FAQ:FAQ General

Questions and Answers

This collection of Questions and Answers refers to the one-dimensional version of WUFI. Some of the program features mentioned here may not apply to WUFI-2D.

1. Sources of Material Data
2. Sources of Climate Data
3. Volume Percent, Mass Percent
4. Relative Humidity in a Building Component

5:<A HREF="#05">Default Moisture Storage Function</A> 6:<A HREF="#06">Water Content of an Air Layer</A> 7:<A HREF="#07">Convergence Failure Caused by Vapor-Permeable Layer</A> 8:<A HREF="#08">Moisture Content at a Monitoring Position</A> 9:<A HREF="#09">Liquid Transport Coefficient of a Paint Layer</A> 10:<A HREF="#10">Rain Absorption Factor of an Unrendered Sandstone Wall</A> 11:<A HREF="#11">Hygric Parameters of Ecological Insulation Materials</A> 12:<A HREF="#12">Typical Initial Moisture Content</A> 13:<A HREF="#13">Water-Repellent Facade</A> 14:<A HREF="#14">Simulation of Absorption Experiment</A> 15:<A HREF="#15">Absorption Experiment with Limited Water Supply</A> 16:<A HREF="#16">Why Never 100% RH in the Component?</A> 17:<A HREF="#17">100% RH at the Facade</A> 18:<A HREF="#18">Criteria for Evaluating Hygrothermal Performance</A> 19:<A HREF="#19">Ventilated Curtain Walls</A> 20:<A HREF="#20">Heat Flow Through Exterior Surface Is Not Realistic?</A> 21:<A HREF="#21">More Recent Weather Data?</A> 22:<A HREF="#22">Averaging of Conductivities at Element Boundaries</A> 23:<A HREF="#23">Conversion of Radiation Data for Other Directions</A>

<A NAME="01">(1):</A>
Where can I find material data for materials which are not included in the database?

Unfortunately, finding material data for hygric simulations can prove difficult since there are no standard collections of such data as yet. While thermal data can be found in many books, hygric data are sparse and hard to come by.

A collection of design values for <A HREF="BasicMaterialData.htm">heat conductivity</A> (including the effect of practical moisture content) and <A HREF="BasicMaterialData.htm">diffusion resistance factors</A> is listed in German standard DIN 4108-4 and numerous textbooks on building physics. The new DIN EN 12524 lists thermal as well as basic hygric design values for building materials.

An extensive list of "NIST Heat Transmission Properties of Insulating and Building Materials" is available on-line at <A HREF="http://srdata.nist.gov/insulation/">http://srdata.nist.gov/insulation/</A>.

<A HREF="MoistureStorageFunction.htm">Moisture storage functions</A> and <A HREF="LiquidTransportCoefficients.htm">liquid transport coefficients</A> may be estimated from the standard parameters <A HREF="MoistureStorageFunction.htm">wf, w80</A> and the <A HREF="LiquidTransportCoefficients.htm">A-value</A> which may also be found in some textbooks (at least for selected materials) and data sheets or can be measured relatively easily.

Occasionally, some data may be found scattered through the specialised literature, but there is no systematic way to retrieve them.

Sometimes the manufacturer may be able to provide material data. Some laboratories (including IBP) can measure the required data if samples are provided.

<A NAME="02">(2):</A>
Where can I find climate data?

Hourly climate data which include rain are even harder to find than material data.

IBP offers one year of hourly weather data with WUFI (the file can also be downloaded from the IBP website). These data from 1991 are considered fairly representative for the climate of the Holzkirchen region.
In addition, weather data for 93 locations in Europe, America and Japan are provided with the professional WUFI version.

Another source of hourly weather data for Germany are the Test Reference Years of the German Meteorological Service DWD which represent typical as well as extreme weather situations. Since they are primarily intended for heating and energy consumption investigations, they have no rain data, however, and thus are of only limited usefulness for hygrothermal investigations.

For other possible climate data sources see <A HREF="SourcesForClimateData.htm">Sources for Climate Data</A>.

If the situation of a specific object is to be investigated, it may be necessary to measure the weather in-situ anyway.

<A NAME="03">(3):</A> WUFI gives me the water content of the simulated wall in units of kg/m³ or in volume percent. However, I need the result in mass percent. How do I convert the results?

WUFI usually gives the water content as "water density", i.e. how many kg of water are in one m³ of building material.
A result given in volume percent tells you how many m³ of water are in one m³ of building material (expressed as percentage).
A result given in mass percent tells you how many kg of water are in one kg of dry building material (expressed as percentage). Please note that the water content in mass percent may easily exceed 100% if the dry material has low density.

With

m_W :	mass of the water in the component
r_W :	density of water (= 1000 kg/m³)
V_W :	volume of the water in the component
m_C :	mass of the component
r_C :	density of the (dry) component
V_C :	volume of the component

we have

water content as expressed by WUFI:
	u	= m_W / V_C [kg/m³]
water content expressed in volume percent:
	u_v	= V_W / V_C * 100
		= (m_W / r_W) / V_C * 100
		= (m_W / V_C) / r_W * 100
		= u * 100 / r_W
		= u * 100 / 1000
		= u / 10
water content expressed in mass percent:
	u_m	= m_W / m_C * 100
		= m_W / (r_C * V_C ) * 100
		= (m_W / V_C) * (100 / r_C)
		= u * (100 / r_C)
		= u / (r_C / 100)

So you get the water content in volume percent if you divide the WUFI result [kg/m³] by 10.
You get the water content in mass percent if you divide the WUFI result by (density of the building component / 100).

<A NAME="04">(4):</A>
I'm trying to make sense of the WUFI results, but I'm confused. What exactly is 'relative humidity' and what is the relative humidity in a building component referred to?

In air the relative humidity is the ratio of the actual water vapor partial pressure p and the water vapor saturation pressure ps. Example: If the air temperature is 20°C (and therefore ps = 2340 Pa) and the actual vapor pressure is 1872 Pa, then the relative humidity is 1872 Pa / 2340 Pa = 0.8 = 80%.

The condition in a porous building material corresponds to a RH of x % if it has been exposed to air with a RH of x % until equilibrium was reached and no moisture was taken up or given off any more.
The moisture in the material is then in equilibrium with the RH of the air in the pore spaces. At RHs less than ca. 50% this means that a molecular layer with a thickness of one or a few molecules has been adsorbed at the surfaces of the pores; at higher RHs capillary condensation occurs.

Here is what happens in detail: the usual formulas for the saturation vapor pressure (such as in German standard DIN 4108) are only valid for plane water surfaces. At concavely curved surfaces, where the water molecules are bound stronger, the saturation vapor pressure is reduced; the more so the stronger the curvature of the surface is.

In a partly filled capillary the interface surface between air and water forms a curved meniscus whose curvature is determined by the surface energies involved and in particular by the radius of the capillary. If the air space in such a capillary is filled with air whose partial water vapor pressure is greater than the saturation vapor pressure at the meniscus (whereas the RH in the air is still less than 100%), then the air in the immediate neighborhood of the meniscus is supersaturated and water condenses from the air onto the meniscus, i.e. the capillary fills up.

In a porous material there exists a wide range of pore sizes. In the smallest pores, any menisci may be curved so strongly that in these pores moisture condenses onto the menisci from 50% RH in the pore air upwards. The smallest pores get filled with water, and subsequently larger and larger pores (with smaller curvatures of the menisci) get filled until a pore size is reached where - because of the larger pore size and the smaller curvature of the meniscus - the saturation vapor pressure at the meniscus is equal to the vapor pressure in the pore air. In this way capillary condensation results in an equilibrium between the moisture content and the relative humidity in the pore air, even if this RH is less than 100%. The amount of water needed to fill the pores up to this point depends on the pore structure and the pore size distribution.

The <A HREF="MoistureStorageFunction.htm">moisture storage function</A> describes the amount of moisture taken up in this manner by the building material if it is exposed to air with a specific RH. Since this relationship between RH and moisture content is largely temperature-independent, the RH is an important and unique parameter describing the moisture content of a material.

<A NAME="05">(5):</A>
When I do not define a moisture storage function for a material, WUFI uses a default moisture storage function instead. What does this function look like?

WUFI needs a well-defined moisture field for each time step, so it must assign a moisture content even to materials which nominally don't have any appreciable moisture content (e.g. water-repellent mineral wool, air layers etc.).

The default <A HREF="MoistureStorageFunction.htm">moisture storage function</A> used by WUFI is described by the function

w = a / (b - phi) + c
w	[kg/m³]:	water content
phi	[-]:	relative humidity

Since phi must be 0 for w=0, it follows immediately that
 c = -a/b
 
The constants a and b are determined as follows:
 
b is set to 1.0105.
 
The moisture content at free saturation, wf, corresponds to a relative humidity of 1 (=100%). Since WUFI also needs a unique relationship between moisture content and RH for moisture contents above free saturation, this oversaturation region is assigned RHs greater than 1, up to phimax = 1.01. This value phimax is reached when the moisture content reaches maximum saturation wmax which is determined by the <A HREF="BasicMaterialData.htm">porosity</A>:
 
 wmax = porosity * 1000 kg/m³
 
Therefore we have
 
 wmax = a / (b-phimax) - a/b.
 
Solving for a yields:
 
 a = wmax * b * (b - phimax) / phimax,
 
and thus:
 
 w / wmax = phi / (b - phi) * (b - phimax) / phimax.
 
 
In particular, for phi=1 we have
 
 wf / wmax = 1 / (b - 1) * (b - phimax) / phimax = 0.047.
 
So this pseudo material has a free saturation of wf = 0.047 wmax.

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<A NAME="06">(6):</A>
I did a WUFI calculation with an assembly that includes an air layer. However, I get completely unrealistic water contents for the air layer. What went wrong?

WUFI was developed to simulate the hygrothermal processes in porous building materials. The detailed simulation of heat and moisture transport in air layers (including convection, turbulence etc.) is much more complicated and is outside WUFI's scope. Furthermore, it does not make much sense to try and implement these inherently two- or three-dimensional processes in a one-dimensional simulation program.

<A HREF="AirLayers.htm">Air layers</A> can therefore only approximately be simulated by treating them as a 'porous' material. It is possible to allow for the amplifying effect of convection on heat and moisture transport by employing appropriate effective <A HREF="BasicMaterialData.htm">heat conductivities</A> and <A HREF="BasicMaterialData.htm">vapor diffusion resistance factors</A>.

However, the <A HREF="MoistureStorageFunction.htm">moisture storage function</A> of an air layer can only very crudely be approximated by the moisture storage function of a porous material. The latter is largely temperature-independent (and implemented as such in WUFI), so that the functional dependence of the moisture content in air on the relative humidity and temperature cannot be reproduced.
Furthermore, the default moisture storage function used by WUFI for materials for which the user has not defined one assumes that capillary condensation will occur in the material already at relative humidities less than 100%, which is not true for an air layer (it has been modeled after the moisture contents of dense mineral wool).

As a result you will get unrealistically large moisture contents for air layers. Note, however, that WUFI uses the relative humidity as the driving potential for moisture transport and computes the water content as a secondary quantity from the resulting relative humidity (using the moisture storage function of the respective material).
So the resulting distribution of relative humidity should in general be quite realistic, its temporal behavior will just be damped much more than in reality (the moisture content acts as a 'capacity term' for moisture transport in the same way the heat capacity acts as a capacity term for heat transport). If short-term fluctuations don't play a major role, the general trend in the behavior of the relative humidity should be tolerably realistic.
This also means that quantities that depend on the relative humidity in or near the air layer (e.g. mould growth rates) can be evaluated more realistically than quantities that primarily depend on the moisture content (e.g. heat conductivity, heat capacity).

Please note that the unrealistically large moisture capacity of an air layer may also affect other layers. If you are interested in the moisture distribution in an assembly that contains an air layer, the air may (or may not) take up more moisture than realistic, so that less moisture remains for distribution among the other layers.

You may mitigate these problems by explicitly defining a slightly more realistic moisture storage function for the air layer. To this end, use a linear function like

phi:	w:
0	0
1	wf

with a low value for wf (the numerics may not be able to cope with very low values, you'll need to experiment a bit) (*). This avoids the spurious capillary condensation.

Also see the next question for a related problem.

(*) Note, however, that the porosity and thus wmax should remain high. If the water content exceeds wf, WUFI reduces the vapor permeability, in proportion to the excess, to reflect the fact that the pore volume gets increasingly filled with water and thus vapor transport decreases. At w=wmax the permeability reaches zero (all pores are filled). For vapor-permeable materials like air layers or mineral wool where moisture transport occurs mainly via vapor transport, wmax should therefore remain at a realistic value.

<A NAME="07">(7):</A>
I tried to perform a WUFI simulation, but the water balance never adds up, regardless whether I make the grid as fine as possible or whether I choose stricter numerical parameters, as suggested in the on-line help. What can I do?

One situation where serious convergence failures tend to occur is a component with a vapor-permeable layer (e.g. air or mineral wool) which has accumulated a lot of moisture (RH ~ 100%) and which is now exposed to a high temperature gradient (e.g. caused by intense solar radiation). WUFI originally wasn't developed to treat these cases which sometimes prove too demanding for the numerics that are mainly tuned to massive porous materials.

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If everything else fails, you may try an alternative <A HREF="MoistureStorageFunction.htm">moisture storage function</A>. In the material database, the moisture storage functions for materials like air or mineral wool are left undefined, so that WUFI uses an internally defined default moisture storage function (see the preceding two questions).

This moisture storage function assumes that for RHs above ca. 50% capillary condensation occurs which leads to increasingly higher moisture contents until free saturation is reached at 100% RH. This is not really realistic for air layers or hydrophobic mineral wool (it may be more appropriate for non-hydrophobic mineral wool).
Since it seems that the problem is mainly caused by the high water content, reduction of the water content by choosing a different moisture storage function often remedies the problem.
Please note that the relative humidity in the material will remain largely unaffected by the specific choice of the moisture storage function, as explained above. So if you are interested in the relative humidity in the layer, your results will be affected only slightly (but please perform a few test calculations with different choices of the moisture storage function to be sure), and if you are interested in the moisture content, you should not rely on the default moisture storage function anyway, but use measured data instead which represent your particular material.

A possible choice for the moisture storage function in these cases is a table like this:

phi:	w:
0	0
1	wf

Use a low value for wf (the numerics may not be able to cope with very low values, you'll need to experiment a bit.) (*).
This linear function is even more realistic than the default function in that it avoids the capillary condensation for RH= 50..100%. The moisture content remains low up to RH=100% (as it should be in air or in hydrophobic insulation materials), and at or above 100% condensation may occur and increase the moisture content beyond wf and up to wmax.

In particular if you are interested in moisture accumulation by condensation in these materials, use such a linear moisture storage function with low wf. Then you know that any moisture content exceeding wf must have been caused by condensation. You can then analyse this excess over wf (test calculations show that this excess is only slightly dependent on the specific choice of wf).

(*) Note, however, that the porosity and thus wmax should remain high. If the water content exceeds wf, WUFI reduces the vapor permeability, in proportion to the excess, to reflect the fact that the pore volume gets increasingly filled with water and thus vapor transport decreases. At w=wmax the permeability reaches zero (all pores are filled). For vapor-permeable materials like air layers or mineral wool where moisture transport occurs mainly via vapor transport, wmax should therefore remain at a realistic value.

<A NAME="08">(8):</A>
How can I get the moisture content at a monitoring position?

WUFI's output includes the temporal behavior of

• temperature and relative humidity at the monitoring positions, and of
• the mean moisture content of each layer.

In order to get the moisture content at a monitoring position, you can either

• calculate it from the relative humidity prevalent at that monitoring position by means of the <A HREF="MoistureStorageFunction.htm">moisture storage function</A>, or
• insert a thin 'diagnostic' layer at the position in question which has the same material properties as the surrounding material. WUFI will output curves for the water contents of each layer, including a separate curve for the diagnostic layer. This is also a useful way to get the water content of, say, the outermost 5 cm of a layer.

<A NAME="09">(9):</A>
I want to examine the effect of driving rain on a painted wall. What liquid transport coefficients Dws do I enter for the paint?

There are no measurements of <A HREF="LiquidTransportCoefficients.htm">transport coefficients</A> or, equivalently, water absorption coefficients for paint layers themselves known to us.

What is measured sometimes is the water uptake for different paint layers by applying the paint on a standard substrate (such as cellular concrete or lime cement mortar) and measuring the water absorption for this composite material.

So the best thing you can do is probably the following:
Don't use a layer of rendering and a layer of paint; instead, use a layer of the 'hybrid' material for which you already know the combined water uptake from the measurements. Use the Dws from the hybrid water uptake (let it generate by WUFI from the measured water absorption coefficient) and use the Dww and other data from the original rendering.

The <A HREF="WaterVaporDiffusion.htm">vapor diffusion resistance</A> of the paint can then be included in the <A HREF="DialogEditSurfaceCoefficients.htm">surface transfer coefficients</A> (as long as it is not markedly moisture-dependent).

Please note some possible problems, though:

• The result of the measurement may (or may not) depend on the substrate material, the details of the application etc. So you should make sure that you are using a water absorption value that has been measured under the same circumstances as the case you consider in your calculations.
• The paint may slowly change its properties when it gets wet (e.g. by swelling). The mean properties over a rain period of two or three hours may be different than the mean properties during a measurement that takes many hours. Again, the measurement should be done close to natural conditions.

<A NAME="10">(10):</A>
What is the right choice for the rain absorption factor for an unrendered natural sandstone wall? When I use the value of 0.7 suggested by WUFI, then the entire wall gets wet like a sponge. When I reduce the absorption factor to 0.5, the same happens, it just takes longer. What's wrong?

This should not happen, but the <A HREF="RainWaterAbsorptionFactor.htm">rain absorption factor</A> is very likely not to blame. It does not depend on the material of the wall (it depends a bit on its surface structure and, of course, on its tilt). After all, it simply expresses the fact that some of the rain water splashes off when it hits the wall surface and is no longer available for absorption.

Are you sure that the amount of rain is okay? Maybe you created your own *.KLI file and used normal rain instead of the correct driving rain?
Several kinds of sandstone have a very high water absorption (e.g. Rüthener) and may accumulate an inacceptable amount of moisture when exposed to a wet climate such as the Holzkirchen weather. Maybe you used one of those?

<A NAME="11">(11):</A>
I want to investigate the hygric behavior of ecological insulation materials, such as flax, hemp or reed. However, these materials consist of fibres, whereas WUFI is mainly designed for capillary-active porous materials. What is the best approach?

The difference between fibres and porous mineral materials is in general not really crucial for the transport equations. The fibre materials may tend to have preferred transport directions, which would have to be allowed for by using appropriate material data for the x and y directions in a two-dimensional calculation.

Determining the <A HREF="LiquidTransportCoefficients.htm">liquid transport coefficients</A>, however, may be difficult or even impossible if they change their consistency upon wetting (e.g. by caking).

On the other hand:
As long as your insulation materials don't become so wet that capillary conduction becomes predominant, you can ignore capillary transport and only consider diffusion transport. That is, you leave the liquid transport coefficients undefined and only enter a <A HREF="BasicMaterialData.htm">µ-value</A>. Surface diffusion phenomena may be allowed for by using a <A HREF="DiffusionResistanceFactorMoistureDependent.htm">moisture-dependent µ-value</A>.

Since you probably only want to assess whether or not the insulation becomes wet by rain or condensation, you will mainly be concerned with water contents in the sorption moisture region of the <A HREF="MoistureStoragefunction.htm">moisture storage function</A>, for which these simplifications should be adequate.
As these materials must be prevented from becoming wetted through anyway, there will be no need to investigate in detail the behavior of an insulation soaked full of water.

<A NAME="12">(12):</A>
I want to find out how long it takes a wall with construction moisture to dry out. Which initial moisture content should I use?

That depends on a number of individual circumstances such as the amount of production moisture (e.g. in cellular concrete or lime silica bricks), the amount of mixing water (in concrete or mortar), the amount of rain hitting the wall while it was unrendered, the season when construction took place (warm/cold) etc., so no general answer is possible here. The table gives examples for typical initial moisture contents:

Material		Water content [kg/m³]
Fresh concrete:
	free water	175
Concrete, 28 days old (at 70% hydratation):
	bound water	85
	dried water	25 ... 45
	free water	65 ... 45
		(sum = 175)
Concrete, 3 to 6 months old (at 90% hydratation):
	bound water	105
	dried water	35 ... 50
	free water	35 ... 20
		(sum = 175)
DIN 4108 "thermal protection"
	practical moisture content of concrete	50
Cellular concrete		180 ... 220
Clay brick masonry		100 ... 150
Calcium silica brick masonry		100 ... 120

 

<A NAME="13">(13):</A>
How can I simulate a wall whose exterior surface has been treated with a water-repellent agent? Is it correct to set the rain water absorption factor to zero? Do I need to change the sd-value of the exterior surface, even though I use a diffusion-permeable treatment?

The <A HREF="RainWaterAbsorptionFactor.htm">rain water absorption factor</A> must be set to zero if the water absorption is indeed completely stopped by the treatment. If water absorption is only reduced, you must determine the <A HREF="LiquidTransportCoefficients.htm">water absorption coefficient</A> for the treated material and replace the part of the wall which corresponds to the penetration depth of the treatment with da layer of the treated material.

If the treatment does not change the <A HREF="WaterVaporDiffusion.htm">diffusion permeability</A> of the material, no <A HREF="DialogEditSurfaceCoefficients.htm">sd-value</A> needs to be specified for the exterior surface.
Many treatments do, however, increase the diffusion resistance factor (µ-value) of the material. In these cases, this additional resistance should be allowed for by an appropriate sd-value. Alternatively, and even better, you can replace the part of the wall which corresponds to the penetration depth of the treatment with a new layer that has the same material properties but an appropriately increased µ-value.

Even if the water absorption is negligible (so that adjusting the rain absorption factor instead of the liquid transport coefficients would be sufficient) and vapor diffusion is not hindered by the treatment (so that no µ-value needs to be adjusted), it might nevertheless be preferable to model the treated part of the wall by defining a separate layer whose liquid transport coefficients have been reduced or even set to zero.
This is because the capillary conduction in this layer does not only determine the amount of absorbed rain water; it also influences the wall's drying behavior.
Drying-out proceeds faster if water from the interior of the wall can be conducted to the surface by capillary transport and can evaporate from there. Drying-out is impeded, however, if capillary transport stops a few centimeters behind the surface and moisture can only dry out after crossing this treated layer by vapor diffusion. So this is another mechanism by which water-repellent treatment may reduce the drying potential of a wall, in addition to a possibly increased µ-value.

<A NAME="14">(14):</A>
WUFI gives me the option to estimate the liquid transport coefficients Dws from the water absorption coefficient. How can I check how good this estimate is?

You can check the <A HREF="LiquidTransportCoefficients.htm">estimated</A> Dws or determine an unknown water absorption coefficient from known Dws by simulating a water absorption experiment.

Define an initially dry layer consisting of the material in question, let it rain on the surface (with a higher rain load than the layer can absorb to be sure that no insufficient rain load is limiting the water uptake) and look at the amount of water absorbed after e.g. 100 or 200 hours.

This may be done with a one-line *.KLI-file such as:

[h:	rain:	rad:	t_ext:	RH_ext:	t_int:	RH_int:]
500	1000	0	20	1	20	0

The number of hours entered for the duration of this one-line climate file does not matter since WUFI re-starts reading from the beginning of the climate file if the simulation period extends beyond the end of the climate file.

Set the vapor diffusion thickness of the interior surface to a very high value to prevent vapor transport through that surface.

You should perform a few test calculations in order to find a suitable thickness of the test specimen which assures that the moisture front traverses most of the specimen (in order to make the most efficient use of the numerical grid) but not all of it.

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<A NAME="15">(15):</A>
I want to evaluate the effect of different material parameters on water absorption by simulating a laboratory experiment in which different specimens are exposed to a limited water supply. I created the corresponding KLI file with a spreadsheet program to avoid typing in all those numbers by hand, but WUFI can't read this KLI file.

The reason why WUFI does not accept your spreadsheet file is probably that you did not write it in ASCII format and/or did not write the header lines in the correct format. Please consult the on-line help for details on creating *.KLI files with your own programs.

Also note that if you want to simulate a simple absorption experiment with a specified constant water supply and constant climate conditions it is sufficient to create a *.KLI file which consists of only one single line, for example the line:

[h:	rain:	rad:	t_ext:	RH_ext:	t_int:	RH_int:]
1000	5	0	20	1	20	0.8

means that for 1000 hours after the starting time of the climate file there is a constant rain load of 5 Ltr/m²h.

An alternative would be a single line like

[h:	rain:	rad:	t_ext:	RH_ext:	t_int:	RH_int:]
1 5	0	20	1	20	0.8

which states that for 1 hour after the starting time of the climate file there is a constant rain load of 5 Ltr/m²h. When WUFI reaches the end of a climate file, it starts reading the file anew from the beginning, so you can simulate an experiment which is running for 100 hours (or whatever) and the climate file will automatically be read 100 times over.

The only difference between these two files is that in the latter case WUFI does not accept a calculation time step greater than 1 hour, whereas in the former case you may also choose any convenient time step greater than one hour.

For a simple absorption experiment I usually make sure the climate file contains a rain load large enough so that it does not limit the absorption of the specimen, for example 100 Ltr/m²h, which ould not be plausible for real rain.

If you want to have a specified limited supply please don't forget that WUFI reduces the amount of rain it reads from the climate file by the <A HREF="RainWaterAbsorptionFactor.htm">rain absorption factor</A> which allows for the fact that some rain splashes off of the wall on impact and is not available for absorption. This factor should be set to 1 during your experiments.

Furthermore, please note a small subtlety involved in using limited rain supply. Let's assume you have a specimen with a water absorption factor of 3 kg/m²h^1/2 and the climate file specifies a rain load of 3 Ltr/m²h. During each time step, WUFI performs a test step with an unlimited supply and subsequently evaluates the amount of water taken up. If this amount of water is less than the amount supplied in the climate file, then the material is the limiting factor and WUFI accepts the result of this time step and proceeds with the calculation.
However, if the amount taken up is more than the amount supplied, WUFI performs additional iteration steps in which a fictitious 'flow resistance' at the specimen surface is adjusted until the amount taken up matches the amount supplied.

If you are using 1-hour steps in your calculation, and the dry specimen absorbs 3 kg/m² of water in the first step and the climate file supplies 3 kg/m²h, then WUFI accepts the trial step done with unhindered absorption and proceeds with the calculation.
But if you are repeating the same calculation with a time step of half an hour, things are different! Since the water uptake is not linear in time, the specimen will absorb more than 1.5 kg/m⊃ in the first half hour, while WUFI compares this with 1.5 kg/m⊃ of rain in the first half hour (assuming the rain is evenly distributed over the hour) and now limits the amount absorbed to 1.5 kg/m².
This is usually of no concern with real rain data and real building materials, but it may be beneficial to be aware of these subtleties if performing test calculations with limited rain supply.

<A NAME="16">(16):</A>
I'm familiar with steady-state water vapor diffusion calculations (in particular, the Glaser method described in German standard DIN 4108). So I knew I had to expect more or less frequent dew conditions in the wall I was simulating. However, when I watched the WUFI film, I could never see the relative humidity reach 100%.

The usual building materials always have some moisture sorption capacity. This sorption capacity buffers changes in relative humidity inside the wall. If you define boundary conditions which would provoke instant condensation in a Glaser calculation, you may nevertheless not get condensation in a realistic case (such as simulated by WUFI).

That's because a relative humidity of 100% would correspond to a <A HREF="MoistureStorageFunction.htm">moisture content</A> equal to free saturation of the material in question, and this amount of water must first be transported into the dew region. The diffusion flows do transport moisture to the location where dew conditions prevail, but the transported amounts of moisture are generally small, and the RH will only slowly rise from the initial value, say 80%, to 81%, 82% etc. It may take days or weeks until sufficient amounts of water have been transported to the dew region so that finally free saturation (i.e. RH=100%) is reached. Meanwhile, boundary conditions may have changed and there are no dew conditions any more.

The Glaser method, on the other hand, simply assumes that 100% RH are reached instantly, it doesn't consider the necessity to actually move water in order to reach the moisture content that corresponds to 100% RH.

Furthermore, real materials (as opposed to Glaser) usually have some <A HREF="LiquidTransportCoefficients.htm">capillary conductivity</A> which tries to dispel any moisture accumulations. This effect actively works against local water build-up, so that 100% RH can't be reached easily.

Of course, you may get water accumulation in your building component if conditions are right (or wrong). But this will rarely be accompanied by 100% RH. If you see relative humidity approaching 100% somewhere in your component, it's probably much too late...

<A NAME="17">(17):</A>
OK, this explains why I didn't see dew conditions in the wall. But shouldn't condensation at least happen at the facade on days with high humidity and little sunshine?

The surface of a normal wall in temperate or cool climate regions will always be somewhat warmer than the surrounding air. By day because of solar radiation (even on foggy or overcast days), by night because of heat flow from indoors (exceptions: air-conditioned dwellings or nightly emission, see below).
Since the RH in the air can't be greater than 100% and the RH at the warmer-than-air wall surface will always be less than the RH of the air, you usually can't reach or surpass 100% there.

You'll have <A HREF="MoistureStorageFunction.htm">free saturation</A> (i.e. 100% RH) at the surface when enough rain is absorbed, but this is not due to dew conditions.

The surface temperature will fall below air temperature when the wall <A HREF="LongWaveRadiationEmissivity.htm">emits</A> more long-wave radiation than it gets back from surrounding surfaces. If it even falls below the dew-point temperature, you will indeed get dew conditions at the surface.
This happens routinely during the night, especially during clear nights, when the long-wave emission of the water vapor in the atmosphere is at a minimum.

In these cases you may get repeated and regular wetting of the surface which may lead to dust accumulation or algae growth, especially with exterior insulations whose surfaces cool down particularly strongly.
Currently, WUFI does not routinely allow for this effect, since the necessary data on atmospheric and terrestrial counterradiation are rarely available. If these data are provided, WUFI can compute nightly emission cooling in principle, but only approximately. Future WUFI versions will have a more sophisticated emission model incorporated.

<A NAME="18">(18):</A>
I used WUFI to compute the water content in a variety of wall assemblies. In order to evaluate their hygrothermal performance, I now need appropriate criteria, e.g. standards that should not be exceeded.

There are no general criteria which are applicable for every case. Different materials and different applications require different criteria. Here are some general hints:

• The most important criterion: the moisture must not accumulate over time. Water condensing in the building component must be able to dry out again. If the moisture content in your component keeps increasing - even slowly - you'll run into problems sooner or later.
• The building materials which come into contact with moisture must not be damaged (e.g. by corrosion or mould growth).
  Mineral building materials are usually not at risk; some of them may be susceptible to frost damage if they contain a lot of moisture.
  Wood should not exceed 20 mass-% of moisture during a prolonged period; otherwise mould growth may result (possible exception: increased moisture while temperatures are low).

German standard DIN 4108-3 adds the following criteria:

• The amount of condensing moisture in roof or wall assemblies must not exceed a total of 1.0 kg/m².
  This is a more or less arbitrary criterion. In order to test it with WUFI, start the calculation with the normal equilibrium moisture (corresponding to 80% RH) and see if the total water content exceeds the starting value by more than 1 kg/m².
• At interfaces between materials that are not capillary-active, no moisture increase exceeding 0.5 kg/m² is permissible.
  This is meant to avoid moisture running or dripping off, which could accumulate elsewhere and cause damage.
• The moisture increase in wood must not exceed 5 mass-%, the moisture increase in materials made of processed wood must not exceed 3 mass-%.
  These are more or less arbitrary numbers.

In addition, special criteria may be applicable in specific cases, for example:

• Are there any materials which are particularly sensitive to moisture damage?
• Does increased heat loss by moist insulation exceed any energy conservation requirements?
• Is the building material at this moisture level sensitive to frost damage?
• Is there salt in the wall which must be kept from crystallizing or from moving around?
• Etc.

Even if you don't have clear criteria which fit your case, you may still perform a ranking of your assemblies by comparing them with each other or with a standard case.

<A NAME="19">(19):</A>
I want to simulate a ventilated curtain wall; how can I do this? I can model the air gap as an air layer in WUFI but it seems these air layers are assumed to be stagnant, which is certainly not the case in my ventilation gap.

If you model the ventilation gap as an <A HREF="AirLayers.htm">air layer</A> in WUFI, it is indeed treated as a closed air layer without connection to the exterior air. The effect of inner convection on heat and moisture transport across the air layer is allowed for (as a first approximation) by use of <A HREF="AirLayers.htm">effective</A> <A HREF="MaterialData.htm">heat conductivities</A> and <A HREF="MaterialData.htm">vapor diffusion resistance factors</A>.

The air flow and air exchange phenomena in a ventilated air layer cannot be simulated with a one-dimensional program like WUFI-1D; WUFI-2D currently does not take air flows into account.
If the air exchange is large enough, it may be justified to assume exterior air conditions in the air gap. That is, you do not model the curtain facade and the air gap, and you consider the surface of the insulation or the wall itself (as the case may be) as the exterior surface in WUFI's component assembly. Rain must be set to zero (simply by setting the <A HREF="RainWaterAbsorptionFactor.htm">rain absorption factor</A> = 0).
It will be advisable to choose appropriate effective values for the exterior heat transfer coefficient and the short-wave solar absorptivity, but this requires calibration by experimental data.

The same problem is encountered in simulations of roofs, either because of a ventilation cavity in the roof or because of the question how to model the covering and the batten space.

The investigations described in [1] used a simplified treatment of a roof. WUFI-1D simulations were carried out to examine the moisture balance in a fully insulated west-facing pitched roof (50° inclination). The covering and the batten space could be omitted from the simulated assembly because measured temperatures in a similar roof on IBP's testing area were available and could be used to determine appropriate effective surface transfer coefficients. The measurements were taken on the waterproofing foil (i.e. directly on the insulation layer) and were compared with the computed temperatures at the outer surface of the modeled insulation layer which sufficed to represent the whole roof for the purpose of a thermal adjustment.
The thermal surface transfer coefficients were adjusted in WUFI until good agreement between measurement and calculation was reached. This was the case with an effective short-wave absorptivity of as=0.6 and an effective heat transfer coefficient of a=19 W/m²K. The effective absorptivity is roughly identical with the real absorptivity (for red roof tiles), while the effective a is slightly higher than the usual standard value of 17 W/m²K. Obviously the covering and the air in the batten space have no major effect on the thermal behavior of the roof, at least in this case. In particular, the amount of heat removed by convection through the ventilated air cavity seems negligible and the entire heat created in the covering by solar radiation is passed on into the underlay.
The question to which extent this isolated result can be generalised could only be answered by more extensive comparisons with measurements.

[1] H.M. Künzel: Außen dampfdicht, vollgedämmt? - Die rechnerische Simulation gibt Hinweise zu dem Feuchteverhalten außen dampfdichter Steildächer. bauen mit holz 8/98, S. 36-41.

<A NAME="20">(20):</A>
I calculated the sum of the heat flows through the exterior and the interior surfaces during one year and I noted that the heat flow out of the building component through the exterior surface is much larger than the heat flow into the component through the interior surface. But shouldn't they be nearly equal? How can more heat flow out of the component than into it? No heat can be created in the wall.

The solar radiation incident on the exterior surface is electromagnetic radiation and not heat flow; it is therefore not included in the heat flow data.
However, after absorption it is converted to heat so that there exists indeed a heat source in the wall. Since the heat source is close to the exterior surface, most of the generated heat flows outward through the exterior surface, only a small amount flows inward through the interior surface. This asymmetric heat flow is superimposed on the usual transmission heat flow (which in colder climates alway goes from the indoor side to the outdoor side of the building element).

Please note that in the film display the heat flow arrow at the exterior surface does include the solar radiation. Otherwise it would look very strange to see the sun shining on the wall surface but a lot of heat flowing out of the wall. This is a concession to the intuitive expectations of the audience.

Also note that there can be a heat source or sink in the wall when water condenses or evaporates. In some cases these latent heat effects can be non-negligible (e.g. drying of a wall wetted by driving rain).

<A NAME="21">(21):</A>
Are there more recent weather data available? The copy of WUFI I downloaded still has those of 1991.

'Recent' weather data would probably not be very useful to you. It is more important to have weather data which are either known to be typical for a specific location or which repesent defined critical conditions (e.g. for design purposes). We consider 1991 to be a fairly typical year for Holzkirchen. 'Critical' weather data, i.e. one particularly cold year and one particularly warm year, are in preparation and may be made available as two 'Hygrothermal Reference Years'.

<A NAME="22">(22):</A> For the numerical solution of the transport equations the component must be divided into a series of grid elements for whose midpoints the resulting temperatures and water contents are computed at each time step, and across whose element boundaries the heat and moisture fluxes required by the equations are flowing. In order to arrive at correct fluxes across the boundaries, effective conductivities have to be assigned to the boundaries which represent the integral effect of the conductivities between the midpoints of the two elements. The investigations reported in [1] show that the results do in fact depend on the way these effective heat and moisture conductivities are determined from the real conductivities of the two elements: obviously linear interpolation between the neighboring conductivities is preferable within a material layer, and a resistance formulation is more appropriate for boundaries between layers with different materials. How does WUFI treat these element boundary conductivities?

It is obvious that at element boundaries where materials with possibly very different conductivities are in contact with each other a simple average of the conductivities (or resistances) cannot result in a realistic effective conductivity to describe the fluxes between the elements. Take as an example a material with very low resistance which borders on a material with very high resistance. The flux flowing between the midpoints of the two elements is determined by the sum of the two successively encountered resistances, not by the arithmetical average of the conductivities.
One might suppose now that this physically motivated reasoning also applies to smaller differences between the neighboring elements and that therefore the resistance formulation (i.e. the harmonic mean of the conductivities) should always be used within the entire component. However, test calculations during the development of WUFI's numerics showed that this is not the case. Within a material the arithmetical mean of the conductivities yielded better results (compared with experimental data), so that WUFI uses harmonic averages at material boundaries and arithmetical averages within a material, in agreement with the cited investigation. The derivation of the resistance formulation assumes equal fluxes in the two element halves, but this need not be the case if transient processes in materials with heat or moisture storage capacities are considered.

[1] Galbraith, G.H. et al.: Evaluation of Discretized Transport Properties for Numerical Modelling of Heat and Moisture Transfer in Building Structures,
Journal of Thermal Env. & Bldg. Sci., Vol. 24, Jan. 2001

<A NAME="23">(23):</A>
I want to perform a hygrothermal simulation of a wall on which every day a shadow is cast for some time by a building on the other side of the street. WUFI does not offer an option to allow for such a shadow, but I could simply use a self-created *.KLI file by converting the measured radiation myself and allowing for the sadow in this process. But, how is the conversion of the radiation data done?

First you need to determine the radiation incident on the surface of your building element from the measured data describing the radiation on a horizontal surface. For this purpose it is necessary to determine the position of the sun in the sky at the time of the measurement.

Position of the Sun:

Let J be the number of the day in the year (1 .. 365 or 366). Then compute the auxiliary quantity x:

x = 0.9856° * J - 2.72°

and the equation of time Z (in minutes):

Z = -7.66*sin(x) - 9.87*sin( 2*x + 24.99° + 3.83°*sin(x) ).

The equation of time describes the variable difference in time between the actual culmination of the sun and noon. Because of the ellipticity of the Earth's orbit and the obliquity of the Earth's axis the sun wanders with slightly irregular speed across the sky. During the course of the year there are thus times where it reaches culmination earlier than a fictitious sun with constant speed (the so-called 'mean' sun) and times where it reaches culmination later.

The local meridian is the great circle that rises from the horizon due north, passes through the point directly above the observer and crosses the horizon again due south. The instant at which the sun crosses the local meridian on its daily path from east to west is also the instant where its position is due south and where it reaches its daily greatest height.

When the apparent sun (i.e. the actually observed sun) crosses the meridian it is 12 noon local apparent solar time (LAT); when the mean sun crosses the meridian it is 12 noon local mean time (LMT). The equation of time is therefore the difference between LAT and LMT (Z = LAT - LMT).

Furthermore, since the place where the measurements were taken is usually not located on the reference meridian of the time zone (15° East for the Central European Time Zone, CET), the difference between local mean time and zone time must be allowed for, which is 4 minutes for 1° difference in geographical longitude L and one hour for 15° difference. If the measurement was timed in Central European Summer Time CEST, convert to CET first by subtracting one hour (CET = CEST - 1h).

In this way you can now compute the corresponding local apparent time LAT from the known measurement time (in CET):

LAT = CET - (15°-L)/(15°/h) + Z/(60 min/h) [h]

and thus determine the position of the sun: at 12 noon LAT the sun is exactly on the meridian, before noon it stands at an appropriate distance to the east of the meridian, after noon, an appropriate distance to the west.
The distance between the sun and the meridian is measured by the hour angle:

w = (LAT - 12h) * 15°/h.

The hour angle w is reckoned perpendicular to the meridian; it is negative before noon, zero at noon and positive after noon; it increases steadily by 15° per hour.

The hour angle gives the distance of the sun from the meridian; the declination d, i.e. the distance of the sun from the celestial equator, then fixes the position of the sun completely. The declination varies between -23°26' at winter solstice, 0° at the equinoxes, and 23°26' at the summer solstice. Since its change during one day is very small, it suffices to compute it once for the day J under consideration:

sin(d) = 0.3978 * sin( x - 77.51° + 1.92° * sin(x) ),
cos(d) = sqrt(1 - sin(d)^2)

where x is the auxiliary quantity introduced above.

The last step is the transformation from the coordinate system determined by w and d into the more familiar coordinates altitude g and azimuth y (=compass direction). The geographical latitude j of the measurement location is needed for this.

sin(g) = cos(d)*cos(w)*cos(j)+sin(d)*sin(j)
cos(g) = sqrt(1 - sin(g)^2)
 
if cos(g)=0 then y = 0
else begin
       sin(y) = cos(d)*sin(w)/cos(g)
       cos(y) = (cos(d)*cos(w)*sin(j)-sin(d)*cos(j))/cos(g)
       y = atn2(sin(y), cos(y))
     end

This formula uses atn2(A,B), the arctangent function for two arguments A and B, which is provided by many programming languages, and which gives the arctangent of A/B in the correct quadrant. If this function is not available to you, you can use the ordinary arctangent and then explicitly determine the correct quadrant (i.e. you compute y=atn(A/B), and in the case B<0 you add 180° if y<=0 or subtract 180° if y>0. If B=0 and A<0 then y=-90°, if B=0 and A>0, then y=+90°.).
The azimuth y is counted from south=0°, positive towards the west and negative towards the east.

Examples for Munich (48.13°N, 11.58°E):

CET	Altitude	Azimuth	Declination
(J=1)
1 Jan. 2001 09:00	6.436°	-44.614°	-22.987°
1 Jan. 2001 12:00	18.836°	-4.209°	-22.977°
1 Jan. 2001 16:00	3.441°	49.590°	-22.962°
1 Jan. 2001 16:25	0.434°	54.316°	-22.961°
(J=79)
20 Mar. 2001 07:00	6.498°	-82.661°	-0.126°
20 Mar. 2001 12:21	41.851°	-0.041°	-0.038°
20 Mar. 2001 16:00	22.726°	62.242°	+0.022°
(J=172)
21 Jun. 2001 08:00	34.501°	-87.522°	+23.437°
21 Jun. 2001 12:00	65.126°	-8.437°	+23.437°
21 Jun. 2001 18:00	19.771°	103.475°	+23.436°

These values were computed with an astronomical ephemeris program. Of course, the simplified method described above cannot reproduce these data exactly, in particular for low altitudes of the sun (1 Jan. 16:25), since it does not allow for atmospheric refraction. On the other hand, the comparison allows you to assess the overall accuracy of this simple method. Your results should agree with these exact positions within a few tenths of a degree. The declinations have been included as well for testing purposes.

 

Converting the Radiation Data:

We assume that your input data are measured hourly values of the global (I_glob) and the diffuse radiation (I_diff) on a horizontal surface.

The radiation incident on the measuring or the component surface is split up into a direct and a diffuse component. The direct component is received directly from the sun and is therefore a directed quantity that depends on the position of the sun. The direct radiation vertically incident on a surface which is facing the sun is the direct normal radiation I_dir_normal. The direct radiation I_dir obliquely incident on a horizontal measuring surface depends on the solar altitude g:

I_dir = I_dir_normal * sin(g).

Since I_dir can be computed as the difference between the measured values of global and diffuse radiation and g can be determined from the measurement location and time by the method given above, the corresponding direct normal radiation is

I_dir_normal = (I_glob - I_diff) / sin(g).

The angle of incidence h, i.e. the angle that the direct normal radiation makes with the normal to the component surface which is tilted by the angle b and oriented in the direction a, is

cos(h) = sin(g)*cos(b) + cos(g)*sin(b)*cos(a-y)
h:	Angle of incidence (vertical=0°)
g:	Altitude of the sun
y:	Azimuth of the sun (south=0°, positive towards west, negative towards east)
b:	Tilt of the component surface (vertical wall=90°)
a:	Azimuth of the normal to the component surface (south=0°, west positive).

The direct radiation incident on the component surface is therefore:

I_dir_in = I_dir_normal * cos(h)
         = (I_glob - I_diffus) * cos(h) / sin(g).

The diffuse component consists of the radiation scattered by the air ("blue sky") and the clouds which comes from all directions and can approximately be treated as isotropic. Diffuse radiation is measured by blocking the direct radiation with a shadow ring around the solarimeter. The measurement gives I_diff, the diffuse radiation incident on the horizontal measuring surface from the entire sky hemisphere. A component surface with arbitrary tilt and orientation receives the same diffuse radiation (since it is isotropic), but for non-horizontal surfaces the fact has to be allowed for that the sky covers a smaller part of its field of view and the total amount of incident diffuse radiation is reduced proportionately (a vertical wall sees sky only in the upper half of its field of view):

I_diff_in = I_diff * ( cos(b/2) )^2.

Additionally, you may add the global radiation reflected from the ground:

I_refl_in = r * I_glob * ( sin(b/2) )^2,

where r is the short-wave albedo of the ground and the reflection is assumed to be isotropic. In the current version, WUFI ignores the reflected component of the radiation.

The total radiation incident on the surface of the building component is the sum of the components:

I_in = I_dir_in + I_diff_in + I_refl_in.

You may now modify or supplement this conversion method according to your needs. For example, you can allow for shadows by setting the direct radiation to zero at times where the sun is behind the obstacle, and by reducing at all times the diffuse radiation in proportion to the reduction of the field of view caused by the obstacle. On the other hand, at times where the sun illuminates the facing side of the obstacle, it may be necessary to add some reflected radiation.

Hint: if the radiation data to be converted have been averaged over some longer interval (e.g. one hour), please note the following:

It is advisable to compute the solar positions for the middle of the measuring interval, i.e. the averaged data measured between 9h and 10h should be converted using the solar position computed for 9:30h.
If the sun has risen or set during such a measuring intervall (which is easy to check for, using the solar altitude), the solar position must be computed for the middle of the visibility interval, not for the middle of the measuring interval.

Independent of the duration of the measuring interval, radiation data obtained at very low solar altitudes should not be used, since under these circumstances the direct normal radiation must be calculated from very small and unreliable values obtained for the direct radiation at grazing angles of incidence.

Details on these conversion methods can be found in:
VDI 3789 Umweltmeteorologie, Blatt 2: Wechselwirkungen zwischen Atmosphäre und Oberflächen; Berechnung der kurz- und der langwelligen Strahlung.

In addition to data on global and diffuse radiation, the weather file IBP1991.WET included with WUFI contains radiation data obtained with a west-facing solarimeter which you can use to test your conversion routines.