example of Nozzle

Fig. 11.10 Schematic of a flow of a compressible substance (gas) through a converging nozzle for example qref{gd:iso:ex:impulse}

Consider a flow of gas into a converging nozzle with a mass flow rate of $1[kg/sec]$ and the entrance area is $0.009[m^2]$ and the exit area is $0.003[m^2]$. The stagnation temperature is $400K$ and the pressure at point 2 was measured as $5[Bar]$. Calculate the net force acting on the nozzle and pressure at point 1.

Solution

The solution is obtained by getting the data for the Mach number. To obtained the Mach number, the ratio of $P_1A_1/A^{*}P_0$ is needed to be calculated. The denominator is needed to be determined to obtain this ratio. Utilizing Fliegner's equation \eqref{gd:iso:eq:fliegnerAir}, provides the following \begin{align*} A^{*} P_0 = \dfrac{\dot{m} \sqrt{R\,T} }{ 0.058} = \dfrac{1.0 \times \sqrt{400 \times 287} }{ 0.058} \sim 70061.76 [N] \end{align*} and \begin{align*} \dfrac{A_2\, P_2 }{ A^{\star}\, P_0} = \dfrac{ 500000 \times 0.003 }{ 70061.76 } \sim 2.1 \end{align*}

Isentropic Flow Input: $\dfrac{A\, P }{ A^{\star} \, P_0} $ k = 1.4
$M$ $\dfrac{T}{T_0}$ $\dfrac{\rho}{\rho_0}$ $\dfrac{A}{A^{\star} }$ $\dfrac{P}{P_0}$ $\dfrac{A\, P }{ A^{\star} \, P_0} $ $\dfrac{F }{ F^{\star}} $
0.27353 0.98526 0.96355 2.2121 0.94934 2.1000 0.96666

With the area ratio of ${A \over A^{\star}}= 2.2121$ the area ratio of at point 1 can be calculated. \begin{align*} \dfrac{ A_1 }{ A^{\star}} = \dfrac{A_2 }{ A^{\star}} \dfrac{A_1 }{ A_2} = 2.2121 \times \dfrac{0.009 }{ 0.003} = 5.2227 \end{align*} And utilizing again Potto-GDC provides

Isentropic Flow Input: $\dfrac{A}{A^{\star} }$ k = 1.4
$M$ $\dfrac{T}{T_0}$ $\dfrac{\rho}{\rho_0}$ $\dfrac{A}{A^{\star} }$ $\dfrac{P}{P_0}$ $\dfrac{A\, P }{ A^{\star} \, P_0} $ $\dfrac{F }{ F^{\star}} $
0.11164 0.99751 0.99380 5.2227 0.99132 5.1774 2.1949

The pressure at point $\bbb{1}$ is \begin{align*} P_1 = P_2 {P_0 \over P_2} { P_1 \over P_0} = 5.0 times 0.94934 / 0.99380 \sim 4.776[Bar] \end{align*} The net force is obtained by utilizing equation qref{gd:iso:eq:NetForce} \begin{align*} F_{net} &= P_2 A_2 {P_0 A^{*} \over P_2 A_2} (1+k) {\left( k+1 \over 2 \right)^{k \over k-1}} \left( {F_2 \over F^{*} } - { F_1 \over F^{*}}\right) \ & = 500000 \times {1 \over 2.1}\times 2.4 \times 1.2^{3.5} \times \left( 2.1949 - 0.96666 \right) \sim 614[kN] \end{align*}

GasDynamics2 GasDynamics1 IdealFlow3 Index TOC
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