written 5.7 years ago by |

*Area velocity relationship for compressible flow*

$\therefore A\times V=\text{constant}$

Above equation proof that as the area increases, velocity decreases.

But incase of compressible fluid, the continuity equation is given by

$\rho AV=\text{constant}..............(1)$

From this it is clear that with change in area, both velocity and density are affected.

$\therefore \rho d(AV)+AVd\rho=0$

$\rho[AdV+VdA]+AVd\rho=0$

$\rho AdV+\rho VdA+AVd\rho=0$

Therefore dividing by $\rho AV$, we get

$\dfrac {dV}V+\dfrac {dA}A+\dfrac {d\rho}\rho=0...................(2)$

Euler's equation for compressible fluid is,

$\dfrac {dP}{\rho}+VdV+gdz=0$

Neglecting the z term, the above equation is written as

$\dfrac {dP}\rho+VdV=0$

(Divide and multiply by $d\rho$)

$\dfrac {dP}{d\rho}\times\dfrac {d\rho}\rho+VdV=0$

But, $\dfrac {dP}{d\rho}=c^2$

Hence the above equation becomes,

$c^2\dfrac {d\rho}\rho+VdV=0$

$c^2\dfrac {d\rho}{\rho}=-VdV$

$\dfrac {d\rho}{\rho}=-\dfrac {VdV}{c^2}$

Substituting the value of $\dfrac {d\rho}\rho$ in equation (2), we get

$\dfrac {dV}V+\dfrac {dA}A-\dfrac {VdV}{c^2}=0$

$\dfrac {dA}A=\dfrac {VdV}{c^2}-\dfrac {dV}V$

$\dfrac {dA}A=\dfrac {dV}V\left[\dfrac{V^2}{c^2}-1\right]$

$\dfrac {dA}A=\dfrac {dV}V\left[M^2-1\right]$

**Important conclusions:-**

For M<1, the flow is sub-sonic and right-hand side equation is negative as ($M^2-1$) is negative for the value M<1.

For M>1, the flow is Super sonic. The value ($M^2-1$) will be positive and hence right hand side will be positive.

For m = 1, the flow is called Sonic flow. The value ($M^2-1$) is zero hence my right hand side of equation will be zero. $\left(\dfrac {dA}A=0\right)$.

This means area is constant.