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Chapter 11 Gd (continue)

11.5 Normal Shock

Shock in a Tube

Fig. 11.11 A shock wave inside a tube, but it can also be viewed as a one–dimensional shock wave.

In this section the relationships between the two sides of normal shock are presented. In this discussion, the flow is assumed to be in a steady state, and the thickness of the shock is assumed to be very small. A shock can occur in at least two different mechanisms. The first is when a large difference (above a small minimum value) between the two sides of a membrane, and when the membrane bursts (see the discussion about the shock tube). Of course, the shock travels from the high pressure to the low pressure side. The second is when many sound waves ``run into'' each other and accumulate (some refer to it as ``coalescing'') into a large difference, which is the shock wave. In fact, the sound wave can be viewed as an extremely weak shock. In the speed of sound analysis, it was assumed the medium is continuous, without any abrupt changes. This assumption is no longer valid in the case of a shock. Here, the relationship for a perfect gas is constructed. In Figure 11.11 a control volume for this analysis is shown, and the gas flows from left to right. The conditions, to the left and to the right of the shock, are assumed to be uniform. The conditions to the right of the shock wave are uniform, but different from the left side. The transition in the shock is abrupt and in a very narrow width. Therefore, the increase of the entropy is fundamental to the phenomenon and the understanding of it. It is further assumed that there is no friction or heat loss at the shock (because the heat transfer is negligible due to the fact that it occurs on a relatively small surface). It is customary in this field to denote $x$ as the upstream condition and $y$ as the downstream condition. The mass flow rate is constant from the two sides of the shock and therefore the mass balance is reduced to \begin{align} \rho_{x} \,U_{x} = \rho _{y}\, U_{y} \label{shock:eq:mass} \end{align} In a shock wave, the momentum is the quantity that remains constant because there are no external forces. Thus, it can be written that \begin{align} P_{x} - P_{y} = \left(\rho_{x} {U_{y}}^2- \rho_{y}\, {U_{x}}^2 \right) \label{shock:eq:momentum} \end{align} The process is adiabatic, or nearly adiabatic, and therefore the energy equation can be written as \begin{align} C_{p} \,T_{x} + \dfrac{{U_{x}}^{2} }{ 2} = C_{p}\, T_{y} + \dfrac{{U_{y}}^{2} }{ 2} \label{shock:eq:energy} \end{align} The equation of state for perfect gas reads \begin{align} P = \rho\, R\, T \label{eq:shock:state} \end{align} If the conditions upstream are known, then there are four unknown conditions downstream. A system of four unknowns and four equations is solvable. Nevertheless, one can note that there are two solutions because of the quadratic of equation qref{shock:eq:energy}. These two possible solutions refer to the direction of the flow. Physics dictates that there is only one possible solution. One cannot deduce the direction of the flow from the pressure on both sides of the shock wave. The only tool that brings us to the direction of the flow is the second law of thermodynamics. This law dictates the direction of the flow, and as it will be shown, the gas flows from a supersonic flow to a subsonic flow. Mathematically, the second law is expressed by the entropy. For the adiabatic process, the entropy must increase. In mathematical terms, it can be written as follows: \begin{align} s_{y} - s_{x} > 0 \label{eq:shock:entropy} \end{align} Note that the greater–equal signs were not used. The reason is that the process is irreversible, and therefore no equality can exist. Mathematically, the parameters are $P, T, U,$ and $\rho$, which are needed to be solved. For ideal gas, equation qref{eq:shock:entropy} is \begin{align} \ln \left(\dfrac{T_y }{ T_x} \right) - \left(k - 1\right)\, \dfrac{P_y }{ P_x} > 0 \label{shock:eq:entropyIdeal} \end{align} It can also be noticed that entropy, $s$, can be expressed as a function of the other parameters. These equations can be viewed as two different subsets of equations. The first set is the energy, continuity, and state equations, and the second set is the momentum, continuity, and state equations. The solution of every set of these equations produces one additional degree of freedom, which will produce a range of possible solutions. Thus, one can have a whole range of solutions. In the first case, the energy equation is used, producing various resistance to the flow. This case is called Fanno flow, and Section 11.8 deals extensively with this topic. Instead of solving all the equations that were presented, one can solve only four (4) equations (including the second law), which will require additional parameters. If the energy, continuity, and state equations are solved for the arbitrary value of the $T_{y}$, a parabola in the $T-s$ diagram will be obtained. On the other hand, when the momentum equation is solved instead of the energy equation, the degree of freedom is now energy, i.e., the energy amount ``added'' to the shock. This situation is similar to a frictionless flow with the addition of heat, and this flow is known as Rayleigh flow. This flow is dealt with in greater detail in Section qref{gd:sec:rayleigh}. Since the shock has no heat transfer (a special case of Rayleigh flow) and there isn't essentially any momentum transfer (a special case of Fanno flow), the intersection of these two curves is what really happened in the shock. The entropy increases from point $x$ to point $y$.

11.5.1 Solution of the Governing Equations

Equations qref{shock:eq:mass}, qref{shock:eq:momentum}, and qref{shock:eq:energy} can be converted into a dimensionless form. The reason that dimensionless forms are heavily used in this book is because by doing so it simplifies and clarifies the solution. It can also be noted that in many cases the dimensionless equations set is more easily solved. From the continuity equation qref{shock:eq:mass} substituting for density, $\rho$, the equation of state yields \begin{align} \dfrac{P_x }{ R\, T_x } \,U_x = \dfrac{P_y }{ R\, T_y } \, U_y \label{shock:eq:continutyNonD} \end{align} Squaring equation qref{shock:eq:continutyNonD} results in \begin{align} \dfrac{ {P_x }^{2} }{ R^{2} \,{T_x}^2}\, {U_x}^{2} = \dfrac{ {P_y }^{2} }{ R^{2} \,{T_y}^2}\, {U_y}^{2} \label{shock:eq:massNonD0} \end{align} Multiplying the two sides by the ratio of the specific heat, k, provides a way to obtain the speed of sound definition/equation for perfect gas, $c^2 = k\,R\,T$ to be used for the Mach number definition, as follows: \begin{align} \dfrac{ {P_x }^{2} }{ T_x \underbrace{k\, R\, {T_x}}_{{c_x}^2} } {U_x}^{2} = \dfrac{ {P_y }^{2} }{ T_y \underbrace{k\, R\, {T_y}}_{{c_y}^2} } {U_y}^{2} \label{shock:eq:massNonD1} \end{align} Note that the speed of sound is different on the sides of the shock. Utilizing the definition of Mach number results in \begin{align} \dfrac{ {P_x}^{2} }{ T_x } {M_x}^{2} = \dfrac{ {P_y}^{2} }{ T_y } {M_y}^{2} \label{shock:eq:massNonD2} \end{align} Rearranging equation qref{shock:eq:massNonD2} results in \begin{align} \dfrac{T_y }{ T_x} = \left( \dfrac{ P_{y} }{ P_{x}} \right)^{2} \left( \dfrac{M_y }{ M_x} \right)^{2} \label{shock:eq:nonDimMass} \end{align} Energy equation qref{shock:eq:energy} can be converted to a dimensionless form which can be expressed as \begin{align} T_y \left( 1 + \dfrac{k-1 }{ 2}\, {M_y}^{2} \right) = T_x \left( 1 + \dfrac{k-1 }{ 2}\, {M_x}^{2} \right) \label{shock:eq:energyDless} \end{align} It can also be observed that equation qref{shock:eq:energyDless} means that the stagnation temperature is the same, ${T_0}_y = {T_0}_x$. Under the perfect gas model, $\rho\, U^{2}$ is identical to $k\, P\, M^{2}$ because \begin{align} \rho U^{2} = \overbrace{P \over R\,T}^{\rho} \overbrace{\left( {U^2 \over \underbrace{k\,R\,T}_{c^2}}\right)} ^{M^2} k\,R\,T = k\, P\, M {2} \label{shock:eq:Rindenty} \end{align} Using the identity qref{shock:eq:Rindenty} transforms the momentum equation qref{shock:eq:momentum} into \begin{align} P_x + k\, P_x\, {M_x}^{2} = P_y + k\, P_y\, {M_y}^{2} \label{shock:eq:Punarranged} \end{align} Rearranging equation qref{shock:eq:Punarranged} yields \begin{align} \dfrac{P_y }{ P_x} = \dfrac{1 + k\,{M_{x}}^2 }{ 1 + k\,{M_{y}}^2} \label{gd:shock:eq:pressureRatio} \end{align} The pressure ratio in equation qref{gd:shock:eq:pressureRatio} can be interpreted as the loss of the static pressure. The loss of the total pressure ratio can be expressed by utilizing the relationship between the pressure and total pressure (see equation qref{gd:iso:eq:pressureDless}) as \begin{align} \dfrac{{P_0}_y }{ {P_0}_x} = \dfrac{P_y\, \left( 1 + \dfrac{k-1 }{ 2}\, {M_y}^{2} \right)^{\dfrac{k}{k-1}} } {P_x\, \left( 1 + \dfrac{k-1 }{ 2}\, {M_x}^{2} \right)^{\dfrac{k}{k-1}} } \label{shock:eq:totalPressureRatio} \end{align} The relationship between $M_x$ and $M_y$ is needed to be solved from the above set of equations. This relationship can be obtained from the combination of mass, momentum, and energy equations. From equation qref{shock:eq:energyDless} (energy) and equation qref{shock:eq:nonDimMass} (mass) the temperature ratio can be eliminated. \begin{align} \left( \dfrac{P_y M_y }{ P_x M_x }\right)^{2} = { 1 + \dfrac{ k-1 }{ 2} {M_x}^{2} \over 1 + \dfrac{ k-1 }{ 2} {M_y}^{2}} \label{shock:eq:combineDimMassEnergy} \end{align} Combining the results of qref{shock:eq:combineDimMassEnergy} with equation qref{gd:shock:eq:pressureRatio} results in \begin{align} \left( \dfrac{1 + k\,{M_{x}}^2 }{ 1 + k\,{M_{y}}^2} \right)^{2} = \left( \dfrac{ M_x }{ M_y }\right)^{2} { 1 + \dfrac{ k-1 }{ 2} {M_x}^{2} \over 1 + \dfrac{ k-1 }{ 2} {M_y}^{2}} \label{shock:eq:toBeSolved} \end{align} Equation qref{shock:eq:toBeSolved} is a symmetrical equation in the sense that if $M_y$ is substituted with $M_x$ and $M_x$ substituted with $M_y$ the equation remains the same. Thus, one solution is \begin{align} M_y = M_x \label{shock:eq:Msolution1} \end{align} It can be observed that equation qref{shock:eq:toBeSolved} is biquadratic. According to the Gauss Biquadratic Reciprocity Theorem this kind of equation has a real solution in a certain range which will be discussed later. The solution can be obtained by rewriting equation qref{shock:eq:toBeSolved} as a polynomial (fourth order). It is also possible to cross–multiply equation qref{shock:eq:toBeSolved} and divide it by $\left({M_x}^2- {M_y}^2\right)$ results in \begin{align} 1 + \dfrac{k -1 }{ 2} \left({M_{y}}^2+ {M_{y}}^2 \right) - k \,{M_{y}}^2\, {M_{y}}^2 = 0 \label{shock:eq:generalSolution} \end{align} Equation qref{shock:eq:generalSolution} becomes

Shock Solution

\begin{align} \label{shock:eq:solution2} {M_y}^2 = \dfrac{ {M_x}^2 + \dfrac{2 }{ k -1} } {\dfrac{2\,k }{ k -1}\, {M_x}^2 - 1 } \end{align}
The first solution qref{shock:eq:Msolution1} is the trivial solution in which the two sides are identical and no shock wave occurs. Clearly, in this case, the pressure and the temperature from both sides of the nonexistent shock are the same, i.e. $T_x=T_y,\; P_x=P_y$. The second solution is where the shock wave occurs. The pressure ratio between the two sides can now be as a function of only a single Mach number, for example, $M_x$. Utilizing equation qref{gd:shock:eq:pressureRatio} and equation qref{shock:eq:solution2} provides the pressure ratio as only a function of the upstream Mach number as \begin{align*} {P_y \over P_x} = \dfrac{2\,k }{ k+1 } {M_x}^2 - \dfrac{k -1 }{ k+1} \qquad \text{or} \end{align*}

Shock Pressure Ratio

\begin{align} \label{shock:eq:pressureMx} \dfrac{P_y }{ P_x} = 1 + \dfrac{ 2\,k }{ k+1} \left({M_x}^2 -1 \right ) \end{align}
The density and upstream Mach number relationship can be obtained in the same fashion to became

Shock Density Ratio

\begin{align} \label{shock:eq:densityMx} \dfrac{\rho_y }{ \rho_x} = \dfrac{U_x }{ U_y} = \dfrac{( k +1) {M_x}^{2} }{ 2 + (k -1) {M_x}^{2} } \end{align}
The fact that the pressure ratio is a function of the upstream Mach number, $M_x$, provides additional way of obtaining an additional useful relationship. And the temperature ratio, as a function of pressure ratio, is transformed into

Shock Temperature Ratio

\begin{align} \label{shock:eq:temperaturePbar} \dfrac{T_y }{ T_x} = \left( \dfrac{P_y }{ P_x} \right) \left( \dfrac{\dfrac{k + 1 }{ k -1 } + \dfrac{P_y }{ P_x}} { 1+ \dfrac{k + 1 }{ k -1 } \dfrac{P_y }{ P_x}} \right) \end{align}
In the same way, the relationship between the density ratio and pressure ratio is

Shock $P-\rho$

\begin{align} \label{shock:eq:densityPbar} \dfrac{\rho_x }{ \rho_y} = \dfrac{ 1 + \left( \dfrac{k +1 }{ k -1} \right) \left( \dfrac{P_y }{ P_x} \right) } { \left( \dfrac{k+1}{ k-1}\right) +\left( \dfrac{P_y }{ P_x} \right)} \end{align}
which is associated with the shock wave.
M_out and P as Function of Min

Fig. 11.12 The exit Mach number and the stagnation pressure ratio as a function of upstream Mach number.

11.5.1.1 The Star Conditions

The speed of sound at the critical condition can also be a good reference velocity. The speed of sound at that velocity is \begin{align} c^{*} = \sqrt{k\,R\,T^{*}} \label{shock:eq:starSpeedSound} \end{align} In the same manner, an additional Mach number can be defined as \begin{align} M^{*} = \dfrac{U }{ c^{*}} \label{shock:eq:starMach} \end{align}

11.5.2 Prandtl's Condition

It can be easily observed that the temperature from both sides of the shock wave is discontinuous. Therefore, the speed of sound is different in these adjoining mediums. It is therefore convenient to define the star Mach number that will be independent of the specific Mach number (independent of the temperature). \begin{align} M^{*} = \dfrac{ U }{ c^{*} } = \dfrac{c }{ c^{*} } \dfrac{U }{ c} = \dfrac{c }{ c^{*} }\, M \label{shock:eq:starMtoM} \end{align} The jump condition across the shock must satisfy the constant energy. \begin{align} \dfrac{c^2 }{ k-1} + \dfrac{U^2 }{ 2 } = \dfrac{{c^{*}}^2 }{ k-1} + \dfrac{{c^{*}}^2 }{ 2 } = \dfrac{ k+ 1 }{ 2\,( k -1) } {c^{*}}^2 \label{shock:eq:momentumC} \end{align} Dividing the mass equation by the momentum equation and combining it with the perfect gas model yields \begin{align} {{c_1}^2 \over k\, U_1 } + U_1 = {{c_2}^2 \over k\, U_2 } + U_2 \label{shock:eq:massMomOFS} \end{align} Combining equation qref{shock:eq:momentumC} and qref{shock:eq:massMomOFS} results in \begin{align} \dfrac{1 }{ k\,U_1} \left[ \dfrac{k+1 }{ 2 }\, {c^{*}}^2 - {k-1 \over 2 } U_1 \right] + U_1 = \dfrac{1 }{ k\,U_2} \left[ {k+1 \over 2 } {c^{*}}^2 - \dfrac{k-1 }{ 2 }\, U_2 \right] + U_2 \label{shock:eq:combAllR} \end{align} After rearranging and dividing equation qref{shock:eq:combAllR} the following can be obtained: \begin{align} U_1\,U_2 = {c^{*}}^2 \label{shock:eq:unPr} \end{align} or in a dimensionless form \begin{align} {M^{*}}_1\, {M^{*}}_2 = {c^{*}}^2 \label{shock:eq:PrDless} \end{align}

11.5.3 Operating Equations and Analysis

In Figure 11.12, the Mach number after the shock, $M_y$, and the ratio of the total pressure, $P_{0y}/P_{0x}$, are plotted as a function of the entrance Mach number. The working equations were presented earlier. Note that the $M_y$ has a minimum value which depends on the specific heat ratio. It can be noticed that the density ratio (velocity ratio) also has a finite value regardless of the upstream Mach number. The typical situations in which these equations can be used also include the moving shocks. The equations should be used with the Mach number (upstream or downstream) for a given pressure ratio or density ratio (velocity ratio). This kind of equations requires examining Table qref{shock:tab:basic} for $k=1.4$ or utilizing Potto-GDC for for value of the specific heat ratio. Finding the Mach number for a pressure ratio of $8.30879$ and $k=1.32$ and is only a few mouse clicks away from the following table.

Mout and P as Function of Min

Fig. 11.13 The ratios of the static properties of the two sides of the shock.

To illustrate the use of the above equations, an example is provided.

Example 11.12

Air flows with a Mach number of $M_x=3$, at a pressure of 0.5 [bar] and a temperature of $0^\circ C$ goes through a normal shock. Calculate the temperature, pressure, total pressure, and velocity downstream of the shock. Assume that $k=1.4$.

Solution

Analysis: First, the known information are $M_{x}=3$, $P_{x}=1.5[bar]$ and $T_{x}=273 K$. Using these data, the total pressure can be obtained (through an isentropic relationship in Table qref{variableArea:tab:basicIsentropic}, i.e., $P_{0x}$ is known). Also with the temperature, $T_x$, the velocity can readily be calculated. The relationship that was calculated will be utilized to obtain the ratios for the downstream of the normal shock. $\dfrac{P_x }{ P_{0x} } = 0.0272237 \Longrightarrow P_{0x} = 1.5/0.0272237 = 55.1 [bar]$ \begin{align*} c_x = \sqrt{k\, R\, T_x} = \sqrt {1.4 \times 287 \times 273} = 331.2 m/sec \end{align*}

Shock Wave Input: $M_x$ k = 1.4
$M_x$ $M_y$ $\dfrac{T_y}{T_x}$ $\dfrac{\rho_y}{\rho_x}$ $\dfrac{P_y}{P_x}$ $\dfrac{{P_0}_y }{ {P_0}_x} $
3.000 0.47519 2.6790 3.8571 10.3333 0.32834

\begin{align*} U_x = M_x \times c_x = 3\times 331.2 = 993.6 [m/sec] \end{align*} Now the velocity downstream is determined by the inverse ratio of \begin{align*} \rho_y/\rho_x = U_x/U_y = 3.85714 \text{.} \end{align*} \begin{align*} U_y = 993.6 / 3.85714 = 257.6 [m/sec] \end{align*} \begin{align*} P_{0y} = {\left( P_{0y} \over P_{0x} \right)} \times P_{0x} = 0.32834 \times 55.1 [bar] = 18.09 [bar] \end{align*}

When the upstream Mach number becomes very large, the downstream Mach number (see equation qref{shock:eq:solution2}) is limited by \begin{align} {M_y}^2 = \dfrac{ 1 + \cancelto{\sim 0}{\dfrac{2 }{ (k -1) {M_x}^{2} } } } { \dfrac{2\,k }{ k -1} - \cancelto{\sim 0}{\dfrac{1 }{ {M_x}^{2}} } } \qquad = \quad \dfrac{k -1 }{ 2\,k} \label{shock:eq:MyLimit} \end{align} The limits of the pressure ratio can be obtained by looking at equation qref{gd:shock:eq:pressureRatio} and by utilizing the limit that was obtained in equation qref{shock:eq:MyLimit}.

The Moving Shocks

In some situations, the shock wave is not stationary. This kind of situation arises in many industrial applications. For example, when a valve is suddenly closed and a shock propagates upstream. On the other extreme, when a valve is suddenly opened or a membrane is ruptured, a shock occurs and propagates downstream (the opposite direction of the previous case). In addition to (partially) closing or (partially) opening of value, the rigid body (not so rigid body) movement creates shocks. In some industrial applications, a liquid (metal) is pushed in two rapid stages to a cavity through a pipe system. This liquid (metal) is pushing gas (mostly) air, which creates two shock stages. The moving shock is observed by daily as hearing sound wave are moving shocks. As a general rule, the moving shock can move downstream or upstream. The source of the shock creation, either due to the static device operation like valve operating/closing or due to moving object, is relevant to analysis but it effects the boundary conditions. This creation difference while creates the same moving shock it creates different questions and hence in some situations complicate the calculations. The most general case which this section will be dealing with is the partially open or close wave. A brief discussion on the such case (partially close/open but due the moving object) will be presented. There are more general cases where the moving shocks are created which include a change in the physical properties, but this book will not deal with them at this stage. The reluctance to deal with the most general case is due to fact it is highly specialized and complicated even beyond early graduate students level. In these changes (of opening a valve and closing a valve on the other side) create situations in which different shocks are moving in the tube. The general case is where two shocks collide into one shock and moves upstream or downstream is the general case. A specific example is common in die–casting: after the first shock moves a second shock is created in which its velocity is dictated by the upstream and downstream velocities.

Stationary Coordinates. Moving Coordinates

Figure 11.14 Comparison between stationary and moving coordinates for the moving shock.

In cases where the shock velocity can be approximated as a constant (in the majority of cases) or as near constant, the previous analysis, equations, and the tools developed in this chapter can be employed. The problem can be reduced to the previously studied shock, i.e., to the stationary case when the coordinates are attached to the shock front. In such a case, the steady state is obtained in the moving control value. For this analysis, the coordinates move with the shock. Here, the prime $\underline{\bbb{'}}$ denotes the values of the static coordinates. Note that this notation is contrary to the conventional notation found in the literature. The reason for the deviation is that this choice reduces the programing work (especially for object–oriented programing like C++). An observer moving with the shock will notice that the pressure in the shock sides is \begin{align} {P_x}^{'} = P_x \qquad {P_y}^{'} = P_y \label{shock:eq:conversionP} \end{align} The temperatures measured by the observer are \begin{align} {T_x}^{'} = T_x \qquad {T_y}^{'} = T_y \label{shock:eq:conversionT} \end{align} Assuming that the shock is moving to the right, (refer to Figure ) the velocity measured by the observer is \begin{align} U_x = U_s - {U_x}^{'} \label{shock:eq:conversionUx} \end{align} Where $U_s$ is the shock velocity which is moving to the right. The ``downstream'' velocity is \begin{align} {U_y}^{'} = U_s - U_y \label{shock:eq:conversionUy} \end{align} The speed of sound on both sides of the shock depends only on the temperature and it is assumed to be constant. The upstream prime Mach number can be defined as \begin{align} {M_x}^{'} = \dfrac{U_s - U_x }{ c_x} = \dfrac{ U_s }{ c_x} - M_x = M_{sx} - M_x \label{shock:eq:conversionMx} \end{align} It can be noted that the additional definition was introduced for the shock upstream Mach number, $M_{sx} = { U_s \over c_x}$. The downstream prime Mach number can be expressed as \begin{align} {M_y}^{'} = \dfrac{U_s - U_y }{ c_y} = \dfrac{ U_s }{ c_y} - M_y = M_{sy} - M_y \label{shock:eq:conversionMy} \end{align} Similar to the previous case, an additional definition was introduced for the shock downstream Mach number, $M_{sy}$. The relationship between the two new shock Mach numbers is \begin{align} \begin{array}{rl} \dfrac{U_s }{ c_x} = \dfrac{c_y }{ c_x} \,\dfrac{U_s }{ c_y} \\ M_{sx} = \sqrt{\dfrac{T_y }{ T_x} }\,M_{sy} \end{array} \label{shock:eq:MsyMsx} \end{align} The ``upstream'' stagnation temperature of the fluid is

Shock Stagnation Temperature

\begin{align} \label{shock:eq:conversionT0T} T_{0x} = T_x \left( 1 + \dfrac{k-1 }{ 2} {M_x}^{2} \right) \end{align}
and the ``upstream'' prime stagnation pressure is \begin{align} P_{0x} = P_x \left( 1 + \dfrac{k-1 }{ 2} {M_x}^{2} \right)^{\dfrac{k }{ k-1} } \label{shock:eq:conversionP0P} \end{align} The same can be said for the ``downstream'' side of the shock. The difference between the stagnation temperature is in the moving coordinates \begin{align} T_{0y} - T_{0x} = 0 \label{shock:eq:stagnationTempDiff} \end{align}

Shock or Wave Drag Result from a Moving Shock

Shock Moving Drag

Fig. The diagram that reexplains the shock drag effect of a moving shock.

It can be shown that there is no shock drag in stationary shock{for more information see ``Fundamentals of Compressible Flow, Potto Project, Bar-Meir any verstion''.}. However, the shock or wave drag is very significant so much so that at one point it was considered the sound barrier. Consider the Figure where the stream lines are moving with the object speed. The other boundaries are stationary but the velocity at right boundary is not zero. The same arguments, as discussed before in the stationary case, are applied. What is different in the present case (as oppose to the stationary shock), one side has increase the momentum of the control volume. This increase momentum in the control volume causes the shock drag. In way, it can be view as continuous acceleration of the gas around the body from zero. Note this drag is only applicable to a moving shock (unsteady shock). The moving shock is either results from a body that moves in gas or from a sudden imposed boundary like close or open valve. In the first case, the forces or energies flow from body to gas and therefor there is a need for large force to accelerate the gas over extremely short distance (shock thickness). In the second case, the gas contains the energy (as high pressure, for example in the open valve case) and the energy potential is lost in the shock process (like shock drag). For some strange reasons, this topic has several misconceptions that even appear in many popular and good textbooks. Consider the following example taken from such a book.

Zucorow Figure

Fig. The diagram for the common explanation for shock or wave drag effect a shock. Please notice the strange notations (e.g. V and not U) and they result from a verbatim copy.

Example

A book (see Figure ) explains the shock drag is based on the following rational: The body is moving in a stationary frictionless fluid under one–dimensional flow. The left plane is moving with body at the same speed. The second plane is located ``downstream from the body where the gas has expanded isotropically (after the shock wave) to the upstream static pressure.'' The bottom and upper stream line close the control volume. Since the pressure is the same on the both planes there is no unbalanced pressure forces. However, there is a change in the momentum in the flow direction because ($U_1 > U_2$). The force is acting on the body. There several mistakes in this explanation including the drawing. Explain what is wrong in this description (do not describe the error results from oblique shock).

Solution

Neglecting the mistake around the contact of the stream lines with the oblique shock(see for retouch in the oblique chapter), the control volume suggested is stretched with time. However, the common explanation fall to notice that when the isentropic explanation occurs the width of the area change. Thus, the simple explanation in a change only in momentum (velocity) is not appropriate. Moreover, in an expanding control volume this simple explanation is not appropriate. Notice that the relative velocity at the front of the control volume {$U_1$} is actually zero. Hence, the claim of $U_1 > U_2$ is actually the opposite, $U_1 < U_2$.

11.6 Qualitative questions

Additional Questions

  1. 1.

    In the analysis of the maximum temperature in the shock tube, it was assumed that process is isentropic. If this assumption is not correct would the maximum temperature obtained is increased or decreased?

  2. 2.

    In the analysis of the maximum temperature in the shock wave it was assumed that process is isentropic. Clearly, this assumption is violated when there are shock waves. In that cases, what is the reasoning behind use this assumption any why?

11.6.1 Tables of Normal Shocks, $k=1.4$ Ideal Gas

$M_x$ 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 1.95 2.00 2.05 2.10 2.15 2.20 2.25 2.30 2.35 2.40 2.45 2.50 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00 5.25 5.50 5.75 6.00 6.25 6.50 6.75 7.00 7.25 7.50 7.75 8.00 8.25 8.50 8.75 9.00 9.25 9.50 9.75 10.00
$M_y$ 1.00000 0.95313 0.91177 0.87502 0.84217 0.81264 0.78596 0.76175 0.73971 0.71956 0.70109 0.68410 0.66844 0.65396 0.64054 0.62809 0.61650 0.60570 0.59562 0.58618 0.57735 0.56906 0.56128 0.55395 0.54706 0.54055 0.53441 0.52861 0.52312 0.51792 0.51299 0.49181 0.47519 0.46192 0.45115 0.44231 0.43496 0.42878 0.42355 0.41908 0.41523 0.41189 0.40897 0.40642 0.40416 0.40216 0.40038 0.39879 0.39736 0.39607 0.39491 0.39385 0.39289 0.39201 0.39121 0.39048 0.38980 0.38918 0.38860 0.38807 0.38758
$\dfrac{T_y}{T_x}$ 1.00000 1.03284 1.06494 1.09658 1.12799 1.15938 1.19087 1.22261 1.25469 1.28720 1.32022 1.35379 1.38797 1.42280 1.45833 1.49458 1.53158 1.56935 1.60792 1.64729 1.68750 1.72855 1.77045 1.81322 1.85686 1.90138 1.94680 1.99311 2.04033 2.08846 2.13750 2.39657 2.67901 2.98511 3.31505 3.66894 4.04688 4.44891 4.87509 5.32544 5.80000 6.29878 6.82180 7.36906 7.94059 8.53637 9.15643 9.80077 10.46939 11.16229 11.87948 12.62095 13.38672 14.17678 14.99113 15.82978 16.69273 17.57997 18.49152 19.42736 20.38750
$\dfrac{\rho_y}{\rho_x}$ 1.00000 1.08398 1.16908 1.25504 1.34161 1.42857 1.51570 1.60278 1.68966 1.77614 1.86207 1.94732 2.03175 2.11525 2.19772 2.27907 2.35922 2.43811 2.51568 2.59188 2.66667 2.74002 2.81190 2.88231 2.95122 3.01863 3.08455 3.14897 3.21190 3.27335 3.33333 3.61194 3.85714 4.07229 4.26087 4.42623 4.57143 4.69919 4.81188 4.91156 5.00000 5.07869 5.14894 5.21182 5.26829 5.31915 5.36508 5.40667 5.44444 5.47883 5.51020 5.53890 5.56522 5.58939 5.61165 5.63218 5.65116 5.66874 5.68504 5.70019 5.71429
$\dfrac{P_y}{P_x}$ 1.00000 1.11958 1.24500 1.37625 1.51333 1.65625 1.80500 1.95958 2.12000 2.28625 2.45833 2.63625 2.82000 3.00958 3.20500 3.40625 3.61333 3.82625 4.04500 4.26958 4.50000 4.73625 4.97833 5.22625 5.48000 5.73958 6.00500 6.27625 6.55333 6.83625 7.12500 8.65625 10.33333 12.15625 14.12500 16.23958 18.50000 20.90625 23.45833 26.15625 29.00000 31.98958 35.12500 38.40625 41.83333 45.40625 49.12500 52.98958 57.00000 61.15625 65.45833 69.90625 74.50000 79.23958 84.12500 89.15625 94.33333 99.65625 105.12500 110.73958 116.50000
$\dfrac{{P_0}_y }{ {P_0}_x} $ 1.00000 0.99985 0.99893 0.99669 0.99280 0.98706 0.97937 0.96974 0.95819 0.94484 0.92979 0.91319 0.89520 0.87599 0.85572 0.83457 0.81268 0.79023 0.76736 0.74420 0.72087 0.69751 0.67420 0.65105 0.62814 0.60553 0.58329 0.56148 0.54014 0.51931 0.49901 0.40623 0.32834 0.26451 0.21295 0.17166 0.13876 0.11256 0.09170 0.07505 0.06172 0.05100 0.04236 0.03536 0.02965 0.02498 0.02115 0.01798 0.01535 0.01316 0.01133 0.00979 0.00849 0.00739 0.00645 0.00565 0.00496 0.00437 0.00387 0.00343 0.00304

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