Consider an inﬁnitesimal CV ﬁlled with ﬂuid as shown in Figure 11.2. Apply a mass balance to this CV to derive the continuity equation in Cartesian coordinates.

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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

[latex]\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 [/latex]

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

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

and

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

The values on the far face are found to be

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

and

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

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 [latex]\int_{CV}[/latex] (∂ρ/∂t)dV + [latex]\int_{CS}[/latex] 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 [latex]\int_{CV}[/latex] (∂ρ/∂t)dV = (∂ρ/∂t)[latex]\sout{V}[/latex] = (∂ρ/∂t)(dx dy dz), while each surface integral provides a term of the form [latex]\int_{CS}[/latex] 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

[latex]ρ(u • n)A = ρ_{near} (u_{near} )(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)[/latex]

[latex]= (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)[/latex]

[latex]= (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[/latex] 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) = [latex]\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)[/latex]

[latex]= (-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[/latex] 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:

[latex]\frac{\partial \rho }{\partial t}(dx \ dy \ dz)+\left[\rho \left(\frac{\partial u}{\partial x} \right)+u\left(\frac{\partial \rho}{\partial x} \right)\right]dx \ dy \ dz+\left[\rho \left(\frac{\partial v}{\partial y} \right)+v\left(\frac{\partial \rho}{\partial y} \right)\right]dx \ dy \ dz[/latex]

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

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

[latex]\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 [/latex]

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

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

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