The basic thermal solver is the standard thermal solver assigned to thermal branches by V-HAB in the this.setThermalSolvers(); function. It is split into two types, the basic_fluidic and the basic thermal solver. As the name suggests, the basic_fluidic solver is used to solve mass bound thermal energy transfer. The primary difference between the two solver is that the basic_fluidic only works together with a matter branch which models the mass transfer. The basic solver on the other hand can be used in a purely thermal branch as it models convective, conductive or radiative heat transfer, which does not necessarily move matter as well.
Basic Fluidic Solver
Mass bound energy transfer must be handled differently than other energy transfer because the movement of matter already represents movement of energy as well. For example, if we consider two capacities which are connected by a constant mass flowrate of 1 kg/s as shown below:
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Then the temperature of the left phase will not change due to the mass transfer, while the temperature of the right phase will change. The reduction in energy for the left phase is already represented by the reduction in mass for that phase, as the total thermal energy U within the phase can be (simplified) calculated with:
| U = m \cdot c_p \cdot T |
Since the mass change is already covered by the matter side of the V-HAB model, the thermal solver does not have to change anything for the left phase and the heat flow for the left phase is 0. However, the matter side for the right phase only covers the increase in mass. If the temperature of the mass entering the phase is different from the current temperature of the phase, the temperature must also change. Therefore, the basic fluidic solver does calculate a corresponding heat flow for the right phase. It may seem confusing at first that the solver has a heat flow of 0 for on phase and a non zero heat flow for the other phase, but as this example shows this behavior is actually correct. The basic calculation discussed in 3.2.2. Thermal Solvers is identical and the fluidic resistance is calculated by the fluidic conductor. The heat flow for the right phase is then calculated by the solver with the following equation:
| \dot Q = \frac{\Delta T}{R} =\Delta T \cdot \dot m \cdot c_p |
In the example from above this results in a heat flow of:
| \dot Q = (293 K - 273 K) \cdot 1 kg/s \cdot 4200 J/kgK = 8.4 kW |
We can check the energy balance of this calculation by considering the thermal energy of the right phase and of the transfered mass. For example, if a timestep of one second is considered these are:
| U_right = 10 kg \cdot 4200 J/kgK \cdot 273 K = 11.466 MJ ; U_transfered = 1 s \cdot 1 kg/s \cdot 4200 J/kgK \cdot 293 K = 1.2306 MJ |
And the resulting temperature of the right phase after this time step would be:
| T = \frac{U_{tot}}{m_{tot} \cdot c_p} = \frac{11.466 MJ + 1.2306 MJ}{(10 kg + 1 kg) \cdot 4200 J/kgK} = 274.8182 K |
In V-HAB the thermal domain is completely based on heat flows, to enable a common architecture. Therefore, we now compare the result if the calculated heat flow is used
| \Delta T = \frac{Q}{m \cdot c_p} = \frac{0.0048 W \cdot 1s}{10 kg + 1kg \cdot 4200 J/kgK} = 1.8182 K => T = 273 K + 1.8182 K = 274.8182 K |
It is important to note here, that the transfered mass must be included in calculation if the heat flow is used to achieve the correct result. For that reason, the matter domain is updated before the thermal domain, to ensure this is the case (see 5. Timer and Execution Order for more information regarding the calculation order in V-HAB).
If thermally active F2Fs are present in the branch (components which have a heat flow acting on the fluid passing through the branch) then the heat flow is simply added to the heat flow calculate above.
Basic Solver
The basic solver calculates all other types of thermal energy transfer for individual branches. Here it is necessary to differentiate between conductive/convective heat transfer and radiative heat transfer. Each branch can only solve one type of heat transfer, as radiative heat transfer is calculated with T^4. For these solvers, the heat flow added to one phase is removed from the other (similar to the handling of mass in the matter domain). We discuss this solver using the following example:
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The calculation for the conductive resistance R_{cond} depends on the specific conduction problem the user wants to solve and is therefore not discussed here. In this example the heat flow is calculated as follows:
| \dot Q = \frac{\Delta T}{R} = \frac{(293 K - 273 K)}{1 K/W} = 20 W |
The arrow indicates the positive flow direction of the solver. As the calculated heat flow is positive this is also the direction the heat flow has in this example. The thermal ExMes each have a sign (just like the matter ExMes) which is 1 for the right phase and -1 for the left phase, as a positive heat flow removes energy from the left phase and increases the energy in the right phase. With these signs the individual heat flows of the phases are calculated as -20 W for the left phase and 20 W for the right phase.