Question #171029

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.


1
Expert's answer
2021-03-15T11:18:10-0400

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

A×V=ConstantA\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


ρ×AV=Constant\rho\times AV = Constant

Differentiating above equation we get,


ρ[AdV+VdA]+AVdρ=0\rho [AdV + VdA] + AVd\rho = 0

or

ρAdV+ρVdA+AVdρ=0\rho AdV + \rho VdA + AVd\rho = 0


Dividing by ρAV\rho AV , we get


dVV+dAA+dρρ=0\dfrac{dV}{V} + \dfrac{dA}{A} + \dfrac{d\rho}{\rho} = 0

The Eulers equation for compressible fluid is given by,


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

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

dpρ×dρdρ+VdV=0\dfrac{dp}{\rho}\times \dfrac{d\rho}{d\rho} + VdV = 0


dpdρ=C2\dfrac{dp}{d\rho} = C^2

Hence the above equation becomes as


C2dρρ+VdV=0C^2\dfrac{d\rho}{\rho} + VdV = 0

dρρ=VdVC2\dfrac{d\rho}{\rho} = -\dfrac{VdV}{C^2}

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


dVV+dVAVdVC2=0\dfrac{dV}{V} + \dfrac{dV}{A} - \dfrac{VdV}{C^2} = 0


dAA=VdVC2dVV\dfrac{dA}{A} = \dfrac{VdV}{C^2} - \dfrac{dV}{V}


dAA=dVV[M21]\dfrac{dA}{A} = \dfrac{dV}{V}[M^2 - 1]


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