Solution:

What is Shock waves?

• A shock wave is a pressure wave of finite thickness, of the order of 10–2 to 10–4 mm in the atmospheric pressure.

• A shock wave takes place in the diverging section of a nozzle, in a diffuser, throat of a supersonic wind tunnel, in front of sharp nosed bodies.

Shock waves are of two types :

(1). Normal shocks which are almost perpendicular to the flow.

(2). Oblique shocks which are inclined to the flow direction.

(1) Normal Shock Wave:

• Consider a duct having a compressible sonic flow (see Fig.).

• Let p1, ρ1, T1, and V1 be the pressure, density, temperature and velocity of the flow (M1 > 1) and p2, ρ2, T2 and V2 the corresponding values of pressure, density, temperature and velocity after a shock wave takes place (M2 < 1).

• For analysing a normal shock wave, use will be made of the continuity, momentum and energy equations.

Assume unit area cross-section, $ A_{1}=A_{2}=1 \\ $.

Continuity equation : $ \quad m=\rho_{1} V_{1}=\rho_{2} V_{2} \\ $

Momentum equation : $ \quad \Sigma F_{x}=p_{1} A_{1}-p_{2} A_{2}=m\left(V_{2}-V_{1}\right)=\rho_{2} A_{2} V_{2}^{2}-\rho_{1} A_{1} V_{1}^{2} \\ $

for, $ A_{1}=A_{2}=1 \\ $, the pressure drop across the shock wave,

$$ \begin{aligned} p_{1}-p_{2} &=\rho_{2} V_{2}^{2}-\rho_{1} V_{1}^{2} \\\\ p_{1}+\rho_{1} V_{1}^{2} &=p_{2}+\rho_{2} V_{2}^{2} \\ \end{aligned} $$

Consider the flow across the shock wave as adiabatic.

Energy equation : $ \left(\frac{\gamma}{\gamma-1}\right) \frac{p_{1}}{\rho_{1}}+\frac{V_{1}^{2}}{2}=\left(\frac{\gamma}{\gamma-1}\right) \frac{p_{2}}{\rho_{2}}+\frac{V_{2}^{2}}{2}....(a) \\ $

$\left(z_{1}=z_{2}\right.$, duct being in horizontal position)

or,

$ \frac{\gamma}{\gamma-1}\left(\frac{p_{2}}{\rho_{2}}-\frac{p_{1}}{\rho_{1}}\right)=\frac{V_{1}^{2}-V_{2}^{2}}{2} \\ $

Combining continuity and momentum equations [refer eqns. (i) and (ii)], we get,

$$ p_{1}+\frac{\left(\rho_{1} V_{1}\right)^{2}}{\rho_{1}}=p_{2}+\frac{\left(\rho_{2} V_{2}\right)^{2}}{\rho_{2}}...(b)\\ $$

This equation is known as Rankine Line Equation.

Now combining continuity and energy equations [refer eqns. (i) and (iii)], we get,

$$ \frac{\gamma}{\gamma-1}\left(\frac{p_{1}}{\rho_{1}}\right)+\frac{\left(\rho_{1} V_{1}\right)^{2}}{2 \rho_{1}^{2}}=\frac{\gamma}{\gamma-1}\left(\frac{p_{2}}{\rho_{2}}\right)+\frac{\left(\rho_{2} V_{2}\right)^{2}}{2 \rho_{2}^{2}}...(c) \\ $$

This equation is called Fanno Line Equation.

Further combining eqns. (i), (ii) and (iii) and solving for $\frac{p_{2}}{p_{1}}$, we get,

$$ \frac{p_{2}}{p_{1}}=\frac{\left(\frac{\gamma+1}{\gamma-1}\right) \frac{\rho_{2}}{\rho_{1}}-1}{\left(\frac{\gamma+1}{\gamma-1}\right)-\frac{\rho_{2}}{\rho_{1}}}....(d) \\ $$

Solving for density ratio $\frac{\rho_{2}}{\rho_{1}}$, the same equations yield,

$$ \frac{\rho_{2}}{\rho_{1}}=\frac{V_{1}}{V_{2}}=\frac{1+\left(\frac{\gamma+1}{\gamma-1}\right) \frac{p_{2}}{p_{1}}}{\left(\frac{\gamma+1}{\gamma-1}\right)+\frac{p_{2}}{p_{1}}}...(e) \\ $$

The eqns. (d) and (e) are called Ranking-Hugoniot equations.

One can also express $\frac{p_{2}}{p_{1}}, \frac{V_{2}}{V_{1}}, \frac{\rho_{2}}{\rho_{1}}$ and $\frac{T_{2}}{T_{1}}$ in terms of Mach number as follows :

$$ \begin{aligned} \frac{p_{2}}{p_{1}} &=\frac{2 \gamma M_{1}^{2}-(\gamma-1)}{(\gamma+1)} \\\\ \frac{V_{1}}{V_{2}} &=\frac{\rho_{2}}{\rho_{1}}=\frac{(\gamma+1) M_{1}^{2}}{(\gamma-1) M_{1}^{2}+2} \\\\ \frac{T_{2}}{T_{1}} &=\frac{\left[(\gamma-1) M_{1}^{2}+2\right]\left[2 \gamma M_{1}^{2}-(\gamma-1)\right]} \\{(\gamma+1)^{2} M_{1}^{2}} \\ \end{aligned} $$

By algebraic manipulation the following equation between $M_{1}$ and $M_{2}$ can be obtained.

$$ M_{2}^{2}=\frac{(\gamma-1) M_{1}^{2}+2}{2 \gamma M_{1}^{2}-(\gamma-1)} \\ $$