Methodology for the Evaluation of Natural Ventilation in ... - Cham

Table 30. Percentage of Heat Loss through Advection for Various Window and StackConfigurations for Single Zone Model7 windows(0.0168 m 2 )5 windows(0.012 m 2 )2 windows(0.0048 m 2 )3 stacks (0.016875 m 2 ) 74.57% 73.45% 71.98%2 stacks (0.01125 m 2 ) 59.84% 67.05% 63.08%1 stack (0.005625 m 2 ) 39.02% 44.04% 41.71%In the three stacks case, the outlet area was always greater than the inlet area, causing the inletarea to restrict the airflow. In the single stack case, it was the outlet area that restricted the flowfor the 3 window through 7 window cases. Only when one opening area was significantlysmaller than the other was a reduction in airflow apparent, as in the 7 windows, 1 stack case.The numerical models created to simulate the experimental case for the single heated zone modelprovided information on some of the limitations with the PHOENICS software package. Initialruns of the simulation modeled the reduced-scale air model, using the exact dimensions andinterior heat loads (heaters) as the physical model. However, as discussed in Chapter 4, there isno conduction through the envelope of the model in PHOENICS. This was apparent in the initialrun when the resulting numerical model had much higher internal temperatures than observed inthe experimental model. Surface temperature measurements were recorded for the single zonemodel, and the resulting heat loss through the main surfaces of the model determined. It wasfound that the main heat loss was through the atrium walls, as much as 69 percent of the totalheat loss in some cases. The heat losses due to conduction through the model envelope for thephysical model are presented in Table 32.The heat losses were calculated using the surface temperatures measured on the exterior of themodel, the floor and ceiling temperatures on the interior of the model, and air temperatures ineach interior zone and the air surrounding the model (the ambient air). The material propertiesobtained from ASHRAE 2001 Fundamentals were used to calculate an overall heat transfercoefficient for each separate wall construction. Two wall constructions were identified and theheat transfer coefficient and surface area determined (Table 31). From the heat transfercoefficient, surface area and temperature measurements, the relative heat loss for each surfacewas calculated.Table 31. Material Properties and Surface Areas for Heat Loss CalculationMaterialsU-Value Total Surface Location(W/m 2 K) Area (m 2 )1-layer Insulation board, MDF 0.38 7.2 All Walls, and Atrium1-layer Insulation board, MDF, batt-insulation 0.37 5.2 Ceiling and FloorThe atrium had the most surface area, primarily due to the north wall, and accounted for 34percent of the total surface area. However, the floors and ceilings of the heated zone were betterinsulated, and so contributed less to the overall heat loss for the model.123