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: Jacket agitated vessel

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.

  • $\begingroup$ This has one minor issue: The first equation is unsteady state. Then, it becomes steady state? The simple LMTD equation is true for shell+tube. The proper approach may need to consider the different geometry. The convective + radiative exchange term counts convection twice over and has nothing to do with radiation. Finally, in the same way that you used $U$ from reactor to jacket, consider using $U_x$ as an external transfer coefficient from jacket to air. $\endgroup$ May 15, 2019 at 0:16
  • $\begingroup$ may be relevant : engineering.stackexchange.com/q/80/10902 $\endgroup$
    – Solar Mike
    May 15, 2019 at 5:41
  • $\begingroup$ It can be unsteady state and I would recommend you to solve it as such but you will need to redefine the equations entirely since this is no longer an ordinary differential equation. This would be a partial one, and you want to solve for T. I honestly am too rusty (and lazy) to solve for this as a partial differential. You can check Trim's Applied Partial differential equations, I would try to tackle it as a Fourier series truncated to the second term, in case that not works redefine the system so it works as a Sturm-Liouville system. $\endgroup$
    – Media
    May 15, 2019 at 15:40
  • $\begingroup$ @JeffreyJWeimer thank you for considering my problem. I reformulating my whole question so that it doesn't seem confusing. I am clearly solving in unsteady state. The system is in still air and I read in the Perry's 8th Ed, that for human body in still air, radiation has half of the effect of convection. And in my equations $h_c$ and $h_r$ refer to $U$ but for convection and radiation (just matter of use). $\endgroup$
    – ParaH2
    May 15, 2019 at 17:40
  • $\begingroup$ @Media thank for your comment. Could you please show me how do you mean using an example to use the Fourier Series. Because to be honest I haven't done maths for years, at least not at that level and I don't really have a strong background on numerical methods as well. $\endgroup$
    – ParaH2
    May 15, 2019 at 17:45

1 Answer 1


Full Analysis

Perfectly Insulated System

The best case is to analyze this system in both the time and position domains. The time domain considers the variation of the reactor temperature $T_r(t)$ and the outlet temperature of the jacket system $T_{jL}(t)$. The position domain considers the variation of the temperature profile through the jacket $T_{j}(z)$. In this approach, you might hope for a set of differential equations that lend themselves to separation of variables for the jacket temperature $T_{j}(z,t) = T_{jz}(z) T_{jt}(t)$. The position domain analysis through the jacket at any instant in time then follows an approach similar to what is posted here for time or posted here for position. The temperature drops exponentially as a function of position along the jacket. The time domain presents a more difficult analysis. In the end, the position domain result is modified by an exponential of time with a respective time constant that contains a ratio of the mass flow and heat capacity of the fluid in the jacket $\dot{m}\tilde{C}_p$ to the overall coefficient and area of the reactor to jacket system $U_r A_r$.

Open to Air

The complexity of the analysis for the above system is increased when the jacket is not perfectly insulated.

Estimate Analysis

Perfectly Insulated System

We can estimate the system with one assumption. Assume the temperature profile of the jacket does not change with time. This is a gross assumption that has serious limitations on the accuracy. We can use it to first approximation to determine the orders of magnitude that parameters have to influence the answer.

To first approximation, the energy balance on the well-insulated system can be written as below to solve for the temperature in the reactor $T_r$ as a function of time $t$.

$$ \sum m \tilde{C}_p \frac{dT_r}{dt} = U_r A_r (T_r - <T>_j) $$

In this, $m$ and $\tilde{C}_p$ are the component masses and specific heat capacities in the reactor, $U_r$ and $A_r$ are the overall transfer coefficient and area of the reactor, and $<T>_j$ is a constant, average temperature for the jacket over its length and over a representative time frame. This equation is immediately solvable with the appropriate boundary condition $T_r(t=0)$. Be aware that this answer may predict that $T_r$ goes to a nonsensical result as $t \rightarrow \infty$.

Open to Air

To first approximation, the energy balance on the un-insulated system can be written as below.

$$ \sum m \tilde{C}_p \frac{dT_r}{dt} = U_r A_r (T_r - <T>_j) + U_j A_j (<T>_j - T_a)$$

In this, $U_j$ and $A_j$ are the jacket overall coefficient and area exposed to air while $T_a$ is the (constant) air temperature. Here again, the equation can be solved immediately with the appropriate boundary condition $T_r(t=0)$.

Making Predictions

Both solutions to the above are exponentials with time. They differ by a factor on the order of $\alpha \equiv U_jA_j(<T>_j - T_a)/\sum m \tilde{C}_p(T_r(t=0) - <T>_j)$. The perfectly insulated system has this factor as ZERO. Increasing values of $\alpha$ go away from perfect insulation. The system open to air has a defined value for this factor. An insightful approach is to plot the dimensionless temperature $\Theta \equiv (T_r - <T>_j)/(T_r(t=0) - <T>_j)$ as a function of time $t$ for variations in the ratio $\alpha$. You can then make slices horizontally (at constant $\Theta$) for given values of $\alpha$ to estimate the effect of improving the insulation.

The best approach might be akin to either of these:

  • I need at least a 20% improvement on time to heat or cool to make the use of insulation worthwhile.
  • The equations predict that I will gain only a 10% improvement on time.
  • $\Rightarrow$ I will make no changes


  • I need at least a 20% improvement on time to heat or cool to make the use of insulation worthwhile.
  • The equations predict that I may gain a 25% improvement on time.
  • $\Rightarrow$ I will plan further experimental bench top studies to validate the approximations

Unfortunately, I cannot immediately see a shortcut to estimate whether this first estimate will be consistently too low or too high compared to a "real" model. At that level, the project is better handed to a (chemical) engineer to tackle in a dedicated manner (rather than as a forum discussion).

  • $\begingroup$ Just to have a complement: with the equation "open to air", did you assume that we calculate <T_J> using the equation with insulation? Otherwise it boils done again to the mish-mash I obtained, doesn't it? $\endgroup$
    – ParaH2
    May 16, 2019 at 19:35
  • $\begingroup$ The system of equations should work either way, cool or heat. The sign of the temperature difference will change, making the reactor cool down or heat up with time accordingly. Keep $<T>_j$ as the same value in both cases (insulated or not insulated). In practice, this will not be true. But, in deeper or longer-term practice, these equations are valid anyway only as first estimates. I might comment in the statement accordingly. $\endgroup$ May 16, 2019 at 20:06

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