# How to Derive the Differential Form of the Contintuity Equation

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.

• one thing is unclear to me. With respect to what do you differentiate each side of the equation? Do you use the grad / $\nabla$? Nov 18, 2020 at 9:41
• You differentiate with respect to a, b, and c, then add these together. Nov 18, 2020 at 9:47
• 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 . Nov 18, 2020 at 10:55
• Hopefully Algo's nicely formatted answer helps? Nov 20, 2020 at 9:11