The infinite thermal solver is used in case an infinite conductivity between the phases shall be assumed (or zero thermal resistance). This assumption results in two seperated phases beeing handled like one capacity in the thermal domain. A thermal resistance of 0 would result in infinite heat flows between the two phases at any temperature difference, and therefore the solver instead spreads all in- and outgoing heat flows of both phases over the two connected phases to ensure that their temperature remains identical. This is especially useful e.g. in the CDRA where a gas flow modeled with flow phases is in close thermal contact with the solid adsorber. In this case the large surface area between the two phases allows this assumption. The solver is discussed using the following rudimentary example:
The condition that no conductor is used in the thermal branch is used to identify infinite thermal solver, which are also assigned by the this.setThermalSolvers() function if this condition is true. In this example, the heat flows for the flow phase are calculated by the basic fluidic solver as discussed in Basic Thermal solver to 8.4 kW entering the flow phase and 0 kW leaving it. An actual flow phase would be handled as if no mass was present at all, but for this example we assume the phase is modeled as a normal phase with 1 kg mass. In this case the infinite solver is calculated after all other thermal solvers and uses the following calculation to spread the heat flows between the two connected phases (flow phase and solid phase):
| \dot Q_{tot} = 8.4 kW - 0 kW = 8.4 KW \\ \displaystyle \dot Q_{flow} = \frac{m_{flow} \cdot c_{p,flow}}{m_{flow} \cdot c_{p,flow} + m_{solid} \cdot c_{p,solid}} \dot Q_{tot} = \frac{1 kg \cdot 1000 J/kgK}{1 kg \cdot 1000 J/kgK + 10 kg \cdot 800 J/kgK} 8.4 kW = 0.9333 kW \\ \displaystyle \dot Q_{solid} = \frac{m_{solid} \cdot c_{p,flow}}{m_{solid} \cdot c_{p,flow} + m_{solid} \cdot c_{p,solid}} \dot Q_{tot} = \frac{10 kg \cdot 800 J/kgK}{1 kg \cdot 1000 J/kgK + 10 kg \cdot 800 J/kgK} 8.4 kW = 7.4667 kW |
If we now calculate the temperature change for both the flow and the solid phase we find that they are identical:
| \displaystyle \Delta \dot T_{flow} = \frac{\dot Q_{flow}}{m_{flow} \cdot c_{p,flow}} = \frac{0.9333 kW}{1 kg \cdot 1000 J/kgK} = 9.3330e-04 K/s\\ \displaystyle \Delta \dot T_{solid} = \frac{\dot Q_{solid}}{m_{solid} \cdot c_{p,solid}} = \frac{7.4667 kW}{10 kg \cdot 800 J/kgK} = 9.3330e-04 K/s\\ |
By inserting \dot Q_{tot}into the equation for the individual temperature change rate, it also becomes obvious that this approach is identical to handling both phases like one:
| \displaystyle \Delta \dot T_{flow} = \frac{\frac{m_{flow} \cdot c_{p,flow}}{m_{flow} \cdot c_{p,flow} + m_{solid} \cdot c_{p,solid}} \dot Q_{tot}}{m_{flow} \cdot c_{p,flow}} = \frac{\dot Q_{tot}}{m_{flow} \cdot c_{p,flow} + m_{solid} \cdot c_{p,solid}} |