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Influence Of Vent Center Elevation Difference On Flow Field Under Transient Thermal Pressure Natural Ventilation I

  • source:metmac;
  • Time:11/19/2020

Natural ventilation is an important means to improve the indoor air environment. This ventilation method does not require any energy consumption and can be used to a certain extent.

To meet people’s psychological needs of being close to nature. Therefore, in HVAC

The industry faces the two major issues of saving building energy consumption and improving indoor air quality

Under the challenge, natural ventilation methods have also received more attention.

 

Based on the existing transient thermal pressure natural ventilation theory, this paper adopts the computational fluid dynamics (CFD) method to change the center spacing of the vents under the condition of the indoor heat source characteristics and the size of the vents. Numerical simulation of ventilated rooms is carried out to obtain the flow field under different working conditions, and the influence of vent center height difference on instantaneous ventilation volume and vertical temperature difference is analyzed, and the airflow characteristics of vent center height difference on transient thermal pressure natural ventilation are summarized. The law of influence.

 

1 Transient thermal pressure natural ventilation model

Due to the complexity of the transient thermal pressure natural ventilation process, there are few theoretical studies on this aspect, and the more representative one is the theoretical model given in the literature [2]. This model studies the transient development process of thermal pressure natural ventilation under the same initial indoor and outdoor temperature, vents are set on the ceiling and bottom of the room, and the heat source is located on the indoor floor. This model is based on the two-zone flow model of PF Linden [3]. The two-zone model believes that during the ventilation process, a rising plume is formed above the heat source, which rises to the ceiling and spreads horizontally, gradually forming hot air in the upper part of the room. Floor. The hot air inside the room is affected by the thermal buoyancy and is discharged from the upper vent, while the outdoor cold air flows in from the lower vent. The interior is divided into two parts, the upper hot air layer and the lower cold air layer. The Kaye & Hunt model assumes that the upper hot air layer is fully mixed during the ventilation process, and the air temperature is evenly distributed. As the ventilation process continues, the hot air layer becomes thicker and thicker, and the thermal stratification interface gradually moves downwards until the steady state.

 

2 Physical model and numerical method

2.1 Physical model

The geometric size of the natural ventilation room simulated in this paper is 4 m × 3 m × 2.8 m, as shown in Figure 1. The air inlet and outlet are on the same side of the room.

Each inch is 1 m and 0.2 m. The indoor heat source is a centralized heat source located in the room

The size of the center position is 0.2 m and 0.2 m, and the heat source power is 100 W. The initial indoor and outdoor temperatures are both 293 K, and the ambient operating pressure is standard air pressure.

--1_1

This paper simulates the thermal pressure flow of 5 kinds of vent of air duct center height difference.

The wind process is used to analyze the influence of the center height difference of the vent of air duct on the ventilation process. The main parameters of the simulated operating conditions are shown in Table 1. Center height difference of inlet and outlet

  (Working condition

Set respectively as 0.8 m (condition 1), 1.2 m (condition 2), 1.6 m

3), 2.0 m (working condition 4), 2.4 m (working condition 5).

--2

2.2 Numerical calculation method

This paper uses FLUENT fluid mechanics calculation software for numerical calculation. The fluid studied is a three-dimensional continuous incompressible fluid. During the research process, the properties of the fluid are considered unchanged. The values of fluid physical parameters are shown in Table 2.

--3_1

The fluid governing equation is Navier Stokes equation, the mathematical model adopts RNG k-impression turbulence model, and the standard wall function is adopted near the wall.

Law processing. Since the flow field studied is a natural convection turbulent flow field, and

The main consideration is the influence of buoyancy caused by temperature difference, so Boussinesq hypothesis [4] is adopted. The surface of the indoor concentrated heat source is a wall with constant heat flow, and the wall and floor of the room are insulated and non-slip walls. Inlet and outlet setting

Is a pressure boundary condition, the relative pressure is 0, allowing the fluid to flow in all directions

In the process of numerical simulation, the inlet and outlet of the calculation area can be

Treated as fully developed turbulence.

The calculation domain uses a non-uniform grid, the grid at the heat source and the air inlet

The grid is denser, and the grid in other computational domains is sparse, which can reduce the grid size.

Divided nodes, saving calculation time. To avoid the low-level format

This simulated mass equation adopts the central difference format [5], the momentum equation adopts the QUICK format [5], the energy equation, k equation and the author

Cheng all adopts the second-order welcome style [5]. At the same time to ensure a good calculation process

Convergence, after repeated attempts, the under-relaxation factor selected in this paper is 0.8.

Numerical simulation of transient flow problems, the early stage of flow, by

If there is a high variable gradient in the flow field, a smaller time step should be used, and

There are a large number of inner loop iterations in each time step. When the flow field is fully developed

A larger time step can be used after the show, and each time step can be appropriate

Reduce inner loop iterations. Therefore, in the simulation process of this study,

The simulation is performed in a time step. The time step of the first 10 steps is 1 s, the time step of steps 11 to 20 is 5 s, the time step of steps 21 to 30 is 10 s, and the time step of the next 42 steps is 20 s, each time step long

There are 150 inner loop iterations in, after 1000 s, the time step becomes 50 s, and there are 50 inner loop iterations in each time step until the calculation is completed.

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