8.2: Mass Conservation

Fig. 8.1 The mass balance on the infinitesimal control volume.

Fluid flows into and from a three dimensional infinitesimal control volume depicted in Figure 8.1. At a specific time this control volume can be viewed as a system. The mass conservation for this infinitesimal small system is zero thus

$\label{dif:eq:RTT} \dfrac{D}{Dt} \int_V \rho dV = 0 \tag{1}$

However for a control volume using Reynolds Transport Theorem (RTT), the following can be written

$\label{dif:eq:RTTe} \dfrac{D}{Dt} \int_V \rho dV = \dfrac{d}{dt} \int_V \rho dV + \int_A U_{rn} \, \rho\, dA = 0 \tag{2}$

For a constant control volume, the derivative can enter into the integral (see also for the divergence theorem in the appendix A.1.2) on the right hand side and hence

$\label{dif:eq:RTTcvM} \overbrace{\int_V \dfrac{d\rho}{dt} \, dV}^{\dfrac{d\rho}{dt}\,dV} + \int_A U_{rn} \, \rho\, dA = 0 \tag{3}$

The first term in equation (3) for the infinitesimal volume is expressed, neglecting higher order derivatives, as

$\label{dif:eq:controlVolRho} \int_V \dfrac{d\rho}{dt} \, dV = \dfrac{d\rho}{dt} \,\overbrace{dx\,dy\,dz}^{\scriptsize dV} + \overbrace{f\left( \dfrac{d^2\rho}{dt^2\dfrac{}{}} \right) + \cdots}^{\sim 0} \tag{4}$

The second term in the LHS of equation is expressed as

$\label{dif:eq:mass2} \displaystyle \int_A U_{rn} \, \rho\, dA = \overbrace{dy\,dz}^{dA_{yz}}\,\left[ \left.\dfrac{}{\dfrac{}{}} (\rho\, U_x)\right|_x - \left.(\rho\,U_x)\right|_{x+dx} \right] + \\ \displaystyle \overbrace{dx\,dz}^{dA_{xz}}\left[\left.\dfrac{}{\dfrac{}{}}(\rho\,U_y)\right|_y -\left.(\rho\,U_y)\right|_{y+dy} \right] + \\ \displaystyle \overbrace{dx\,dy}^{dA_{xz}}\,\left[\left.\dfrac{}{\dfrac{}{}}(\rho\, U_z)\right|_z - \left.(\rho\,U_z)\right|_{z+dz} \right] \tag{5}$

The difference between point $$x$$ and $$x+dx$$ can be obtained by developing Taylor series as

$\label{dif:eq:mass3} \left.(\rho\,U_x)\right|_{x+dx} = \left. (\rho\, U_x)\right|_x + \left.\dfrac{\partial \left(\rho\, U_x\right) } {\partial x}\right|_x dx \tag{6}$

The same can be said for the $$y$$ and $$z$$ coordinates. It also can be noticed that, for example, the operation, in the $$x$$ coordinate, produces additional $$dx$$ thus a infinitesimal volume element $$dV$$ is obtained for all directions. The combination can be divided by $$dx\,dy\,dz$$ and simplified by using the definition of the partial derivative in the regular process to be

$\label{dif:eq:mass4} \int_A U_{rn} \, \rho\, dA = - \left[ \dfrac{\partial (\rho\, U_x) }{\partial x} + \dfrac{\partial (\rho\, U_y) }{\partial y} + \dfrac{\partial (\rho\, U_z) }{\partial z} \right] } \tag{7}$

Combining the first term with the second term results in the continuity equation in Cartesian coordinates as

$\label{dif:eq:continuityCart} \dfrac{\partial \rho }{\partial t} + \dfrac{\partial \rho\,U_x }{\partial x} + \dfrac{\partial \rho\,U_y }{\partial y} + \dfrac{\partial \rho\,U_z }{\partial z} = 0 \tag{8}$

Cylindrical Coordinates

Fig. 8.2 The mass conservation in cylindrical coordinates.

The same equation can be derived in cylindrical coordinates. The net mass change, as depicted in Figure 8.2, in the control volume is

$\label{dif:eq:cyl:dRhodt} d\,\dot{m} = \dfrac{\partial \rho}{\partial t} \overbrace{dr\,dz\,r\,d\theta}^{\scriptsize dv} \tag{9}$

The net mass flow out or in the $$\mathbf{\widehat{r}}$$ direction has an additional term which is the area change compared to the Cartesian coordinates. This change creates a different differential equation with additional complications. The change is

$\label{dif:eq:cly:massR} \left( \, \dfrac{}{\dfrac{}{}} \text{flux in $$\pmb{\mathbf{r}}$$ direction} \right)= d\theta\,dz\,\left( r\,\rho\, U_r - \left(r\,\rho\, U_r + \dfrac{\partial \rho\,U_r\,r}{ \partial r \dfrac{}{} } dr \right) \right) \tag{10}$

The net flux in the $$r$$ direction is then

$\label{dif:eq:cly:massRnet} \text{ net flux in the \pmb{\mathbf{r}} direction} = d\theta\,dz\,\dfrac{ \partial \rho\,U_r\,r}{ \partial r} dr \tag{11}$

Note that the $$r$$ is still inside the derivative since it is a function of $$r$$, e.g. the change of $$r$$ with $$r$$. In a similar fashion, the net flux in the $$z$$ coordinate be written as

$\label{dif:eq:cly:massZnet} \mbox{net flux in $$z$$ direction} = r\,d\theta\,dr\,\dfrac{ \partial \left( \rho\,U_z\right)}{ \partial z} dz \tag{12}$

The net change in the

$$\theta$$ direction is then $\label{dif:eq:cly:massThetanet} \mbox{net flux in \theta direction} = dr\,dz\,\dfrac{\partial\rho\,U_{\theta}}{\partial \theta} d\theta \tag{13}$

Combining equations (11) and dividing by infinitesimal control volume, $$dr\;r\,d\theta\,dz$$, results in

$\label{dif:eq:cly:massCombP} \text{total net flux } = -\left( \dfrac{1}{r}\dfrac{ \partial \left( \rho\,U_r\,r \right) }{ \partial r\dfrac{}{}} + \dfrac{ \partial \rho\,U_z\,r}{ \partial z} +\dfrac{\partial\rho\,U_{\theta}}{\partial \theta} \right) \tag{14}$

Combining equation (14) with the change in the control volume divided by infinitesimal control volume, $$dr\;r\,d\theta\,dz$$ yields

Continuity in Cylindrical Coordinates

$\label{dif:eq:continuityCyl} \dfrac{\partial \rho }{\partial t} + \dfrac{1}{r}\dfrac{\partial \left(r\,\rho\,U_r\right) }{\partial r} + \dfrac{1}{r} \dfrac{\partial \rho\,U_{\theta }}{\partial \theta} + \dfrac{\partial \rho\,U_z }{\partial z} = 0 \tag{15}$

Carrying similar operations for the spherical coordinates, the continuity equation becomes

Continuity in Spherical Coordinates

$\label{dif:eq:continuitySph} \dfrac{\partial \rho }{\partial t} + \dfrac{1}{r^2}\dfrac{\partial \left(r^2\,\rho\,U_r\right) }{\partial r} + \dfrac{1}{r\,\sin\theta} \dfrac{\partial \left(\rho\,U_{\theta } \,\sin\theta\right)} {\partial \theta} + \dfrac{1}{r\,\sin\theta} \dfrac{\partial \rho\,U_{\phi} }{\partial z} = 0 \tag{16}$

The continuity equations (8) and can be expressed in different coordinates. It can be noticed that the second part of these equations is the divergence (see the Appendix A.1.2 page Hence, the continuity equation can be written in a general vector form as

Continuity Equation

$\label{dif:eq:continuityV} \dfrac{\partial \rho }{\partial t} + \boldsymbol{\nabla} \cdot \left( \rho \,\pmb{U}\right) = 0 \tag{17}$

$\label{dif:eq:massIndexNotation} \dfrac{\partial \rho }{\partial t} + \dfrac{\partial \left(\rho\,U\right)_i }{\partial x_i} = 0 \tag{18}$
Where $$i$$ is is of the $$i$$, $$j$$, and $$k$$. Compare to equation (8). Again remember that the meaning of repeated index is summation.