I am currently trying to derive a model to help estimate as precisely as possible how much time can be saved by insulating the jacket of a reactor used for eating and cooling.
EDIT: I edited my post so that it is clearer for everybody, I understand it could have be quite confusing. I also corrected a few typos.
Presentation of the system
Let's consider the following reactor:
where, $T_{iJ}$ denotes the temperature at the jacket's inlet, $T_{oJ}$ denotes the temperature at the jacket outlet, $T_{surf}$ is the temperature at the surface of the jacket in contact with the ambient air and $T_{film} is the temperature very close to the jacket.
The temperature of the air far away from the jacket (compare to the width of the film at the surface of the jacket) will be denoted by $T_{\infty}$.
Energy balance with insulation of the jacket
The reactor is supposed to be perfectly adiabatic and operating as batch, therefore:
$$\left(\sum_i m_i c_{p_{i}}\right)\frac{dT_r}{dt} = UA \Delta T_{ln} = UA \frac{T_{iJ}-T_{oJ}}{\ln \left(\frac{T_{iJ}-T_r}{T_{oJ}-T_r}\right)} \tag 1$$
The temperature in the jacket can be averaged knowing the temperature at both inlet and outlet. If the temperature at the outlet is not know on the process and if the jacket is insulated, thus:
$$F_j c_{p_j}\left(T_{iJ} - T_{oJ}\right) = UA \Delta T_{ln} \tag 2$$
where, $F$ is the mass flow rate of water into the jacket.
which leads to,
$$T_{oJ} = T_r + \left(T_{iJ}-T_r\right)\exp\left(-\frac{UA}{F_j c_{p_j}}\right) \tag 3$$
Then $(1)$ can be directly solved. It can also be solved as equation $(4)$ using $T_j = (T_{iJ} + T_{oJ})/2$.
$$\left(\sum_i m_i c_{p_{i}}\right)\frac{dT_r}{dt} = UA (T_j - T_r) \tag 4$$
Energy balance without insulation of the jacket
If the jacket is not insulated, therefore $(2)$ becomes the following:
$$F_j c_{p_j}\left(T_{iJ} - T_{oJ}\right) = \underbrace{UA \Delta T_{ln}}_{\text{exchange jacket/reactor}} + \underbrace{h_c A_c \left(T_{film}-T_{\infty}\right) + h_r A_r \left(T_{surf}-T_{\infty}\right)}_{\text{convective and radiative exchange between the jacket and the room}} \tag 5$$
Rearranging the terms it leads to equation $(6)$ as follow,
$$\ln \left(\frac{T_{iJ}-T_r}{T_{oJ}-T_r}\right) = \frac{UA}{F_j c_{p_j}} + \frac{h_c A_c}{F_j c_{p_j}}\cdot\frac{T_{film}-T_{\infty}}{T_e - T_s} + \frac{h_r A_r}{F_j c_{p_j}}\cdot\frac{T_{surf}-T_{\infty}}{T_e - T_s} \tag 6$$
where $T_{film} = (T_{surf} + T_{\infty})/2$ and $T_{surf} = f(T_j)$ but can be set for this problem as $T_j - 3$.
Here comes the tricky part as the temperature we need to solve, $T_{oJ}$, depends on partially itself, $T_j$, and must be used during the numerical solving to calculate so solve $(1)$.
I have no idea on how to proceed to solve the whole system of equations $(1)$ and $(6)$.
I would be glad if you could bring some hints on either how to proceed to solve this messy equations or another method much easier and sufficiently precise to forecast the savings on the process.