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# Experiment of Building Health Loss

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Published: Thu, 31 Aug 2017

Chapter 3

CASE STUDY

For this case study a residence is selected in Kocaeli province of Turkey in order to determine the heat loss due to ventilation, infiltration and transmission by taking into consideration design conditions such as indoor and outdoor temperatures, U-values of the wall, window and floor.

3.1 Overview

The internal and external temperature differences of a building are critical parameters affecting the heat transfer. The heating and cooling energy requirements of buildings vary in parallel with the instantaneous change in indoor and outdoor conditions. Since the energy requirement alters depending on the ambient conditions, it is necessary to use a practical and applicable calculation method in building design. The heating energy requirement for a building is the minimum required energy for the heating system to maintain its internal environment at a specified comfort temperature within a year. The calculations to be made for heat loss are utilized to determine outward heat flow of the hot air inside the all spaces of dwelling. Accordingly, a heater (floor heating, combi, heat pump, etc.) is selected to meet the total heat loss according to the construction of the building.

3.2 Fundamental Assumptions for Case Study

The basic assumptions considered in the calculations are as follows:

• All calculations have been carried out considering the methodology described in the following sections.
• Outdoor air temperature, which is one of the design conditions, must be low enough to cover the worst weather conditions, but should not be selected from the low temperatures encountered for a short period of time. Meteorological records demonstrated that the most severe climatic conditions are not repeated every year. Therefore, outdoor air temperature is taken as -4 °C for Kocaeli provinence.
• Indoor air temperature is taken as 20°C for all rooms in a residence including, three bedrooms, two bathrooms, one kitchen and one living room. Because thermal comfort conditions are the determinants of the internal climate condition, it must be provided in terms of human health in the indoor environment and for the sake of not consuming excessive energy.
• Inner surface resistance and outer surface resistance of the building for wall surfaces, windows, and floor is 0.123 and 0.055 (m²âˆ™ °C /W).
• Thermal conductivities and thicknesses of the construction materials are determined for an insulated building. R- and U-values of walls, windows and floor is calculated because the residence is located on the ground floor.

3.3 Overall Building Heat Loss

The total heat loss () of a building is determined by the sum of transmission (fabric) heat loss () by conduction and convection, heat loss by infiltration () and heat loss () by ventilation.The general formula used for calculating of total heat loss and the sources of the heat losses are indicated through the equation (3.1) and Figure 1 below, respectively.

Figure 1 : House with arrows of heat loss

3.3.1 Transmission (Fabric) Heat Loss

Heat will flow through the structure toward lower temperatures in such a way that conduction and convection co-exist when a temperature difference exists between the internal and external sides of a structure. Total fabric heat loss is the sum of the heat losses through the building enclosure (i.e., the walls, roofs, ceilings, windows, doors, and floors) and it is expressed in equation (3.2) below.

where is transmission heat loss [W], is surface area [m2], is overall heat transfer coefficient [W/(m²·°C)], is internal air temperature [°C], is external air temperature [°C].

3.3.2 Heat Loss through Ventilation

Natural or mechanical ventilation is used to create a comfortable and healthy environment in buildings. The number of air changes must be determined to calculate heat loss through ventilation. The number of air changes in the building differ between natural and mechanical ventilation because of differences in components, tightness, and construction. To calculate the heat loss through ventilation, two different calculation methods are employed according to whether the ventilation is natural or mechanical. Ventilation heat losses are estimated for doors and windows in the rooms by use of the following equation (3.3).

where is ventilation loss through windows [W], is spesific heat capacity of air [kj/kg°C], is density of air [kg/m3], is air change rate [h-1], is internal air temperature [°C] and is external air temperature [°C].

The values for and are 20°C and 100 kPa, respectively, for calculating the heat loss through natural ventilation; however, these parameters can vary slightly, depending on the temperature and pressure. The enhancement in enthalpy between entering and leaving the system is also neglected.

According to the above explanations, the heat loss through natural ventilation in equation (3.5) can be calculated with the help of the following equation (3.4) by substituting = 1.184 kg/m3 and = 1.005 kJ/kg°C into equation (3.3).

=

3.3.3 Heat Loss through Infiltration

Outside air leaking into a dwelling causes the same amount of hot air to leak out. In this case, the cold outside air leaking into the room needs to be heated up to room temperature. Infiltration (air leaking) is called heat loss to the amount of heat required to heat the cold leaking air. Infiltration heat loss can be easily calculated by the following formula (3.6).

where is infiltration loss [W], is spesific heat capacity of air [kj/kg°C], is  number of air changes per hour, is volume of the room [m³], is internal air temperature [°C], and is external air temperature [°C].

3.4 Overall Heat Transfer Coefficient ( – value)

-value or -factor indicates the level of insulation of all insulation materials and varies with each material. The resulting value shows how much heat is transferred to the material to be used. It is the most important property expected from the insulation products, and the fact that it is low is one of the reasons of preference of the material. Being an important criterion in comparing different materials, it may not always possible to give the correct results considering the construction materials of the building to be implemented. The heat resistance () value must also be calculated in order to evaluate the performance of the application made or to be made (see Figure 2). Since the thermal insulation performance is also related to the thermal resistance () value, it is calculated by the ratio of the thickness () and the thermal conductivity value () of each material. -value can be obtained by using equations (3.6 and 3.7). -value has an inverse relationship to -value as can be seen in the equation 3.8.

Figure 2 : The thermal resistance network

where is the total resistance to heat transfer of the combination, expressed as

or the following formula is obtained by replacing with and with

where and are inside and outside surface resistances

3.4.1 Overall -Values for the Enclosure Sections

The cross-section view of the wall is depicted in Figure 3. The thicknesses and thermal conductivities of each layer forming the wall are shown in Table 2 below.

Figure 3 : Schematic cross-section of walls

Table 2 : Thermal and physical properties of the wall

 Material Thickness Thermal Conductivity R – value () () Surface Resistance Outside 0.055 Cement plaster with sand aggregate 0.02 0.72 0.028 Brickwork 0.075 0.84 0.089 Insulation (Glass wool) 0.05 0.034 1.47 Brickwork 0.075 0.84 0.089 Cement plaster with sand aggregate 0.02 0.72 0.028 Surface Resistance Inside 0.123

The cross-section view of double glazed window with argon filled is displayed in Figure 4. The thicknesses and thermal conductivities of each layer forming the window are shown in Table 3 below.

Figure 4 : Schematic cross-section of windows

Table 3 : Thermal and physical properties of window

 Material Thickness Thermal Conductivity R – value () () Outside Surface Resistance 0.055 Glass 0.004 0.96 0.004 Argon 0.010 0.016 0.625 Glass 0.004 0.96 0.004 Inside Surface Resistance 0.123

The cross-section view of floor is illustrated in Figure 5. The thicknesses and thermal conductivities of each layer forming floor are shown in Table 4 below.

Figure 5 : Schematic cross-section of the floor

Table 4 : Thermal and physical properties of the floor

 Material Thickness Thermal Conductivity R – value () () Surface Resistance Outside Laminate 0.007 0.13 0.054 Cement mortar 0.05 1.73 0.029 Extrude polystren foam 0.10 0.035 2.86 Unreinforced concrete 0.20 1.65 0.12 Surface Resistance Inside 0.123

3.5 Energy Loss Calculations

Before calculating the energy loss, the dimensions (length (L) and height (H)) of all facades of the building are identified as in Table 5 and 6. Since the temperatures of all the rooms are selected as the same (20°C), the dimensions of the walls and windows that do not lose heat are not subject to evaluation. Table 8 to 10 express transmission heat losses due to windows, walls, and floor. Table 11 and 12 show ventilation and infiltration heat losses through windows. Table 13 is created to give a general review of heat losses through all rooms.

Table 5 : Lengths and heights of walls by facades

 Room North South East West L H L H L H L H Kitchen 4,61 2,8 4,01 2,8 0 0 4,01 2,8 Living room 0 0 6 2,8 0 0 5,49 2,8 Bedroom 1 0 0 2,75 2,8 0 0 0 0 Bedroom 2 0 0 2,75 2,8 0 0 3,23 2,8 Bedroom 3 0 0 3,51 2,8 5,75 2,8 0 0 Bathroom 1 1,2 2,80 0 0 0 0 0 0 Bathroom 2 1,2 2,80 0 0 0 0 0 0

Table 6 : Lengths and heights of windows by facades

 Room North South East West L H L H L H L H Kitchen 0 0 0 0 0 0 1,4 1,4 Living room 0 0 2,4 1,4 0 0 2,4 1,4 Bedroom 1 0 0 1,4 1,4 0 0 0 0 Bedroom 2 0 0 1,4 1,4 0 0 0 0 Bedroom 3 0 0 1,4 1,4 0 0 0 0 Bathroom 1 0,4 0,6 0 0 0 0 0 0 Bathroom 2 0 0 0 0 0 0 0 0

Table 7 : Net area of walls by facades

 Room North South East West Kitchen 12,91 11,23 0 9,27 Living room 0 13,44 0 12,01 Bedroom 1 0 5,74 0 0 Bedroom 2 0 5,74 0 9,05 Bedroom 3 0 7,87 16,11 0 Bathroom 1 3,12 0 0 0 Bathroom 2 3,36 0 0 0

Table 8 : Transmission heat loss through windows

 Room U-value Area Ti To ΔT Qt (W) Kitchen 1,23 1,96 20 -4 24 57,86 Living room 1,23 6,72 20 -4 24 198,37 Bedroom 1 1,23 1,96 20 -4 24 57,86 Bedroom 2 1,23 1,96 20 -4 24 57,86 Bedroom 3 1,23 1,96 20 -4 24 57,86 Bathroom 1 1,23 0,24 20 -4 24 7,08 Bathroom 2 1,23 0 20 -4 24 0,00 Total (W) 436,90

Table 9 : Transmission heat loss through walls