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Chapter 11 Gd (continue)
11.5 Normal Shock
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This assumption is no longer valid in the case of a shock.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 $Ts$ 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{k1 }{ 2}\, {M_y}^{2} \right) =
T_x \left( 1 + \dfrac{k1 }{ 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{k1 }{ 2}\, {M_y}^{2} \right)^{\dfrac{k}{k1}} }
{P_x\, \left( 1 + \dfrac{k1 }{ 2}\, {M_x}^{2} \right)^{\dfrac{k}{k1}} }
\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{ k1 }{ 2} {M_x}^{2} \over 1 + \dfrac{ k1 }{ 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{ k1 }{ 2} {M_x}^{2} \over
1 + \dfrac{ k1 }{ 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}{ k1}\right) +\left( \dfrac{P_y }{ P_x} \right)}
\end{align}
which is associated with the shock wave.
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 }{ k1} + \dfrac{U^2 }{ 2 } =
\dfrac{{c^{*}}^2 }{ k1} + \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
 {k1 \over 2 } U_1 \right] + U_1 =
\dfrac{1 }{ k\,U_2} \left[ {k+1 \over 2 } {c^{*}}^2
 \dfrac{k1 }{ 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 PottoGDC 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.
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.
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.
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{k1 }{ 2} {M_x}^{2} \right)
\end{align}
and the ``upstream'' prime stagnation pressure is
\begin{align}
P_{0x} = P_x \left( 1 + \dfrac{k1 }{ 2} {M_x}^{2} \right)^{\dfrac{k }{ k1} }
\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
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, BarMeir 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.
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.
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.
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.
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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