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4.4 Shock Waves

Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics
stationary normal shock

Figure 4.4:Stationary normal shock

The starting point is to set up the governing equations for one-dimensional steady compressible flow over a control volume enclosing the normal shock (Fig. Figure 4.5).

normal-shock control volume

Figure 4.5:Normal-shock control volume

continuity:

ρ1u1=ρ2u2\rho_1 u_1=\rho_2 u_2

momentum:

ρ1u12+p1=ρ2u22+p2\rho_1 u_1^2+p_1=\rho_2 u_2^2+p_2

energy:

h1+12u12=h2+12u22h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2

Divide the momentum equation by ρ1u1\rho_1 u_1

1ρ1u1(ρ1u12+p1)=1ρ1u1(ρ2u22+p2)={ρ1u1=ρ2u2}=1ρ2u2(ρ2u22+p2)\frac{1}{\rho_1 u_1}\left(\rho_1 u_1^2+p_1\right)=\frac{1}{\rho_1 u_1}\left(\rho_2 u_2^2+p_2\right)=\left\{\rho_1 u_1=\rho_2 u_2\right\}=\frac{1}{\rho_2 u_2}\left(\rho_2 u_2^2+p_2\right) \Rightarrow
p1ρ1u1p2ρ2u2=u2u1\frac{p_1}{\rho_1 u_1}-\frac{p_2}{\rho_2 u_2}=u_2-u_1

For a calorically perfect gas a=γp/ρa=\sqrt{\gamma p/\rho}, which if implemented in Eqn. (4.55) gives

a12γu1a22γu2=u2u1\frac{a_1^2}{\gamma u_1}-\frac{a_2^2}{\gamma u_2}=u_2-u_1

The energy equation (Eqn. (4.53)) with h=CpTh=C_p T

CpT1+12u12=CpT2+12u22C_p T_1 + \frac{1}{2}u_1^2=C_p T_2 + \frac{1}{2}u_2^2

Replacing CpC_p with γR/(γ1)\gamma R/(\gamma-1) gives

γRT1γ1+12u12=γRT2γ1+12u22\frac{\gamma RT_1}{\gamma-1} + \frac{1}{2}u_1^2=\frac{\gamma RT_2}{\gamma-1} + \frac{1}{2}u_2^2

With a=γRTa=\sqrt{\gamma RT} this becomes

a12γ1+12u12=a22γ1+12u22\frac{a_1^2}{\gamma-1} + \frac{1}{2}u_1^2=\frac{a_2^2}{\gamma-1} + \frac{1}{2}u_2^2

Eqn. (4.59) can be set up between any two points in the flow. Specifically, we can use the relation to relate the flow velocity, uu, and speed of sound, aa, in any point to the corresponding flow properties at sonic conditions (u=a=au=a=a^*).

a2γ1+12u2=γ+12(γ1)a2\frac{a^2}{\gamma-1} + \frac{1}{2}u^2=\frac{\gamma+1}{2(\gamma-1)}{a^*}^2

If Eqn. (4.60) is evaluated in locations 1 and 2, we get

a12=γ+12a2γ12u12a22=γ+12a2γ12u22\begin{aligned} &a_1^2 = \frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_1^2\\ &a_2^2 = \frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_2^2 \end{aligned}

Since the change in flow conditions over the shock is adiabatic (no heat is added inside the shock), critical properties will be constant over the shock. Especially aa^* will be constant.

Eqn. (4.61) inserted in (4.56) gives

1γu1(γ+12a2γ12u12)1γu2(γ+12a2γ12u22)=u2u1\frac{1}{\gamma u_1}\left(\frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_1^2\right)-\frac{1}{\gamma u_2}\left(\frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_2^2\right)=u_2-u_1 \Rightarrow
(γ+12γ)a2(1u11u2)=(γ+12γ)(u2u1)\left(\frac{\gamma+1}{2\gamma}\right){a^*}^2\left(\frac{1}{u_1}-\frac{1}{u_2}\right)=\left(\frac{\gamma+1}{2\gamma}\right)\left(u_2-u_1\right) \Rightarrow
a2(1u11u2)=(u2u1){a^*}^2\left(\frac{1}{u_1}-\frac{1}{u_2}\right)=\left(u_2-u_1\right) \Rightarrow
a2(u2u1u2u1u1u2)=(u2u1){a^*}^2\left(\frac{u_2}{u_1 u_2}-\frac{u_1}{u_1 u_2}\right)=\left(u_2-u_1\right) \Rightarrow
1u1u2a2(u2u1)=(u2u1)\frac{1}{u_1 u_2}{a^*}^2\left(u_2-u_1\right)=\left(u_2-u_1\right) \Rightarrow
a2=u1u2{a^*}^2=u_1 u_2

Eqn. (4.67) is sometimes referred to as the Prandtl relation. Divide the Prandtl relation by a2{a^*}^2 on both sides gives

1=u1au2a=M1M21=\frac{u_1}{a^*}\frac{u_2}{a^*}=M^*_1M^*_2

or

M2=1M1M^*_2=\frac{1}{M^*_1}

The relation between MM^* and MM is given by

M2=(γ+1)M22+(γ1)M2{M^*}^2=\frac{(\gamma+1)M^2}{2+(\gamma-1)M^2}

from which is can be seen that MM^* will follow the Mach number MM in the sense that

Eqn. (4.70) inserted in Eqn. (4.69) gives

(γ+1)M122+(γ1)M12=2+(γ1)M22(γ+1)M22\frac{(\gamma+1)M_1^2}{2+(\gamma-1)M_1^2}=\frac{2+(\gamma-1)M_2^2}{(\gamma+1)M_2^2}
M22=1+[(γ1)/2]M12γM12(γ1)/2M^2_2=\frac{1+\left[(\gamma-1)/2\right]M^2_1}{\gamma M^2_1-(\gamma-1)/2}

The Mach number relations above effectively show that if the Mach number upstream of the shock is greater than one, the downstream Mach number must be less than one and vice versa. We can also see that a sonic upstream flow M1=1.0M_1=1.0 gives sonic flow downstream of the shock. So, apparently the relation as such holds for both supersonic and subsonic upstream flow mathematically. The question is if it is also physically correct. For a supersonic upstream flow we will get a discontinuous compression and if the flow upstream of the control volume is subsonic we will instead get a discontinuous expansion inside the control volume but, again, is this physically correct? We will get the answer by analyzing the entropy change over the control volume.

Analyzing the energy equation and the second law of thermodynamics shows that there is a direct relation between entropy increase and total pressure drop.

s2s1=CplnT2T1Rlnp2p1s_2-s_1=C_p\ln\dfrac{T_2}{T_1}-R\ln\dfrac{p_2}{p_1}
s2s1=CplnT2To2To1T1To2To1Rlnp2po2po1p1po2po1s_2-s_1=C_p\ln\dfrac{T_2}{T_{o_2}}\dfrac{T_{o_1}}{T_1}\dfrac{T_{o_2}}{T_{o_1}}-R\ln\dfrac{p_2}{p_{o_2}}\dfrac{p_{o_1}}{p_1}\dfrac{p_{o_2}}{p_{o_1}}

using the isentropic relations we get

s2s1=CplnTo2To1Rlnpo2po1s_2-s_1=C_p\ln\dfrac{T_{o_2}}{T_{o_1}}-R\ln\dfrac{p_{o_2}}{p_{o_1}}

and since the process is adiabatic and thus To2=To1T_{o_2}=T_{o_1} the change in entropy is directly related to the change in total pressure as

s2s1=Rlnpo2po1s_2-s_1=-R\ln\dfrac{p_{o_2}}{p_{o_1}}

or

po2po1=e(s2s1)/R\dfrac{p_{o_2}}{p_{o_1}}=e^{-(s_2-s_1)/R}

Figure Figure 4.6 shows the entropy change over a normal shock. As can be seen in the figure, a subsonic upstream Mach number leads to a reduction of entropy, which once and for all rules out all such solutions as non-physical and thus the question about the upstream conditions can now be considered answered. This in turn implies that the Mach number downstream of a normal shock will always be subsonic, which can be seen in Fig Figure 4.7 below.

entropy change over a normal shock

Figure 4.6:Entropy change over a normal shock (Δs\Delta s) as function of upstream Mach number (M1M_1)

downstream Mach number as function of upstream Mach number

Figure 4.7:Downstream Mach number (M2M_2) as function of upstream Mach number (M1M_1)

By rewriting the right-hand side of Eqn. (4.72), it is easy to realize that the downstream Mach number M2M_2 approaches a finite value for large values of the upstream Mach number, M1M_1.

M22M1=2/M12+(γ1)2γ(γ1)/M12M1=γ12γ\left.M_2^2\right|_{M_1\rightarrow\infty}=\left.\dfrac{2/M_1^2+(\gamma-1)}{2\gamma-(\gamma-1)/M_1^2}\right|_{M_1\rightarrow\infty}=\dfrac{\gamma-1}{2\gamma}