The optimal decision of insulation in cladding structures for energy efficient buildings

Execution of thermal insulation for enclosing structures of residential properties is a sophisticated part of the thermal protection. The choice of insulation materials for external walls is one of the main ways to increase the performance of heat insulation. This article contains thermotechnical calculations of two enclosing structures due to conditions of Saint-Petersburg. According to the results, the thickness of the insulating layer and a total wall thickness of the exterior walls composed of concrete, insulation and sand-cement rendering are smaller than the same characteristics of wall construction of hollow tile, insulation and sandcement rendering.


Introduction
The implementation of thermal insulation for enclosing structures of residential facilities is the major area of thermal protection.Firstly, insulation allows using fuel and energy resources efficiently because of the ability of the supporting structures to keep heat inside the house.The right choice of insulation increases the comfort of living in any period of time [1].
Contemporary design and advanced technology allow engineers to meet all the requirements of thermal insulation process ant to provide heat-saving efficiency.
There are the key methods of thermal protection of cladding structures: insulation of external walls, sandwich panels for exterior structures and internal insulation.The choice of insulation materials for external walls is an essential part of the design of the thermal protection [5][6][7].
The purpose of this article is to compare the most commonly used wall constructions with using of extruded foam polystyrene as insulation with the aid of thermo-technical calculations of outer walls [8][9][10][11].
The insulation layer in these constuctions is the material "PENOPLEX WALL".It is an eco-friendly, water-resistant heater with the highest ability for a heatshield.This material has the nominal size 1200х600х (20; 30; 40; 50; 60; 80; 100) mm.
These are the following tasks of the article: • To make the thermo-technical calculations of two outer walls constructions; • To carry out a comparative analysis of wall constructions; • To conclude about the efficiency of cladding structures.We made the thermo-technical calculations in accordance with the requirements of the following regulations [12,13]: • Building code 23-02-2003 "Thermal protection of buildings"; • Building code 23-01-99 "Construction climatology"; • Code of conduct 23-101-2004 "Design of thermal protection of buildings".

Experimental section
The data: residential building in Saint-Petersburg.Relative humidity φ int = 55%.Mean air temperature in the building: t int =20 °C.The type of enclosing structure: external walls.The task is to determine the required thickness of the insulation layer.Humidity conditions during the cold period of the year depends on the relative humidity φ int = 55% and the inside air temperature t int = 20 °C.According to table 1 [2], they are normal.
It is necessary to determine heat transmission resistance of cladding structure to calculate the required thickness of the insulating layer due to the requirements of sanitary code and energy efficiency [14]. 1) Firstly, we should determine the required heat transmission resistance R req on the basis of sanitary-hygienic conditions according to the formula: where t int -the inside air temperature, °C, t int = 20 °C; t ext -the average outside air temperature, °C, for Saint-Petersburg t ext = -26 °C [3]; n -coefficient which takes into account the dependence of the position of the enclosing structure to the outer air [2], n= 1; α int -coefficient of heat transfer of the internal surface of enclosing structures, W/(m•°С) [2], α int = 8.7 W/(m•°С); Δt n -temperature gradient, °C [2], Δt n = 4.0 °C.Then R req = 1.32 m 2 °С/W 2) Next step is to define required reduced total thermal resistance R req due to the conditions of energy efficiency [15][16][17]: SPbWOSCE-2016 6020 where t htthe average outside air temperature, °C [3].For the season with the average daily temperature of outer air not above 8°С t ht = -1.8°С.
z htthe duration of the heating period, days [table 1, 3].For the season with the average daily temperature of outer air not above 8 °С z ht = 220 days.

D b = 4796 °С•days
Then the required reduced total thermal resistance R req : R req = 3.079 m 2 °С/W 3) The normative heat protection.We use the maximum value of reduced total thermal resistance 3.079 m 2 •°С/W.4) The area of humidityhumid [18,19], we choice the thermotechnical characteristics of materials of cladding structures under conditions B [2].We should calculate the thermal resistance for each layer of wall construction using formula: The first layer (concrete D400): R 1 = 2 m 2 •°С/W.The third layer (sand-cement rendering): R 3 = 0.00753 m 2 •°С/W.The evaluation of minimal required thermal resistance of insulation [4]:

Results and Discussion
The value of reduced total thermal resistance is bigger than the value of required resistance.That is why this enclosing structure meets the heat transmission requirements [20].
The thickness of insulation is sizable for this wall construction.

Conclusion
The calculations showed that the wall construction composed of concrete, insulation and cement-sand plaster has smaller thickness of the insulating layer and lower total wall thickness and greater thermal resistance than the wall construction of the hollow brick, insulation and cement-sand plaster.In this case, it is recommended to use the first option while the design of the building in Saint-Petersburg.

Table 1 .
The layers of construction no.1.

Table 2 .
The layers of construction no.2.

Table 3 .
The results of analogical calculations for the second construction are in table3.matecconf/201 106 Characteristics of wall constructions no.1, no.2.