Normal Shocks

As previously described, there is an effective discontinuity in the flow speed, pressure, density, and temperature, of the gas flowing through the diverging part of an over-expanded Laval nozzle. This type of discontinuity is known as a normal shock. Let us investigate the properties of such shocks.

Our fundamental equations are the mass conservation equation [see Equation (14.30)],

$$\frac{D\rho}{Dt} = -\rho\,\nabla\cdot{\bf v},$$ (14.80)

the momentum conservation equation [see Equation (14.31)],

$$\frac{D{\bf v}}{Dt}= -\frac{\nabla p}{\rho},$$ (14.81)

and the energy conservation equation [see Equation (1.75)],

$$\frac{D{\cal E}}{Dt} =-\frac{p}{\rho}\,\nabla\cdot{\bf v}.$$ (14.82)

In writing the previous equations, we have neglected viscosity, heat conduction, and potential energy. Note that we have used the more fundamental energy conservation equation, (1.75), rather than Equation (14.32), because the latter equation incorporates the ideal gas law, and this law is not valid inside the shock, because the gas there is not in thermodynamic equilibrium.

Consider a compressible gas flowing steadily down a duct of constant cross-sectional area $A$ . Let $x$ measure distance along the duct. The gas's temperature, $T(x)$ , density, $\rho(x)$ , pressure $p(x)$ , specific internal energy ${\cal E}(x)$ , and normal velocity $u(x)=v_x$ , are all assumed to be constant across any cross-section, and independent of time. Let the shock be situated at $x=0$ . Suppose that in the region upstream of the shock, $x<0$ , the temperature, density, pressure, specific internal energy, and flow speed of the gas take the constant values $T_1$ , $\rho_1$ , $p_1$ , ${\cal E}_1$ , and $u_1$ , respectively. Likewise, suppose that in the region downstream of the shock, $x>0$ , the temperature, density, pressure, specific internal energy, and flow speed of the gas take the constant values $T_2$ , $\rho_2$ , $p_2$ , ${\cal E}_2$ , and $u_2$ , respectively. Of course, in the immediate vicinity of the shock, $T(x)$ , $\rho(x)$ , $p(x)$ , ${\cal E}(x)$ , and $u(x)$ are all rapidly-varying functions of $x$ . We wish to find the relationship between the upstream and downstream gas parameters. We can achieve this goal by invoking conservation of mass, momentum, and energy across the shock--in other words, by making use of Equations (14.80)-(14.82).

The mass continuity equation (14.71) yields

$$u\,\frac{d\rho}{dx} = -\rho\,\frac{du}{dx},$$ (14.83)

which can be rearranged to give

$$\frac{d}{dx}\,(\rho\,u) = 0.$$ (14.84)

The momentum conservation equation (14.72) yields

$$u\,\frac{du}{dx} =-\frac{1}{\rho}\,\frac{dp}{dx},$$ (14.85)

which can be rearranged to give

$$\rho\,u\,\frac{du}{dx} + \frac{dp}{dx} = 0,$$ (14.86)

or

$$\frac{d}{dx}\left(\rho\,u^{\,2}+p\right) = 0,$$ (14.87)

where use has been made of Equation (14.84). Finally, the energy conservation equation (14.73) yields

$$u\,\frac{d{\cal E}}{dx}=-\frac{p}{\rho}\,\frac{du}{dx},$$ (14.88)

which can be rearranged to give

$$\rho\,u\,\frac{d{\cal E}}{dx}+\frac{d}{dx}\,(p\,u) = u\,\frac{dp}{dx},$$ (14.89)

or

$$\frac{d}{dx}\left(\rho\,u\,{\cal E} +p\,u\right)= -\rho\,u^{\,2}\... ...o\,u\,\frac{du^{\,2}}{dx}=-\frac{d}{dx}\left(\frac{1}{2}\,\rho\,u^{\,3}\right),$$ (14.90)

where use has been made of Equations (14.84) and (14.86). Thus, we obtain

$$\frac{d}{dx}\left(\rho\,u\,{\cal E} + p\,u+\frac{1}{2}\,\rho\,u^{\,3}\right)=0,$$ (14.91)

which reduces to

$$\frac{d}{dx}\left({\cal E} + \frac{p}{\rho}+\frac{1}{2}\,u^{\,2}\right)=0$$ (14.92)

with the aid of Equation (14.84). The previous equation can also be written

$$\frac{d}{dx}\left({\cal H} +\frac{1}{2}\,u^{\,2}\right)=0,$$ (14.93)

where ${\cal H}={\cal E}+p/\rho$ is the specific enthalpy of the gas. [See Equation (14.18).]

Integrating Equations (14.84), (14.87), and (14.93) across the shock, we obtain

$$\rho_1\,u_1 = \rho_2\,u_2,$$ (14.94)

$$\rho_1\,u_1^{\,2}+p_1 = \rho_2\,u_2^{\,2}+p_2,$$ (14.95)

$${\cal H}_1+\frac{1}{2}\,u_1^{\,2} = {\cal H}_2+\frac{1}{2}\,u_2^{\,2},$$ (14.96)

where ${\cal H}_1={\cal E}_1+p_1/\rho_1$ , et cetera. We shall assume that the gas upstream and downstream of the shock obeys the ideal gas law, and has the same ratio of specific heats, $\gamma$ . It follows that

$$p_{1,2} = {\cal R}\,\rho_{1,2}\,T_{1,2},$$ (14.97)

$${\cal H}_{1,2} = \left(\frac{\gamma}{\gamma-1}\right){\cal R}\,T_{1,2}.$$ (14.98)

where use has been made of Equations (14.1), (14.18), and (14.60). Let ${\rm Ma}_1$ and ${\rm Ma}_2$ be the Mach numbers upstream and downstream of the shock, respectively. Thus,

$${\rm Ma}_{1,2} = \frac{u_{1,2}}{\sqrt{\gamma\,{\cal R}\,T_{1,2}}},$$ (14.99)

where use has been made of Equations (14.45) and (14.57). Equations (14.96), (14.98), and (14.99) can be combined to give

$$\frac{T_2}{T_1}=\frac{1+(1/2)\,(\gamma-1)\,{\rm Ma}_1^{\,2}}{1+(1/2)\,(\gamma-1)\,{\rm Ma}_2^{\,2}}.$$ (14.100)

Equations (14.94) and (14.99) yield

$$\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\sqrt{\frac{T_1}{T_2}}\,\frac{{\rm Ma}_1}{{\rm Ma}_2},$$ (14.101)

or

$$\frac{p_2}{p_1} = \sqrt{\frac{T_2}{T_1}}\,\frac{{\rm Ma}_1}{{\rm Ma}_2},$$ (14.102)

where use has been made of Equation (14.97). Finally, Equations (14.95), (14.97), and (14.99) give

$$\frac{p_2}{p_1} = \frac{1+\gamma\,{\rm Ma}_1^{\,2}}{1+\gamma\,{\rm Ma}_2^{\,2}}.$$ (14.103)

Eliminating $T_2$ , $\rho_2$ , and $p_2$ between Equations (14.100), (14.101), and (14.103), we obtain

$${\rm Ma}_2=\left[\frac{2+(\gamma-1)\,{\rm Ma}_1^{\,2}}{1-\gamma+2\,\gamma\,{\rm Ma}_1^{\,2}}\right]^{1/2}.$$ (14.104)

Note that

$$\frac{d{\rm Ma}_2^{\,2}}{d{\rm Ma}_1^{\,2}}=-\left(\frac{1+\gamma}{1-\gamma+2\,\gamma\,{\rm Ma}_1^{\,2}}\right)^{\,2}<0.$$ (14.105)

Moreover, it is easily verified that ${\rm Ma}_2=1$ when ${\rm Ma}_1=1$ . Hence, we deduce that

$${\rm Ma}_2\lesseqgtr 1 {\rm Ma}_1\gtreqless 1.$$ (14.106)

In other words, if the downstream flow is subsonic then the upstream flow is supersonic, and vice versa. Equations (14.1), (14.100), (14.101), (14.103), and (14.104) can be combined to give

$$\frac{p_2}{p_1} = \frac{1-\gamma+2\,\gamma\,M_1^{\,2}}{\gamma+1},$$ (14.107)

$$\frac{\rho_2}{\rho_1} =\frac{u_1}{u_2}=\frac{(\gamma+1)\,{\rm Ma}_1^{\,2}}{2+(\gamma-1)\,{\rm Ma}_1^{\,2}},$$ (14.108)

$$\frac{T_2}{T_1} = \frac{p_2}{p_1}\,\frac{\rho_1}{\rho_2}.$$ (14.109)

The previous three equations completely describe the conditions downstream of the shock in terms of the upstream conditions. We can rearrange these equations to give

$${\rm Ma}_1^{\,2} = \frac{\gamma-1+(\gamma+1)\,(p_2/p_1)}{2\,\gamma},$$ (14.110)

$$\frac{\rho_2}{\rho_1} = \frac{u_1}{u_2} =\frac{\gamma-1+(\gamma+1)\,(p_2/p_1)}{\gamma+1+(\gamma-1)\,(p_2/p_1)},$$ (14.111)

$$\frac{T_2}{T_1} =\left(\frac{p_2}{p_1}\right)\left[\frac{\gamma+1+(\gamma-1)\,(p_2/p_1)}{\gamma-1+(\gamma+1)\,(p_2/p_1)}\right].$$ (14.112)

The latter two equations are known as the Rankine-Hugoniot relations.

According to Equation (14.25), the jump in specific entropy across the shock is

$${\cal S}_2-{\cal S}_1={\cal C}_V\ln\left[\frac{p_2}{p_1}\left(\frac{\rho_1}{\rho_2}\right)^\gamma\right].$$ (14.113)

It follows from Equations (14.107) and (14.108) that

$${\cal S}_2-{\cal S}_1 = {\cal C}_V\ln\left[G({\rm Ma}_1^{\,2})\right],$$ (14.114)

where

$$G(x) = \left(\frac{1-\gamma+2\,\gamma\,x}{\gamma+1}\right)\left[\frac{2+(\gamma-1)\,x}{(\gamma+1)\,x}\right]^{\,\gamma}.$$ (14.115)

It is easily demonstrated that

$$\frac{dG}{dx} = 2\,\gamma\,(\gamma-1)\,(1-x)^{\,2}\,\frac{[2+(\gamma-1)\,x]^{\,\gamma-1}}{[(\gamma+1)\,x]^{\,\gamma+1}}\geq 0.$$ (14.116)

Moreover, $G(1)=1$ . Hence, we deduce that

$${\cal S}_2-{\cal S}_1 \lesseqgtr 0 {\rm Ma}_1\lesseqgtr 1.$$ (14.117)

Now, the second law of thermodynamics forbids a spontaneous decrease in the specific entropy of the gas as it passes through the shock (Reif 1965). In other words, the second law of thermodynamics demands that ${\cal S}_2-{\cal S}_1\geq 0$ . It follows that ${\rm Ma}_1\geq 0$ . However, it is easily seen from Equations (14.107)-(14.109) that $p_2/p_1=\rho_2/\rho_1=u_2/u_1=T_2/T_1=1$ when ${\rm Ma}_1=1$ . In other words, there is no shock (i.e., no discontinuity in the properties of the gas at $x=0$ ) when the upstream Mach number is exactly unity. Thus, we conclude that the only type of normal shock that is consistent with the second law of thermodynamics is one in which the upstream flow is supersonic, and the downstream flow subsonic--that is, ${\rm Ma}_1>1$ and ${\rm Ma}_2<1$ . It is apparent from Equations (14.107)-(14.109) that if ${\rm Ma}_1>1$ then $p_2/p_1>1$ , $\rho_2/\rho_1>1$ , and $T_2/T_1>1$ . In other words, the passage of the gas through the shock front leads to both compression and heating.

The dimensionless parameter

$$\frac{{\mit\Delta}p}{p_1} =\frac{p_2-p_1}{p_1}$$ (14.118)

is a convenient measure of shock strength. Thus, a shock is said to be weak if ${\mit\Delta}p/p_1\ll 1$ , and strong if ${\mit\Delta}p/p_1\gg 1$ .

Consider a weak shock. According to Equations (14.111), (14.112), (14.114), and (14.116),

$$\frac{{\mit\Delta}\rho}{\rho_1} \simeq \frac{1}{\gamma}\,\frac{{\mit\Delta}p}{p_1},$$ (14.119)

$$\frac{{\mit\Delta}T}{T_1} \simeq \left(\frac{\gamma-1}{\gamma}\right)\frac{{\mit\Delta}p}{p_1},$$ (14.120)

$$\frac{{\mit\Delta}{\cal S}}{{\cal R}} \simeq \frac{(\gamma+1)}{12\,\gamma^{\,2}}\left( \frac{{\mit\Delta}p}{p_1}\right)^{\,3},$$ (14.121)

where ${\mit\Delta}\rho =\rho_2-\rho_1$ , et cetera, and use has been made of Equation (14.59). It can be seen that the flow across the shock is isentropic (i.e., $p/\rho^{\,\gamma}$ and $T^{\,\gamma}/p^{\,\gamma-1}$ are constant) to first order in the shock strength. This is the case because the increase in specific entropy across the shock is only third order in the shock strength.

Consider a strong shock. It can be seen from Equations (14.111) and (14.112) that

$$\frac{\rho_2}{\rho_1} \rightarrow \frac{\gamma+1}{\gamma-1},$$ (14.122)

$$\frac{T_2}{T_1} \rightarrow \left(\frac{\gamma-1}{\gamma+1}\right)\frac{{\mit\Delta}p}{p_1},$$ (14.123)

as ${\mit\Delta}p/p_1\rightarrow\infty$ . Clearly, there is a finite limit to the degree of compression of gas passing through a strong shock. Thus, in the case of a strong shock, the large increase in the pressure across the shock front is predominately caused by a large increase in the temperature. In other words, the gas flowing through a strong shock is subject moderate compression, but intense heating.

Finally, let us transform to a frame of reference that is co-moving with the gas upstream of the shock. In this reference frame, the shock appears to propagate through a stationary gas of pressure, density, and temperature, $p_1$ , $\rho_1$ , and $T_1$ , respectively, at the speed $u_1$ . In other words, in the new reference frame, our stationary shock is transformed into a shock wave. (See Section 14.4.) The pressure, density, and temperature behind the wave are $p_2$ , $\rho_2$ , and $T_2$ , respectively. Also, the gas behind the wave follows the shock front at the speed $u_1-u_2$ . We can regard ${\rm Ma}_1$ as the Mach number of the shock wave in the unperturbed gas. It follows from Equation (14.110) that

$${\rm Ma}_1^{\,2} = 1+\left(\frac{\gamma+1}{2\,\gamma}\right)\frac{{\mit\Delta} p}{p_1}.$$ (14.124)

Thus, for a weak shock wave (i.e., ${\mit\Delta}p/p_1\ll 1$ ),

$${\rm Ma}_1\simeq 1.$$ (14.125)

In other words, a weak shock wave propagates through the unperturbed gas at the local sound speed. Indeed, such a wave is essentially indistinguishable from a conventional sound wave. On the other hand, for a strong shock wave (i.e., ${\mit\Delta}p/p_1\gg 1$ ),

$${\rm Ma}_1\simeq \sqrt{\left(\frac{\gamma+1}{2\,\gamma}\right)\frac{{\mit\Delta} p}{p_1}}.$$ (14.126)

We conclude that a strong shock wave propagates through the unperturbed gas at a speed that greatly exceeds the local sound speed.