rezistente termice in camp curent

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II. Thermal insulation design for the envelop elements The equation used for the thermal resistance estimation in the hypothesis of one direction thermal flux is: e k k i e k i ef o d  R  R  R  R α λ α 1 1 ,  + + = + + =   [m 2 K/W] (1) Thermal characteristics of the materials used in the construction of the envelope: Nr. Material Density ρ Thermal conductivity λ Thermal receptivity S m  Vapour permeability resistance 1 /K D  Kg/m 3  W/(mk) W/(m 2 k) - 0 1 2 3 4 5 1 Reinforced concrete 2500 1,74 16,25 21,63 2 Reinforced concrete 2400 1,62 15,36 21,3 3 Cement/lime plaster 1700 0,87 4,47 8,5 4 Plaster over insulation 1200 0.52 3.15 4.7 5 Mineral wool 100/130 0,04 0,50 1,3 6 Polystyrene 20 0,04 0,29 30,0 7 Cardboard plaster 700 0,23 3,13 3,4 8 Sand infilling 1600 0,58 7,50 3,9 10 Gravel infilling 1800 0,70 8,74 2,4 11 Aerated concrete (b.c.a.) masonry with slim joints 675 0,27 3,38 3,8 12 Aerated concrete strips 625 0,25 3,13 3,7 13 Fir timber 550 0,17 4,12 10,4 14 Oak and beech timber 800 0,23 5,78 11,3 15 PFL board 300 0,07 1,95 2,7 16 Bituminous cardboard 600 0,17 3,28 -

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Page 1: Rezistente Termice in Camp Curent

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II. Thermal insulation design for the envelop elementsThe equation used for the thermal resistance estimation in the

hypothesis of one direction thermal flux is:

ek 

i

ek ief o

d  R R R R

α λ α 

11,

  ++=++=   ∑∑   [m2K/W] (1)

Thermal characteristics of the materials used in the construction of theenvelope:

Nr. Material

Density

ρ 

Thermalconductivity

λ 

Thermalreceptivity

Sm 

Vapourpermeabilityresistance

1/KD 

Kg/m3  W/(m⋅k) W/(m2⋅k) -

0 1 2 3 4 5

1 Reinforced concrete 2500 1,74 16,25 21,63

2 Reinforced concrete 2400 1,62 15,36 21,3

3 Cement/lime plaster 1700 0,87 4,47 8,54 Plaster over insulation 1200 0.52 3.15 4.75 Mineral wool 100/130 0,04 0,50 1,3

6 Polystyrene 20 0,04 0,29 30,0

7 Cardboard plaster 700 0,23 3,13 3,4

8 Sand infilling 1600 0,58 7,50 3,9

10 Gravel infilling 1800 0,70 8,74 2,4

11 Aerated concrete(b.c.a.) masonry withslim joints

675 0,27 3,38 3,8

12 Aerated concrete strips 625 0,25 3,13 3,7

13 Fir timber 550 0,17 4,12 10,4

14 Oak and beech timber 800 0,23 5,78 11,315 PFL board 300 0,07 1,95 2,7

16 Bituminous cardboard 600 0,17 3,28 -

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Other necessary information :

Surface heat loss factors: αi and αe:

Working condition factor:

One requirement for the thermal resistance is:  Ro,ef ≥ Rnec 

Where the corrected mean resistance for building envelope are :R’o,nec= 1.4 for external wallsR’o,nec= 3 for terace roofR’o,nec= 1.65 for over basement

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 1. External walls type I (cross shear wall):

ek 

i

ek ief o

d  R R R R

α λ α 

11,

  ++=++=   ∑∑  

αi=8, αe=24

Nr. MaterialLayer

Thickness

Therm.conduct.

Therm.asim.factor

Heaeloss

resist.

Vapourconducti

v.

d λ  Sm  R 1/KD 

(m) (W/mK) (W/m2K) (m2K/W) -

1Internal Cement / limeplaster

2 Reinforced concrete

3 Polystyrene

4 External plaster

Total on element

External walls: type II

Nr. Material

Layer

Thickness

Therm.

conduct.

Therm.a

sim.factor

Heae

lossresist.

Vapour

conductiv.

d λ  Sm  R 1/KD 

(m) (W/mK) (W/m2K) (m2K/W) -

1Internal Cement / limeplaster

2

Cellular concrete(b.c.a.) or brickmasonry

3 Polystyrene

4 External plaster

Total on element

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2. Top floor (terrace roof) :

1- waterproofmembrane protection(mortar 3 cm + gravelor concrete boards);2 – waterproof

membrane;3 –diffusive layer;4 – waterproofmembrane support(mortar 3 cm);5 – plistyrene termo-insulation 20 cm;6 – vapor barrier;6’ – membrane support(mortar 3 cm)7 –diffusion layer;

7’ –membrane support(mortar 3 cm)8 – sloping layer(b.c.a.) 8 cm;9 – reinforced concretefloor slab – 12…15 cm;10 – ceiling plaster -1,5 cm;

3. Floor over the basement:

11

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floor over basement:1- ceiling plaster - 1,5 cm;2- thermal insulation ofpolystyrene – 10 cm;

3- reinforced concrete floor slab – 12…15 cm;4- mortar bead – 3 cm5 – Ceramic floor tile – 1.2 cm