I Why is first TD law different for chemical reactors

[ussername]

I've learned that first thermodynamic law for some open system is in the form of:

where total change of system energy $\frac{\partial E}{\partial \tau }$ is equal to the transferred heat and work.
Total change of system energy $\frac{\partial E}{\partial \tau }$ is equal to the energy transferred via mass flux (surface integral) plus energy change within the system (volume integral).

Now I've seen energy balance of CSTR reactor and it is in the form of:

Full derivation is here or here.

I think $\frac{\partial E}{\partial \tau }$ in the bottom picture means total change of system (internal) energy, because further in the derivation it is replaced with total enthalpy differential and it is substituted with CSTR mass balance:
$$\frac{\partial n_{i}}{\partial \tau }=F_{i}^{0}-F_{i}+V\cdot \upsilon _{i}\cdot r_{V}$$
which makes sense only if $\frac{\partial E}{\partial \tau }$ is total change of energy.

If $\frac{\partial E}{\partial \tau }$ in the bottom picture means total change of system energy then it is not the same equation as in the first picture.

Can anybody explain?

 

[Chestermiller]

I have correctness problems with both the insets in your post, and one of the links. In my judgment and experience, the development that is done correctly is this one: http://jbrwww.che.wisc.edu/home/jbraw/chemreacfun/ch6/slides-enbal.pdf. If you wish to learn more about correctly-done energy balances (both for mechanical energy and thermal energy), see Bird, Stewart, and Lightfoot, Transport Phenomena.

 

[ussername]

My second picture is from the link you posted.

I had basically problem to understand where to put enthalpy total differential H=f(T,p,n):
$$\frac{\partial H}{\partial \tau}=\dot{H}_{out}-\dot{H}_{in}+\frac{\partial }{\partial \tau}\left (\int_{V}^{ }\frac{\partial H}{\partial m}\,dm \right )$$
First I thought that $\frac{\partial H}{\partial \tau}$ is the total enthalpy change of the system so the total differential H=f(T,p,n) should belong to there.
But the derivation says that total differential belongs to this term:
$$\frac{\partial }{\partial \tau}\left (\int_{V}^{ }\frac{\partial H}{\partial m}\,dm \right )$$and it seems partially logical because this integral adds elements of system enthalpy $dH$ along the system volume, so this integral is equal to the system enthalpy.

 

[Chestermiller]

My second picture is from the link you posted.

I had basically problem to understand where to put enthalpy total differential H=f(T,p,n):
$$\frac{\partial H}{\partial \tau}=\dot{H}_{out}-\dot{H}_{in}+\frac{\partial }{\partial \tau}\left (\int_{V}^{ }\frac{\partial H}{\partial m}\,dm \right )$$
First I thought that $\frac{\partial H}{\partial \tau}$ is the total enthalpy change of the system so the total differential H=f(T,p,n) should belong to there.
But the derivation says that total differential belongs to this term:
$$\frac{\partial }{\partial \tau}\left (\int_{V}^{ }\frac{\partial H}{\partial m}\,dm \right )$$and it seems partially logical because this integral adds elements of system enthalpy $dH$ along the system volume, so this integral is equal to the system enthalpy.

Are you saying that the equations in the second inset you posted are incorrect? Are you familiar with the derivation of the open system (control volume) version of the first law of thermodynamics?