Isentropic expansion waves

Assumptions:

1. Steady flow.
2. Negligible body forces.
3. Adiabatic flow.
4. No work terms.
5. Negligible gravitational effect.

The continuity equation is $${\displaystyle {\frac {\partial }{\partial t}}\int \limits _{CV}\rho dV+\int \limits _{CS}\rho {\bar {v}}\,d{\bar {A}}=0\qquad \qquad (1.1)}$$

First term is zero by assumption (1). Now, $${\displaystyle {-\rho v\sin \alpha A}+{(\rho +d\rho )(v+dv)\sin(a-d\theta )A}=0}$$ which can be rewritten as $${\displaystyle \rho v\sin \alpha =(\rho +d\rho )(v+dv)\sin(\alpha -d\theta )\qquad \qquad (1.2)}$$

Now we consider the momentum equation for normal and tangential to shock. For y-component, $${\displaystyle F_{S_{y}}+F_{B_{y}}={\frac {\partial }{\partial t}}\int \limits _{CV}v_{y}\rho dV+\int \limits _{CS}v_{y}\rho {\bar {v}}\,d{\bar {A}}\qquad \qquad (1.3)}$$

Second term of L.H.S and first term of R.H.S are zero by assumption (2) and (1) respectively. Then,

$${\displaystyle 0=v\cos \alpha (-\rho v\sin \alpha A)+(v+dv)\cos(\alpha -d\theta ){(\rho +d\rho )(v+dv)\sin(\alpha -d\theta )A}}$$

Or using equation 1.1 (continuity), $${\displaystyle v\cos \alpha =(v+dv)\cos(\alpha -d\theta )}$$

Expanding and simplifying [Using the facts that, to the first order, in the limit as ${\displaystyle d\theta \rightarrow 0}$, ${\displaystyle \cos {d\theta }\rightarrow 1}$ and ${\displaystyle \sin {d\theta }\rightarrow d\theta }$], we obtain

$${\displaystyle d\theta ={\frac {-dv}{v\tan \alpha }}}$$

But, $${\displaystyle \sin \alpha ={\frac {1}{M}}}$$

so, $${\displaystyle \tan \alpha ={\frac {1}{\sqrt {M^{2}-1}}}}$$

and $${\displaystyle d\theta =-{\frac {{\sqrt {M^{2}-1}}\ dv}{v}}\qquad \qquad (1.4)}$$

We skip the analysis of the x-component of the momentum and move on to the first law of thermodynamics, which is

$${\displaystyle {\dot {Q}}-{\dot {W}}_{s}-{\dot {W}}_{shear}-{\dot {W}}_{other}={\frac {\partial }{\partial t}}\int \limits _{CV}e\rho dV+\int \limits _{CS}h\rho {\bar {v}}.d{\bar {A}}\qquad \qquad (1.5)}$$

First term of L.H.S, next three terms of L.H.S and first term of R.H.S are zero due to assumption (3), (4) and (1) respectively.

where, $${\displaystyle e=u+{\frac {v^{2}}{2}}+gz}$$

For our control volume we obtain

$${\displaystyle 0=\left(h+{\frac {v^{2}}{2}}\right)(-\rho v\sin \alpha A)+\left[(h+dh)+{\frac {(v+dv)^{2}}{2}}\right]{\bigl (}(\rho +d\rho )(v+dv)\sin(\alpha -d\theta )A{\bigr )}}$$

This may be simplified as

$${\displaystyle {h+{\frac {v^{2}}{2}}}=(h+dh)+{\frac {(v+dv)^{2}}{2}}}$$

Expanding and simplifying in the limit to first order, we get

$${\displaystyle dh=-vdv}$$

If we confine to ideal gases, ${\displaystyle dh=c_{p}dT}$ , so

$${\displaystyle c_{p}dT=-vdv\qquad \qquad (1.6)}$$

Above equation relates the differential changes in velocity and temperature. We can derive a relation between ${\displaystyle M}$ and ${\displaystyle v}$ using ${\displaystyle v=Mc=M{\sqrt {kRT}}}$. Differentiating (and dividing the left hand side by ${\displaystyle v}$ and the right by ${\displaystyle {\sqrt {kRT}}}$ ),

$${\displaystyle {\frac {dv}{v}}={\frac {dM}{M}}+{\frac {dT}{2T}}}$$

Using equation (1.6)

$${\displaystyle {\begin{aligned}{\frac {dv}{v}}&={\frac {dM}{M}}-{\frac {vdv}{2c_{p}T}}\\[2pt]&={\frac {dM}{M}}-{\frac {dv{\frac {v^{2}}{c_{p}T}}}{2v}}\\[2pt]&={\frac {dM}{M}}-{\frac {dv{\frac {M^{2}c^{2}}{c_{p}T}}}{2v}}\\[2pt]&={\frac {dM}{M}}-{\frac {dv{\frac {M^{2}kRT}{c_{p}T}}}{2v}}\\[2pt]&={\frac {dM}{M}}-{\frac {dv[M^{2}(k-1)]}{2v}}\end{aligned}}}$$

Hence,

$${\displaystyle {\frac {dv}{v}}={\frac {2}{2+M^{2}(k-1)}}{\frac {dM}{M}}\qquad \qquad (1.7)}$$

Combining (1.4) and (1.7)

$${\displaystyle {\frac {d\theta }{2{\sqrt {M^{2}-1}}}}=-{\frac {1}{2+M^{2}(k-1)}}{\frac {dM}{M}}\qquad \qquad (1.8)}$$

We generally apply the above equation to negative ${\displaystyle d\theta }$, let ${\displaystyle d\omega =d\theta }$. We can integrate this between the initial and final Mach numbers of given flow, but it will be more convenient to integrate from a reference state, the critical speed (${\displaystyle M=1}$) to Mach number ${\displaystyle M}$, with ${\displaystyle \omega }$ arbitrarily set to zero at ${\displaystyle M=1}$,

$${\displaystyle \int \limits _{0}^{\omega }d\omega =\int \limits _{1}^{M}{\frac {2{\sqrt {M^{2}-1}}}{2+M^{2}(k-1)}}{\frac {dM}{M}}}$$

Leading to Prandtl-Meyer supersonic expansion function,

$${\displaystyle \omega ={\sqrt {\frac {k+1}{k-1}}}\tan ^{-1}\left[{\frac {\sqrt {k-1}}{\sqrt {k+1}}}(M^{2}-1)\right]-\tan ^{-1}(M^{2}-1)}$$

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