2. Benchmark solution#

2.1. Calculation method#

2.1.1. Calculating vapour pressure from temperature#

We assume the linear saturation curve. So it is written:

eq 2.1.1-1

The equation [éq 2.2.3.3-2] from the reference document [R7.01.11] then gives:

Eq 2.1.1-2

We write that the total mass of water and the total mass of air are preserved (because there is no water or gas flow at the edge) and we get:

Eq 2.1.1-3

Eq 2.1.1-4

[R7.01.11] [éq 4.1.4-1] also gives:

Eq 2.1.1-5

The coupling of equations [éq 2.1.1-3], [éq 2.1.1-4] and [éq 2.1.1-5], to which must be added the ideal gas equation for steam, dry air and dissolved air as well as Henry’s law is a highly non-linear system that we will solve in small disturbances, which allows it to be linearized.

All calculations done, we get:

Eq 2.1.1-6

2.1.2. Temperature calculation#

Equation [éq 3.2.4.3-1] from the reference document [R7.01.11] gives:

Eq 2.1.2-1

(since the other expansion coefficients are zero).

Equation [éq 3.2.4.3-2] gives:

eq 2.1.2-2

So we get:

Eq 2.1.2-3

In this problem,

is nothing but the heat provided per unit volume.

By calling

the total volume of the room and

its lateral surface and

the application time of the flows:

Eq 2.1.2-4

2.1.3. System to be solved#

Eq 2.1.3-1

					(calculated)



5,00E-01	-1,00E-12	3,00E+02	3,70E+03	2.50E+06	2,67E-02	1.00E+03

		(calculated)		l		(calculated)



2.20E+03	3,00E-01	2,93E+03	1.05E+03	4,18E+03	1.90E+03	2.78E+06





1.00E+06	10	6,00E+04	1.00E+06