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Insulation Theory

Building Insulation2040BIEN1003December 2003

Insulate for life

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Making Far-reaching ChoicesInsulation makes good economic sense as it reduces

energy consumption in buildings. Insulation as a single

investment pays for itself many times over during the

life cycle of a building. Reduced energy consumption

also benefits the environment.

This information is common knowledge in the EU.

The EU has already passed the Energy Performance

Directive that requires the member states to adopt new

practices to improve the energy efficiency of buildings.

We are well prepared for the increasing

requirements. The unique properties of stone and the

extreme Nordic weather conditions have inspired us to

develop economically and ecologically rewarding, safe

and sound insulation solutions. Stone wool, made

from nature’s own stone, is extremely fire resistance,

durable and reliable choice for protecting lives and

property.

We are happy to share our knowledge in this

Insulation theory, which has become something of an

institution and has already guided generations of

builders in the field.

We hope that you find this material useful in

designing high quality insulation solutions, which will

meet the requirements in the years to come.

Contents

Energy saving and economics ofinsulation 4-5

Insulation Economy 6-8

Structural design 9-12

Heat 13-18

Moisture 19-26

Frost 27-28

Ground insulation 29-33

Fire 34-43

Acoustics 44-57

CE marking 58

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Insulation TheoryDesigning safe and sound insulation structures requires knowledge from all fieldsof building physics. Here the concepts and fundamental functions related toinsulation material are presented. The content is divided into the following:

Energy saving and economics ofinsulation

Insulation is a powerful mean in reducing energy consumption. Economicallyoptimal insulation can be calculated and it is often much thicker than is stipulatedin the regulations.

Constructive designFour principles to follow in carrying out the construction to avoid the risk ofunnecessarily high energy consumption and damp damage.

HeatFundamental definitions of heat followed with the heat transfer mechanism ofinsulation material and the calculation of thermal transmittance for buildingcomponents

MoistureFundamental definitions and terms and a number of calculation examples inaddition to a list of practical tips in order to minimise the risk of moisture inbuildings.

FrostThe occurrence of frost and how to reduce or prevent its build up by usinginsulation.

Ground insulationThe use of stone wool as insulation of constructions to the ground with one sidewarmed up. The chapter is referring to experiences from different follow up ofdamages in buildings.

FireFundamental terms in addition to the classification system applied to the materialsand buildings. Finally, the section includes graphs, which facilitate theestablishment of dimensions for fire insulation in different types of constructions.

AcousticsThis part gives a theoretical description of acoustics and the function of stone woolfor different usage in buildings. Typical values for stone wool are given. It alsorefers to common European standards regarding absorption and sound reduction.

CE markingIn this part a short description of the European regulations for marking of mineralwool is given. What does it mean and what kind of quality control is promised?

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Energy saving and economics of insulation

Energy economising and thermal insulation make up the“Principal requirement” of the EU BuildingCommodities Directive. The EU Commission hastherefore chosen to set out the importance ofconstructing buildings with a view to the need for good

Even if a better insulation standard than that required bythe regulations were to increase costs, it is still a veryinexpensive measure in relation to its efficiency. The extracosts can be viewed as a very inexpensive insurance policyagainst what will occur in the future. It makes goodeconomic sense to bear in mind rising energy prices, sothat making further investments in additional insulation inthe future can be avoided. It can be said that choosing ahigh insulation standard is a favourable insurance policyagainst future rises in energy prices.

Energy savingNational regulations stipulate the minimum requirementsso as to limit the need for heating energy in buildings. Theaim is to achieve good energy economy. But what is goodenergy economy? Who is it good for? For the homeowner?For the tenant? For the community?

There is no conflict here. If the most personallyfavourable thickness is chosen (the thickness that incommon terms is called optimal thickness), it appears thatit is generally thicker than is stipulated in the regulations.Furthermore it provides a more comfortable indoor climateand above all is more energy efficient as seen from society’sperspective when environmental considerations areincreasingly being taken into account.

The most favourable insulation must be calculated basedon a particular lifetime for the building. The insulationdoes not wear out, does not require maintenance and doesnot require replacing. A lifetime of 50 years is normallyreckoned for insulation that is to match the estimatedworking life for the building. However, this is much tooshort. If the structure has been correctly designed, therewill be nothing to affect the insulation when it is in place.It will have the intended insulating effect for as long as it isin place and we do not know the age of a piece ofinsulation. In practice, the lifetime is unlimited. Therefore,the lifecycle analysis for stone wool insulated structureproves that a significantly greater amount of insulationshould be used than that which is stipulated in theregulations.

Therefore, it is important that proper insulation is fittedwith a view to the future when new construction orrenovation will be carried out. Seen over the lifetime of thebuilding, there are hardly any measures that increase energyefficiency that are as favourable to the homeowner aseffective insulation.

energy saving and consequently to a good standard ofthermal insulation. This has been done after seeing theconsequences of the escalating energy prices of recentyears and with increased awareness of our globalenvironmental problems.

■ I N S U L A T I O N T H E O R Y – E N E R G Y S AV I N G

Total environmental balance for stone woolinsulation

Positive contribution Negative contribution

Manufacture Usage stage Demolition

>100

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Figure 1

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Optimal insulationTraditionally we have calculated the economical optimalthickness for different structures in a building. In this case,the fact that building costs will rise when insulation isincreased is taken into consideration. But at the same timethe annual energy consumption will decrease without therebeing any outlay for maintenance. A schematic example isshown here for a traditional attic joist floor where theeconomic thickness is normally around 0.50 m. This isfound where the curve has its minimum point, simplyexpressed as the lowest annual expenditure for building andenergy costs.

Environmental concernsHow does insulation affect our environment?Does the manufacture and transportation of insulation leadto significant environmental damage?How is the insulation finally dealt with after demolition?

The use of thermal insulation has a very positive effecton the environment. Manufacture, including the extractionof raw materials, transportation and assembly have anegative environmental effect that is compensated forduring the first year in which the insulation is used. It isusually said that the environmental usefulness is severalhundred times greater than the environmental stress.

If the total lifetime of the building is considered,operation and maintenance equate to approx. 85 % of thetotal environmental strain. Totally overwhelming is theamount energy required for heat and hot water. Approx.15 % comes from the manufacturing process and less than1 % from the demolition. It is easy to show that investingin extra insulation will pay for itself many times over whentaking into account the environmental strain over theentire lifecycle. And this additional insulation is motivatedon purely personal economic grounds.

0 0.5 1.0 1.5 2.0 2.5 3.0

The total environmental strain of a building

Normal insulation Increased insulation

Manufacture Usage stage DemolitionFigure 2

Figure 3

Sacrifice

Economy

Environment

Optimal thickness in m

The diagram also suggests the location of anenvironmentally optimal insulation thickness. The curverepresents the sum of the environmental influences uponconstruction and for annual operation. Theenvironmentally optimal insulation thickness for the atticjoist floor is an unrealistic 2.5 m.

It is mainly due to the fact that the additional effects onthe environment when the thickness increases originatealmost entirely from the insulation itself. The calculationshows that an investment in insulation thicknesses in excessof normal thicknesses will also lead to an investment in ourenvironment.

The house as an energy systemIf we were to construct a house that satisfied therequirements for optimal energy economising we should belooking at the bigger picture and not just the individualparts. It often proves to be the case that extra insulationwill allow a slightly simpler heating system to be chosen.For example, it must be appreciated that the variouscomponents in a building have different lifetimes. Werecommend that the effects that different insulationstandards have on the choice of heating system in thehouse be examined at an early stage.

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The important conditionBy 2006 each Member State of the EU must implementthe Energy Performance Directive for Buildings into thenational legislation. This significantly changes the way thebuilding regulations stipulate for the energy use inbuildings. The new regulation will be based on the totalenergy consumption of a building taking into conside-ration the thermal losses through the envelope, ventilationlosses and heat recovery gains, solar and other gains as wellas tap water generation, cooling by means of airconditioning systems etc.

However, reduced energy consumption in a privatebuilding may only take place under the conditions of aproperly insulated climate shield. Only then, it will bepossible to have full usage of efficient installations for theproduction of energy.

A high insulation standard for floors, walls, roof andwindows does not only mean lower energy consumption. Itwill also reduce the power need and makes the heatingperiod shorter. It improves the conservation of existing freeenergy and creates conditions for simpler heating systems.

A high insulation standard is an investment with verygood profitability for unlimited time. There is norequirement for running costs or maintenance.

B = ∆U · Q

Insulation Economy

SC MethodSC stands for saving costs. The SC method means that youcompare the costs of saving energy - the saving cost - withthe current price of energy. Step by step, the thickness ofthe insulation is increased and you can calculate themarginal saving cost. As long as it is lower than the currentenergy price, the insulation measure is profitable.

The savings cost - SC – may be calculated using thefollowing formula:

– ∆I = increased investment cost (EUR/m2)– B = energy saving per year (kWh/m2)

The energy saving may be calculated

where ∆U is the improvement of the U value and Q is thethermal consumption figure for the actual area inapproximately 1000 degree hours/year.

– α = is the correction figure, which takes intoconsideration the life span (n), the real energy priceincrease (q), desired real return on the investment (r)and is calculated using the following formula.

– n = life span (years)

B · α∆ I

1 – t

1 + q

1 + rα = ; t =

1 – tn

Heating and running of buildings contribute toapproximately 40 % of the total energy consumption inEurope. There is therefore great potential to reduce theenergy consumption. Especially important is that at timesof new construction adequate insulation standards areused. Even when renovating, you should consider whichconstructions ought to be additionally insulated.

The following includes a calculation to estimate theeconomical insulation thickness. This method is used foreconomically optimal insulation thickness in the tables that

show the association to the construction solutions in thetable.

The SC method provides the insulation level that leadsthe lowest annual expense with the chosen calculationprerequisites. The minimum point is fairly flat that meansthat the annual costs only increase marginally if you chosea slightly higher insulation standard. You can say that thereis a low premium to ensure against future extreme energyprice increase.

■ I N S U L AT I O N T H E O R Y – I N S U L AT I O N E C O N O M Y

SC = EUR/kWh

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d1

0.042

Note that if the formula should be applied for a measurethat demands maintenance, for example waste heatrecovery, the ∆I also contains the current value of theannual expenses for maintenance.

Average and marginal savingcostsThe financial benefit of each extra centimetre of insulationdecreases with thickness. It is profitable to increase thethickness of the insulation until the last centimetre fulfilsthe return requirement. That is to say, until the savingsexceed the cost.

Using the marginal saving cost – SCmarg – you cancalculate the consequences of gradual increases of thickness.

For the maximum financial benefit, the SCmarg shouldbe the same as the current energy price. This determinesthe financial optimal thickness of the insulation. Since theoptimal thickness is decided, you may find the totalprofitability of the measure by calculating the averagesaving cost - SCaverage.

NOTE! For profitable insulation measures, the SCaverage isalways lower than the SCmarg. This generally also applies to:SCmarg dimensions the insulation thicknessSCaverage describes profitability.

ProfitabilityExample. Below we indicate how to calculate economicalthermal insulation for a building. As an example, we havechosen a detached house in the middle of Sweden with aloft made of wood, insulated with Paroc loose wool. It isnormally performed using an optimising calculation on thecomputer.

Figure 4: Wooden joists.

Conditions for calculation:– Current energy price 0.60 SEK/kWh– Real interest, r 4 %– Real annual energy price increase 2 %– Life span, n 50 years– Heat consumption figure, Q 110 · 103 K h/year– Investment in order to increase the

thickness of the insulation by 1 cm 3 SEK– α, calculated using the above rate

of interest energy price increaseand life span 32.4

– λD for insulation 0.042 W/m K

The following thermal resistance may be applied in thecalculations

* Increased successively until economical thickness isachieved.

We have chosen to start the calculation from an insulationthickness of 170 + 50 mm. It provides an Up value of0.205 W/m2 K. With an increase in insulation standard by20 mm the investment increases by 6.00 SEK/m2. At thesame time, the U value drops by 0.018 W/m2 K.

Construction R m2 K/W

Surface transfer resistance+ outer roof + inner covering 0.47

d2 = 170 mm Loose wool, 5 % ruleshare 3.85

d1* Loose wool without wood frame

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The energy saving will be as follows:

B = ∆U · Q = 0.018 · 110 = 1.98 kWh/m2

With the aid of the energy saving, the investmentdifference and other conditions, the above will be:

The measure is profitable since the SCmarg < 0.60 SEK/kWh

The corresponding calculation is completed for everyincrease in insulation thickness for the SCmarg, until itexceeds the applicable energy price. The increase inthickness is chosen according to the standard thickness sothat the step will be 20 – 30 mm.

According to this calculation, it will become apparentthat the optimal insulation thickness today isapproximately 650 mm for the construction in question.

The average savings cost is determined by calculating thetotal step from 220 to 650 mm insulation thickness.

B = ∆U · 110 = (0.205 – 0.093) · 110 = 12.3 kWh/m2

The measures are very profitable since

SCaverage < 0.60 SEK/kWh.

Temperature inside the house

Area (Sweden) 18 °C 20 °C 22 °C

Kiruna 156 175 194Arjeplog 150 166 182Piteå 140 155 170Lycksele 131 146 161Östersund 122 136 150Härnösand 111 123 135Falun 109 121 133Gävle 103 114 125Örebro 98 109 120Nyköping 95 106 117Visby 89 99 109Kalmar 88 98 108Göteborg 82 91 100Malmö 79 88 97

Table 5: Heat consumption figure Q at different temperaturesinside the house.

∆I · 100

1.98 · 32.4

6.0 · 100

B · α0.0935 SEK/kWh

SCaverage = = = 0.29 SEK/kWhB · α 12.3 · 32.4

38 · 3 · 100∆I · 100

■ I N S U L AT I O N T H E O R Y – I N S U L AT I O N E C O N O M Y

SCmarg = = =

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Structural design

When planning, it is important that the house is looked atin its entirety and not just by the performance of theindividual components.

Even if the calculation of the heat losses has been carriedout correctly in theory, there is no guarantee that the resultwill agree with the actual outcome. Construction must becarried out in a professional way. This means that the workmust be performed both correctly and accurately.

It is important that these four principles are followed whenconstruction is carried out, otherwise there will be a risk ofunnecessarily high energy consumption and in the worstcase damp damage may result. There now follows someadvice and tips for each of the points. If you follow therecommendations the building will function correctly!

Air and vapour barriersA modern house must be airtight in order for the ventila-tion to function as intended. Therefore, an air and vapourbarrier is required, this will operate during the entirety ofthe lifetime of the house. Normally a plastic sheeting isbuilt into the structure, which is placed on the “warm side”of the insulation. Other materials, such as concrete, canprovide airtightness.

• Air and vapour barriersThe building must have an airtight layer, a so-called airand vapour barrier, on the inside of the structure. Thelayer must not only prevent the transfer of moisture fromthe inside to the outside, but should above all make thestructure airtight. A structure that is not airtight willresult in higher energy consumption and there will be arisk of damage due to damp and mould within thestructure. In addition, draughts can cause discomfort.

• Installation of insulationThermal insulation must fill up the whole of its space.There must be no air gaps. It is particularly importantto avoid air gaps on the “warm” side of the insulation. Ifthe insulation does not fill up the whole of its space, aircan begin to circulate, a convection that can seriouslydecrease the intended insulation efficiency.

• Wind protectionWhen the air moves behind the facade, it is importantthat it cannot penetrate the primary insulation or thegaps around the insulation. Therefore, there must bewind protection in place to prevent this. The windprotection must be adapted to the insulation material,the façade material and the entire structure.

• Ventilated air spaceThere should normally be an air space that is ventilatedby outdoor air behind a façade layer or under a largenumber of roof coverings. The air space allows themoisture that comes in from the outside to be ventilatedaway. It also functions as an extra safety device if anypart of the inside of the structure has not been madeairtight. Certain structures with totally airtight exteriors- e.g. warm roofs and sandwich structures - do notrequire an air space.

If the material and the construction are both perfect, therewill be a certain safety margin in relation to the calculatedvalue. But any errors in the execution of the work or faultsin the finished structure can affect both the insulatingefficiency and durability.

The following points are particularly important toconsider in order for the work to be performed correctlyand accurately:

■ I N S U L AT I O N T H E O R Y – S T R U C T U R A L D E S I G N

Overpressure inside: Air isforced out

Underpressure inside: Air istaken in

Figure 6

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■ I N S U L AT I O N T H E O R Y – S T R U C T U R A L D E S I G N

THINK ABOUT THIS IN PARTICULAR:• Place an airtight layer, e.g. a 0.2 mm PE foil on the insideof the insulation to prevent air leakage and vapour diffusion

• If possible place the installations on the inside of the plasticfoil

• Think about making all joins and routings airtight. Usedurable tape, adhesive, caulking compound or other specialarrangements

• Pack and seal large gaps

• On roofs made of concrete or light concrete the barrier isto protect against building moisture.

• Low-sloping unventilated roofs on supporting sheets shouldalways be constructed using an air and vapour barrier on thesheet. This becomes particularly important if the activity on thepremises markedly increases the moisture content of the air

• Plastic foil must not be used in structural components thatare in direct contact with the ground, e.g. basement outerwalls, basement floors or slabs on the groundThe ground usually has higher moisture content than theinternal air. Therefore, the majority of the insulation should belaid on the outside and the underside respectively. If insulationis used on the inside, an effective vapour barrier should beplaced under the insulation for slabs on the ground orbasement floors. A vapour barrier is not to be used for aninverted insulated basement wall. Wood must not be incontact with the supporting elements.

A correctly functioning air and vapour barrier isparticularly important when there is too much pressureindoors. This occurs nearly always at the top of thebuilding during the winter. If the attic joist floor is notairtight, heat and damp air can penetrate their way into thestructure and condense.The consequences can be seriousmould damage. In addition, if the insulation is not keptdry, its insulating properties will be reduced.

Moisture convection, moisture that accompanies airwhen it penetrates into a structural component, is muchmore dangerous than moisture diffusion, that is moisturewhich is transferred due to differences in vapour content.

Airtightness is therefore very important. But the barriershould also prevent vapour diffusion into the structure.Otherwise water vapour can condense and cause damage.The driving force for diffusion is highest during the wintersince moisture will flow into the building from people andfrom activities. The barrier must then be placed on theinside in order to be effective. If it is placed on the outside,it will have almost the opposite effect to that intended. Inthis case the moisture will condense on the barrier.

It is sometimes stated that a vapour barrier on the insidecan cause damage during warm, rainy summer days whenthe diffusion drives the moisture from the outside to theinside of the structure. However a large number ofinvestigations show that these fears are exaggerated. It is thedriving forces during the winter that must be guardedagainst.

The air and vapour barrier is usually a 0.2 mm PE foilthat satisfies national standards for ageing resistance. Jointsmust be kept to a minimum and be as well sealed aspossible.

A lot of damage has been reported from buildings wherethe construction has been made knowingly permeable inorder to allow it to “breathe”. Paroc would most definitelywarn against such solutions. Be careful when constructingthe air and vapour barrier. The most critical points are

– Connections between different building components

– Routings for pipes, electrical points or ventilation devices

– Joins in the barrier

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Wind protectionThe wind protection must prevent air that moves behind afaçade or an external wall from ruining the thermalinsulation capabilities of the insulation. Therefore the airthat moves parallel to the insulation is to be protectedagainst using the wind protection, the air and vapourbarrier will deal with air movement through the structure.

The requirement for wind protection depends on thesize of the air movements to be expected behind the façadelayer. A well-walled brick façade will provide significantlylower air movements than a wooden panel, for example.High buildings provide greater air movements than lowbuildings and buildings exposed to the wind providegreater air movements than buildings protected against thewind. Particularly exposed are the corners of the buildingswhere the difference in wind pressure between both sidescan be great.

THINK ABOUT THIS IN PARTICULAR:• Wind protection must not be so airtight against vapourdiffusion that it prevents moisture that has come into thestructure from evaporating outwards.

• Be particularly careful with wind protection at the corners ofthe building. There must be no unnecessary joins here.

• Follow the instructions in the recommendations, which canbe gathered from the figures below. They are based on manyyears’ experience and are a guarantee for correct functioning.If an alternative solution is chosen, this will be at your ownrisk. If this should be the case, consult the manufacturer or arecognised expert.

Installation of insulationThe construction of an insulating material with cells or alattice of fibres causes the air to move and the heat transferwill thereby be significantly reduced. Therefore, it isimportant that the insulation completely fills out theintended space. Otherwise the air can begin to movethrough the gaps and spaces.

Since warm air is lighter than cool air, the air worksalong an outer wall after rising along the warm side of theinsulation and sinking down the cool exterior. Thesedriving forces increase as the temperature difference acrossthe insulation increases. In a roof the air will move throughthe structure.

It is therefore important to avoid spaces, cavities, gaps orother imperfections in the insulation and in particular onthe warm side. If cool external air is allowed to reach theinside of the insulation, the insulation has been short-circuited.

THINK ABOUT THIS IN PARTICULAR:• Be careful to cut the insulation so that it fits. Be careful whenassembling so that the insulation completely fills out the spacefor which it is intended.

• Stone wool must be cut to be slightly larger in length andwidth. No air spaces between the insulation and thesurrounding surfaces.

• Insulation in several layers must be assembled with offsetjoints where this is possible.

• The air and vapour barrier and the insulation must lie closeagainst each other. If there is a thin panel in the roof, forexample, the barrier is to be placed on top of this.

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Ventilated air spacesBehind the façade layer and under the roof coverings thereshould be a ventilated air space. The purpose of an airspace is to ventilate (and in walls also to drain) away anyrain water that has penetrated and to prevent it fromreaching other moisture sensitive constructioncomponents. Furthermore, the space must ventilate awayany moisture that comes from within the building.

The air space should be at least 20 mm wide and mustnot be packed with lath or mortar remains.

Sandwich structures, concrete elements or so-calledindustrial roofs or low-sloping roofs do not normallyrequire an air space.

THINK ABOUT THIS IN PARTICULAR:• If the façade material has a smooth rear side, nailingbattens or similar must not seal the air space.

• If the edge of the joist must be sealed in order to preventthe risk of fire spreading, air permeable stone wool should bechosen

• Ensure that you build in good ventilation at the eaves of theattic floor joist and supplement it with ridge ventilation orgable ventilation.

• Follow the instructions in the recommendations. They arebased on many years’ experience and are a guarantee forcorrect functioning. If an alternative solution is chosen, this willbe at your own risk. If this should be the case, consult themanufacturer or a recognised expert.

Light stone wool like PAROC UNS 37

Stone wool, minimum PAROC WAS 50

Stone wool, minimum PAROC WAS 25 (30 mm) or PAROC WAS35 (50, 80 mm)

Optional façade material

Façade layer of facing stone, concrete, etc.

Wind protection of plaster, board, foil or paper

Figure 7

■ I N S U L AT I O N T H E O R Y – C O N S T R U C T I V E D E S I G N

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Thermal ConductivityThermal transmission may be theoretically calculatedstarting from the laws of physics, however in practice thecalculation is difficult to carry out. Therefore, the thermalconductivity is measured in the various materials. This isdefined as a heat amount in Wh per hour – h – passingthrough a 1 metre thick layer with an area of 1 m2 and thedifference in temperature across the material is 1 °C. Figure8 illustrates this definition and it may be written as amathematical formula:

this may be shortened to W/m K.Using this formula, thermal conductivity can be

expressed as a figure. This is signified by the Greekcharacter λ (lambda).

λ valueλ may be used when calculating the amount of heat thathas been transported through a certain material over acertain period of time.

Example: If two bodies, both with an area of A and withthe temperatures t1 and t2 are separated by an insulatedmaterial with the thermal conductivity of 1 and thethickness of d, in the time of h, a heat amount of Q istransported through the insulation material. See figure 9.

The lower the λ value, the better the insulation quality ofthe material. Normal insulation materials carry a value ofapprox. λ = 0.03 – 0.04 W/m K (measurements are takenin laboratory conditions where the average temperature isapproximately 10 °C).

W · h · mh · m2 · K

Thermal transmission, or the transfer of heat from awarmer body to a colder body may in principal take placein the following ways:

1) Conduction – transfer of heat through solid/liquidmaterial.

2) Convection – the moving of heat through moving fluidor gas.

3) Radiation – transfer of heat by means ofelectromagnetic waves.

Thermal transmission through fixed opaque material onlytakes place by conduction. Convection and radiationtransfer heat in liquids and gases. Thermal transmission ina vacuum is only possible by means of radiation.

The materials that are applied as thermal insulation areall porous; part of the material is filled with gas and mostoften with air. Thermal transmission through traditionalinsulation material takes place as a result of theconduction, convection and radiation.

The thermal insulation capacity of a material isdesignated by thermal conductivity signified by λ.

Figure 9

Figure 8

Heat

■ I N S U L AT I O N T H E O R Y – H E AT

Q = · (t2 – t1) · A · hλd

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■ I N S U L A T I O N T H E O R Y – H E AT

How the structure of the material influencesthermal conductivityThe insulation properties of a material depend on thematerial density, which in turn is influenced by theporosity of the material. Porosity can be achieved indifferent ways.

When cellular plastics such as XPS, Extruded polystyrene,are used the spaces become sealed due to the porosity. Thebasic material is entirely connected and rigid. Stone woolhas a completely different porosity. The pore volume iscontinuous and the basic material, the fibres, are in contactwith each other only in certain points. In all the otherporous materials, the porosity exists in the form somewherebetween these two extremes.

Figure 10: Composition of thermal conductivity in Paroc stone wool and the air column.

q = (t2 – t1)λ air

λradiation

λconductivity

d

λconductivity

λradiation

λconvection

Thermal transportIf an air gap separates two surfaces of different temperatu-res, heat will transfer from the warmer surface to thecolder. The thermal flow through the air gap per m2 andper hour may be expressed in the following way:

λ air consists of three parts:λcd= the actual thermal conductivity of airλcv = contribution from convectionλra = contribution from radiation

The value highlighted in Figure 10 applies to a 100 mm aircolumn between wood surfaces or brickwork (not shinymetallic surfaces).

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If a porous insulation material is placed in the air gap, theair gap is divided by either the cell walls or fibres. Theconvection in the air gap stops almost completely as the airmovements are strongly inhibited by dividing the volume.Radiation contribution, which used to be the dominatingtransmission form, is significantly reduced and there arenow many small cells or fibres which transfer radiation atvery small differences in temperature. If the insulationmaterial is filled with air, the actual thermal radiationthrough the air does not change. In addition to the otherfactors present in the three aforementioned transmissionforms, there is now also a contribution to the thermaltransfer in the form of thermal conduction through thebasic insulation material – cell or fibre material.

Similar to the case with the convection contribution, thefewer fibres or cells that there are in the insulation material,the smaller the radiation contribution. In order to reducethe total thermal radiation, fibres or cells may be used. Onthe other hand, a large quantity of insulation material willlead to increased thermal conductivity through the basicmaterial. Therefore, the more fibres there are per mass unitin the fibre material, the better.

Insulation W/m K Air gap W/m K

λcd: 0.028 (82 %) 0.025 (4%) λcv: 0.000 (0 %) 0.116 (19%) λra: 0.006 (18 %) 0.476 (77%)

0.034 (100 %) 0.617 (100 %)

λ W/m K10

0.040.030.020.010.00

0 100 200 γ kg/m3

λ W/m K10

0.040.030.020.010.00

0 100 200 γ kg/m3

The influence of density on the insulationproperties of Paroc stone woolThe following five figures indicate the way in which theinsulation properties of Paroc stone wool vary based ondensity with regard to the four abovementionedcontributions to the thermal transfer as well as the overallthermal conductivity.

ConvectionAs a result of the temperature difference, convection maytake place in stone wool insulation. It is clear from thefigure that the influence of convection is marginal, whenthe density is 20 kg/m3 or greater. All Paroc products onthe market are above this density.

Figure 11: The effect of convection.

Thermal conduction through airThermal conduction through stationary air makes thelargest contribution to the total thermal transfer throughthe insulation material and only varies very little accordingto density. This is due to the fact that the fibrous materialunder all conditions represents only a small part of thetotal volume

Figure 12: The effect of conductivity in air.

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■ I N S U L A T I O N T H E O R Y – H E AT

λ W/m K10

0.040.030.020.010.00

0 100 200 γ kg/m3

λ W/m K10

0.040.030.020.010.00

0 100 200 γ kg/m3

λ W/m K10

0.040.030.020.010.00

0 100 200 γ kg/m3

RadiationThe radiation contribution is strongly dependent on thedensity of the stone wool and may at lower densitiesbecome dominant. Furthermore, it should be noted that aradiation contribution is also dependent on temperatureand the radiation contribution increases in higher tempe-ratures.

Figure 13: The effect of radiation.

Thermal conductivity through fibrous materialsAs the amount of fibre grows proportionally with thedensity (for the same fibre diameter), so does the fibreconductivity contribution.

Figure 14: The effect of convection in the fibres.

The graph will look different if the fibre consistency of thematerial changes.

Total thermal transferIf you add the abovementioned contributions to a totalthermal transfer, the result will be a graph that presents thethermal conductivity of Paroc stone wool as a function ofdensity. The graph illustrates that the thermal conductivityis at its minimum when the density is approximately80 kg/m3. However, the minimum level varies dependingon the temperature.

The above described connection between the thermalconductivity and the density applies to the majority ofinsulation materials, nevertheless, with different figures.

It should be noted that this relation regarding stonewool can also vary, as the fibre orientation can be altered inorder to obtain optimal properties for the installation andthe compressive strength.

Figure 15: The relationship between thermal conductivity anddensity.

λ value for calculationIt is the λ10 value, with statistical variations, that is usedwhen calculating the thermal insulating status of aconstruction. In EN-regulations it is called λD, where Dstands for declared. EN 10456 describes the way ofcalculating the λD in detail. There may also be nationaladding, depending on local constructions, for the λcalculation. Paroc always introduce current λD values forthe specific products.

17

Thermal conductivity’s dependence on temperatureThe thermal conductivity of stone wool increases inaccordance with the average temperature. The increase isapproximately 0.5 % per °C in relation to lighter products,and approximately 0.3 % per °C in relation to heavierproducts, within the temperature range of 0–100 °C.

This depends on the fact that thermal transmission bymeans of radiation and thermal conductivity throughstationary air increase with raising temperature.Convection and fibre conductivity are much lessdependent on temperature.

The dependence of thermal conductivity on tempera-ture at different densities is illustrated in Figure 16.

Figure 16: The dependence of thermal conductivity ontemperature at different densities.

Figure 17 shows thermal conductivity as a function of bothdensity and average temperature. It is clear from the figurethat the λ-minimum is moved towards higher densitywhen the thermal transfer takes place at a higher tempera-ture. This means that it is advantageous to use the productswith higher densities for insulation work in high tempera-ture applications.

The dependence of thermal conductivity on thewater content of the insulation materialIf insulation material contains water, this will naturallyaffect the thermal conductivity of the material. In themanufacturing of Paroc stone wool, water-repellentproperties are added to the wool and in practice this has asignificant effect on how the stone wool absorbs water. Thematerial will only absorb water when it is pressed in.Experience shows that it is very difficult in any other wayto reach a water content that exceeds 0.5 % volume.

The amount of water absorbed is minimal. At 95 % ofrelative air moisture there is hygroscopic water content instone wool of only 0.004 % volume.

The material is open to diffusion and the value, watervapour transfer coefficient is approximately 0.5 mg m/hN.This low figure means that when vapour passes through theinsulation layer and cooles down, no condensation takesplace.

The properties of Paroc stone wool mean that thematerial may be used as a capillary breaking layer.

Figure 17: λ value as the function of density and averagetemperature.

Furthermore, it ought to be noted that the λ value shouldalways be used in conjunction with a figure showing theaverage temperature it is measured at. For buildinginsulation thermal conductivity λ10 is used, which meansthat the λ value is measured at 10 °C.

λ W/m K

0.20

0.10

0.00 100 200 300 400 500 600 Average temperature °C

PAROC UNS 37

PAROCFPS 10

0.05

γ kg/m 50 100 150 200

λ W/m K

0.15

0.00

3

Average temperature °C

350300250200150

4504000.10

18

For further information of material properties and ourproducts see www.paroc.com

Limiting heat lossesThermal resistanceWhen the material property, λD is known, the thermalresistance over the insulating layer can be determined. It iscalculated in accordance with EN 6946 from the formula:

R = d/ λD ; m2 K/W,d = insulation thickness in metres.

EN 6946 stipulates how the thermal resistance is calculatedfor different types of structures. Rsi and Rse, the inner andouter transition resistances in different directions, are alsogiven. Furthermore, the standard describes how differentair spaces and other specific details are dealt with.

The end result for a structural component is RT

The thermal transmittance can then be calculated from

U = 1/RT ; W/m2 K

Corrected thermal transmittancefor building componentsThe corrected thermal transmittance, Uc, for a buildingcomponent is calculated according to the equation below:

Uc = U + ∆Uf + ∆Ug + ∆Ur

To calculate Uc EN ISO 6946 is used.

In order to calculate the thermal transmittance for astructural part (not a window), access to the following isrequired:

• EN ISO 6946 “Building components and structuralparts – Thermal resistance and thermal transmittance -Calculation methods”

• EN 12524 “Building materials and building products –Moisture technical and thermal technical properties –Tables with calculated values.”

• EN ISO 10456 “Building materials and buildingproducts – Procedures for the determination of declaredand calculable thermal values.”

• Material data from the manufacturer of the thermalinsulating material. Note that these data are manufacturerspecific.

The majority of manufacturers state the corrected thermaltransmittance, Uc for the structures that are recommended.If it is decided to use these values, there is no need to carryout your own calculation. However, you will need to checkthat the manufacturer refers to the correct standard.

On the market there also are computerised calculationprogrammes that simplify the task of calculation for thosewho wish to carry out their own calculations.

When following EN ISO 6946, a U value in W/m2 Kwill be arrived at. Then a correction is carried out usingthe three terms that are stated as ∆U values. Informationon these is to be found in Appendix D to the standard.

∆Uf is a correction term for extra thermal flow caused bysmaller metallic attachments in the structure. The term isoften insignificant especially in wooden structures.

∆Ug is a correction term that takes account of normalconstruction errors incurred when assembly takes place.The standard is general and does not provide sufficientguidance for national structures.

∆Ur is a correction for precipitation and wind that have anadditional influence on the heat losses for inverted roofs.

Figure 18: λ value as the function of water content

λ W/m K

0.100

0.050

2 4 6 8 10 Moisture content (vol.-%)

■ I N S U L A T I O N T H E O R Y – H E AT

19

This section firstly deals with a number of terms relatingto moisture followed by the moisture properties of stonewool. Moreover, an outline will be given of the differenttypes of moisture that may appear in addition to moisturetransport.

A few special moisture problems will also be addressedin this section including air tight housing, the relationshipbetween insulation and ventilation, cellar outer wall as wellas slab on the ground floor.

As much of the international research regardingbuildings relates of natural reasons to moisture problems,and hence there is a vast amount of information availableon the subject. Information may for example be found inthe Moisture handbook from the Moisture group at LundInstitute of Technology in Sweden. There are also largequantities of computer programs that facilitate moisturecalculations.

DefinitionsMoisture diffusionThe transportation of water vapour as a result ofcompensation of steam content or steam pressure. Diffu-sion is a relatively slow course of events.

Moisture convectionThe transportation of water vapour as a result of airmovement is a result of differences in air pressure.Convection is a relatively quick process.

Moisture contentThe relationship between the total mass of steam and thetotal volume of the gaseous mixture. Expressed in kg/m3.

Saturated water vapour pressureThe partial pressure for the water vapour in the air may ata certain temperature amount to the certain highest value.This is called the saturated steam pressure and may only bevaried by change of temperature. The higher the tempera-ture, the higher the saturated steam pressure.

Saturated water vapour contentThe steam content at a certain temperature corresponds tothe saturated steam pressure called the saturated steamcontent. It is also the greatest amount of steam that airmay contain at a certain temperature.

Moisture

Relative humidityThe relative humidity (RH) is also referred to as therelative moisture and is measured using the relationshipbetween the actual moisture content (water vapourpressure) and saturated moisture content (saturated watervapour pressure). RH is expressed as a percentage.

The relative humidity is of great significance whendetermining origins of moisture damage.

The moisture properties of stonewoolThe water vapour permeability of stone wool is high incomparison to other building materials. This means thatcondensation does not take place inside an insulated layerof Paroc stone wool despite the possible drop in tempera-ture across the insulation and even if the so-called dewpoint falls inside the insulation.

Figure 19

■ I N S U L AT I O N T H E O R Y – M O I S T U R E

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■ I N S U L AT I O N T H E O R Y – M O I S T U R E

Water resistance and moisture stabilityParoc stone wool products have a very high resistancetowards water and moisture. They are made of fibrousmaterial resistant against moisture and have a hardenedbinder – phenol resin, which displays very good moisturestability.

CorrosionAny insulation material that is in contact with metal maycontribute both passively and actively towards corrosion ifthere is a presence of water or moisture.

A passive contribution to corrosion provides insulationmaterial if it binds the water against the exterior of themetal. Since Paroc stone wool is water repellent and lacksboth hygroscopic and capillary absorbing tendencies, it ispossible to reduce the corrosion contribution to a mini-mum. Therefore, the lowest diffusion resistance facilitatesdrying when the conditions are favourable. Conversely, thelower diffusion resistance leads to a situation in which thestone wool cannot contribute towards preventing moisturevaporisation from a cold surface. If there is air in theinsulation, corrosion may take place on corrosive materialif the moisture does not dry out.

The insulation material that is water-soluble mayincrease water’s electrolytic capacity or significantly alterthe water’s pH value and by that means contribute activelytowards corrosion. The high moisture resistance of Parocstone wool means that the solubility is very low. Theelectrolytic capacity and pH value do not change.

Certain other types of insulation material may containmaterials that directly contribute to the events causingcorrosion, such as fire retardant salts. Paroc stone wool isincombustible and contains no such materials.

Water repellents and hygroscopicityEvery Paroc stone wool product is manufactured in such away that makes it water repellent.

The purpose of repelling water means that water willrun off the outside of the Paroc slabs, water will not soakthe fibres and will not be absorbed in the wool either.

Only if the water is exposed to pressure it may press inthe slabs. In this situation, the fibres do not absorb anywater. Therefore, drying will take place quickly, not leastbecause of the high water vapour permeability.

Paroc stone wool products do not absorb water in acapillary action. Furthermore, they do not absorb moisturefrom the air other than in small amounts at extremehumidity.

Figure 20

21

Figure 21

Air moisture

Rain moisture

Building moisture

Ground moisture

Runningwater

Sources of moisturePart of a building may be subject to moisture throughprecipitation, condensation of water vapour in the air,absorption of ground moisture or leakage. Furthermore, allmaterials come into contact with the water vapour in theair and absorb a certain amount of water. During the timeof building, the construction may also be subject to greatamounts of water, known as building moisture.

Normally, the main sources of moisture are:– air moisture– building moisture– rain moisture– ground moisture (vapour content of 100 %)– running water

Air moistureAir contains water vapour and the content level is denotedby RH.

The relative humidity level outdoors may be assumed to85 % during the winter and 70 % during the summer.

The relative humidity level of the air inside the house isdetermined by the outside air temperature and the vapourcontent, the inside air temperature, production of moistureinside the house in addition to the ventilation intensitybelow the stationary circumstances. That is to say, if thereis an even production of moisture and level ventilationintensity, the correlation may be written as vapour contentinside the house = vapour content outside the house + themoisture contribution. The full value of this moisturecontribution during the winter months may be; 3 g/m3 forthe office and 4 g/m3 for the normal dwellings.

Building moistureBuilding moisture is moisture to which constructions aresubjected during the building stage or during themanufacturing of the building materials.

After the building phase, building moisture should dryout in order that the construction comes into equilibriumwith the surrounding relative vapour content.

22

■ I N S U L AT I O N T H E O R Y – M O I S T U R E

Ground moistureThe influence of ground moisture is largely dependant onthe level of the ground water, but also the type of land, theground level, the cause of the water and the ground’sdrainage properties.

Ground moisture may be divided into the followingcategories:– Surface water– Infiltration water (i.e. surface water penetrating into

construction)– Ground water– Fracture water– Capillary absorbed water

Above the highest surface of the ground water (HSGW),the ground moisture should always be assumed as 100 %RH.

ExampleA building is heated to +20 oC and ventilated with 0.5 airchanges per hour (outside air). The volume of the buildingis 300 m3 and the moisture production inside the house asa result of people, animals and plants etc is 0.6 kg/h.Outside the house, the air temperature is +- 0 oC and therelative humidity is 90 %, that is to say the vapour contentis 0.0043 kg/m3. The saturation vapour content at +20 oCis 0.0173 kg/m3. The relative humidity within the house is:

0.5 · 300

The vapour content of the air inside the building =

0.0043 +

(The amount of vapour in the outside air per m3 plus atotal vapour production inside the house through theventilation level).

0.6

RH =

0.0043 +0.5 · 300

0.6

0.0173= 48 %

RH = saturation vapour content

vapour content

Temperature(°C)

Saturation vapourcontent

cm (10–3 kg/m3)

Saturation vapourpressure

(mmHg)Pm (Pa)

The reduced ventilation level inside the house increases therelative humidity. This is a problem that may in some casesbe acute in today’s well-insulated and air tight houses withpoor ventilation.

Moisture transportThe most important moisture transport mechanisms are:– Diffusion– Convection (as water vapour)– Capillary absorption– Force of gravity (as liquid)

DiffusionMoisture diffusion strives to level out the differences invapour content in the air through molecule movements.The moisture flows from an area with higher vapourcontent to an area with lower vapour content. Diffusionmay in practice be regarded nondependent of the tempera-ture.

– 12 1.81 217.3 1.63– 10 2.15 259.9 1.95

– 8 2.54 309.3 2.32– 6 3.00 367.9 2.76– 4 3.53 437.2 3.28– 2 4.15 517.2 3.88

0 4.86 610.5 4.58

1 5.18 657.2 4.932 5.57 705.2 5.293 5.96 758.2 5.694 6.37 813.1 6.105 6.79 871.8 6.54

6 7.26 934.4 7.017 7.74 1001.0 7.518 8.27 1073 8.059 8.83 1148 8.61

10 9.40 1228 9.21

11 10.03 1312 9.8412 10.67 1402 10.5213 11.38 1494 11.2314 12.05 1598 11.9915 12.83 1705 12.74

16 13.66 1817 13.6317 14.45 1937 14.5318 15.36 2063 15.4819 16.29 2197 16.4820 17.3 2338 17.54

21 18.3 2486 18.6522 19.4 2643 19.8323 20.6 2809 21.0724 21.8 2983 22.3825 23.0 3167 23.76

26 24.4 3360 25.2127 25.8 3564 26.7428 27.2 3779 28.3529 28.7 4004 30.0430 30.4 4242 31.82

Table 22: The correlation between temperature – saturationvapour content and saturation vapour pressure.

23

Figure 25

Figure 23: Diffusion from high to low vapour content.

High – Low vapour content

Outsidetemperature +10 °CRH 50 %Vapour pressure 614 Pa

Insidetemperature +20 °CRH 40 %Vapour pressure 935 Pa

Figure 24: Diffusion through untreated walls.

The outside temperature is +10 oC. The relative humidityis 50 %. According to the table, this provides a vapourpressure of 614 Pa.

The temperature inside the house is +20 oC and the RH40 %. This provides a vapour pressure of 935 Pa.

Diffusion takes place from higher vapour pressure to alower vapour pressure i.e. inside to outside.

ConvectionMoisture convection refers to the fact that the water vapourcontent of the air follows the air travelling through aconstruction. If the air travels from a warmer area to acolder area, the water vapour in the air will condense onthe cold exterior. If the air travels from a cold to a warmarea, condensation will not take place; the air flow dries thestructure. Thus it is dangerous to have over pressure insidethe house for normal applications.

Air movement and therefore convection is reduced ifthere is an airtight layer anywhere in the construction.

Capillary suctionCapillary suction attempts to level out the moisturecontent in a material through moisture travel in the fluidphase.

Capillary suction may normally be neglected on drymaterial but if certain critical moisture content is found,there will be a continuous water mass in the material andmoisture transport through capillary suction will besignificant. This type of capillary water transport rarely needsto be taken into consideration. However it occurs aroundinsulation on the ground and by oncoming pelting rain.

Convection predominantMoisture diffusion and moisture convection may existsimultaneously and either cooperate or counteract.Previously, the importance of having a vapour barrier in theconstruction has always been noted. If you observe theamount of moisture that may be transported from oneplace to another during a certain amount of time as a resultof diffusion, you will always find small in comparison withthe amount that is transported via convection. Therefore, avapour barrier in an outer wall or roof is primarily effectiveas a convection or air barrier.

The amount of moisture transported via convection is,in addition to the air pressure difference across theconstruction, dependent on the total area of perforation. Itis important to be aware that one large hole permits agreater moisture transport than many small holes withexactly the same area. It is therefore most important toavoid larger leaks.

Transportedamount of water

Convection

Diffusion

Void with

24

■ I N S U L AT I O N T H E O R Y – M O I S T U R E

Air tight housingThis applies to efficient building insulation in order to saveenergy. Simultaneously, the building has to be constructedair tight in order for the ventilation to be controllable. Agood air tightness means that the air is transmitted to allthe largest parts via the ventilation system. The ventilationamount may be adjusted to the requirements of thebuilding irrespective of wind pressure and similar.

The requirement for ventilation may exist for manyreasons: to remove odours (from people, tobacco smoke,cooking etc), provide people with vital oxygen, avoid highCO2 levels; prevent dangerous levels of radon gas andformaldehyde, as well as to remove moisture (avoidingcondensation on windows and walls, mould etc).Therefore, the requirement for ventilation significantlyvaries between different types of buildings.

Reduced ventilation raises relative humidityFrom the point of view of the moisture prevention, wellfunctioning ventilation is of great importance. The level ofventilation affects the relative air humidity, which is themain determinant on the existence of mould.

In an average residence, a lot of moisture is generated.The human being emits moisture even at temperaturesbelow +20 oC. At +20 oC, the average person emits 40 gwater per hour and this amount increases by 7 g/per ºC.Cooking, washing, laundry and plants provide their owncontribution to this moisture load.

The generated moisture transforms into water vapourand is absorbed by the air inside the house, the moisture

Figure 27: The effect of moisture content.

content level of which may be described by the relativehumidity. That is to say, the relative humidity is a quotebetween the actual vapour content and the saturationvapour content of the actual inside air temperature.

A house equipped with furniture, walls etc tends tomoisture equilibrium with air in the house. With goodventilation, the vapour content will remain at the normallevel.

On the other hand, with poor ventilation, the moistureenrichment will take place in any material that can absorbmoisture. The concentration will proceed and the risk ofthe formation of mould will increase.

It is easy to see that as the level of ventilation is reduced,the relative air humidity increases.

The risk of mould formation will increase. Even at arelative humidity of 75 %, certain types of mould fungusthrive at average room temperature.

Mould fungus may form on woodand other organic materialsWood-based materials are subjected toincreased moisture movement

Wood rot

Adhesive on plastic flooring is broken down

Plastic based materials are subjected togreater moisture movements

Figure 26: The effect of low and high ventilation.

25

Cellar wallsCellar walls are susceptible to different sources of damp. Inthe cellar walls there is building moisture, gaps within thewalls contain air moisture and in the ground outside thewall, there is ground moisture. Furthermore, the area maybe subjected to obtaining local water pressure against thewall as a result of rain, melted water or water currents inthe ground. Moisture may also be absorbed via capillaryaction through the lower plate in walls.

Therefore, damp in cellar constructions must be driedout. The design engineer has to presuppose that one is ableto provide the interior with dense material for example,vinyl tape or tight acrylic paint. The scientific way is toprevent the moisture problem in cellar walls thus making itpossible for the construction to dry out from the outside.

If the cellar wall is insulated from the outside with acapillary breaking, vapour permeable material must theoutside forthcoming moisture be diverted. Buildingmoisture may dry out through the vapour permeableinsulation. This means that it does the same irrespective ofthe material on the inner jacket. It is also advised to have ataut coating on the inner jacket as well.

Slab on the ground and cellarfloorsSlab on the ground floor and cellar floors may be insulatedabove and below concrete. Many complaints in recent yearshave focused on the moisture problem of the slab on theground foundations. Most cases refer to wood framedflooring above the concrete slab. Therefore, this constructionsolution is applied to hardly any constructions today.

Insulation above concreteOne reason in support of thermal insulation on top of theslab is that it feels more comfortable to walk on than onewith plastic carpet fixed directly on the concrete. Anotherreason is that the surface of the concrete slab requires lessaccuracy.

The disadvantage of having insulation above the slab isthat the transport of moisture in the vapour phase upthrough the slab must be stopped with a vapour barrier. Ifthis is not done, the floor may get damaged. Cellularplastics may not replace the vapour barrier since there willalways be cracks between the plates. Cellular plastic is noteven sufficient to be used as diffusion seal.

If you are at all unsure about achieving a lasting tautvapour barrier you should place the insulation below theconcrete instead.

Figure 28: The capillary breaking insulation of cellar walls. Figure 29: Insulation below the concrete slab.

Building moisture

Ground moisture

Drainage Capillary breaking layer

Moi

sture

tran

spor

t

26

Insulation below concreteThe best and safest way to build a slab on the ground flooris to insulate the under side of it with open insulation. Thistype of thermal insulation incorporates a moisturemechanical advantage and allows moisture transport fromthe slab to the ground instead of from the ground to theslab. How is this possible?

Well, the ground has 100 % RH in certain temperature,say 17 oC. This provides a vapour pressure of 1937 Pa. Theinsulation means that you receive a temperature on theunderside of the plate that is higher than in the ground, forexample 20 oC. The saturated vapour pressure in the plates,i.e. at 100 % RH will be 2338 Pa at that temperature.

Since the vapour pressure attempts equilibrium, itresults in vapour transport in a downward direction. Thiscontinues until the vapour pressure is the same in theground and the plate. In the above example, the slab willreach 83 % RH. The level of humidity will not influencethe plastic carpet or the adhesive.

To ensure that the vapour transport is downwards, atemperature difference of at least 2 oC is required, whichcan be achieved by using 30 – 40 mm thick stone wool forthe slab widths up to 15 metres. From the energy efficientpoint of view, people often choose a significantly thickerinsulation. This provides even better protection againstmoisture damage.

Insulation must be laid under the entire floor. If theinsulation is only placed on the edge, the inner parts of thefloor will not be protected against ground moisture.Insulation also has to be taken from beneath the edgesection, stiffening etc. It is advised that an insulationmaterial that may bear a higher load than the rest of theinsulation be used.

Figure 30: Floors without insulation.

Floors without insulationWhat happens if you have laid a taut floor covering on aconcrete slab which already had time to dry to an averagehumidity of 90 % RH?

Beneath the slab lies a drained and capillary breakinglayer. The humidity of the ground is 100 % RH. Since thetemperature of the slab will be the same as the ground, youwill get vapour transport from the ground to the dry plate.Vapour transport will continue until the vapour pressurereaches equilibrium i.e. 100 % RH.

The result will be saponification of the adhesive on thefloor coating or mould growth on organic material.

Moi

sture

tran

spor

t

■ I N S U L AT I O N T H E O R Y – M O I S T U R E

For further information of material properties and ourproducts see www.paroc.com

27

When heat is transformed from the ground and thetemperature is lower than 0 oC, it transforms the watercontent in the ground to ice and the ground freezes. Thereare two types of frost: ice stripes frost (discontinuous) andhomogenous frost. Only the ice stripes create the frostelevation.

One term of significant importance in connection withfrost is capillaries. If you place a small pipe in a bowl withwater, the water will rise into the pipe. Exactly the samehappens in the ground, i.e. a dry area of land has the abilityto absorb water provided there are small holes in theground which serve as pipes for transport by the groundwater from a lower level to a higher.

The origin of frostEvery type of land has a certain purpose to bind water. Awater coating surrounds every grain of earth and itsthickness depends on the size of the grain.

When the heat leaves the ground, the grains of earth inthe water coating are transformed to ice crystals – frost. Ifthe frost now stops at a certain depth known as the frostboundary and the conditions are suitable for new watermolecules to be transported there, they will also be trans-formed to ice and join the existing mass of ice crystals.With that, the ice layer receives an increase in volume that,according to the smallest resistance layer, straightens out inan upward direction and consequently raises the layer ofearth above.

Figure 31: The frost elevation mechanism.

Frost

■ I N S U L AT I O N T H E O R Y – F R O S T

Frost elevation

Directed resultant force

Frost boundary: ice crystalformation provides anincrease in volume

Capillary moisture transport

Ground water

The conditions for the ice layer growing are that theremust be a capillary connection with the ground water. Theice layer will appropriate the water molecules from thegrains of earth that are nearest under the frost boundary.The resolved grain of earth will in turn appropriate fromthe nearest grain lying beneath and wave of thefts willcontinue in this way as long as the ice layer has capillariesconnected to the ground water. In the event that theconnection is lost, there will be no more ice volume andthere will be no space for the frost elevation.

Different kinds of ground provide differenttypes of frostThe more fine-grained the area of land, the thicker thewater holes resulting in the individual grain of earth willbe. This means that the water molecules may betransported easier and even quicker when the grain of earthis small and the transport route short. In a fine-grainedarea of land however, the frost elevation will be easier sincethe number of contact points between the ice sheets andthe amount of grains is significantly greater (the load ateach point is smaller). Clay is an exception since it has alow capillary path speed.

On an area of land with coarse particles, the watertransport will be complicated because of a long transportway and more narrow water holes. The load in the contactpoints is so great that the ice crystals will not be able toraise the layer above and instead fill the space of the holebetween the grains. In order for the frost elevation to takeplace, the following conditions must be met simultaneously.

– The area of land needs to be prone to frost.– Water should be able to transport to the frost boundary.– Enough significant amount of heat should disappear

from the area of land.– The load on the ground must be less that the lifting

power of the frost.

28

Insulation in the ground prevents frost damageFrost damage may be prevented in different ways. You can:

– Change the frost prone area of land for one less prone tofrost.

– Lower the ground water level in order that the earthcannot absorb water.

– Foundations for frost free depth.– Lay a layer of thermal insulation in the ground.

From an economical point of view, the most interestingalternative is to position a thermally insulated layer. Theadvantage with ground insulation is that the thermalcurrent can be limited from the ground during insulation.As a result there is less frost depth since the temperaturebeneath the insulation layer seldom falls short of 0 oC. Thereduced frost depth in turn:

– Much smaller risk for frost damage– Less foundation depth for houses etc– Less disposition depth for water and sewer

■ I N S U L AT I O N T H E O R Y – F R O S T

In order to obtain functioning ground insulation, thefollowing requirements are also demanded in the insulationmaterial:

– It must not rot– It must withstand acid found in the ground– It must have high pressure strength– It must have good thermal insulation power

Extruded cellular plastics fulfil these requirements.Stone wool is not recommended as insulation against frostin roads, railways or other cold constructions. In the longtime run stone wool will get wet and the thermalconductivity reduced. If one side of the construction iswarmed up, stone wool works very well.

29

Ground insulation

■ I N S U L AT I O N T H E O R Y – G R O U N D I N S U L AT I O N

There are many different recommendations for insulationmaterials that are to be used in the ground and forstructures on the ground.

The old tradition of observing natural geography isunfortunately not always followed. Today, houses are builton old swamps or dried seabed or lakebeds, or on othertypes of ground that are less suitable for the purpose. Theconditions of the locality should be taken into conside-ration when deciding on a solution.

The various ground insulation solutions are more or lessresistant to moisture load. The functioning of the variousmaterials and the major differences are presented below.

Statements were obtained from The Swedish NationalTesting and Research Institute (SP) and from MunthersTorkteknik AB with regard to the structures presentedbelow. Both organisations have had to deal with multipledifficulties and have thus learned how not to build. Theyare called SP respectively MT in the following.

Ground SlabGeneral descriptionA ground slab supporting a heated building must always beprovided with heat insulation. Its main purpose is to limitrelative humidity in the floor to a level that does notdamage the flooring material. The insulation shall alsoreduce heat losses along outer parts of the floor. If theinsulation of the slab is very thick, ground frost insulationmay be necessary on the outside.

Thermal insulation can be placed either underneath oron top of the concrete slab.

In order for the ground slab to function properly, highlyreliable capillary barrier and drainage between ground andthe concrete slab is required. When a concrete ground slabis used, the slab must be protected from contact with watersucked up by capillary action. The insulation shall be dryso as to ensure that excessive humidity does not reach thefloor.

A correctly designed ground slab is theoretically safewith regard to moisture, and it is also considerably lessexpensive than other solutions.

Critical factors– The structure shall be such that it prevents ground

water, capillary water or water seeping in from theoutside from reaching the thermal insulation or flooringmaterial susceptible to moisture.

– The structure shall also reduce the relative humidity, i.e.the thermal insulation function shall hold the RH solow that flooring materials susceptible to moisture areprotected.

– The structure must also be durable and non-deformableso as to bear moving loads.

– The structure shall consist of materials that withstandany moisture loads without being ruined or releasinghazardous materials.

Important details– The slab of concrete must be able to dry out upwards or

downwards before tight flooring materials are laid.Thick parts of the slab may be particularly problematicin this respect.

– The moisture barrier separates the moisture sensitivematerial, e.g. wood on the slab, from the slab.

– When the insulation is underneath the slab, a moisturebarrier is placed on the slab, if required.

– Diffusion-proof flooring with moisture sensitive gluerequires a moisture barrier.

– When the insulation is above the slab, a moisture barrieris placed between the slab and the insulation. Thisrequires that the slab be carefully cleaned.

– The recommendations are generally valid for smallhouses. Where larger slabs - width over 10 m - are used,a special moisture solution is required, such as amoisture barrier between insulation and the slab.

– When floor heating is used, the insulation should becompleted with moisture barrier between the slab andthe insulation underneath.

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■ I N S U L AT I O N T H E O R Y – G R O U N D I N S U L A T I O N

Slab on insulation Insulation above slab

Figure 32: The entire slab is always insulated. Minimum 100 mmdraining material under insulation. Edges are insulated, ifinsulation of the slab is over 120 mm thick.

Insulation with Paroc stone wool under the slab is conside-red to be a dry application; no extras when calculating thethermal isolating capacity.

The drying time for a slab with stone wool isapproximately 40 days. The slab continues dryingdownwards even after installation of the flooring.

If EPS or other plastic insulation is used, the dryingtime will be 60 days. Plastic insulation can give extraprotection against moisture from below.

COMMENT:SP: Little damage on this floor. If stone wool is replaced withcell plastic insulation, drying will take longer after installationof tight flooring. Before laying the flooring the slab will drymainly upwards irrespective of the insulation material.Adequate drying time is necessary. When the direction ofmoisture movement is reversed, e.g. when turning off floorheating, dense insulation is better than open stone wool.Moisture barrier is required in such cases.

MT: With regard to protection against moisture this is the bestground slab structure. Sills and the like should also beprotected from concrete. The moisture content in the concreteslab will usually be so low that the surface layer can be selected

quite freely.

Figure 33: A PE vapour barrier or a moisture barrier is placedbetween the insulation and the slab. The layer breaking down thecapillary phenomenon should be at least 150 mm thick.

This solution gives a soft and warm floor. No problemswith moisture from the structure, if the concrete surface isperfectly clean.

Requires careful work. The structure is susceptible tomoisture from above.

Keeping the vapour barrier during building work can bea problem.

COMMENT:SP: Frequent damage above all on floor structures with studs.Plastic foil does not guarantee freedom from damage. Therewill be moisture under the plastic sheets, and microorganismswill flourish. If smell occurs, a floating floor is more difficult tofix. Vapour movement will make the slab moist in spite of alayer stopping capillary action.

MT: As for moisture, this structure should be avoided. Theinsulation layer may not include organic material. Sills andthe like are placed above the upper edge of the insulation layer,in warm room air, and should be sufficiently insulated againstmoisture. The vapour resistance of the insulation layer shouldbe taken into consideration with regard to other layersincluding the flooring.

31

Insulation above and below the slab Basement wallGeneralThe basement is subject to various sources of moisture.The wall structure contains moisture that must be allowedto dry out. There is moisture from the ground outside thebasement wall. Rain, water from melting snow and ice orwater currents in the ground can also cause local waterpressure against the basement wall. Capillary action cancause water to be sucked through the slab and up the wall.

Moisture in structureThe moisture in a basement structure must be allowed todry out, either outwards or inwards. If the inside of thewall is coated with a non-breathable material, e.g. a vinylwallpaper or plastic paint, the moisture can in practice dryout only outwards.

A correctly constructed basement wall therefore allowsdiffusion on the outside.

The moisture dries out over a period of several years,after which the inside of the wall can be sealed againstdiffusion without a risk of damages, even if the outside wallis diffusion-proof.

Ground moistureGround moisture is the main cause of problems. Youshould count with a relative humidity (RH) of 100 % inthe ground, even if the value is occasionally lower.

Ground moisture in the form of free water can bedrained away. The wall should for the sake of safety havean anti-capillary layer so that the water current in thedrainage does not damage the wall.

Surface moisturePlan a slope leading away from the building.

Critical factors– Moisture in the structure must be able to dry outwards

if the inside of the basement wall is sealed.– The wall must be protected against moisture from the

outside.

Figure 34: Combination of insulation below and above the slabas per previous alternatives.

The main insulation underneath the slab can be combinedwith a thin layer of comfort insulation.

If the slab is insulated on the underside with PAROCGRS 30, the slab can continue drying downwards after thedense flooring has been laid.

A plastic foil is placed between concrete and stone woolon the topside.

Optimal moisture protection and comfort:PAROC GRS 30 under the slab and 17 or 25mm PAROCSSB 2 on the topside. Double floor sheets for evencompression on the topside. Plastic foil between theinsulation and the slab.

COMMENT:SP: Good, comfortable floor. No damage. An alternativesolution is ventilated sealing layer, moisture protection mat,which allows the drying of moisture in the structure, and atthe same time the floor is warm.

MT: With regard to moisture insulation this is only betterwith insulation underneath. With regard to comfort thefooting should be separated from the concrete slab so that floorsurface temperature will not be too low. This can occasionallybe achieved with correct flooring material. The insulationlayer should never have a load-bearing frame made of organicmaterial. Place sill and the like above the top edge of the

insulation layer.

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■ I N S U L AT I O N T H E O R Y – G R O U N D I N S U L A T I O N

Important details– The concrete slab extends outside the basement wall,

and a cove is poured on the edge. The cove must bepoured carefully in order to cover the gap between thewall and the slab. To be on the safe side, the wall shouldbe insulated against moisture by bitumen 0.5 m abovethe cove.

– One way to interrupt the capillary action is to set up aground insulation slab, PAROC GRS 30, against thebasement wall.Any water in the ground will flow parallel to the slab,as the flow resistance of the slab is usually higher thanthat of the ground. In dense ground the opposite ispossible. To prevent water from running in through theslab into the wall there must be a drainage layer outsidethe slab made of suitable material. The slab is a partof the drainage system and must therefore contact withthe part leading the water away.

– A grooved insulation slab or a moisture protection matare open at the bottom and thus increase the risk ofwater penetration.

– If there is a high risk of water pressure on the wall, fit anbitumen mat on the wall irrespective of type ofinsulation.

– Draining filler shall lead to the drainage line.

External insulation

Figure 35: Down: Insulation contacting with drainage.Up: Plaster on insulation or base element.Wall: Bitumen coating approximately 0,5 m up.

A draining layer of a minimum thickness of 200 mm isplaced closest to the insulation.

Stone wool stops capillary action, but has an openstructure that allows the wall to dry outwards. Water is ledoff along the insulation surface.

Stone wool is sensitive to high pressure, but a flat layerunderground is ok.

The thermal resistance is reduced, as per instruction inthe Swedish building codes, with 0.20 m2K/W when stonewool contacts with ground.

COMMENT:SP: Good structure irrespective of insulation material. Waterpressure in the ground, resulting from blocked drainage, causesproblems. It is important to drain surface water away from thehouse.

MT: Clearly the best alternative with regard to humidity.Paint the inside wall with a silicate or KC paint that will pass

through vapour. You can also leave the wall untreated.

33

Internal insulation

Figure 36: N.B.! Wooden studs must not be placed inside thewall – RISK OF ROTTING. Do not use a plastic sheet as vapourbarrier.

If wooden studs are used, fit first a 20 mm stone woolboard directly on the entire wall.

The wall can to a degree dry inwards, when stone woolis used as insulation.

COMMENT:SP: Great risk of mildew on the wall. This is a riskconstructors should be aware of.

MT: The basement wall will be colder and therefore moisterthan without insulation or with external insulation. Theinsulation layers should never have a frame of inorganicmaterial. Pay attention to the vapour resistance of theinsulation layers with regard to other materials in the wall. Inrelation to comfort, the structure has the same benefits as

external and internal insulation.

External and internal insulation

Figure 37: Recommendations for external and internal insulationcan be combined. The insulation is placed mainly on the outsideof a wall.

The external insulation is the main insulation and itcompleted with a thin insulation layer on the inside forincreased comfort.

COMMENT:SP: The point is the external insulation. On the inside aboard, such as gypsum, is enough, since a wall is not touchedin the same way as the floor is.

MT: With regard to moisture this version is worse thaninsulation on the outside only. The moisture permeabilitycharacteristics of the insulation layer must be taken intoconsideration when “closing in” parts of building. Theinsulation layers should never have a frame of organicmaterial. As for comfort, this alternative has benefits ascompared with externally insulated walls. Counter radiationfrom the wall is reduced, and the basement is experienced asless ”raw”.

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■ I N S U L AT I O N T H E O R Y – F I R E

Fire

Fire can be defined as a destructive thermal process whichincreases damage on its own until all combustible materialexposed to fire is burned out.

The following are required for a fire to ignite and to sustain:– Combustible material– Sufficient amount of oxygen– Heat that causes a material to reach ignition temperature

AirThe oxygen required for combustion usually comes fromair. The intensity of the fire depends on the oxygen supply.A reduction in the amount of oxygen can thereforeextinguish or dampen a fire.

HeatWhen the temperature of combustible material reachesignition temperature, rapid combustion will result. Theheat required for ignition can be produced by:– open flame (e.g. match)– a heated body (e.g. welding spark)– optical phenomenon (e.g. burning lens)– electrical phenomenon (e.g. arc)– friction (e.g. over-heating of a bearing)

The combustion temperature reached depends of severalfactors, such as the heat value of the burning materials(MJ/kg), rate of combustion (depends on thepulverisation), air supply and the amount of flue gasesgenerated.

A fire in a room can be divided into three mainphases: ignition, combustion and cooling down.

Figure 38: Fire triangle shows the requirements for a fire.

FuelWhen combined with oxygen, combustible materialsgenerate more heat than is required for the chemicalreaction. Combustibility is graded as inflammability.

• Self-igniting materialsMaterials that can start burning without the influence ofan external heat source, e.g. linseed oil soaked in cottonwaste.

• Flammable materialsMaterials, which finely dispersed can be ignited with amatch and which continue burning in air, such as paper,wood splinters and most textiles.

• Materials that do not ignite easilyMaterial that will ignite when heated locally and willburn as long as heated, but which will not continue toburn after the heat source has been removed. Someexamples: wood-wool cement boards and certainplastics. Test methods exist for determining whethermaterial is difficult to ignite.

Non-combustible materials include the common buildingmaterials such as cement, concrete, aerated concrete,gypsum, brick and stone wool. The non-combustibility isusually tested.

OXYGEN H

EAT

FUELTemperature

Time

Figure 39: The various stages of fire on time/temperature curve.

1. IgnitionFlammable interior material, such as textiles or otherupholstery material, catch fire as a result of careless handlingof a heat source. A fire can be caused by cigarettes, matches,radiators, welding equipment or something similar. Fault inelectrical equipment and arson are other possible reasons.The ignition phase can be up to several hours long if the firebegins as glowing combustion. The process can, however, beextremely rapid if flammable materials, liquids or gasesignite. Smoke and gases are generated in a room whenglowing starts and the atmosphere can be life endangeringlong before the temperature in the room starts increasing.

Ignition Flame phase Cooling down

35

2. Flame phase, fully developed fireIgnition becomes a full fire with the so-called flashover.This is a critical phase in the development of a fire. Afterthe flashover it is no longer possible to stay in the roomand constraining and extinguishing the fire will bedifficult. All combustible surfaces in the room are now onfire, and the temperature rises rapidly. The maximumtemperature in a room during the fire phase is 800-1,100 °C. The structure is subject to fire stress, and the firecan spread to other rooms and other parts of the buildingthrough the spread of flames, heat radiation, thermalconduction or convection of combustible gases. Thegeneration of smoke and gases can be extreme.

3. Cooling downDuring this phase the amount of combustible materiallessens and the temperature drops. The combustion processceases gradually.

The most important factors determining the shape ofthe time/temperature curve in Figure 39 are the amountand distribution of combustible material, ventilation,oxygen supply, and the characteristics of buildingcomponents (such as thermal capacity etc).

Radiation

Radiation

Ignition protectedcoating

Wooden floor

Steel beamCeiling

Conduction

Figure 40 A. Fire spread through radiation.

RadiationHeat is radiated from warm bodies to cold ones. For themain part, heat radiation is invisible infrared radiation.The radiation intensity is reduced as a square of distance.

Spreading of fireA fire is spread mainly through radiation, conduction andconvection.

Figure 40 B. Fire spread through conduction.

ConductionHeat is conducted in a material (solid body, liquid or gas)or from one object being in direct contact with another.Metals are the best conductors of heat. Liquids conductlittle heat, and gases even less.

In case of a fire, heat can also be conducted throughnon-combustible materials and structures. A thin concretewall, for example, is no sure obstacle to a spreading fire. Asmetals are good heat conductors, pipes and other suchstructures that penetrate walls can be a risk.

36

Figure 40 C. Fire spread through convection.

ConvectionThe generated flue gases and the surrounding air are heatedin a fire. Since warm gases are lighter than cool ones, therewill be convection or thermal radiation of hot gas mixtures.

In case of an indoor fire, such convected heat can causesecondary fires at long distances from the main fire, partlyby heating combustible materials to ignition temperature,partly because gases that have not burned due to lack ofoxygen are ignited when sufficient oxygen is available.

■ I N S U L AT I O N T H E O R Y – F I R E

Fire protection of buildingsThe purpose of the national building code is to buildhouses in a way that prevents fires. On the other hand, thespreading of fire within a building or to other buildingsshould be prevented. Here is a simple summary of the goalsof fire protection regulations

– To save lives– To save property

This can also be expressed in the following way:

• Preventing firesRequirements concerning non-combustible material,sufficient distance to combustible materials, surfacelayers etc.

• Allowing a secure escape in case of fireRequirements concerning two exits or the maximumlength of escape routes etc.

• Ensuring the durability of structures in case of fireThe frame should consist of fire-engineered structuresetc.

• Reducing the risk of spreading fireRequirements concerning surface materials, fire cells inattics etc.

• Facilitating the putting out of firesRequirements concerning ventilation, fire posts andaccessibility for rescue vehicles etc.

These requirements are the basis for the fire engineering ofa building.

Con

vect

ion

37

Fire insulationFire insulation of buildings is designed to prevent thecommencement and spreading of fire as well as ensuringthat the potential fire can be restricted in such a way that isbeneficial for personal safety.

The construction performance and fire techniqueassessments are in the majority of cases based on full-scaletesting, type tests of all the constructions or testing of aparticular construction. We are currently investing in thedevelopment of theories, which enable the dimensioning offlammable constructions using calculations. It is thereforeimportant to have knowledge regarding the characteristicsof materials at high temperatures.

Fire loadThe term fire load refers to the relationship between thetotal amount of heat that is released on total combustion ofthe entire inflammable material in a building, inclusive ofgoods in the warehouse, inventories etc and the total area ofthe building.

The fire load is expressed as MJ/m2, however in writing,it is often referred to as kg wood/m2. This unit isinternationally recognised and useful, as a significantnumber of fires are related to wood. Other combustiblematerials are calculated from wood in ratio withcombustion heat. The fire load may therefore be calculatedas a sum of the total combustion heat of the material,divided by the combustion heat of one kg tree (pine) that is19000 kJ/kg.

Type of building Fire load in kgWood m2 floor area

House 25 – 45Hospital 20 – 40Hotel 15 – 25Office 20 – 95School 15 – 30Library 200 – 400Storage premises 30 – 125Warehouse 15 – 50

Table 42: The table shows the results of an investigation of thesize of fire load in different buildings with mainly concreteconstructions.

The table indicates that the fire load does not only varybetween different categories of buildings, but also withinthe same category. This depends on the different types andextent of furniture.

A broad knowledge of existing fire loads in differenttypes of building will also provide a foundation for therequirements for the building constructions (bearingconstructions, outer walls, joists etc.) and for the firetechnical dimensions of these.

Figure 43

Fire courseThe fire course – that is to say the temperature/length oftime, largely depends on the fire load and the geometry ofthe firecells.

A complete description of the fire course in a firecell iscomplicated due to the number of influencing factors.Therefore, it is advised to use the normal fire curve (referto Figure 44) for all tests and calculations where there areno particular reasons to veer from it. Such a reason maypose a risk for petrol fires for which there is another curve,the hydrocarbon curve, which portrays another time andtemperature course.

The unique fire properties of stone wool are clear whencompared to the normal fire curve.

Figure 41

Fire load = kg wood/m2

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1200

1000

800

600

400

200

00 30 60 90 120 150 180 210 240 270 300

■ I N S U L AT I O N T H E O R Y – F I R E

Natural stone starts shrinking and thestrength of the concrete is significantlyreduced (120 mins)

Wood gas ignites (approx 5 mins)

Rubber and plastic materials melt and ignite (approx 3 mins)

Figure 44: The normal fire curve

FirecellA firecell may include a room or a group of rooms in abuilding and is designed in such a way that a fire may beprevented from spreading to another area of the buildingwithin a time frame.

Figure 45: A building with two firecells.

The time frame is determined in terms of the function ofthe building and the number of floors.

In the final construction of the firecell, there are nobuilding parts, such as windows and doors with less fireresistance than that which corresponds to the firecell.Providing that the fire can still be prevented fromspreading to these parts of the building, via intervention ofthe fire defence within the normal effort time or in anotherway.

Fire temperature

Length of fire - minutes

Fire technical building classesIn national building codes it is common to devide the typeof buildings into classes. In Sweden, as an example,buildings are divided into three different fire technicalclasses. When dividing the classes, the attic should becalculated as a floor if it is used as a residence or main partof a home.

Buildings in class Br1Buildings with three or more floor plans should be classifiedin class Br1.

The following buildings with two floors should beclassified in class Br1:– Buildings designated to be slept in by people, who are

not expected to have good knowledge of thesurroundings.

– Buildings designated for persons, who are not able toeasily move to safety by themselves.

– Buildings with assembly premises on other floors.

Buildings in Br2 classThe following buildings with two floor plans should beclassified as lowest in the class Br2:– Buildings designated for more than two flats and where

the living area or workroom is in the attic.– Buildings with assembly premises on the ground level.– Buildings which have a building area greater than

200 m2 and which are not divided by fire walls intounits bigger in size than in the lowest class REI 60-M.

Glass melts (approx 7 mins)

39

Buildings with a one floor plan with assembly premises in orunder the ground level should, as lowest, be grouped in theclass Br2.

Since other buildings may depend on a particular typeof usage, there are special guidelines to be followed. Otherbuildings, not mentioned above, are classified as Br 3.

Detailed guidelines for different parts of a building anddifferent types of buildings may be found in provisions infor example The Swedish National Board of Housing,Building and Planning or the Swedish Insurance Sector.

Euro classesAs part of the EU, a new common system for testing andclassifying the fire properties of building materials has beentaken into use. This means that the number of fire testingmethods is reduced and replaced by a fewer number ofharmonised measures. All earlier national classifications arereplaced by the new Euro classes A1, A2, B – F. A1 is thebest class. During the transitional period, the old classesapply within parentheses. The Euro classes describe thebuilding materials’ contribution to fire and the risk of overignition. No over ignition takes place in classes A1, A2 andB. Classes A2-D should be combined with additionalclasses which describe the building material releasingsmoke from itself (s1, s2, s3) or emits burning drops (d0,d1, d3) when affected by the fire. Class E may only becombined with the additional class d2. For the lowest classF, the performance has not been established or it meansthat the product burns easily. The National Buildingguidelines should outline how the different materials maybe used. Paroc stone wool without the outer coating orcovered by glass fibre tissue is classified in the top class A1.

CoatingsIgnition protected coating is a coating made from fire-resistant or another suitable material, which is fixed safelyand which, when fire tested in accordance with standardi-sed methods for a period of at least 10 minutes, preventsthe ignition of the flammable material it covers.

In specific construction solutions, Paroc stone wool isclassified as an ignition protected coating.

Class division of the coatingThe coating is decided upon in terms of smoke progressionand fire spreading in a room during the fire’s preliminaryphase. Therefore, the fire exit requirements are very strict.The regulation divided the coating into three requirementlevels: Euroclass B-s1, d0 (previously coating class I), C-s2,d0 (previously class II) and D s2, d0 (previously class III).Coatings of a lesser quality D-s20, d0 are not utilised.

Classification of the building sectionsThe regulations set requirements for building sectionsdepending on the function of the classes:– R (load-bearing capacity)– E (integrity)– I (insulation)The term integrity refers to the constructional capacity toprevent the entire building burning. Insulation limits thetemperature of the areas not exposed to the fire – normallya maximum temperature of 140 °C.

The class designations that are usually a combination oftwo or three classes are connected to a time frame inminutes such as RE 30 or REI 60.

The classification may be combined with an additionalclassification– M with special consideration paid to mechanical

damage– C for doors with locking device fitted

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■ I N S U L AT I O N T H E O R Y – F I R E

Fire resistant capacity of Paroc stone woolParoc stone wool is manufactured from Diabase stone thatis heated to its melting point of approximately 1500 °Cand is then transformed into fibre by a special process.

Paroc stone wool is therefore to a high degree aninorganic product, apart from a very small content ofbinder and oil.

Resistance to temperatureParoc stone wool is an insulation material with a highmelting point appropriate for constructions with high firerequirements. Insulation does not increase the fire load andthe protected insulation capacity during a fire. The binderin Paroc stone wool products melts at a temperature ofapprox 200 °C. Stone wool fibres reach up to 1000 °C.Paroc stone wool may therefore be used at temperatureshigher than 200 °C – the fibres remain in tact and protectthe underlying material from the effects of the flame.Insulation should be placed in the construction so that themechanical influence may not change form when thebinder leaves.

As a rule, only one side is exposed to high temperaturesfollowed by the breaking down of the binder. Paroc stonewool has good thermal insulation properties even in thehigh temperatures that are incurred during fire. Thetemperature drop of the outermost insulation layer is sogreat that the rest of the insulation remains intact.

Efficiency of the insulationThe following factors conclude the efficiency of the fireinsulation:1) Thermo stability

Insulation material should be able to endure hightemperatures that exist during a fire without melting orshrinking significantly in size.

2) Insulation powerThe insulation power is dependant on the temperature.The insulation material should have good thermalinsulation properties even at high temperatures.

3) Thermal capacityThe thermal capacity of Paroc stone wool is somewhatdependant on the low density. Concrete for examplehas a high thermal capacity.

DimensioningThe aim of fire technical dimensioning is to bring about aconstruction that with safety withstands the influence offire it may be exposed to without collapsing. Furthermore,it often applies to preventing the ignition of material onthe side not exposed to fire.

However, the calculations are complex. In many cases,the dimensions that lead to tests have already beenconducted. For certain constructions, it is possible tocarryout data calculations such as fire insulation of steelconstructions with PAROC FPS 14. The Addition methodis adopted for divided wooden constructions. A summarydescription may be found in the following.

41

Position coefficient (kn) for various material layers in a wallwith a single board layer.

Addition methodGeneralThe following method is not 100 % exact and does notcompensate real fire tests. There are still too many unsureparameters. The result of such a calculation can be used ascomplement to the testing or a reference test can completethe calculation.

Divided Timber StructuresThe addition method is a method for calculating the fireresistance of timber structures with wooden studs and thehighest fire-engineering class of EI 60. By adding the fireresistance of the structure’s various material layers it ispossible to get an estimate of the entire structure’s fireresistance (btot). The calculations are based on a largenumber of fire tests. The starting point is the so-called basevalue (bn). This is the fire resistance of a material layer. Thelocation of the layer in the structure can be established bymultiplying the base value with a position coefficient (kn).The following formula can also be used.

btot = b1k1 + b2k2 + ... = Σbnkn (formula 1)

Examples of a material’s base value (b) and positioncoefficient (k) are shown in the table below.

Base value (bn) of various materials

Type Density Thickness Base value(kg/m3) (mm) (min)

Wood based boards 450–590 12 11.1and all plywood 20 18.7

Chip boards and 600–800 12 13.6fibre boards 22 24.6

Gypsum boardNormal 680–780 13 18.0Protect F ≥830 15 22.0

Glass wool 19 45 5.095 10.0

120 12.0195 20.0

Stone wool 28 45 9.095 19.0

120 24.0195 39.0

Air gap 45–195 5.0

*) With regard to thickness of the exposed board.

**) 0.8 when stud spacing is ≥ 70 mm.

When Protect F or equivalent is used as the exposed board,i.e. the board resists fire ≥ 60 minutes, the followingposition coefficients can be applied:1) The same as for stone wool, however max. 2.9.2) 1.5 for wood-based boards.3) 1.8 for gypsum board and fibre cement boards.4) 2.0 for glass wool

(A) Air gapbtot=(13.6 x 0.8)+(5.0 x 1.0)+(13.6 x 0.6)= 24.0 min

(B) Glass woolbtot=(13.6 x 0.78)+(10.0 x 1.0)+(13.6 x 0.67)=29.7 min

(C) Stone woolbtot=(13.6 x 0.78)+(19.0 x 1.0)+(13.6 x 2.9)= 69.0 min

12 mm chip board95 mm timber stud95 mm (A) air gap, (B) glass wool, (C) stone wool12 mm chip board

A sample calculation.Here is an example of fire resistance (btot) calculatedaccording to the Addition method (formula 1).

Type Thick- Position coefficients

ness

(mm) Exposed board Non-exposed board

cover on rear with covered on front with

Glass wool/ Air Ins/ *)1)Glass Stone wool (mm) 2)3)

stone wool gap air gap wool Air

(mm) (mm) gap

45–195 45–195 45 70 95 145

Wood based 12 0.78 0.8 1.0 0.67 1.9 2.4 2.9 3.9 0.6board and 20 0.94 0.8 1.0 1.23 1.9 2.4 2.9 3.9 0.6**)

all plywood

Chip and 12 0.78 0.8 1.0 0.67 1.9 2.2 2.9 3.9 0.6fibre boards 22 0.98 0.8 1.0 1.37 1.9 2.4 2.9 3.9 0.6**)

Gypsum boardNormal 13 0.80 0.8 1.0 0.74 1.9 2.4 2.9 3.9 0.7Protect F 15 0.84 0.8 1.54) 0.88 1.9 2.4 2.9 3.9 0.7

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■ I N S U L AT I O N T H E O R Y – F I R E

Divided Timber Structures

Addition method -Complementary data for Paroc stone wool

The base values for materials were obtained through anextensive test programme, in which samples were testedboth in full and reduced scale. In order to complement thestudy and mainly to study the influence of stone wooldensity we initiated a test project at the Swedish Institutionfor Technical Education in Norrköping. The methods usedwere the same as the first ones used by Trätek. Some testswere carried out in parallel, and they proved to becompatible. The results are shown in the tables below.

Base values (bn) for various Paroc stone wool products andfor Gyproc Normal gypsum board

Product number Density Thickness Base value(kg/m3) (mm) (min)

Gypsum board 730 13 18.7

PAROC UNS 37 26 45 7.770 10.995 11.6

145 20.3

PAROC FPS 4 45 45 10.470 16.895 20.2

PAROC COS 10 80 50 12.680 24.5

100 32.3

PAROC FPS 14 140 30 11.950 23.570 38.080 43.7

Position coefficients (kn)

Position coefficient Stone Position coefficients

Type Thick- Exposed lined wool Stone Non-exposed lined board with

ness board with stone density wool stonewool on front side, thickness (mm)

(mm) wool on back side (kg/m3) 30 45 50 70 80 95 100 145

Gypsum 13 0.9 26 1.0 – 2.1 – 2.6 – 2.9 – 3.5board 0.9 45 1.0 – 2.3 – 2.7 – 3.3 – –

0.9 80 1.0 – – 3.1 3.5 – 4.1 4.3 – 0.9 140 1.0 2.6 – 3.6 4.1 4.3 – – –

NB. (-) = no data available.

Position coefficients (kn) for various materials layers inwalls with two board layers.

Exposed board Isolation/ Non-exposed boards

Structure2) board 1 board 2 air gap board 3 board 4

Exposed/non-exposed board exposed closest closest non-

+ board closest to stud to stud to stud exposed

2 x wood-based board 1.0 0.6 1.0 0.5 0.7air gap

2 x gypsum board 1.0 1.0 1.0 1.0 0.73)

air gap

Gypsum+ wood-based board 1.0 1.0 1.0 0.8 0.73)

air gap

Wood-based board+ gypsum 1.0 0.6 1.0 1.0 0.73)

air gap

2 x wood-based board 1.0 0.6 1.0 1.01) 2.01)

stone wool, 28 kg/m3

2 x gypsum board 1.0 1.0 1.0 1.01) 3.51)

stone wool, 28 kg/m3

Gypsum+ wood-based board 1.0 1.0 1.0 1.01) 2.01)

stone wool, 28 kg/m3

Wood-based board+ gypsum 1.0 0.6 1.0 1.01) 2.51)

stone wool, 28 kg/m3

1) The value is clearly on the safe side. To obtain highervalues more base layer is required.2) Total board thickness max 26 mm per side of wall.3) 1.0 when stud spacing ≥ 70 mm.

43

A sample calculation.Here are sample calculations of the structure describedabove as per (formula 1) on page 41.

(D) PAROC UNS 37 (26 kg/m3)btot = (18.7 x 0.9)+(11.6 x 1.0)+(18.7 x 2.9)= 82.7 min

(E) PAROC FPS 4 (45 kg/m3)btot = (18.7 x 0.9)+(20.2 x 1.0)+(18.7 x 3.3)= 98.7 min

Addition method –Fire resistance period over 60 minsAccording to the above calculations long fire resistanceperiods are obtained when using stone wool as insulation.These periods are usually longer than the fire-engineeringclasses given in other information material from Paroc.Why? In a fire the inner lining either burns or falls down –usually after the fire has been burning for 15 to 25minutes. After that the studs and the isolation are exposedto fire. Withstanding temperatures of over 1,000 °C stonewool stays in place and protects the non-exposed skinplate. The studs are charred at a rate of 0.7–1.0 mm perminute, and will thus be consumed in about 100 minutes.Oversized insulation units are commonly clamped betweenthe studs, but as the studs burn off, the insulation fallsdown. After that the fire will penetrate the non-exposedboard. As this process can vary from case to case, it isdisregarded when calculating the fire resistance. This iswhy certain care is required when making suchcalculations. This is also the main reason why the Additionmethod may not be used when the burning time exceeds60 minutes.

Another reason for lower fire-engineering classes is that thefire resistance period is rounded down to the closest value.In example D above 82.7 minutes is reduced to 60minutes.

13 mm gypsum board95 mm PAROC slab45 x 95 mm stud c 600 mm13 mm gypsum board

For further information of material properties and ourproducts see www.paroc.com

Other Paroc information regarding fireInstruction for how to insulate steel structure against fireare to be found on our web pages. There will also othertested and approved constructions be found, depending onhow far the implementation of the EN-regulations havegone in different countries.

44

Acoustics

Sound pressureWe are not able to hear the sound intensity directly, as ourhearing experience is based on the perceived sound pressurein the sound wave. The relationship between the intensityand sound pressure may generally be written as follows:

I = p2/(nZ) = p2/(n c) (3)

p = sound pressure, PaZ = c = characteristic impedance of the medium = 400 for

normal air= density of the medium, kg/m3

c = speed of sound in the medium, m/sn = a constant, varying between 1 and 4 depending on the

character of the sound field.

Free plane sound waveWhen the sound source is located at a great distance and ina space where there are no reflective surfaces, a free planesound wave is propagating.n = 1 in the formula (3). Formula (1) applies.

Diffuse sound fieldIn a diffuse sound field, the sound waves arrive from allpossible directions.n = 4 in the formula (3).

In a reverberation room, a diffuse sound field prevails. If asound source is placed in such a room, balance is achievedwhen the supplied and emitted sound power are the same,i.e. in the reverberant field:

P = I · A (4)

P = supplied sound power, WI = intensity, W/m2

A = absorption of the room, m2

Sound power and sound intensitySound is energy and treating sound as energy can solvemany acoustic problems. This is a simple way of looking atsound issues.

A sound source may be considered as a power pointsource emitting P (watt). A point source emits equally in alldirections and at the distance r (m) the intensity I (watt/m2) may be calculated using the following formula. Wepresume no reflective surfaces in the space.

I = P/(4πr2) (1)

P = sound power, Wr = distance, mI = sound intensity, W/m2

For hemispherical conditions, i.e. a sound source on theground, the result will be:

I = P/(2πr2) (2)

Figure 46: The distribution of sound from a point source on the ground.

NOTE! The intensity is inversely proportional to the(distance)2 from the source (Reduced 6 dB every time thedistance is doubled).

Table 47: Sound power output of some sound sources.

Example 1: A crowded Ullevi Stadium in Gothenburg,Sweden (40,000 people) shouts goal! What is the soundpower produced?

Answer: 40,000 · 10-3 W = 40 Watt

P I

r

■ I N S U L AT I O N T H E O R Y – A C O U S T I C S

Source of sound Power, W LW dB

Whispering 10-9 30Conversation 10-5 70Scream 10-3 90Lorry 10-2 100Trumpet, grand piano 10-1 110Compressed air, riveting hammer 100 120Large orchestra 101 1304-engine propeller plane 103 150Jet plane 104 160Rocket 107 190

45

Example 2: The output of a machine is 0.1 W soundpower. It is placed:a) Outdoors on a hard groundb) Indoors in a reverberation room with the absorption

= 6 m2.What is the sound intensity and the sound pressure at100 m distance outdoors and in the reverberant fieldindoors?

Answer: a) I = P/(2πr2) = 0,1/2 · π · 1002 = 1,6 · 10-6 W/m2.p2 = I · (n c) = 1,6 · 10-6 · 1 · 400.p = 0,025 Pa.b) I = P/A = 0,1/6 = 0,017 W/m2.p2 = I · (n c) = 0,017 · 4 · 400.p = 5,2 Pa.

TransmissionWhen a sound wave is incident upon a partition separatingtwo spaces, some of it will be reflected and a small amountwill be transmitted through the partition. We disregard thefact that the wall may absorb sound power.

Iin

Iin

Ir

Itr

Ia

Ir

AbsorptionWhen a sound wave strikes a surface of a room, a propor-tion of the sound will be reflected. The remaining soundwill be absorbed.

Figure 48: Transmission through a partition.

Iin = Itr + Ir (5)τ = Itr / Iin (6)ρ = Ir / Iin (7)

Iin = Incident sound intensity, W/m2

Itr = Transmitted sound intensity, W/m2

Ir = Reflected sound intensity, W/m2

τ = Transmission coefficientρ = Reflection coefficient

Figure 49: Absorption of a surface.

Iin = Ir + Ia (8)α = Ia / Iin (9)ρ = Ir / Iin (10)

Iin = Incident sound intensity, W/m2

Ia = Absorbed sound intensity, W/m2

Ir = Reflected sound intensity, W/m2

α = Absorption coefficientρ = Reflections coefficient

Note: In reality, of course, absorption and transmissionoccur at the same time in most cases.

46

dB – quantityThe terms dB and Bel (=10 dB) are purely mathematicalterms and not special measures for sound. If you compare twomagnitudes such as A and B, it can be said that A is a certainnumber of times greater (older, heavier etc.) than B, or acertain number of dB in relation to B. If a great grandfather is100 years old and his great granddaughter is 1 year old, he is100 times = 102 = 2 Bel older than her. To be precise, grand-father is 20 dB old (compared to his great granddaughter). Belis logarithmic to a relationship between two magnitudes.

Within acoustics, we talk about a level (L), which ispresented in dB, regarding the sound power as well asthe sound intensity and the sound pressure (plusvibrations etc.). Therefore, the different kinds of dB valuesand their reference values must always be kept in mind.

Reference values:Sound power W0 = 10-12 WSound intensity I0 = 10-12 W/m2

Sound pressure p0 = 20 · 10-6 Pa

Levels:Power level Lw = 10 · log(W/W0) dBIntensity level LI = 10 · log(I/I0) dBSound pressure level Lp = 10 · log(p/p0)2 dB

The reference value for sound pressure corresponds to thehearing threshold (at 1000 Hz). The intensity level andsound pressure level are about the same in free plane soundwave in “normal” air.

Adding dBdB is a logarithmic quantity. One has to revert to linearquantities in order to add or subtract and then go back tothe logarithm.

Example 3: Five different dB values are to be added: 43,45, 33, 32 and 38 – what is the sum?

Answer: Pass to the Bel values first: 4.3, 4.5, 3.3, 3.2 and3.8 Bel are to be added. Now go to the linear figures andadd. 104.3 + 104.5 + 103.3 + 103.2 + 103.8 = 20000 + 30000+ 2000 + 1600 + 6300 = 60000 = 104.8. So the sum will be4.8 Bel = 48 dB.

Sound reduction indexThe term sound reduction index is actually the same as thetransmission coefficient but expressed in dB. In order toobtain a positive value of dB, you have to invert thecoefficient.

R = 10 · log(1/τ) (11)

R = Sound Reduction Index, dB

τ = Transmission Coefficient

Example 5: The transmission coefficient at a certainfrequency isa) for 160 mm concrete = 0.000003b) 13 mm plasterboard = 0.001.What is the Sound Reduction Index?

Answer: a) R = 10 · log 1/(3 · 10-6) = 55 dBb) R = 10 · log 1/0.001 = 30 dB

The resultant sound reduction indexIf a wall consists of two or more elements (windows, doorsetc) with different sound reduction index, we may calculatethe resultant sound reduction index for the entire wall byusing the following formula:

τ0 · S0 = τ1 · S1 + τ2 · S2 + τ3 · S3 + ... (12)

Example 4: Express the answer for the above examples 1and 2 in dB!

Answer: 1) Lw = 10 · log(W/W0) dB = 10 · log(40/10-12)dB = 136 dB.2 a) LI = 10 · log(I/I0) dB = 10 · (1,6 · 10-6/10-12) dB = 62 dBLp = 10 · log(p/p0)

2 dB == 10 · log(1,6 · 10-6 · 1 · 400/400 · 10-12) dB = 62 dB(Lp similar to LI in a free plane sound wave)2 b) LI = 10 · log(I/I0) dB = 10 · (0,017/10-12) dB = 102 dBLp = 10 · log(p/p0)

2 dB == 10 · log(0,017 · 4 · 400/400 · 10-12) dB = 108 dB(Lp is 6 dB greater than LI in a diffuse sound field).

■ I N S U L AT I O N T H E O R Y – A C O U S T I C S

47

If a wall with the area of S m2 divides two rooms, thesound reduction index can be calculated using thefollowing formula:

R = L1 - L2 - 10 · log(A2/S) (14)

L1 = sound pressure level in the source room, dBL2 = sound pressure level in the receiving room, dB

A2 = absorption of the receiving room

S = area of the dividing partition, m2

The sound pressure levels L1 and L2 are measured directly,whereas A2 is calculated using the formula (17), aftermeasuring the reverberation time T.

Note: R represents the sound reduction index of a partition(measured in a lab).R' represents the apparent sound reduction index betweentwo rooms (measured in the field).

L1 S

R

L2

A2

Figure 50: Sound insulation between two rooms.

Measurement and calculation of the soundreduction index

or expressed in dB for two surfaces:

R0 = R1 - 10 · log[S1/S0 + S2/S0 · 100,1(R1-R2)] (13)

τ0 = resultant transmission coefficient

R0 = resultant sound reduction index

S0 = total areaτn = transmission coefficient of element n

Rn = sound reduction index of element n

Sn = area of element n

Example 6: A door with R = 30 dB is placed in a wall withR = 52 dB. The area of the door = 2 m2 and the wall area(including the door) = 25 m2. What is the resultant soundreduction index?

Answer: R0 = R1 - 10 · log[S1/S0 + S2/S0 · 100,1(R1-R2)] == 52 - 10 · log[23/25 + 2/25 · 100,1(52-30)] = 52 - 11 = 41 dB

Example 7: Practising a piano in the living room produces95 dB. There is a bedroom in an apartment next to theliving room with a common dividing area of 8 m2. Assumethat the wall’s sound reduction index is 55 dB and that theabsorption in the bedroom is 12 m2. What is the soundlevel in the bedroom?

Answer: L2 = L1 - R - 10 · log(A2/S) == 95 - 55 - 10 · log(12/8) = 38 dB

48

Room acousticsSound propagationThe sound pressure level from a point sound source with aknown output power may be achieved by:

Lp = LW + 10 · log(1/4πr2 + 4(1-αm)/A) (15)

Lp = Sound pressure level at a distance of r, dB

LW = Sound power level from the point sound source, dB

r = Distance, m

αm = Average absorption coefficient of the surfaces of the room

A = Absorption of the room, m2

The first term describes the sound pressure level in thedirect field and the second term describes the soundpressure level in the reverberant field.

A reduction of 6 dB in the sound pressure level in directfield corresponds to a doubling of distance from a pointsound source. Compare with a linear sound source (forexample, traffic on a road) that provides 3 dB/doubling ofdistance.

AbsorptionA fraction of the sound intensity is absorbed when the soundis incident to a surface. Each surface has a certain absorption.

A = S · α (16)

α = absorption coefficient

S = area of the surface, m2

A = absorption of the surface, m2

Reverberation timeThe reverberation time is used to characterise the absorp-tion properties of a room. It is defined by the size and theabsorption of the room.

T = 0.16 · V/A (17)

T = Reverberation time, s

V = Volume of the room, m3

A = Absorption of the room, m2

The reverberation time is defined as the time for the soundpressure level to decay by 60 dB, after the sound from aloudspeaker has been turned off or a gun has been fired. Astraight line may generally approximate the decay of thesound pressure level.

The absorption of a room can be calculated as follows:

A = A1 + A2 + A3 +... =∑Sn · αn = S · αm (18)

A = Total absorption of the room, m2

A1 = Absorption of the surface 1, m2

S1 = Area of the surface 1, m2

α1 = Absorption coefficient for the surface 1

Sn = Area of surface n, m2

αn = Absorption coefficient for the surface n, m2

αm = The room’s average absorption coefficient

S = The total area of the room, m2

Example 8: The desired reverberation time in a classroomis max. 0.8 seconds. Assuming the dimensions of theclassroom are 6 · 10 · 3 m and that you want to achieve thementioned reverberation time with just an absorbentceiling. What is the absorption coefficient required for theabsorbent ceiling?

Answer: A = 0.16 · V/T = 0.16 · 180/0.8 = 36 m2.α = A/S = 36/ 60 = 0.6

Sound as a wave motionIn many cases, the sound cannot be treated in terms ofenergy but has to be considered as a wave motion. Airmolecules move and vibrate around their equilibrium in asound wave. The distance between two particles in thesame motion phase constitutes a wavelength. The numberof oscillations per second makes up the frequency. Thefollowing equation describes the relationship:

c = f · λ (19)

c = speed of sound, m/s

f = frequency, Hz

λ = wavelength, m

In air, the speed of sound is approximately 340 m/s(depending on the temperature).

In gases, there are only longitudinal waves and the speeddoes not depend on the frequency.

In plates (sheet material such as plaster and chipboard)there are also bending waves. The speed of sound inbending waves is dependant on the frequency and increasesas the frequency increases.

■ I N S U L AT I O N T H E O R Y – A C O U S T I C S

49

Weighting dB

Figure 51: Equal loudness curves according to ISO 226.

FilterWhen measuring, attention must be taken to the sensitivity ofthe ear by using a filter connected between the microphoneand the measuring instrument. Usually, the measuring is doneusing an A filter, dBA, when there is a risk of hearing damageand discomfort.

In apartments, C filters are often used in order toestimate the low frequency noise annoyance.

Hearing and frequency weightingfiltersThe human ear responds differentially throughout theaudio frequency spectrum. Our hearing threshold is about0 dB in the frequency bands of 1 – 4 kHz, but is muchhigher at low frequencies. It is 40 dB at 50 Hz.

As a rule of thumb, the lowest change in sound pressure levelthat can be heard is 3 dB and a change of 10 dB is subjectivelyheard as a doubling of the loudness. As highlighted in thediagram showing curves representing equal loudness for puretones, these rules of thumb are very simplified.

Figure 52: Standardised weighting curves.

Octave Hz A-filter C-filter

16 -56,7 -8,531 -39,4 -3,063 -26,2 -0,8

125 -16,1 -0,2250 -8,6 0500 -3,2 01000 0 02000 1,2 -0,24000 1,0 -0,88000 -1,1 -3,0

Table 53: Weighting values for filters A and C.

By noise control the sound is preferable measured in octavebands and by sound insulation problems the measurementsare made in third octave bands (1/3 octaves). These bandshave a constant relative bandwidth. The following applies:

B = fu - fl. (20)

Octave band: fu = 2 · fl and fc = √2 · flThird octave band: fu = 21/3 · fl and fc = 21/6 · fl

B = Bandwidth, Hz

fu = Upper band edge, Hz

fl = Lower band edge, Hz

fc = Centre frequency, Hz

50

Answer:

125 250 500 1000 2000 4000 Octave

69 62 55 52 48 45 Sound pressure level

-16 -9 -3 0 +1 +1 A-filter

53 53 52 52 49 46 A-weighted level in fanroom

20 30 50 55 60 60 Sound reduction index

33 23 2 - - - A-weighted level inbedroom

The correction term 10 · log(A2/S)= 0, so L2 = L1 - RThe total level in the fan plant room (logarithmic

addition of 53 + 53 + …) = 59 dBAThe total level in the bedroom (logarithmic addition of

33 + 23 + 2) = 33 dBA

Sound insulationThere are two types of insulation related to buildings.

Airborne sound insulationAirborne sound insulation is the expression used when thesound is produced directly to the air, like speech, singingand sound from the radio and TV.

Airborne sound insulation is determined frommeasurements of the sound reduction index (definedearlier) in 21 third octave bands from 50 to 5000 Hz.

A high R or R' indicates good insulation.

Impact sound insulationImpact sound insulation stands for insulation againstsound created when someone walks on a floor close to theflat.

Impact sound insulation is determined frommeasurements of the sound pressure level, when the floor ismechanically worked on with a standardised hammer. Thelevel is measured in 21 third octave bands from 50 – 5000Hz.

Example 9: In a fan plant room, the sound pressure levelswere measured in octave bands as shown in the table. Thefan plant room shares a partition with a bedroom with anarea of 10 m2, and the absorption of the bedroom is10 m2.The sound reduction index of the wall in octave band isshown in the table. What is the sound level in dBA in thefan plant room and in the bedroom?

■ I N S U L AT I O N T H E O R Y – A C O U S T I C S

Ln = Li + 10 · log(A/10) (21)

Li = Impact sound pressure level in the receiving room, dB

A = Absorption in the receiving room, m2

Ln = Impact sound pressure level for a floor (lab.

measurements)

L'n = Impact sound pressure level (field measurements)

A low value for Ln or L'n means good insulation.

51

Calculation of RwA sound insulation measurement is presented by a curve ofR or R' from 50 to 5000 Hz according to the figure. R andR' are given with one decimal.

When the weighted value Rw or R'w is to be calculated,the curve has to be weighted in a suitable manner. Theweighting performance means that the measured curve iscompared with a standardised reference curve between 100and 3150 Hz. See figure 54.

Figure 54: Measured sound reduction index, R’, and thereference curve.

The reference curve is moved in steps of 1 dB towards themeasured curve, until the sum of the deviations below thereference curve is as large as possible, however, notexceeding 32.0 dB.

The value of the reference curve at 500 Hz after it hasbeen moved is defined as Rw or R'W for the measured curve.

Sound reduction index dB

Frequency Hz

C termsThe Rw concept is taken to pass the situation when the humanvoice is the main source of sound. As we have noise sourcessuch as stereos in flats today, the term is insufficient. When itcomes to the insulation of outside walls where the source ofsound is made up by traffic, the Rw term is also unsuitable.

Therefore, the ISO 717 standard incorporates the Rwcompleted with the spectrum adoption terms or C terms.The complete single-number quantity reads:

Rw (C; Ctr; C50-3150; Ctr,50-3150)

C and Ctr include the frequency range 100 – 3150 Hz.

The single-number quantities

Rw + C and Rw + C50-3150

express the sound insulation in dBA for a noise spectrumwith identical levels in all third octave bands.

The single-number quantities

Rw + Ctr and Rw + Ctr,50-3150

express the sound insulation in dBA for a standardisedtraffic noise spectrum.

Floor with wooden joistsFigure 54 and Table 56 show the results from a fieldmeasurement of a floor with wooden joists according toFigure 55. The floor covering consist of 14 mm parquetflooring and 3 mm plastic foam.

Figure 55: Floor with wooden joists.

52

Impact sound pressure level dB

Frequency Hz

Calculation of Ln,wImpact sound insulation is calculated from measurementsof the sound pressure level that a standardised hammerproduces in the room below. The result can be presented asa curve of Ln or L'n from 50 – 5000 Hz. Ln and L'n aregiven with one decimal.

When calculating the single-number quantity Ln,w orL'n,w, you proceed in a similar way from the impact soundlevels for the 16 frequencies 100 – 3150 Hz and comparethe curve with a standardised reference curve.

The same conditions as for Rw apply regarding totaldeviations max. 32.0 dB in addition to the reading of thereference curve after moving at 500 Hz. Observe that thedeviations this time lies above the reference curve. Seefigure 57.

f R' Ref. ∆R' Pink A-filter A- L-R'Hz dB curve noise weighted

pinknoise L

50 21,0 -11 -30,2 -41 -61

63 27,9 -11 -26,2 -37 -64,9

80 29,8 -11 -22,5 -34 -63,8

100 37,5 39 1,5 -11 -19,1 -30 -67,5

125 42,7 42 0 -11 -16,1 -27 -69,7

160 43,4 45 1,6 -11 -13,4 -24 -67,4

200 48,5 48 0 -11 -10,9 -22 -70,5

250 45,4 51 5,6 -11 -8,6 -20 -65,4

315 47,3 54 6,7 -11 -6,6 -18 -65,3

400 50,5 57 6,5 -11 -4,8 -16 -66,5

500 54,0 58 4 -11 -3,2 -14 -68

630 57,3 59 1,7 -11 -1,9 -13 -70,3

800 59,7 60 0,3 -11 -0,8 -12 -71,7

1000 62,6 61 0 -11 -0 -11 -73,6

1250 64,7 62 0 -11 0,6 -10 -74,7

1600 65 62 0 -11 1 -10 -75

2000 64,3 62 0 -11 1,2 -10 -74,3

2500 64,9 62 0 -11 1,3 -10 -74,9

3150 69,7 62 0 -11 1,2 -10 -79,7

R’w = Σ∆R’ = logΣ = logΣ =

58 dB 27,9 dB 0 dB -55 dB

Table 56: Calculation of R’w and C50-3100 for the floor in Figure54. When the reference curve is moved as high as possiblewithout Σ∆R’ >32.0 dB, the R’w value will be read at 500 Hz.The pink noise stands for similar levels in each third band. Thelevel is chosen so that the A-weighted level (logarithmic sum of L)is 0 dB (normalised to 0 dB).The logarithmic sum of L-R’ will be -55 dB that is to say theconstruction insulates 0- (-55) = 55 dB for the selected pink noisesound spectrum.Therefore, C50-3100 will be 55-58 = -3 dB

■ I N S U L AT I O N T H E O R Y – A C O U S T I C S

The complete single-number quantity for the airbornesound insulation is as follows:

R'w = 58 (-2; -6; -3; -15) dB, so

R'w + C = 58-2 = 56 dB, and

R'w + C50-3150 = 58-3 = 55 dB

The result means that the single-number quantity will dropwhen care is taken to the poor insulation at lowerfrequencies.

Figure 57: Measured impact sound pressure level, L’n and thereference curve.

53

f L'n Reference ∆L

Hz dB curve

50 61,863 64,880 59,7

100 56,4 54 2,4125 58,0 54 4,0160 60,0 54 6,0200 58,0 54 4,0250 58,3 54 0,3315 58,3 54 4,3400 54,6 53 1,6500 52,4 52 0,4630 47,6 51 0800 44,6 50 01000 44,4 49 01250 41,4 46 01600 40,2 43 02000 38,4 40 02500 35,1 37 03150 28,6 34 0

logΣ = L'n,w = Σ∆L70,1 dB 52 dB 27,0 dB

Table 58: Calculation of L’n,w and C50-2500 for the floor in Figure55. When the reference curve is moved downwards as far aspossible without ∑∆L> 32.0 dB, L’n,w is read at 500 Hz.L’n -values for the frequency range 50 – 2500 Hz are summarisedlogarithmically and provide a total level of 70.1 dB from thehammer.C50-2500 is calculated from C50-2500 = log∑L -15 - L’n,w = 70 - 15 -52 = 3 dB

C termsThe Ln,w concept does not provide a completely correctpicture of the impact sound insulation for different typesof floors, especially with wooden joists. Above all, you needto pay attention to higher impact sound levels atfrequencies below 100 Hz for light wood beams.Therefore, ISO-717 complements Ln and L'n with twospectrum adoption terms or C terms:

Ci,100-2500 and Ci,50-2500

These terms take into consideration the measured soundlevel from the hammer for the entire emitted frequencyrange and the single-number quantity becomes:

Ln,w (C; Ci,50-2500)

Flats separated by floor with wooden joistsFigure 57 and Table 58 show the results from an impactsound level measurement (in the field) on the same lightfloor with wooden joists shown in Figure 55.

The single-number quantity for the impact sound level is:

L'n,w = 52 (0; 3) dB, so

L'n,w + C = 52 + 0 = 52 dB, and

L'n,w + Ci,50-2500 = 52 + 3 = 55 dB

The result means that the single-number quantity increases(= poor impact insulation), when the impact sound levelsbelow 100 Hz are taken into consideration.

54

CoincidenceThe speed of sound for free bending waves in the panel isdependent on the frequency. At a critical frequency, thesound velocity in the panel will be the same as the speed ofsound in air. At this coincidence frequency, the wavepattern of the panel comes close to the sound waves in theair and the panel´s insulation capacities decrease.

The coincidence frequency is defined by the stiffness ofthe sheets. In the coincidence area and at higherfrequencies, the sound reduction index is affected by thepanel’s internal damping.

As an example it may be noted that the coincidence of13 mm plaster occurs at approximately 3000 Hz. Therefore,if 26 mm plaster should be used, the coincidence frequencymoves to lower frequencies, which is unfavourable. 2 times13 mm plaster is therefore a better solution.

Double leaf partitions and cavity wallsCavity walls are a suitable alternative when high insulationis required for a light construction. Cavity wallconstructions are defined due to their resonance frequency.Below the resonance frequency, the wall behaves as a singlewall in terms of the wall mass, near the resonancefrequency the sound reduction curve dips and above theresonance frequency the insulation will be very high.

Figure 59: Cavity wall with a steel studded frame.

■ I N S U L AT I O N T H E O R Y – A C O U S T I C S

FaçadesIf a wall or a window shall insulate against traffic noisefrom the street, the single-number quantity for insulationagainst traffic noise needs to be as high as possible.

Typical value for a highly (thermal) insulated woodenfaçade is:

Rw = 48 (-2; -7; -2; -12) dB

The insulation against traffic noise for such a woodenfaçade would be:

Rw + Ctr,50-3150 = 48 - 12 = 36 dB

Wall constructions insulatedagainst soundSingle leaf panelSingle leaf partitions are one-layer constructions, one sheetof plaster, glass, brick, concrete etc. The sound reductionindex is described by the mass of the wall, the stiffness, theinternal damping and interaction with flanking walls aswell as the area. Looking only at the mass of the wall panelcan make a good first approximation. The sound reductionindex according to the mass law is as follows:

R = 20 · log m + 20 · log f - 49 dB (22)

m = mass/area, kg/m2

f = frequency, Hz

Note: The sound reduction index increases by 6 dB whenthe mass is doubled and with 6 dB when the frequency isdoubled, 6 dB/octave.

Example 10: Calculate the sound reduction index for 1 mmsteel plate at 500 Hz.

Answer: R = 20 · log m + 20 · log f - 49 == 20 · log 8 + 20 · log 500 - 49 = 23 dB

55

AbsorberWhen a highly absorbent material as stone wool is used inthe air cavity, the sound insulation increases. The greaterthe cavity, the greater the benefit obtained from theabsorber. Generally you can expect an increase of about 5 –10 dB of R with a filled wall compared to an empty wall.

Rigid connectionsA rigid connection between two cavity walls may have adreadful effect when the cavity wall panels consist of stiffmaterials, such as concrete or light concrete. For resilientpanels such as plaster sheets, the deterioration is not so severe.

”Resilient Skin”The radiation-decreasing resilient skin is a typical way toimprove the insulation of a heavy wall. It consists of aresilient panel, for example 13 mm of plaster, which ismounted to a heavy wall such as light concrete with stonewool behind. The mounting may be carried out withordinary studs, or according to special mounting methodswithout direct contact to the existing wall. In order toreach a good effect at lower frequencies, the thickness ofthe stone wool must be increased.

Resonance frequencyThe resonance frequency is calculated as follows:

fr = 60 · √(m1+m2)/(m1 · m2 · d) (23)

fr = Resonance frequency, Hz

m1 = Mass of partition wall 1, kg/m2

m2 = Mass of partition wall 2, kg/m2

d = Distance between partition walls, m

Example 11: A cavity wall with a panel of 13 mm plaster(9 kg/m2) on both sides. If you want a resonance frequencyof 63 Hz, what distance would you need between theplaster panels?

Answer: fr = 60 · √(m1+m2)/(m1 · m2 · d) = 63 == 60 · √18/(81 · d). d = (60/63)2 · (18/81) = 0.2 m

Sound insulated floorsFloors should satisfy both the demand of high airbornesound insulation and low impact sound levels.

As a rule, it is more difficult to meet the impact soundrequirement.

Airborne sound insulationThe principles for single leaf walls and cavity walls are alsoapplicable to floors.

A homogenous concrete solid floor of 16 cm can beexpected to fill the demands of sound reduction indexoutlined for example in the Swedish regulation. Hollowconcrete floors should give the same effect when the weightcorresponds to 16 cm of homogenous concrete.

Floors with wooden beams may not be expected to meetthe standard requirements. In order to achieve highinsulation with the wooden beams, the cavity wall principleneeds to be applied, i.e. using a free hanging ceiling and afloating floor.

In practice you can achieve a good effect by hanging theceiling (plaster sheets etc.) in resilient clips or studs.

A so-called floating floor consisting of a board etc on aresilient layer gives a very good effect.

AbsorberLike in cavity walls, stone wool absorbers are effective infloors with wooden joist structure if the ceiling issuspended. But the effect is poor if the beams connect thefloor on the upper side with the ceiling on the lower side.

Impact sound insulationThe impact sound requirement can be satisfied by using oneof two main principles, floating floors or soft floor covering.

Soft floor coveringThis solution to the impact sound insulation problem canonly be used when the floor itself satisfy the requirementsfor airborne sound insulation and it is the normal solutionfor concrete solid floors.

The soft floor covering may consist of a soft fitted carpetor linoleum with a soft underside. The Swedish NationalTesting and Research Institute (SP) have approved impactsound tested carpets.

The carpets have been tested on a concrete floor andreceived a weighted reduction of impact sound pressurelevel, ∆Lw dB. If you assume that for a concrete floor Ln,wis about 75 dB, you must have a soft carpet with ∆Lw atleast 17 dB in order to reach Ln,w = 58 dB.

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■ I N S U L AT I O N T H E O R Y – A C O U S T I C S

Vibration isolationA machine such as a fan placed directly on a floor maytransmit the vibration as sound throughout the entirebuilding. In order to avoid this, the machine must beplaced on vibration isolators made of steel or rubber or astone wool slab. In order to get an effective isolation, thefloor under the springs must be heavy or solid. As a rule ofthumb, the weight of the ground floor should exceed thatof the machine by approximately 4 times. A wooden joistmay have to be fitted with rigid steel joists in order tofunction as a ground.

Figure 61: Vibration isolation.

The advantage with a stone wool slab as an isolator is thatthe internal damping is high, for which reason theamplitude will not be great at resonance. It is importantthat the vibration insulation is correctly calculated. Theresonance frequency for the system(sometimes referred toas natural frequency), fr, Hz should be much lower thanthe lowest disturbing frequency from the machine fs, Hz.

Floating floorA floating floor consists of a board, a slab, etc on a resilient layer.

An effective floating floor should have as heavy sheetsand as soft resilient layers as possible. If the board consistsof a concrete slab and the elastic layer of stone wool, theeffect will be excellent. It is extremely important that theconcrete slab has no contact to the basic floor. Thereforethe resilient layer also must divide the concrete slab fromthe surrounding walls.

Figure 60: Floating floor.

However, you can achieve a fairly good effect with a lightand dry floating floor made from chipboard and plaster onlayers of stone wool. The effect may be improved by layingthe floor on a sheet of sand above the resilient layer, whichwill increase the mass.

If you would try to achieve a high sound insulation, youmay need to use a floating floor system.The floating floor isused for both airborne and impact sound insulation.

Sound insulation in buildingsIn order to achieve the desired insulation in the buildingfrom the chosen constructions, all non-desired soundtransport must be avoided. These are of two types:

Flanking transmissionIn a building, a fraction of the sound transmission betweentwo rooms may go by a flanking building element, such asthe outer wall or the ceiling. In order to avoid this, themanufacturer’s instructions must be followed carefully.There are often requirements for a safety margin on thedifferent sound data of the elements in order to avoid theflanking transmission.

LeakageSlits, ventilation channels, common tubes for the TV-cables, are all examples of objects that may result in soundleakage. This can be avoided by good planning and jobperformance.

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Elastic facilities of stone wool slabsParoc stone wool slabs are a heterogeneous material. Thismeans that the dynamic elasticity may not be deducedfrom the measured results of the static compression, butmust be measured separately.

The principal appearance of the dynamic stiffness sdrelated to the load, is presented in Figure 62. Observe thatsd is constant above a certain load of approximately500 kg/m2.

Figure 62: The principal appearance of the dynamic stiffnessrelated to the load for Paroc stone wool slabs.

The resonance frequency for the vibration isolation systemcan be calculated from the following formula:

fr = (1/2π) · √s/m Hz (24)

fr = Resonance frequency, Hz

s = Dynamic stiffness, N/m3

m = Load, kg/m2

When stone wool and similar materials are used as”springs” the dynamic stiffness, s, consists of twocomponents - sd is the stiffness of the material and sa thestiffness of the enclosed air. sa can be calculated fordifferent thickness to the following values.

h mm sa MN/m3

5 2210 1120 630 450 2

100 1

Choice of slab/thicknessWhen choosing the slab and the thickness, the followingfactors need to be taken into consideration:• Compression of the slab under static load

• Recommended static maximum load of the slab

• The dynamic stiffness of the slab, sd according to Table 63.

In floating floor constructions, the elastic layers should be assoft as possible. According to the test standards, the dynamicstiffness of stone wool shall be presented by a load of200 kg/m2, when the stone wool is used as a resilient layerunder a concrete slab in a floating floor construction.

Dynamic stiffness, MN/m3

Thickness mm 17 25 30 50

ProductPAROC SSB 1 12 10PAROC SSB 2t 20 15

Table 63. Dynamic stiffness, MN/m3 for Paroc slabs. The valueat a load of 200 kg/m2 load shall be used in floating floorconstructions with concrete slabs. The value at a load of> 500 kg/m2 in vibration isolation of machines etc. To thesevalues, the dynamic stiffness of the enclosed air must be added.

Example 12: A machine weighs 100 kg and is to be placedon a concrete slab of 1 times 2 metres. It is then placed ona 100 mm stone wool slab with sd = 10 MN/m2 at a load> 500 kg/m2 on top of a concrete floor. If you need aresonance frequency of 30 Hz for the system, how thickmust the concrete slab be?

Answer: fr = (1/2π) · √sd/m = 30 = (1/2π) · √10 · 106/mm =[1/(30 · 2π)]2 · 10 · 106 = 280 kg/m2. The machineweighed 100/2 = 50 kg/m2, i.e. the concrete slab shallweigh 230 kg/m2 and should therefore be approximately10 cm thick. (According to the diagram, sd and fr areprobably lower at this lower load. This is only favourablefor the vibration isolation).

30

25

20

15

10

5

0

Load kg/m2

sd MN/m3

0 500 1000 1500 2000

For further information of material properties and ourproducts see www.paroc.com

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CE marking

Why CE marking?In order to facilitate trade within Europe, harmonizedstandards have been produced for a number of goods to befreely sold within the whole EU without nationalrestrictions. The standards for thermal insulation productscontain relevant product properties. Reference is made totesting methods and the designations and levels of theproperties are fixed, sometimes in the form of limiting

■ I N S U L AT I O N T H E O R Y – R E G U L A T I O N S

values, but mostly in the form of classes. CE marking is theway to ensure that the product properties are tested andreported in the same way within the whole of the EU.

Standards for thermal insulationproductsThe European standard EN13162 is applicable to mineralwool that is intended to be used as thermal insulation inbuildings. It is called “Thermal insulation products forbuildings – Factory manufactured mineral wool products(MW) – Statement of properties”.

There are 11 materials included in the standard packagefor building insulation, all with lower declared thermalconductivity, λD, than 0.06 W/mK. Common propertydesignations and class limits have now been introduced forall of these 11 materials. A minimum level of internaltesting routines has been introduced for manufacturers,sometimes supplemented by external manufacture testingof certain properties.

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PAROC OY ABNeilikkatie 17P.O.Box 294FIN-01301 VANTAA, FinlandTel. +358 204 55 4868Fax +358 204 55 4738

A M E M B E R O F P A R O C G R O U P

PAROC GROUP is one of the leading manufacturers of mineralwool insulation products and solutions in Europe. Paroc productsand solutions include building insulation, technical insulation,marine insulation, structural stone wool sandwich panels andacoustics products. We have production facilities in Finland,Sweden, Lithuania, Poland and Great Britain. We have sales andrepresentative offices in 13 countries in Europe.

Paroc building insulation is a widerange of products and solutions for alltraditional building insulation. Thebuilding insulation is mainly used for thethermal, fire and sound insulation ofexterior walls, roofs, floors andbasement, intermediate floors andpartitions.

2040BIEN1003

Paroc technical insulation is usedfor thermal, fire and sound insulation inbuilding techniques, industrial processesand piping, industrial equipment andship structures.

Paroc Fire Proof Panels are steel-faced lightweiht panels with a corematerial of stone wool. Paroc Panels areused for façades, partition walls andceilings in public, commercial andindustrial buildings.

Warranty: Our recommendations are based on our most up-to-date knowledge and experience. As theproducts are used outside our control we cannot take responsibility for any damage which may be causedwhen using the product. This brochure replaces all earlier ones. Because of constant development allinformation is subject to change without notice.