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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?

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

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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. enter image description here

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  • $\begingroup$ one thing is unclear to me. With respect to what do you differentiate each side of the equation? Do you use the grad / $\nabla$? $\endgroup$ – NMech Nov 18 '20 at 9:41
  • $\begingroup$ You differentiate with respect to a, b, and c, then add these together. $\endgroup$ – Jonathan R Swift Nov 18 '20 at 9:47
  • $\begingroup$ so in turn you take partial derivatives in both sides of equation like $\frac{\vartheta}{\vartheta\rho}\frac{\vartheta}{\vartheta A}\frac{\vartheta}{\vartheta V}$ for both sides of the equation? That makes some sense, but still (simpler example to fit the comment but you can extend) if you do $\frac{\vartheta}{\vartheta\rho} \left(\frac{\vartheta}{\vartheta A} \rho A\right)=\frac{\vartheta}{\vartheta\rho} \left(\rho \right)=1$. Apologies if I am being pedantic, I see how the partial derivatives work, but I truly don't see the missing link which makes all fit together . $\endgroup$ – NMech Nov 18 '20 at 10:55
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    $\begingroup$ Hopefully Algo's nicely formatted answer helps? $\endgroup$ – Jonathan R Swift Nov 20 '20 at 9:11

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