Isentropic flow with area change

Speed of sound

$$a^{2}=\frac{\partial P}{\partial \rho}\Bigg|_{x}^{y}$$

 

As our flow condition is all isentropic, \(ds=0\)

$$dP=a^{2}d\rho\cdots(b)$$

 

Substitute equation (a) with (b), Then

$$\frac{dP}{\rho}+vdv=0$$

$$\frac{a^{2}d\rho}{\rho}+vdv=0$$

 

And

$$\frac{a^{2}d\rho}{\rho}+vdv=0$$

$$\frac{d\rho}{\rho}+\frac{v}{a^{2}}dv=0$$

$$\frac{d\rho}{\rho}+\frac{v^{2}}{a^{2}}\frac{dv}{v}=0$$

$$\frac{d\rho}{\rho}+M^{2}\frac{dv}{v}=0$$

 

Therefore,

$$\frac{d\rho}{\rho}=-M^{2}\frac{dv}{v}$$

 

Here, \(~\vec{v}=v\)

$$\frac{d\rho}{\rho}+\frac{dv}{v}+\frac{dA}{A}=0$$

$$-M^{2}\frac{dv}{v}+\frac{dv}{v}+\frac{dA}{A}=0$$

$$(1-M^{2})\frac{dv}{v}=-\frac{dA}{A}$$

$$\frac{dv}{v}=-\frac{1}{1-M^{2}}\frac{dA}{A}\cdots(a)$$

 

Apply this results on other equation (b)

$$\frac{dP}{\rho \vec{v}^{2}}=-\frac{v}{{v}^{2}}dv=-\frac{dv}{v}=\frac{1}{1-M^{2}}\frac{dA}{A}$$