Expanded polystyrene (EPS), extruded polystyrene (XPS), glass wool and rock wool are still used intensively for insulation purposes in Turkey. The heat transfer coefficient varies depending on the density of the material, the pore structure, humidity and temperature. In our country, 4 different climatic zones have been determined according to TS 825. The provinces are divided into regions by taking into account the outdoor temperature, sunshine time and humidity values. Konya is located in the 3rd region and Sivas in the 4th region. The correct boundary conditions determined for the area where the building is located ensure the correct determination of the optimum insulation thickness. For this study, EPS, XPS, glass wool and rock wool plates with various density values were used as insulation.These samples were prepared in accordance with a 300x300 mm measurement equipment to determine their thermal conductivity (k). In accordance with the standard, the k value was measured on samples produced at an average temperature of 10 ºC. The thermal conductivity coefficient was determined using the "Heat Flow Meter, (HFM) Fox 314" equipment (Fig. 1). The thermal conductivity coefficient was measured using the standards TS EN 12664 (2009), TS EN 12667 (2003), and ASTM C518 (2003).
A Fox-314 HFM according to EN 12664 and 12667 standards was used to assess thermal conductivity. Starting at -10°C and working up to 50°C in 5°C intervals, measurements were taken. The samples used in the studies had a density ranging from 8.9 to 60 kg/m3. The thermal conductivity of all closed cell insulating materials increases as the temperature rises, like the findings in the referred literature [22]. The thermal conductivity of samples with lower densities increased more quickly as the temperature rose. In other words, low density means wider pores and a lot more air, which leads to higher k values. According to the findings, increasing density lowers k values for the same types of materials, which is consistent with previous research [22]. The LaserComp Heat Flux Transducer, created and produced exclusively for thermal conductivity measurements, is the key component of the measurement. To provide an accurate measurement of the total heat flow, the Heat Flux Transducer integrates throughout the whole active surface (100 × 100mm). Heat flow distortion is not possible because the overall transducer thickness is less than 1 mm. Each transducer has a type E thermocouple bonded in the middle, and both are sealed to maintain accuracy over the instrument's lifetime. The thermocouples give accurate readings of the sample's surface temperature (with a resolution of 0.01°C) and heat flow since they are all placed within 0.1 mm of the sample surface. The control of plate temperature likewise employs the same thermocouples. The uncertainty of the HFM type FOX-314 thermal conductivity measuring device is ± 3 to 5% and the reproducibility value is ± 0.5%.
In addition, the 3ω method is used to measure the thermal properties of building materials [23–24]. The thermal conductivity coefficients of the samples will be used to determine the optimum insulation thickness. The k values and average unit prices measured by the HFM method are given in Table 1. Optimum insulation thicknesses were determined by using these properties of the measured samples. For this purpose, an externally insulated wall structure consisting of 2 cm plaster, 20 cm thick brick and insulation material on its outer and inner surfaces is considered.
Table 1
Measured thermal conductivity coefficients and unit prices of insulation materials
| EPS | XPS | Glass wool | Rock wool |
ρ (kg/m3) | 32 | 32 | 100 | 150 |
Price [$/ m3] | 42 | 91 | 47 | 72 |
k (W/mK) | 0,0306 | 0,026 | 0,032 | 0,033 |
Air infiltration through exterior walls, windows, ceilings, and floors is the most common source of heat gain and loss in buildings. However, only the heat gains and losses on the outside walls were considered in this study, and the ideal insulation thickness was computed using the heat transmission coefficient, which changes based on densities. The heat gain and loss from the outside wall's unit surface are as follows [9, 20, 21].
\(\text{Q}=\text{U}.\varDelta \text{T}\) [1]
Here, U is the total heat transfer coefficient ΔT, the difference between the outdoor temperature changing throughout the day and the constant indoor temperature. In this case, the annual heat gain and loss from the unit surface depending on the degree-day number (DD) are given [9, 20, 21].
\({\text{q}}_{\text{A}}=86400. \text{D}\text{D}.\text{U}\) [2]
The total heat transfer coefficient of the wall can be written as:
U=\(\frac{1}{\text{R}\text{i}+{\text{R}}_{\text{w}}+{\text{R}}_{\text{y}}+{\text{R}}_{\text{o}} }\) [3]
Here, Ri and Ro are the thermal resistances of the indoor and outdoor environment, and Rw is the thermal resistance of the non-insulated wall layers. Ry is the thermal resistance of the insulation material and can be written as follows.
\({\text{R}}_{\text{y}}=\frac{{\text{x}}_{\text{y}}}{{\text{k}}_{\text{y}}}\) [4]
Here xy is the thickness of the insulation material and ky is its thermal conductivity. The total heat transfer coefficient, where Rwt is the total thermal resistance of the wall excluding the insulation material, is as follows [20–26].
\(\text{U}=\frac{1}{{\text{R}}_{\text{w}\text{t}+\frac{{\text{x}}_{\text{Y}}}{{\text{k}}_{\text{y}}}}}\) [5]
The heat transfer coefficient on the inner and outer surfaces of the wall was taken as 7,69 and 25 W/m2K, respectively, and Rwt = 0.6 m2K/W. By applying insulation to the outer walls of the buildings, heat gain and loss are significantly reduced. In this case, it is necessary to know the optimum thickness of the insulation in terms of energy saving. The optimum thickness of the insulation is the value that provides the minimum total cost, including the cost of insulation and the cost of energy consumption over the life of the building. Therefore, the optimum insulation thickness should be determined by performing a cost analysis. The annual energy costs of heating and cooling are as follows, respectively:
\({C}_{A,I}=\frac{86400.HDD{C}_{f}}{\left({R}_{wt}+\frac{{x}_{y}}{{k}_{y}}\right).{H}_{u}.\eta }\) [6]
\({C}_{A,S}=\frac{86400.CDD{.C}_{e}}{\left({R}_{wt}+\frac{{x}_{y}}{{k}_{y}}\right).COP }\) [7]
Here, HDD and CDD are heating degree-days and cooling degree-days, respectively, and are given in Table 2 for Konya and Sivas.
Table 2
Heating and cooling degree day value according to Konya and Sivas
Climate zone and city | Heating degree day value (HDD) | Cooling degree day value (CDD) |
3rd Zone-Konya | 2507 | 480 |
4th Zone- Sivas | 3003 | 538 |
\({\text{C}}_{\text{f}}, {\text{C}}_{\text{e}}, {\text{H}}_{\text{u}}, {\eta }\) and COP are the fuel price ($/kg), the price of electricity ($/kWh), the lower heating value of the fuel (J/kg), the efficiency of the heating system and the cooling performance coefficient, respectively. In this case, the total annual energy cost is written as follows.
\({C}_{A}= {C}_{A,I}+{C}_{A,S}\) [8]
The total cost of an insulated building is calculated using the following equation.
\({\text{C}}_{\text{T}}={\text{C}}_{\text{A}}.\text{P}\text{W}\text{F}+{\text{C}}_{\text{y}}.{\text{x}}_{\text{y}}\) [9]
Here \({\text{C}}_{\text{y}} \text{v}\text{e} {\text{x}}_{\text{y}}\) are the price ($/m3) and thickness of the insulation, respectively. CA is the sum of the annual heating and cooling costs per unit surface. When determining the optimum insulation thickness, the total heating cost over N-year life should be evaluated together with the present value factor (PWF). The PWF is calculated as follows depending on the interest rate (i) and the inflation rate (g) [11, 18, 22].
\(PWF=\frac{{\left(1+\text{r}\right)}^{\text{N}}-1}{\text{r}.{\left(1+\text{r}\right)}^{\text{N}}}\) , i>g \(\text{r}=\frac{\text{i}-\text{g}}{1+\text{g}}\) [10]
The insulation thickness that will minimize the total cost gives us the optimum insulation thickness. Accordingly, the optimum insulation thickness is obtained by taking the derivative of the equation that gives the total cost according to the insulation thickness (x), as follows.
\({x}_{opt}=293.94\sqrt{\frac{PWF.k}{{C}_{y}}\left(\frac{{C}_{f}.HDD}{{H}_{U}.\eta }+ \frac{{C}_{e}.CDD}{{\text{3,6.10}}^{6}.COP}\right)}-{k}_{y}.{R}_{wt}\) [11]
The payback period is calculated with the help of the following equation:
\({P}_{b}=\frac{\text{ln}\left[1-\left(\frac{i-g}{1+g}\right).\left(\frac{{C}_{y}.{x}_{y}}{{S}_{A}}\right)\right]}{\text{ln}\left(\frac{1+g}{1+i}\right)}\) [12]
Here, SA is the annual savings and is calculated from the difference between the annual energy costs of the uninsulated wall and the insulated wall as follows [9, 20, 21].
\({S}_{A}={\left({C}_{A}\right)}_{yalsz}-{C}_{A}\) [13]
The properties of the fuels used in the calculations and other parameters are given in Table 3.
Table 3
Parameters used in the calculations
NATURAL GAS | COAL | ELECTRICITY | |
Price [$/m3] | 0,214 | Price [$/kg] | 0,249 | Price [$/kWh] | 0,102 | RWT [m2 K/ W] | 0,6 |
Hu [J/m3] | 34,526*106 | Hu [J/kg] | 29,307*106 | HU [J/kWh] | 36,00648*105 | i [%] | %9,75 |
Yield [%] | 93 | Yield [%] | 65 | Yield [%] | %99 | g [%] | %8,5 |
| | COP | 2,5 | | | N [year] | 10 |