# Flow term in Reynolds transport theorem

The basic form of Reynolds transport theorem for a fixed control volume can be written as $$\frac{D B_{sys}}{Dt}=\frac{\partial B_{CV}}{\partial t}-\dot{B_{in}}+\dot{B_{out}}$$

Why is the $\dot{B_{in}}$ term negative and the $\dot{B_{out}}$ term positive instead of the other way around? Doesn't a flow of $B$ into the system increment that property?

My textbook says "In the limit of time $\mathrm dt\to 0$, instantaneous change of $B$ in the system is the sum of change within, plus the outflow, minus the inflow." That would mean $\frac{D B_{sys}}{Dt}$ represents the rate of change of $B$ in the system as opposed to an external source term. The exact equation given there is $$\frac{\mathrm d(B_{sys})}{\mathrm dt}= \frac{\mathrm d}{\mathrm dt} \left(\,\int\limits_{CV} \beta \rho \,\mathrm d \upsilon\right) + \int\limits_{CS} \beta \rho V \cos \theta \,\mathrm dA_{out} - \int\limits_{CS} \beta \rho V \cos \theta \,\mathrm dA_{in}$$

Is the textbook wrong then?

It's just a matter of convention. The in and out terms are based on the dot product of the velocity vector and the surface normal, $\vec{v}\cdot\vec{n}$. These normals are always taken as the outward facing normals.

So, if you have a 1D control volume for example, the left side has a normal vector in the $-x$ direction and the right side has one in the $+x$ direction. If your velocity is in the $+x$ direction (meaning flow goes into the domain), then $\vec{v}\cdot\vec{n}$ on the left face is negative and on the right face is positive, so you get the signs you expect.

The other thing that would make it line up with your intuition is how you write it. The expression really is (ie. you didn't write an equation, you wrote a definition -- you need something on the RHS):

$$\frac{D B_{sys}}{Dt}=\frac{\partial B_{CV}}{\partial t}-\dot{B_{in}}+\dot{B_{out}} = S$$

where S is some source term (could be zero), which can be rearranged to be:

$$\frac{\partial B_{CV}}{\partial t} = \dot{B_{in}}-\dot{B_{out}} + S$$

and now it matches your intuition. Things flowing in will increase B in time, things flowing out will decrease it.

• My textbook says "In the limit of time $dt\rightarrow 0$, instantaneous change of $B$ in the system is the sum of change within, plus the outflow, minus the inflow. That would mean $\frac{D B_{sys}}{Dt}$ represents the rate of change of$B$ in the system as opposed to an external source term. The exact equation given there is $\frac{d(B_sys)}{dt}= \frac{d}{dt} (\int\limits_CV \beta \rho d \upsilon) + \int\limits_CS \beta \rho V cos \theta dA_out - \int\limits_CS \beta \rho V cos \theta dA_in$ – Skawang Jun 18 '18 at 13:31
• Latex doesn't seem to be working in comments so I've edited my question in reply – Skawang Jun 18 '18 at 13:39
• @Skawang I didn't include any external source terms in anything -- the $S$ I mentioned is the "change within". What you have written is just the definition of the substantial derivative -- $D/Dt = \partial /\partial t + \vec{v}\cdot \nabla$. The $S$ on the RHS in my equation is that "change within" they talk about. There's no inconsistency between what I have written and what the book says, as far as I can tell. – tpg2114 Jun 18 '18 at 14:07
• Isn't the substantial derivative used on points in a velocity field and particles? What do you mean when you use it on $B_{sys}$? Also, the textbook seems to use $\frac{\mathrm d(B_{sys})}{\mathrm dt}$ instead of $\frac{D B_{sys}}{Dt}$. So i thought it was referring to the normal derivative ,which had me confused. – Skawang Jun 25 '18 at 10:40
• Also, if particles flow into the control volume, that would mean $\frac{\mathrm d(B_{sys})}{\mathrm dt}$ is positive assuming steady state ($\frac{\mathrm d}{\mathrm dt} \left(\,\int\limits_{CV} \beta \rho \,\mathrm d \upsilon\right)=0$) but RHS is $- \int\limits_{CS} \beta \rho V \cos \theta \,\mathrm dA_{in}$ which is negative. How is that possible? – Skawang Jun 28 '18 at 9:18