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Consider a 1D adiabatic room (of volume V) with a wall conducting heat on its left side. The wall is solid and heat conduction in it is driven by the common heat equation. There is convection at the interface between the wall and the room.

  1. How to follow the room temperature over time using finite difference method?

  2. Does it make sense to consider the temperature uniform in the room?

  3. If so, how to use the convection coefficient as the difference $ T_{interface}(t) - T_{room}(t) $ would be zero?

  4. Using Tomáš Létal (asnwer) method and discretising it, would it be: $\frac{dT_r(t)}{dt}\cdot c_p = \left(T_w-T_r(t)\right)\cdot \alpha\cdot A \rightarrow T_r^{t+1} = T_r^{t} + \frac{dt\cdot\alpha\cdot A}{c_p}\cdot(T_N^t - T_r^t)$

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    $\begingroup$ Convection is driven by a temperature difference. Where is one? $\endgroup$
    – Solar Mike
    Oct 30, 2022 at 19:00
  • $\begingroup$ You should list the common heat equation, but will there be any heat flow if the spaces on either side of the wall are at the same temperature? $\endgroup$ Oct 30, 2022 at 19:29
  • $\begingroup$ @StainlessSteelRat Yes, that's my problem. But if the room is "long enough" so that we can consider the temperature drop (or rise) near the wall to be relatively small (so that the temperature of the room is uniform), when using the equation ... = h(T(interface, t+dt) - T(room, t+dt)) for a time step, as T(room, t+dt) is not a constant (this is what I want to solve) and T(interface, t+dt) is unknown, what should I do? $\endgroup$ Oct 31, 2022 at 10:50
  • $\begingroup$ You need to put that in your question. The site supports mathjax. Put your equations in. The more speciic you are the easier it will be for someone to help you. $\endgroup$ Oct 31, 2022 at 15:42

1 Answer 1

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You can simplify this, so you don't need finite difference method. If you start with room temperature $T_{r0}$, constant wall temperature $T_w$ and constant convection coefficient $\alpha$, the instant heat flux from room to wall $\dot{Q}_{r\rightarrow w}$ should be:

$$\dot{Q}_{r\rightarrow w} = \left(T_r(t)-T_w\right)\cdot \alpha\cdot A$$

The room has to have heat capacity $c_p$, which will make its temperature dependent on the transferred heat:

$$\frac{dT_r}{dt}\cdot c_p = -\dot{Q}_{r\rightarrow w}$$

Combining these 2 equations: $$\frac{dT_r(t)}{dt}\cdot c_p = \left(T_w-T_r(t)\right)\cdot \alpha\cdot A$$

This is simple differential equation, you can separate the variables and integrate:

$$\int\limits_{T_{r0}}^{T_r(t)} \frac{1}{T_w-T_r(t)} dT_{room} = \frac{\alpha\cdot A}{c_p}\cdot \int\limits_{0}^t 1 dt$$

Resulting in:

$$\ln\left(\frac{T_w-T_r(t)}{T_w-T_{r0}}\right) = -\frac{\alpha\cdot A}{c_p}\cdot \left(t-t_0\right)$$

Function for temperature:

$$T_r(t) = T_w+\left(T_{r0}-T_w\right)\cdot \exp\left(-\frac{\alpha\cdot A}{c_p}\cdot t\right)$$

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