Tell me, how is the thickness of insulation in the blind area and the floors on the ground calculated?

Blind area is concrete (or asphalt) strip, built around the perimeter of the building (on the outside of the walls) to protect its foundation from rainwater. The blind area must be at least 1 meter with an obligatory slope in the direction from the wall.

As you can see, blind area is not a walling with heat insulating properties. Therefore, investing insulation in it is not always advisable. However, in cases where there is a significant depth of soil freezing in winter, a deep basement room and soils swelling under the influence of negative temperatures, it is still better to warm the blind area, for example, with extruded polystyrene foam 5 cm thick. In this case the blind area will look like this :

But the floor laid on the ground, should be warmed only if the room in which it will be, is heated. In an unheated room, the floor on the ground should have only a flat surface and good waterproofing so that the ground water does not penetrate the basement or the ground floor. Recommended layers of such a “cake” for the floor you can see in SP 31-105-2002 section 5 “Foundations, basement walls, floors along the ground”.

Calculation of insulation for the floor on the ground can be performed as follows:

• Determine how deep (relative to ground level) the floor of your room will be.
• Determine the temperature of the soil in the winter at this depth, based on the data in the table (source: http://kotly-minsk.by/index.php?option=com_content&view=article&id=189:2012-06-11-10-11-36&catid= 3: 2012-03-08-22-20-08 & Itemid = 4). If the depth of the floor is between the values ​​indicated in the table, use the interpolation method to determine the temperature.
• Since there is no official method for calculating the thickness of the heat-insulating layer of the floor on the ground, you can use the 9.9., Given in SP 23-101-2004 "Design of thermal protection of buildings", making a few changes to it:
  • First determine the rate of thermal protection: 1. First the value we need is GOSOP (degree-day of the heating period) = (tvn - tgr.z.) * Zot.p., that is: (indoor temperature minus the winter temperature of the soil at the required depth) multiplied by the duration of the heating period ( see column 11 of table 1 SNiP 23-01-99 *);
    2. calculate the standard value of the heat transfer resistance according to the formula: Rн = 0.00045 * GSOP + 1.9 (the coefficients used in this calculation were taken as to overlap over the unheated underground). Rн is measured in m2 * C / W and is the norm of thermal protection for the floor over the ground.

• • Determine the thickness of the layers of your floor on the ground (for example, a layer of reinforced concrete, cement-sand screed on insulation, linoleum, or any other option), except for the layer of insulation (its thickness, we still need to determine). Determine the thickness and material of the basement wall or basement below ground level and separating the floor along the ground from the ground outside the building.
  • For each of these materials, find in the table D1 SP 23-101-2004 coefficient of thermal conductivity.
  • Calculate the thermal resistance for each layer (as well as for the building wall below ground level) using the formula: Rlayer = layer thickness (in square meters) / material thermal conductivity coefficient (W / (m × ° С)).
  • Fold all the resulting thermal resistance of the layers.
  • Add two factors to the calculated value: the heat transfer coefficients of the inner and outer floor surfaces. We take them equal, respectively, 1 / 8.7 = 0.115 (according to table 7 of SNiP 23-02-2003), and 1/23 = 0.043 (according to table 8 SP 23-101-2004), taking into account the most unfavorable conditions. As a result of these calculations, we obtain the value Ro - thermal resistance of all layers of the floor except for the layer of insulation.
  • Now it is necessary to determine the minimum value Rout (thermal resistance) of the insulation we need. To do this, from the value of Rн we subtract the value of Rо, that is, Rout = Rn - Rо.
  • Decide what kind of insulation you will use in this floor construction and determine its thermal conductivity coefficient using the table D1 SP 23-101-2004.
  • It remains the last action - to determine the required thickness of insulation: multiply the value of Rout by the coefficient of thermal conductivity of the selected material.

According to the above calculation, it is possible to determine the approximate thickness of the insulation around the perimeter of the building directly next to its outer walls (basement or basement walls). When moving from external walls to the center of the room, the temperature difference inside the room and under the building will decrease and, consequently, the required thickness of insulation will decrease.

An example of calculating the floor on the ground for Krasnoyarsk.

Initial data:

• the depth of the floor relative to ground level - 40 cm
• soil temperature in winter at a depth of 0.4 meters - -8.2 degrees,
• the heating period for Krasnoyarsk is 234 days,
• indoor temperature + 20 degrees
• floor and wall materials:

The name of the material	Layer thickness, m	Heat conductivity coefficient, W / (m × ° С)
Reinforced concrete screed	0.1 m	1,69
Insulation (Styrofoam)	Need to calculate	0,030
Cement-sand screed	0.05 m	0,58
Linoleum	0,004 m	0,38
Wall concrete	0.6 m	1, 69
Extruded polystyrene foam wall	0.05 m	0,030

Payment:

1. GOSP = (18 - (- 8.2)) * 234 = 6130.8
2. Norm: Rn = 0.00045 * 6130.8 + 1.9 = 4.6589 m2 * C / W.
3. Thermal resistance for wall and floor:
- concrete screed R1 = 0.1 / 1.69 = 0.0592,
- DSP R2 = 0.05 / 0.58 = 0.0862,
- linoleum R3 = 0.004 / 0.38 = 0.0105,
- wall, concrete R4 = 0.6 / 1.69 = 0.355,
- wall, polystyrene foam R5 = 0.05 / 0.030 = 1.6667.

4. Total thermal resistance: Ro = 0.0592 + 0.0862 + 0.0105 + 0.355 + 1.6667 + 0.115 + 0.043 = 2.3356 m2 * C / W.
5. The difference between the norm of thermal protection and the resulting value of the total thermal resistance will be: Rout = 4.6589 - 2.3356 = 2.3233 m2 * C / W
6. Insulation thickness (extruded polystyrene foam): 2.3233 * 0.030 = 0.0697 m.

According to the calculation, for Krasnoyarsk, the thickness of insulation (expanded polystyrene) for the floor over the ground at a depth of 40 cm from the ground level should be 7 cm. Increasing the insulation of the walls of a room can reduce the thickness of the insulation of the floor.