This example consists of a volume with a carbon dioxide concentration that corresponds to 1.519E-3 kg/kg, which is equal to 1000 PPM. There is a fresh air stream with a carbon dioxide concentration of about 300 PPM. The fresh air stream is such that the air exchange rate is about 5 air changes per hour. After 1 hour of ventilation, the volume's carbon dioxide concentration is close to the concentration of the fresh air.
The nominal value for the trace substance is set to C_nominal={1.519E-3}.
This scales the residual equations that are used by the solver to the right order
of magnitude.
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