Answer to Question #171029 in Math for Md Asif Iqbal

Question

Derive the following relation for flow through variable area duct:

( ) 2 = −1 dA dV M

A V

Where A is cross sectional area, V is velocity of flow and M is Mach number. Also discuss the

significance of this equation.

Answer 1

The area velocity realtionship for incompressible fluid is given by the continuity equation as

$A\times V = Constant$ .

From the above equation it is clear that with the increase of area, velocity decreases. But in case of compressible fluid equation is given by

$\rho\times AV = Constant$

Differentiating above equation we get,

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

or

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

Dividing by $\rho AV$ , we get

$\dfrac{dV}{V} + \dfrac{dA}{A} + \dfrac{d\rho}{\rho} = 0$

The Eulers equation for compressible fluid is given by,

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

Neglecting the Z term, Hence we can write the equation.

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

$\dfrac{dp}{d\rho} = C^2$

Hence the above equation becomes as

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

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

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

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

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

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