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The book "Gas Turbine Engineering Handbook" by Boyce, on p. 117 states, the continuity equation,
$ \dot m = \rho A V $,
has as its differential form,
$ 0 = \frac{d\rho}{\rho} + \frac{dA}{A} + \frac{dV}{V}$.
How did the second equation get derived from the first?
for $y = f(x_1, ..., x_n)$, the sum of the partial differentials with respect to all of the independent variables is the total differential: $$ dy = \frac{\partial y}{\partial x_1}dx_1+...+\frac{\partial y}{\partial x_n}dx_n $$
For our case: $$ \dot{m} = \rho Av$$ $$ d\dot{m} = \frac{\partial \dot{m}}{\partial \rho}d\rho + \frac{\partial \dot{m}}{\partial A}dA+ \frac{\partial \dot{m}}{\partial v}dv$$ $$\frac{\partial \dot{m}}{\partial \rho} = vA$$ $$\frac{\partial \dot{m}}{\partial A} = \rho v$$ $$\frac{\partial \dot{m}}{\partial v} = \rho A$$
Substituting: $$ d\dot{m} = Av \ d\rho + \rho v \ dA+ \rho A \ dv $$
for steady-state case, $\dot{m} = \text{const}$, so $d\dot{m} = 0$, then: $$ Av \ d\rho + \rho v \ dA+ \rho A \ dv = 0 $$
Divide by $\rho A v$: $$ \boxed{\frac{d\rho}{\rho} + \frac{dA}{A}+ \frac{dv}{v} = 0}$$
the conservation of mass equation $ \dot m = \rho A V $ may be differentiated to obtain
$A V d\rho + \rho V dA + \rho A dV = 0 $
Which upon division by $ \rho A V $ yields
$ 0 = \frac{d\rho}{\rho} + \frac{dA}{A} + \frac{dV}{V}$.
The method for the first step is "partial derivatives". Explaining this method is outside the scope of the question I believe, but see this screenshot from Wolfram as proof. 
