Hygrothermal Assessment

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    Hygrothermal assessment 1. Global thermal insulation coefficient (G)

    I. Calculus of geometrical characteristics of the buildingII. Calculus and verification of the overall thermal resistance (R-value)

    III. Calculus and verification of global thermal insulation coefficient (G)

    The building is placed in Suceava City, climatic area IV, having the external winter temperatureTe=-21 0 C (according to C107/3-2005, Annex D) and T i=20 0C

    The global thermal insulation coefficient G is computed according to C107/1-2005.

    The calculation method is based on the appropriate design thermal conductivities or design

    thermal resistances of the materials and products involved.

    Performance Requirements Criteria and Levels for Outer Building Elements:

    Overall Thermal Resistance (R) is higher than the Minimum Overall Thermal Resistance(R min -value) that is given for each type of element under inspection, indoor and outdoor environments, building destination, and placement.

    min R R

    The overall thermal transmittance (U), in W/(m 2*K), as a reciprocal value of the overallthermal resistance, is a measure of heat transfer through an element. Overall ThermalTransmittance (U) is not higher than the Maximum Overall Thermal Transmittance (U max -value) that is given for each type of element under inspection.

    ma x

    1U

    RU

    Minimum Overall Thermal Resistance and Maximum Overall Thermal Transmittance for building elements according to C107 for buildings designed after January 1 st 2011:

    Building element R min

    [m *K/W] Umax

    [W/(m *K)]External wall 1.80 0.56Woodwork 0.77 1.30Slab over last floor 5.00 0.20Slab over unheated semi-

    basement2.90 0.35

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    Symbols and units

    Symbol Quantity UnitA area mR design thermal resistance m *K/W

    R se external surface resistance m *K/WR si internal surface resistance m *K/W linear thermal transmittance W/( m*K) relative humidity of the air %

    i(e) indoor (outdoor) air temperature C

    drop (difference) of temperature CU thermal transmittance W/(m *K)

    d thickness m

    h heat transfer coefficient W/(m *K) design thermal conductivity W/( m*K)U Average Thermal Transmittance including

    Thermal Bridge LossesW/(m *K)

    R Average Thermal Resistance includingThermal Bridge Effect

    m *K/W

    I. Calculus of geometrical characteristics of the building

    1. The area of the envelope elements

    i. The area of the superior slab A roof = 376.3 m 2 ii. The area of the slab over the semi-basement A sb=376.3 m 2

    iii. The area of the external woodwork Type b h No. Area Perimeter

    m m b*h (m2) 2(b*h) (m)W1 0.6 0.8 16 7.68 44.8W2 0.8 1.15 16 14.72 62.4

    W3 1.6 1.55 8 19.84 50.4D1 1.6 2.15 16 55.04 120

    97.28 277.6Total for East Wall

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    2. The interior heated volume

    3. The compactness of the building (A total /V interior )

    A/V=3172.2/9896.7 =0.32 m2

    /m3

    II. Calculus of the overall thermal resistance (R-value)Calculating the air-to-air thermal resistance of a multilayer element, the resistances of the

    layers and those of the surfaces will be summarized:

    min

    11 R

    h

    d

    h R R R R

    e

    s

    i

    se s si

    where,

    R : overall thermal resistance, in m2

    .K/Wi and e and s indices : inner and outer surfaces, and material layer surface including any airspaceR si(se) : thermal resistances of a surface, in m 2*K/Wh i(e) : surface heat transfer coefficients, in W/(m 2*K).

    A thermal bridge is a part of the building envelope where the otherwise uniform thermalresistance is significantly changed by :

    full or partial penetration of the build ing envelope by materials with a different thermalconductivity and/or a change in thickness of the fabric and/or

    a difference between internal and external areas, such as occur at wall -floor-ceiling junctions.Heat loss due to the thermal bridges makes 15% increment on the U-value. Extra

    heat losses due to the linear thermal bridge effect are almost 5% of the losses calculated.

    1. External walls

    No. Material Layer Thickness ThermalConductivityThermal

    Resistance

    d (mm) [W/( m*K)] R=d/

    [m2*K/W]1 Interior Plaster 0.015 0.870 0.017

    2 BCA Masonry 0.250 0.700 0.357

    3Thermal Insulation

    (expanded polystyrene)

    0.100 0.047 2.128

    4 Faade Plaster 0.025 0.870 0.029

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    125.08

    11

    i

    sih

    R [m2*K/W]

    042.024

    11

    e

    seh

    R [m2*K/W]

    698.224

    1029.0128.2357.0017.0

    8

    111

    e

    s

    i

    se s sih

    d

    h R R R R

    R = 2.698 m 2*K/W > 1.80 = R min Taking into account the heat losses due to thermal bridges :R= 2.698 * 0.8 =2.1584 > 1.80 = R min

    2. Superior slab (under the non-pedestrian terrace)

    No. Material Layer ThicknessThermal

    ConductivityThermal

    Resistance

    d (m) [W/( m*K)] R=d/ [m2*K/W]1 Ceiling Plaster 0.015 0.87 0.017

    2 Reinforced concreteslab 0.15 1.74 0.086

    3 Sloping concrete layer 0.03 1.28 0.023

    4Vapour

    Barrier(cardboard-

    bitumen)

    - - 0.029

    5Thermal Insulation

    (extruded polystyrene)

    0.20 0.03 6.667

    6 Double layer roof skin 0.01 0.17 0.059

    1.010

    11

    i

    sih

    R [m2*K/W]

    042.024

    11

    e

    se

    h R [m2*K/W]

    995.624

    1059.0667.6029.0023.0086.0017.0

    10

    111

    e

    s

    i

    se s sih

    d

    h R R R R

    R = 6.995 m 2*K/W > 5 =R min Taking into account the heat losses due to thermal bridges :R = 6.995 * 0.8 =5.596 > 5 =R min

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    3. Inferior slab (slab over the semi-basement)

    No. Material Layer ThicknessThermal

    ConductivityThermal

    Resistance

    d (m) [W/( m*K)] R=d/ [m2*K/W]1 Floor tiles 0.02 1 0.020

    2 M100 euqalizinglayer 0.03 0.93 0.032

    3 Reinforced concreteslab 0.18 1.74 0.103

    4 Thermal Insulation(Glass Wool ) 0.15 0.032 4.688

    5 Ceiling Plaster 0.015 0.87 0.017

    167.06

    11

    i

    sih

    R [m2*K/W]

    083.012

    11

    e

    seh

    R [m2*K/W]

    110.512

    1017.0688.4103.0032.0020.0

    6

    111

    e

    s

    i

    se s sih

    d

    h R R R R

    R = 5.110 m 2*K/W >2.90 =R min Taking into account the heat losses due to thermal bridges :R = 5.110 * 0.8 =4.088 > 2.90 =R min

    III. Calculus and verification of global thermal insulation coefficient (G)C107/1-2005 contains the verification methodology for the global thermal insulation

    coefficient G for dwellings. This is defined as the amount of heat losses made by direct transmissionthrough the building`s envelope surface, for a temperature difference of 1 C (or 1 K), relative to the

    volume of building, plus the heat losses associated with refreshing of the internal air, and those dueto additional infiltration of cold air. " Heat Loss Coefficient per Volume (G-value) in W/(m 3*K) must be below the Nominal

    Heat Loss Coefficient per Volume (GN-value ).According to C107/1-2005 Annex 2, for A/V=0.32 and 9 floors, GN will be considered:

    GN = 0.39 W/(m 3*K)In order to compute G, we use the following relation:

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    n- the natural ventilation speed (expressed in the number of air exchanges per hour in a givenspace) n=0.5 h -1 ,according to C107/1-2005, table 1

    non-dimensional factor of correction V the heated volume of the building (m 3)R j - the revised specific thermal resistance of the element j [m 2*K/W];A j the surface of element j with a function of thermal insulation [m 2];

    G=0.351 < GN=0.39 [m 3*K]

    2. Condensation check

    With increasing levels of insulation, the use of impermeable cladding materials and newmethods of construction, can often come increased risk of condensation in buildings.Condensation can give rise to a range of problems from surface staining of internal finishes tolong term degradation of the struct ure. Recent health scares from toxic mold growth in

    buildings have also highlighted the dangers that may be caused if condensation persists.

    The Condensation Risk Analysis according to the latest standards and assessment of therisk of both surface and interstitial condensation occurring for different building elements needsto be realized. If a potential risk is indicated the design can be reconsidered and if necessarymodified to minimize the risk.

    When heat flows out of the envelope, the temperature of the internal surfaces is slightlylower than the internal air temperature due to the surface thermal resistance. The size of thistemperature drop is related to the overall thermal resistance of the construction. Should the air carrying moisture vapor drop below its dew-point at the wall surface -excess moisture will bedeposited. If this situation persists then surface mold will begin to form.

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    I. Climate parameters

    a) Interior climateT i=20 0C (the interior calculus temperature)i=60% (the relative humidity of interior air)R si=0.125 m 2*K/W (the interior superficial thermal resistance)

    b) Exterior climateTe=-21

    oC (external winter temperature corresponding to climate zone 4 according toAnnex D, C107-2005)e=60% (the relative humidity of exterior air)R se=0.042 m 2*K/W (the exterior superficial thermal resistance)

    II. Risk of condensation on the interior surface of the building element (external wall)

    In order to avoid surface condensation, the temperature of the inner surface of the building element in the most unfavorable zone from heat transfer resistance point of viewshould be bigger than the dew point temperature( .

    In order to determine the temperature on the surface of the element, in the unidirectionaltransfer hypothesis, the following relation is used:

    Taking into account the values for the interior humidity i=60% and for the interior temperature i=+20

    0C, the dew point temperature can be obtained, in our case: 12.0 0C.

    ( ) The condition being satisfied, no condensation will appear on the interior surface of the

    external wall.

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    III. Condensation checking within the building elements structure

    Interstitial condensation occurs as a result of moisture diffusing through the permeable building fabric from the high vapor pressure side to the low vapor pressure side. As the air ismoving through the building fabric, it can be cooled, in the process increasing the moisture

    saturation of the air. Where the temperature drops sufficiently to cause the air to reachsaturation, water vapor in the air can condense out. The point at which this occurs is called thedew point which is the point where condensation occurs.

    The condition for avoidance of condensation is that at any point within the element the partial pressure does not reach the saturation pressure value p spv.

    a) Establishing the temperature variation in the structure of the element (considering theannual average temperature)

    , where