2. Reference 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
[R7.01.11] equation [éq 3.2.1-2] then gives:
Eq 2.1.1-2
We write that the total mass of water is preserved (because there is no water flow at the edge) and we get:
Eq 2.1.1-3
[R7.01.11] equation [éq 4.4-1] also gives
Eq 2.1.1-4
The coupling of equations [éq 2.1.1-3] and [éq 2.1.1-4], to which must be added the ideal gas equation for steam, is a highly non-linear system that we will solve in a small disturbance, which allows it to be linearized.
All calculations done, we get:
Eq 2.1.1-5
2.1.2. Temperature calculation#
[R7.01.11] equation [éq 3.2.4.3-1] 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 2.1.2-5
2.2. Benchmark results#
The value of the temperature, liquid pressure and vapor pressure, solution of the system (10) is given with the data summarized in paragraph [§1.2] and recalled below. To calculate the calorific capacities, the following relationships are used:
, the latter relationship being true because the grain expansion coefficient is zero.
(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 |
100 |
6,00E+04 |
1.00E+06 |
After resolution, the following results are obtained:
425 |
|
l |
-1,4E+06 |
2E+00 |
|
2.3. Uncertainties#
The uncertainties are quite large because the analytical solution is an approximate solution due to the linearization of the equations.