# Building Physics: Heat, Air and Moisture

Hugo Hens

Building Physics Heat, Air and Moisture Fundamentals and Engineering Methods with Examples and Exercises 2nd Edition

132

2  Mass Transfer

∂w div ( gx ) ± S x =x (2.9) ∂t where Sx is the source or sink, in kg/(m3 · s). The flows can be either diffusive or convective. In the first case, they depend on the gradient of a driving potential. In the second case, they depend on the driving potential. The relationships with the driving potential deliver the vector equations needed.

2.2

Air transfer

2.2.1 Overview Dry air, suffix da, is a mixture of 21% m3/m3 oxygen (O2), 78% m3/m3 nitrogen (N2) and traces of other gases (CO2, SO2, Ar, Xe). Generally, dry air is assumed to behave as an ideal gas, with as equation of state: pda V = mda Rda T, where pda is the (partial) dry air pressure in Pa, T the temperature in K, mda the mass of (dry) air in kg filling volume V (m3), and Rda the gas constant for dry air, equal to 287.055 J/(kg · K). In reality, dry air is a non-ideal gas, obeying the following ‘exact’ equation of state: pda V B C = 1 + aa + aaa 2 nda Ro T V  V   n    l  nl  with n da the number of moles filling volume V (m 3); R o the general gas constant (8314.41  J/(mol · K)) and Baa, Caaa the viral coefficients: B = 0.349568 ⋅ 10−4 − aa

0.668772 ⋅ 10−2 2.10141 92.4746 − + [m3 /mol] T T2 T3

C= 0.125975 ⋅ 10−2 − aaa

0.190905 63.2467 + [m6 /mol2 ] 2 T T

At ambient temperature, they are almost zero. For the dry air concentration the ideal gas equation of state gives: = rda

mda pda (2.10) = V Rda T

For moist air, a mixture of dry air and water vapour, suffix a, the ideal gas equation of state, also called the ideal gas law, and the concentration become: = Pa V m= ra a Ra T

Pa Ra T

where Pa is the air pressure in Pa, equal to the sum of the partial dry air and the partial water vapour pressure (Pa = pda + pv, large P’s indicating total pressures, small p’s partial pressures). With respect to dry air, the presence of water vapour modifies the air mass ma, the gas constant Ra and the concentration ra. However, for temperatures under 50 °C, the effect is so small the following holds: Ra ≈ Rda, Pa ≈ pda and ra ≈ rda or: Pa / (Ra T) ≈ pda / (Rda T).

133

2.2  Air transfer

Air transfer only occurs when wind, stack and fans create air pressure differences over assemblies, between rooms and between buildings and the environment. Of course, for this to happen, assemblies and building fabrics must be air permeable, a consequence of applying air permeable materials, having cracks in and between assemblies and overlaps in layers made of planks, small elements and boards and, requiring trickle vents as air inlets in ‘living’ rooms, flow through grids in all inside doors and air outlet pipes in ‘wet’ rooms.

2.2.2 Air pressure differences 2.2.2.1 Wind Wind pressure is given by: = Pw Cp

ra v 2 ≈ 0, 6 Cp v 2 (2.11) 2

The equation follows from Bernoulli’s law applied to a horizontal wind, whose velocity drops from a value v to zero after colliding with an infinite obstacle (conversion from kinetic to potential energy), corrected by a pressure factor Cp that cares for the difference between hitting a finite instead of an infinite obstacle. Finiteness deflects the air flow to the upper and side edges of the obstacle where it forms vortexes, while a lee zone develops behind. Consequently the pressure factor Cp links the real pressure at a specific point on a surface to the reference velocity, its value being a function of that reference. In principle this concerns the velocity measured in an open field at a height of 10 m. Of course, another reference can be used depending on the situation. For buildings, pressures on the façades are often correlated to the wind velocity above the ridge. A change in reference alters the pressure factors. Also wind direction, the location of the building (open, wooded, urban area, close to high-rise buildings), the geometry of the building and the façade spot considered alter the pressure factor. Cp positive means overpressure, Cp negative depressurization. Overpressure is found at the weather side, depressurization at the leeside and along all surfaces more or less parallel to the wind direction (Figure 2.2). Wind pressure differences are extremely variable over time and may change sign regularly.

Figure 2.2.  Bernouilli’s law and the wind pressure field around a building.

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2  Mass Transfer

2.2.2.2 Stack effects Stack or buoyancy in gases and liquids has two causes: differences in temperature and differences in composition. Confined to atmospheric air, the decrease of air pressure at a height h is given by dPa = –ra g dh, where g is the acceleration of gravity. Height zero stands for the sea level. Inserting the ideal gas law gives: dPa g dh (2.12) = − Pa Ra T With temperature T constant, the solution is:  gh  = Pa Pa0 exp  − (2.13)  Ra T  This result is known as the barometric equation. Air pressure apparently depends on the height. However, temperature differences (T variable) and different gas composition (Ra variable) induce pressure gradients at the same height, called ‘stack’. For the pressure difference between two points, one at height h1, the other at height h2, we have: DPa = g Pa,ho

(h2 − h1 ) ( Ra T )

where the suffix ho denotes the average height over the interval h2 – h1. The ‘stack potential’ in a point at height h is then given by the pressure difference between that point in an air mass with variable temperature and composition, and a point at identical height in an air mass at constant temperature and composition:  h  1 dh h  1  = − = − PStack g Pa,ho  ∫  g Pa,ho h    ( Ra T )m ( Ra T )o   h = o Ra ( h ) T (h ) ( Ra T )o  with (Ra T)m the harmonic average of the product Ra(h) T(h) at height h:  ( Ra T )m =  

h

h

h=o

  dh  Ra ( h ) T (h ) 

The resulting stack between two points at same height h, but in a column of air at different temperature and composition, becomes:   1 1 pT,1− 2 = PStack2 − PStack1 = g Pa,ho h  −   ( Ra T )m2 ( Ra T )m1  If between height zero and height h, the temperatures T1 and T2 are different but constant while the composition of the air is the same, the equation simplifies to: pT,1− 2 ≈

g Pa,ho ( Ra1 T1 − Ra2 T2 ) h Ra1 T1 Ra2 T2

≈ ra g  (q1 − q 2 ) h

ra0 g ( Ra1 q1 − Ra2 q 2 ) h Ram12 Tm12 (2.14)

2.2  Air transfer

135

where  (=  1 / Tm12) is compressibility, here of air. If, on the contrary, temperature is constant but the composition differs, stack becomes:  1 1  pT,1− 2 ≈ ra Ra g  z  (2.15) −  Ra2 Ra1  In case air volumes are connected by leaks, a neutral plane with stack zero develops somewhere between the lowest and highest leak. Without other causes of pressure difference, air goes from warm to cold or, in the case of different vapour concentration, from higher to lower concentration above that plane. Below, it moves from cold to warm or from lower to higher concentration. Where the neutral plane sits, depends on leak distribution and leak sizes. Unless for high-rise buildings, air pressure differences by thermal stack are small. Concentration stack is even negligible. If during winter the inside temperature is 20 °C and the outside temperature 0 °C, then over the height of a room (z = 2.5 m) thermal stack reaches pt = 1. 2 · 9.81 · 1 / 273.15 · 20 · 2.5 = 2.15 Pa (ra0 = 1.2 kg/m3, g = 9.81 m/s2,  = 1 / 273.16 K–1, qm12 – q0 = 20 °C, z = 2.5 m). Instead, for a 250 m high-rise building, thermal stack between the lowest and highest floor equals 215 Pa, which makes it as important as the pressure difference between windward and leeside for a wind velocity of 62 km/h. Thermal stack anyhow is much more stable over time than wind pressure differences are. 2.2.2.3 Fans Fans are part of any air heating, air conditioning or forced ventilation system. The pressure differences they create are range from slightly to quite higher than those by stack and wind. Fan induced pressure differences are also very stable over time.

2.2.3 Air permeances We have to differentiate between: 1. Open porous materials, such as no-fines concrete, mineral fibre and wood-wool cement 2. Air permeable layers (layers made of small elements with joints or overlaps in between), leaks, cracks, cavities and intentional or unintentional gaps. Air open All coverings, sidings and finishings made of scaly or plate elements (tiles, slates, corrugated sheets, lathed finishes, wooden laths) and all layers composed of board-like or strip formed materials (plywood, OSB, insulation layers, underlay). Even masonry is air open (cracks between bricks and joints!) (Figure 2.3).

Figure 2.3.  Examples of air-open layers: tiles, slates and a lathed ceiling.

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2  Mass Transfer

Joints and cracks Joints are found between construction parts and between boards and other elements. Cracks form spontaneously when tensile strength is superseded. Leaks Each nail hole, electricity box in a wall, light spot in a ceiling Cavities, voids They are found in a variety of outer wall, party wall and roof solutions Openings in the envelope Occur intentionally (trickle vents) or unintentionally (damage, bad joint filler, loose connection, etc.).

In an open porous material, air displacement follows Poiseuille’s law of proportionality between air flow rate and driving force, in the case being the change of air pressure per unit length. The proportionality factor is called air conductivity or permeability ka, units s: ga = − ka grad Pa (2.16) The minus sign in the equation indicates air flows from points at higher to points at lower air pressure. The permeability ka is a scalar for isotropic and a tensor with three main directions and a value differing per direction for anisotropic materials. Permeability increases with open porosity and the number of macro pores in a material. For air permeable layers, leaks, cracks, joints, cavities and openings, the flow equations become: Air permeable layers (per m 2 ) g a = − K a DPa , Ga in kg/(m 2 ⋅ s), K a in s/m Joints, cavities (per m)

Ga = − K aψ DPa , Ga in kg/(m ⋅ s), K aψ in s

Leaks, openings (per unit)

Ga = − K aχ DPa , Ga in kg/s, K aχ in m ⋅ s

(2.17)

where Ka, K aψ and K aχ are the air permeances and DPa the air pressure differentials. Because the flow is not necessarily laminar, the air permeance depends on the pressure differential, in an exponential relationship is expressed as: K ax = a ( DPa )

b −1

where a is the air permeance coefficient for a pressure differential of 1 Pa, and b is the air permeance exponent. For laminar flow, b is 1, for turbulent flows it is 0.5. For transition flows, b lies between 0.5 and 1. Normally, the air permeance is only quantifiable by experiment. For joints, leaks, cavities and openings with a known geometry the value can sometimes be calculated using hydraulics. Pressure losses to take into account are first of all frictional losses: = DPa

f

L ra v 2 L 2 ≈ 0, 42 f g (2.18) dH 2 dH a

where f is the friction factor, dH the hydraulic diameter of the section with length L, v the average flow velocity in that section (Figure 2.3). Local losses near bends, widenings, narrowings, entrances and exits add (Figure 2.4): ra v 2 ≈ 0, 42  g a2 (2.19) 2 with  the factor of local loss. = DPa 

137

2.2  Air transfer

Figure 2.4.  Friction losses, local losses and continuity.

Values for the friction factor are given in Table 2.2. Table 2.3 shows some factors of local loss. The calculation of air permeances then stems from the continuity principle, stating that in each section, the same air flow passes (= mass conservation, Figure 2.5). In other words, along the flow line, the air flow Ga is constant. Further on, total pressure loss – the sum of the frictional and all local losses – must equal the driving force. Both allow calculating the air flow Ga as function of pressure difference DPa. Table 2.2.  Friction factor ‘f’. Reynolds number(1)

Flow

Friction Factor f

Re ≤ 2500

Laminar

96 Re

2500 ≤ Re ≤ 3500

Critical

Re &gt; 3500

Turbulent

fT(2)

Re  3500

Stable turbulent

fT = Cte, single function of roughness

0,038 (3500 − Re) + f T,Re = 2500 (Re − 2500) 1000

v dH where v is average flow velocity,  kinematic viscosity and dH hydraulic diameter:  – for a circular section: the diameter of the circle 2ab – for a rectangular section: , where a and b are the sides of the rectangular a+b – for a cavity: 2 b, where b is the width of the cavity

Re can also be written as: Re ≈ 56,000 ga dH with ga air flow rate

(1)

Re =

    10  + 0, 2   4,793 log      Re  (2)  2 log  − + 0, 2698    fT = Re         

−2

with  relative roughness (see Figure 2.5).

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2  Mass Transfer

Figure 2.5.  Definition of roughness for a circular tube. Table 2.3.  Factor of local loss . Local resistance

Entering an opening

0.5

Leaving an opening

1.0 Re ≤ 1000

–0.036  +  9.6 · 10–5 Re + D

1000 &lt; Re ≤ 3000

1.28 · 10–5 Re1.223 + D

Re &gt; 3000

0.21 Re0.012 + D

Widening  = A0 / A1 A0 small section A1 large section

with

0.98 Re–0.03 + A

1000 &lt; Re ≤ 3000

10.59 Re–0.37 + A

Re &gt; 3000

0.57 Re–0.01 + A

where A = 0.0373 s2 – 0.067 

Leak

2.85 kRe0

Angle or curve 0 b0 b1 f0

refers to the inlet channel width of the inlet channel channel width after the curve friction factor in the inlet channel

D = 0.78 – 1.56 s D = 0.48 – 0.96 

Re ≤ 1000

Narrowing  = A0 / A1 A0 small section A1 large section

 ≤ 0.5  &gt; 0.5

g

( )

e

k0

b  where g = 0.885  1   b0 

3000 ≤ Re &lt; 40,000

−0,86

and

Re ≥ 40,000

kRe0

k0 / (dH)0

kRe0

k0 / (dH)0

0

45 f0

1

1.1

1

0–0.001

45 f0

1

1.0

1  +  0,5 · 103 a

&gt; 0.001

45 f0

1

1.1

1

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2.2  Air transfer

2.2.4 Air transfer in open-porous materials 2.2.4.1 Conservation of mass For air, we have (neither source nor sink): ∂ (2.20) ∂ Air content wa depends on total open porosity Ψo, saturation degree Sa, air pressure Pa and temperature T: Wa = Ψ0 Sa ra = Ψ0 Sa Pa / (Ra T). Whereas the gas constant Ra and the open porosity Ψo are invariable, saturation degree Sa, air pressure Pa and temperature T may vary. Therefore, the differential of air content with respect to time turns into: div

= −

Ψ0 ∂wa = ∂t Ra T

∂S S P ∂T   ∂Pa + Pa a − a a  Sa  ∂t ∂t T ∂t 

If the saturation degree is constant, then the equation simplifies to: ∂wa Ψ 0 Sa = ∂t Ra T

 ∂Pa Pa ∂T  −   ∂t T ∂t 

∂T = 0 (that of course excludes thermal stack): ∂t  Ψ S  ∂P div ga = −  0 a  a (2.21)  Ra T  ∂t

In isothermal conditions,

The ratio Ψ0 Sa / (Ra T) is called the isothermal volumetric specific air content, symbol ca, units kg/(m3 · Pa). Both the open porosity and saturation degree for air now cannot pass 1. Replacing Ra by its value, 287 J/(kg · K) so gives: Ψ0 Sa / (Ra T) &lt; 0.00348 / T, a very small number. 2.2.4.2 Flow equation See equation (2.16). 2.2.4.3 Air pressures Inserting equation (2.16) in the isothermal mass balance (2.21) gives: ∂Pa ∂t In case the air permeance ka is constant, we can write: div ( ka grad Pa ) = ca

ca ∂Pa ∇ 2 Pa = (2.22) k a ∂t In (2.22) the ratio ka / ca stands for the isothermal air diffusivity of the material, symbol Da, units m2/s. That value is generally so high each adjustment in air pressure happens within a few seconds. Only for pressures that fluctuate over extremely short periods, such as wind gusts, will the air pressure response show damping and time shift. With respect to periods of some minutes or longer, damping turns to one and the time shifts to zero. Take for example mineral wool with a density of 40 kg/m3. Properties:

140

2  Mass Transfer

Thermal

Air

  =  0.036  W/(m · K)

ka  =  8 · 10–5 s

 c = 33,600 J/(m3 · K)

ca  =  1,17 · 10–5 kg/(m3 · Pa)

This gives a thermal diffusivity 1,1 · 10–6 m2/s, and an isothermal air diffusivity 6,8 m2/s. That is the latter is 6,355,000 times larger! Thus, when combining air transfer with much slower heat and moisture transfers, the isothermal equation (2.22) reduces to steady-state: ∇ 2 Pa = 0 With the flow equation (ga = –ka grad Pa)), it forms a copy of steady state heat conduction. In non-isothermal condition, i.e. included stack, the same conclusion holds: steady state suffices. Yet, the isobaric volumetric specific air capacity is larger than the isothermal one (≈ 3525 / T 2), resulting in more inertia but for periods beyond minutes, air pressure damping remains one. As thermal stack interferes now, the equations are best explicitly written: ∇ 2 ( Pao − ra g z ) = 0

ga = − ka grad ( Pao − ra g z )

showing that a solution is only possible in combination with the thermal balance. The same holds for non-isothermal flow through air permeable layers, joints, cavities, openings, etc. 2.2.4.4 One dimension: flat assemblies The results are analogical as for steady state heat conduction. Thus, in a single-layered assembly, the air pressures form a straight line (Figure 2.6) while the air flow rate is given by: DPa DPa = g a k= (2.23) a d d ka

Figure 2.6.  Airflow across a single-layered flat assembly.

141

2.2  Air transfer

The ratio d / ka is called the (specific) air resistance of the assembly; symbol W, units m/s. The inverse gives the air conductance or permeance; symbol Ka, units s/m. For the air flow rate across a composite assembly, we have: ga =

DPa (2.24) di ∑ i =1 ka,i n

where Σ di / ka, i is total air resistance of the composite assembly, symbol WT, units m/s. The inverse 1 / WT is called the total air permeance KaT. di / ka,i represents the air resistance of layer i, symbol Wi. The inverse 1 / Wi = ka,i / di is called the air permeance Ka,i of layer i. Air pressure on a random interface in the wall is given by: j

∑ Wi

Pa, j =Pa1 + i =1 WT

( Pa2 − Pa1 ) =Pa1 + ga W1 j (2.25)

This result is also found graphically, after transposing the composite assembly into the [W, Pa] plane, where it behaves as if single-layered, and transferring the intersections with the interfaces there to the assembly in the [d, Pa] plane (Figure 2.7). As air pressure difference DPa, j over any layer j we have: = DPa, j

Wj WT

( Pa2 − Pa1 )

showing that the largest air pressure difference in a composite assembly is found over the most air-tight layer (Figure 2.8). Including such a layer diminishes air permeance substantially, which is why it is called an ‘air barrier’ or ‘air retarder’. Its purpose is to limit the air flow across the assembly. Of course, an air barrier must be able to withstand wind load.

Figure 2.7.  Airflow across a composite assembly. The graph on the left gives the air pressure in a [Wa, Pa]-plane, the graph on the right in the [da, Pa]-plane.

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2  Mass Transfer

Figure 2.8.  Effect of an airtight layer on the air pressure distribution in an assembly. Attracts total pressure difference.

Is such flat assembly approach of practical relevance? The answer is no. The prerequisite to one-dimensional flow excludes stack, which by definition gives a pressure profile along the height. Small air layers in the interfaces may additionally activate stack flow along the height. Also wind pressure is not uniform along the height. And, unintentional cracks can disrupt one-dimensionality. More than a theoretical value should not be attributed to the terms ‘flat’ and ‘one dimension’. 2.2.4.5 Two and three dimensions For isotropic open porous materials the calculation is a copy of two and three dimensional heat conduction. Of course, mass conservation substitutes energy conservation: the sum of air flows from the neighbouring to each central control volume is zero: Σ Ga, i +j = 0. The algorithm then becomes:

(

)

K a,′ i + j Pa, i + j − Pa, l , m, n K a,′ i + j = 0 i l ,= m, n i l , m, n j= ±1 j= ±1 with K a,i ′ + j the air permeance between each of the neighbouring and the central control volumes. Value within a material: ka A / a, with A the contact surface of any control volume with each of the neighbouring control volumes and a the distance along the mesh between their central points (Figure 2.9). For p control volumes, the result is a system of p equations with p unknown:

Figure 2.9.  Conservation of mass: sum of airflows from adjacent to central control volume zero.

143

2.2  Air transfer

[ Ka′ ] p ⋅ p [ Pa ] p =  Ka,′ i, j , k Pa,i, j , k  p (2.26) where [ K a′ ] p ⋅ p is a p rows, p columns permeance matrix, [Pa]p a column matrix of the p unknown air pressures and [ K a,′ i , j , k Pa, i , j , k ] p a column matrix containing the p known air pressures and/or air flows. Once the air pressures are calculated, then the air flows follow from:

(

)

= Ga, i , i + j K a,′ i , i + j Pa, i + j − Pa, i (2.27) For anisotropic materials the same algorithm applies, though per mesh line the direction-related K a′ -value has to be taken. Besides, it is implicitly understood the mesh lines coincide with the main directions of the air permeability tensor. Under non-isothermal conditions, thermal stack intervenes and the thermal and air balances have to be solved simultaneously. Most of the time this requires iteration.

2.2.5 Air flow across permeable layers, apertures, joints, leaks and cavities 2.2.5.1 Flow equations See above. For operable windows and doors, the air permeance exponent b is typically set at 2/3: Ga = a L DPa2 3 with L the casement length in m and a the air permeance per meter run of casement, units kg/(m · s · Pa2/3). 2.2.5.2 Conservation of mass: equivalent hydraulic circuit Most constructions combine air-open layers with open porous materials, joints and cavities. In such a case, expressing the conservation law as a partial differential equation is impossible. An approach consists of transforming the construction into an equivalent hydraulic circuit, i.e., a combination of well chosen points, connected by air permeances (Figure 2.10). In each point the sum of the airflows from the neighbouring points must be zero, or: Σ Ga = 0. As each of the air flows is given by Ga = K ax DPa , insertion in the conservation law results in:

(

)

K a,x i + j Pa, i + j − Pa, l , m, n K a,x i + j = 0 (2.28) i l ,= m, n i l , m, n j= ±1 j= ±1

Figure 2.10.  Equivalent hydraulic circuit.

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2  Mass Transfer

With p points, a system of p non-linear equations ensues. The p unknown air pressures are found by solving for the known boundary conditions. Insertion of the air pressures found into the flow equations then gives the airflows between points. However, non-linearity makes iteration necessary, starting by assuming values for the p air pressures, calculating related air permeances and solving the system. The p new air pressures allow recalculating the air permances and solving the system again. Iteration that way continues until standard deviation between preceding and present air pressures reaches a suitable value (for example &lt; 1/1000). As soon as flat assemblies contain air permeable layers, leaks, joints and cavities, that hydraulic circuit method applies. In the exceptional case we nevertheless have one-dimensional flow, and the assembly dissolves in a series circuit of linear and non-linear air permeances. As the same airflow rate ga migrates across each, the following holds: 1

= Layer 1 g a K= a1 DPa1

a1 DPab1 1

 g a  b1 DPa1 or =  a  1

= Layer 2 g a K= a 2 DPa 2

a2 DPab2 2

 g a  b2 DPa 2 or =  a  2

b an DPa n n

 g a  bn DPa n or =  a  n

1

 1

n g a K= Layer = a n DPa n

Sum:

1  −1  bi  ga  g a  ∑ 1  = DPa  b  ai i  

(2.29)

where DPa is the air pressure difference over the flat assembly. Equation (2.29) requires iteration. Once the air flow rate ga determined, the air pressure distribution in the flat assembly then follows from the layer equations.

2.2.6 Air transfer at building level 2.2.6.1 Definitions In buildings, a distinction is made between inter-zone and intra-zone airflow. Inter-zone considers the air transfer between spaces and between spaces and the outside. To model, we replace the building and its environment by a grid of zone points, linked by flow paths, and write the mass equilibrium per point. Intra-zone considers the air movements within a space. Is air looping from the ceiling along the windows to the floor or vice versa? How does the zone and ventilation air mix? Are there corners, which lack air washing? All this demands application of conservation of energy, mass and momentum to the air filling the space, which demands applying CFD. Here, we limit the discussion to inter-zone airflow.

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