# 11.5.3: Operating Equations and Analysis

- Page ID
- 809

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 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.

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 11.12

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: \(M_z\) | 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 (21)) is limited by

\[

{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} \tag{34}

\]

The limits of the pressure ratio can be obtained by looking at equation (15) and by utilizing the limit that was obtained in equation (34).

### 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.

Shock Stagnation Temperature

\[

\label{shock:eq:conversionT0T}

T_{0x} = T_x \left( 1 + \dfrac{k-1 }{ 2} {M_x}^{2} \right) \tag{42}

\]

and the "upstream'' prime stagnation pressure is

\[

P_{0x} = P_x \left( 1 + \dfrac{k-1 }{ 2} {M_x}^{2} \right)^{\dfrac{k }{ k-1} }

\label{shock:eq:conversionP0P} \tag{43}

\]

The same can be said for the "downstream'' side of the shock. The difference between the stagnation temperature is in the moving coordinates

\[

T_{0y} - T_{0x} = 0

\label{shock:eq:stagnationTempDiff} \tag{44}

\]

### Shock or Wave Drag Result from 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.

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, \(U_1 < U_2\).

### Contributors

Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.