Question 11.2: Consider an infinitesimal CV filled with fluid as shown in Figu...

Figure 11.2 serves as the appropriate sketch. If the value of density at the cube center located at (x, y, z) is ρ, then the value on the near face is given by a Taylor series as

\rho _{near}=\rho +\left(\frac{\partial \rho }{\partial x} \right)\left(\frac{dx}{2} \right)+\frac{1}{2!}\left(\frac{\partial^2 \rho }{\partial x^2} \right)\left(\frac{dx}{2} \right)^2+\cdot \cdot \cdot

To the first order (since the cube is infinitesimal, we can safely assume that higher order terms are negligible), we have ρ_near = ρ + (∂ρ/∂x)(dx/2). Similarly, for velocity we find

u_{near}=u +\left(\frac{\partial u }{\partial x} \right)\left(\frac{dx}{2} \right),          v_{near}=v +\left(\frac{\partial v }{\partial x} \right)\left(\frac{dx}{2} \right)

and

w_{near}=w +\left(\frac{\partial w }{\partial x} \right)\left(\frac{dx}{2} \right)

The values on the far face are found to be

\rho _{far}=\rho -\left(\frac{\partial \rho}{\partial x} \right)\left(\frac{dx}{2} \right),          u _{far}=u -\left(\frac{\partial u}{\partial x} \right)\left(\frac{dx}{2} \right),          v _{far}=v -\left(\frac{\partial v}{\partial x} \right)\left(\frac{dx}{2} \right),

and

w_{far}=w-\left(\frac{\partial w}{\partial x} \right)\left(\frac{dx}{2} \right)

The same procedure is applied to the other four faces, using the appropriate spatial derivative and differential element in each case.

To apply a mass balance, we write \int_{CV} (∂ρ/∂t)dV + \int_{CS} ρ(u • n)dS = 0 and use the appropriate values of density and velocity to evaluate the integrals. Note that for an infinitesimal cube, the integrand in each volume or surface integral is constant. For example, the volume integral yields \int_{CV} (∂ρ/∂t)dV = (∂ρ/∂t)\sout{V} = (∂ρ/∂t)(dx dy dz), while each surface integral provides a term of the form \int_{CS} ρ(u • n)dS = ρ(u • n)A, where A is the area of the face. The value of ρ (u • n) A on each face involves only one velocity component, and it is convenient to consider faces in pairs. For example, consider the near and far faces. On the near face we have (u • n) = u_near , and the mass flux is

= (ρu)dy \ dz+\frac{1}{2} \left[\rho \left(\frac{\partial u}{\partial x} \right)+u\left(\frac{\partial \rho}{\partial x} \right)\right]dx \ dy \ dz+\left[\frac{1}{4} \left(\frac{\partial \rho }{\partial x} \right) \left(\frac{\partial u }{\partial x} \right)dx\right](dx \ dy \ dz)

= (ρu)dy \ dz+\frac{1}{2} \left[\rho \left(\frac{\partial u}{\partial x} \right)+u\left(\frac{\partial \rho}{\partial x} \right)\right]dx \ dy \ dz          to first order

On the far face we have (u • n) = − u_far, and the mass flux is

ρ(u • n)A = ρ_far(−u_far)(dy dz) = \left[\rho -\left(\frac{\partial \rho }{\partial x} \right)\left(\frac{dx}{2} \right)\right]\left[-u +\left(\frac{\partial u}{\partial x} \right)\left(\frac{dx}{2} \right)\right] (dy \ dz)

= (-ρu)dy \ dz+\frac{1}{2} \left[\rho \left(\frac{\partial u}{\partial x} \right)+u\left(\frac{\partial \rho}{\partial x} \right)\right]dx \ dy \ dz          to first order

The two faces combine to give a net mass flux of [ρ(∂u/∂x)+u(∂ρ/∂x)]dx dy dz. The other face pairs yield similar net mass fluxes of [ρ(∂v/∂y)+v(∂ρ/∂y)]dx dy dz and [ρ(∂w/∂z) + w(∂ρ/∂z)]dx dy dz.

Completing the mass balance we find:

+\left[\rho \left(\frac{\partial w}{\partial z} \right)+w\left(\frac{\partial \rho}{\partial z} \right)\right]dx \ dy \ dz=0

Dividing by (dx dy dz) and rearranging yields the continuity equation in the form of Eq. 11.2a:

\left(\frac{\partial \rho }{\partial t}+u\frac{\partial \rho }{\partial x}+v\frac{\partial \rho }{\partial y}+w\frac{\partial \rho }{\partial z} \right)+\rho \left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z} \right)=0

We see that it is possible to derive this equation in different ways.