Single Slope Solar Still, Paper Numerical Validation, ANSYS Fluent CFD Simulation Training

Description

Description

The present problem simulates the surface evaporation process within a solar desalination system using ANSYS Fluent software. We perform this CFD project and investigate it by CFD analysis.

This simulation is based on the information of a reference article [Productivity estimation of a single-slope solar still: Theoretical and numerical analysis], and its results are compared and validated with the results in the article.

In general, evaporation is a surface process that occurs at any temperature on the liquid surface. In this simulation, single slope solar still is designed in which a certain amount of water with a certain height is present.

The sloping surface of the solar still is made of glass and is responsible for heat transfer from the sun to the water surface. The surface of the water rises as it receives heat from the solar radiation passing through the glass, resulting in surface evaporation.

The present model is designed in three dimensions using Design Modeler software. The present model is related to single slope solar still.

The length of the chamber is equal to 0.98 meters, and the height of the vertical wall of the chamber is equal to 0.47 meters so the angle of the sloping surface of the chamber is equal to 19 degrees.

The meshing of the model has been done using ANSYS Meshing software. The mesh type used for this geometry is structured, and the element number is 8200. Also, due to the nature of this project, the transient solver has been used.

Surface Evaporation Methodology

Initially, there is a certain water level inside the chamber where the static airflow is located. As the system starts, the water surface evaporates, and water vapor mixes into the air inside the solar still.

Therefore, to define three phases simultaneously, including liquid water, air, and water vapor, the multiphase mixture model must be used; Because in this three-phase current, the boundary of separation and differentiation between different phases is unclear.

Mass transfer is used to define the phase change process between liquid water and water vapor; A UDF function is used to define the surface evaporation process.

Also, to define the amount of heat caused by solar radiation, it is assumed that the glass wall has a constant temperature due to solar radiation. According to the article information, this simulation has been done in two different modes, and its results have been compared with the results of the article.

In the first case, the initial temperature of the water is equal to 40 ° C, and the constant temperature of the glass is equal to 30 ° C. In the second case, the initial temperature of the water is equal to 50 ° C, and the constant temperature of the glass is equal to 40 ° C.

Surface Evaporation Conclusion

At the end of the solution process, two-dimensional contours related to temperature, velocity, and mass transfer rate between two phases (surface evaporation rate), air volume fraction, liquid water volume fraction, and water vapor volume fraction were obtained.

As seen from the contours, the water surface in the solar still chamber evaporates over time. As a result, water vapor occupies part of the space above the water surface inside the chamber.

The validation of the present simulation is based on the results of Table 3 of the mentioned paper. The amount of surface evaporation rate in the two simulations is compared and validated with the model evaporation rate presented by the paper.

This value of the evaporation rate represents the maximum value of the mass transfer rate between the two phases of liquid water and water vapor.

In the first case, the glass temperature equals 30 ° C. The water temperature is equal to 40 ° C, while in the second case, the glass temperature is equal to 40 degrees Celsius, and the water temperature is equal to 50 degrees Celsius.

Error (%)	Present Simulation	Proposed Model @ Paper	Surface Evaporation
12.21	2.4053614e-5	2.74e-5	Case 1 : Tg = 30 ºC & Tw = 40 ºC
0.57	4.5052254e-5	4.53e-5	Case 2 : Tg = 40 ºC & Tw = 50 ºC