# Continuity equation

In order to obtain the differential formulations, it is necessary to apply the divergence theorem from vector calculus. For a general vector quantity ${\mathbf{\Omega }}$, which is continuously differentiable, and for a control volume V, enclosed by a piecewise smooth control surface A, the divergence theorem states that

$\int_A {{\mathbf{\Omega }} \cdot {\mathbf{n}}dA} = \int_V {\nabla \cdot {\mathbf{\Omega }}dV} \qquad \qquad(1)$

The desired differential equations may be obtained by applying this relationship to the integral form of the basic laws.

Furthermore, since the control volume shape and size are fixed in time, Leibniz’s rule for the specific general quantity φ is also valid:

$\frac{d}{{dt}}\int_V {\rho \phi } dV = \int_V {\frac{{\partial (\rho \phi )}}{{\partial t}}} dV \qquad \qquad(2)$

The surface integral in eq. (4) may be converted to a volume integral by applying eq. (1) as follows:

$\int_A {\rho ({{\mathbf{V}}_{rel}} \cdot {\mathbf{n}})dA = \int_V {\nabla \cdot \rho {{\mathbf{V}}_{rel}}dV} } \qquad \qquad(3)$

Substituting eq. (3) into eq. (4) and considering eq. (2), the entire left-hand side of eq. (4) is included in a single volume integral, i.e.,

$\int_V {\left( {\frac{{\partial \rho }}{{\partial t}} + \nabla \cdot \rho {{\mathbf{V}}_{rel}}} \right)dV = 0} \qquad \qquad(4)$

The only condition that ensures that eq. (4) is true for any arbitrary shape and size of the control volume is that the integrand must be equal to zero, i.e.,

$\frac{{\partial \rho }}{{\partial t}} + \nabla \cdot \rho {{\mathbf{V}}_{rel}} = 0 \qquad \qquad(5)$

which is the differential form of the law of the conservation of mass.

Equation (5) can also be rewritten as

$\frac{{D\rho }}{{Dt}} + \rho \nabla \cdot {{\mathbf{V}}_{rel}} = 0 \qquad \qquad(6)$

where D / Dt is the substantial derivative (or material derivative) defined by

$\frac{D}{{Dt}} \equiv \frac{\partial }{{\partial t}} + {{\mathbf{V}}_{rel}} \cdot \nabla \qquad \qquad(7)$

For a stationary reference frame, Vrel = V, and eq. (6) becomes

$\frac{{D\rho }}{{Dt}} + \rho \nabla \cdot {\mathbf{V}} = 0 \qquad \qquad(8)$

For incompressible flow, in which the density of the fluid is constant ($\rho \equiv {\rm{const}}$), eq. (8) simplifies as

$\nabla \cdot {\mathbf{V}} = 0 \qquad \qquad(9)$

For steady-state compressible flow, the continuity equation can be obtained by simplifying eq. (5), i.e.,

$\nabla \cdot \rho {{\mathbf{V}}_{rel}} = 0 \qquad \qquad(10)$

## References

Faghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Elsevier, Burlington, MA

Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.