transport of water vapour...– Winterday -10°C 90% – Summerday 25°C 40% • On which day is the air more humid (absolute humidity)? 32 . vv v v. mp V RT. ρ= = To remember •

TRANSPORT OF WATER VAPOUR Chapter 6

1

Quantities for moist air Section 6.1

2

3 http://www.atmosedu.com/meteor/Animations/chapter23/

http://www.atmosedu.com/meteor/Animations/chapter23/

saturated vapor pressure

• imagine a closed container, partially filled with liquid water

vice versa, all gases have a tendency to condense

like all liquids, water has the

tendency to evaporate

the corresponding vapor pressure in the gas is called “saturated vapor pressure” (𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 [Pa])

at equilibrium Gevaporation = Gcondensation

4


• now increase the temperature

since the vapor pressure increases, the condensation rate increases as well

molecules have more energy to escape

from the liquid and evaporation increases

a new equilibrium is found, characterized by a higher “saturated vapor pressure” (𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 [Pa])

5


• how to find 𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 for a given temperature? – from charts (not sufficiently accurate) – from tables – from mathematical expressions

these three methods have a different accuracy for exercises, you can choose between tables (book) or mathematical expression

6


• determination of 𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 based on a chart

𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 ≈ 1100 Pa

𝜃𝜃 ≈ 8.4 °C

7

For building physics applications, pv,sat lies between ≈ 100-10’000 Pa !!!


• determination of 𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 based on a table

𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 = 1102 Pa at 8.4 °C 8


• determination of 𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 based on a formula

𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 = 1101.8 Pa at 8.4 °C

𝜃𝜃 < 0℃

𝜃𝜃 ≥ 0℃

9

22.5273

, 610.5v satp eθθ+

=

17.27237.3

, 610.5v satp eθθ

+ =

relative humidity

• now we remove the lid of the container

since there are less molecules above the solution, the condensation rate decreases

evaporation will still continue

some molecules escape

the vapor pressure in the air is lower than 𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠

10

relative humidity

• relative humidity (RH) is the ratio of the vapor pressure to the saturated vapor pressure – symbol: 𝜑𝜑

– formula: 𝜑𝜑 = 𝑝𝑝𝑣𝑣𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠

– unit: dimensionless, sometimes in %

by definition, 𝜑𝜑 has to have a value between 0 and 1! so between 0% – 100%!

11

relative humidity

12

For building physics applications, pv lies between ≈ 0-10’000 Pa !!!

Warm and humid air => high pv

Cold and dry air => low pv

dew point

• the dew point is the temperature to which humid air must be cooled, at constant pressure, for vapor to condense into water – symbol: 𝜃𝜃𝑑𝑑

– unit: °C

• the dew point can be found using (i) a chart,

(ii) a table or (iii) a formula

13

relative humidity

14 Dewpoint θd

Temperature at which pv,sat(θd)= pv(θamb,RHamb)

What is the dew point at 25°C and RH = 50%?

condensation

• condensation occurs when the temperature at a certain location is equal to or below the dew point of the humid air at that location:

• condensation occurs when the vapor pressure at a certain location becomes equal to the saturated vapor pressure at that location:

dθθ ≤

satvv pp ,=15

,

100%v

v sat

pp

ϕ = =

vapor concentration

• the vapor concentration is the mass of vapor per unit volume of air – symbol: 𝜌𝜌𝑣𝑣

– formula: 𝜌𝜌𝑣𝑣 = 𝑚𝑚𝑣𝑣𝑉𝑉

• 𝜌𝜌𝑣𝑣 depends on 𝑝𝑝𝑣𝑣 and 𝑇𝑇!

– unit: kg/m3

Example: at 99% RH and 20°C

→ 𝜌𝜌𝑣𝑣 =𝑝𝑝𝑣𝑣𝑅𝑅𝑣𝑣𝑇𝑇

16 ( )

3

0.99 2337 0.0171462 20 273.15

.

v vsatv

v v a

p p Pa kgJR T R T mK

kg K

ϕρ ×= = = =

+

Also called absolute humidity

vapor ratio

• the vapor ratio is the mass of vapor per unit mass of dry air – symbol: 𝑥𝑥 – formula:

• often used in HVAC product documentation

– unit: kg/kg Example: at 99% RH and 20°C

d

v

dv

vd

d

v

d

v

pp

pRpR

mmx 62.0====

ρρ

17

3

3

0.01421.

0.0171

2

v

a

v

d

a

v

dd

m

m

kgkgx kg kg

ρρ

= = =

Amount of water vapour that can be contained in the air is small!

31.2d akgm

ρ ρ≈ = 101325d ap P Pa≈ =

Question humidity

• The air in a closed room (30°C) has an absolute humidity of 18 g/m3. When the air is cooled, how do the following parameters change: – Relative humidity – Absolute humidity – Dew point temperature

18

unit equation

Ideal gas law -

Relative humidity %

Vapor concentration kg/m³

Vapor ratio kg/kg

Air pressure Pa

overview

v v

d d

mxm

ρρ

= =

19

v vv

v

m pV R T

ρ = =

,

v

v sat

pp

ϕ =

v v vp V m R T=

a d vP p p= +

exercise

• you place some food in a box (15 x 15 x 5 cm3) in the kitchen (20°C, 50% RH). You close the lid of the box, and put the box in the fridge (2°C). – what happens? – express initial and final

state in 𝑝𝑝𝑣𝑣, 𝜑𝜑, 𝜌𝜌𝑣𝑣, 𝑥𝑥, 𝑚𝑚𝑣𝑣

(𝑅𝑅𝑣𝑣 = 462 J/(kg.K); 𝜌𝜌𝑠𝑠 = 1.2 kg/m3)

20

exercise - solution

• requested

21

initial final with condens

θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50%

𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 [Pa]

𝑝𝑝𝑣𝑣 [Pa]

𝜌𝜌𝑣𝑣 [kg/m3]

𝑥𝑥 [kg/kg]

𝑚𝑚𝑣𝑣 [kg]

exercise - solution

• requested

22


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 100%

𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 [Pa] 2337 Pa 705 Pa 705 Pa

𝑝𝑝𝑣𝑣 [Pa]

𝜌𝜌𝑣𝑣 [kg/m3]

𝑥𝑥 [kg/kg]


exercise - solution

• requested

23

, , , 0.5 2337 Pa 1169 Pav ini ini v sat inip pϕ= = × =

( ),

, 3

1169 Pa 0.0086462 20 273.15 K

v ini vv ini

v ini a

p kgJR T m

kgK

ρ = = =+


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 100%


𝑝𝑝𝑣𝑣 [Pa] 1169 Pa

𝜌𝜌𝑣𝑣 [kg/m3] 0.0086 kg/m3

𝑥𝑥 [kg/kg]


exercise - solution

• requested

24

3,

3

0.00860.0072

1.2

v

v ini a v

aa a

a

kgm kgx kg kg

m

ρρ

= = =


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 100%


𝑝𝑝𝑣𝑣 [Pa] 1169 Pa 705 Pa


𝑥𝑥 [kg/kg] 0.0072 kg/kg


31.2d aa

a

kgm

ρ ρ≈ =

exercise - solution

• requested

25

( )3 6, 30.0086 0.15 0.15 0.05 9.71 10v

v v ini box va

kgm V m kgm

ρ −= = × × = ×


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 100%




𝑥𝑥 [kg/kg] 0.0072 kg/kg

𝑚𝑚𝑣𝑣 [kg] 9.71x10-6 kg

exercise - solution

• requested


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 100%



𝜌𝜌𝑣𝑣 [kg/m3] 0.0086 kg/m3 0.0086 kg/m3

𝑥𝑥 [kg/kg] 0.0072 kg/kg 0.0072 kg/kg

𝑚𝑚𝑣𝑣 [kg] 9.71x10-6 kg 9.71x10-6 kg

26

exercise - solution

• requested


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 100%


𝑝𝑝𝑣𝑣 [Pa] 1169 Pa 1097 Pa 705 Pa

𝜌𝜌𝑣𝑣 [kg/m3] 0.0086 kg/m3 0.0086 kg/m3

𝑥𝑥 [kg/kg] 0.0072 kg/kg 0.0072 kg/kg

𝑚𝑚𝑣𝑣 [kg] 9.71x10-6 kg 9.71x10-6 kg

27

( ),fin ,fin 30.0086 462 2 273.15K 1097vv v v fin

a

kg Jp R T Pam kgK

ρ= = + =

exercise - solution

• requested


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 156% 100%


𝑝𝑝𝑣𝑣 [Pa] 1169 Pa 1097 Pa 705 Pa

𝜌𝜌𝑣𝑣 [kg/m3] 0.0086 kg/m3 0.0086 kg/m3

𝑥𝑥 [kg/kg] 0.0072 kg/kg 0.0072 kg/kg

𝑚𝑚𝑣𝑣 [kg] 9.71x10-6 kg 9.71x10-6 kg

28

( ),fin ,fin 30.0086 462 2 273.15K 1097vv v v fin

a

kg Jp R T Pam kgK

ρ= = + =

,finfin

, ,fin

1097 156%705

v

v sat

p Pap Pa

ϕ = = = Condens until RH = 100%

exercise - solution

• requested


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 156% 100%


𝑝𝑝𝑣𝑣 [Pa] 1169 Pa 1097 Pa 705 Pa

𝜌𝜌𝑣𝑣 [kg/m3] 0.0086 kg/m3 0.0086 kg/m3

𝑥𝑥 [kg/kg] 0.0072 kg/kg 0.0072 kg/kg

𝑚𝑚𝑣𝑣 [kg] 9.71x10-6 kg 9.71x10-6 kg

29 Condens until RH = 100%

exercise - solution

• requested


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 156% 100%


𝑝𝑝𝑣𝑣 [Pa] 1169 Pa 1097 Pa 705 Pa

𝜌𝜌𝑣𝑣 [kg/m3] 0.0086 kg/m3 0.0086 kg/m3 0.0055 kg/m3

𝑥𝑥 [kg/kg] 0.0072 kg/kg 0.0072 kg/kg 0.0046 kg/kg

𝑚𝑚𝑣𝑣 [kg] 9.71x10-6 kg 9.71x10-6 kg

30

( ),cond

,cond 3

705Pa 0.0055462 2 273.15K

v vv

v fin a

p kgJR T m

kgK

ρ = = =+

,cond 0.0046v v

a a

kgxkg

ρρ

= =

exercise - solution

• requested


θ [°C] 20°C 2°C 2°C

𝜑𝜑 [%] 50% 156% 100%


𝑝𝑝𝑣𝑣 [Pa] 1169 Pa 1097 Pa 705 Pa

𝜌𝜌𝑣𝑣 [kg/m3] 0.0086 kg/m3 0.0086 kg/m3 0.0055 kg/m3

𝑥𝑥 [kg/kg] 0.0072 kg/kg 0.0072 kg/kg 0.0046 kg/kg

𝑚𝑚𝑣𝑣 [kg] 9.71x10-6 kg 9.71x10-6 kg 6.24x10-6 kg

31

( )3 6, ,cond 30.0055 0.15 0.15 0.05 6.24 10v

v cond v box va

kgm V m kgm

ρ −= = × × = ×

6, , ,cond 3.46 10cond liq v ini v lm m m kg−= − = × Amount of condensed liquid water

Question humidity

• 2 cases – Winterday -10°C 90% – Summerday 25°C 40%

• On which day is the air more humid (absolute humidity)?

32

v vv

v

m pV R T

ρ = =

To remember

• Moisture quantities

• 𝑝𝑝𝑣𝑣,𝑠𝑠𝑠𝑠𝑠𝑠 from table or formula 33

Additional exercises Vapour - part 1

34

Exercise 1: quantities for moist air • Air at 34°C and 39% relative humidity?

– What is the partial vapor pressure (pv) 2073 Pa – What is the vapor concentration (ρv) 0.0146 kg/m3

– What is the vapor ratio (x) 0.0122 kg/kg – What is the dew point (θd) ≈ 18.1°C

• The air is heated to 40°C (in a closed container)

– What is the partial vapor pressure (pv) 2114 Pa – What is the relative humidity 29 % – What is the vapor concentration (ρv) 0.0146 kg/m3

– What is the vapor ratio (x) 0.0122 kg/kg

use the formula to calculate pv,sat use the table to determine dew point

35

Vapor transport Section 6.2

36

advection ↔ diffusion

advection • air moves • water vapor is transported

together with the air • driven by a pressure

difference (ΔPa)

diffusion • air is stagnant • water vapor is transported

through the air • driven by a concentration

difference (Δpv)

37 High concentration (pv >>>)

Low concentration (pv <<<)

diffusion

• Applications in this course – Diffusion through building materials/components – Interstitial condensation

38

diffusion

• imagine a closed container, consisting of two compartments

water vapor dry air

vapor concentration is different in both compartments

39

diffusion

• now take away the barrier and wait until a new equilibrium is found

water vapor dry air

this transport mechanism is called diffusion

40

vapor diffusion in air

• diffusion in air is described by

a

vvdav Dg

ρρρ ∇−=

with ρa [kg/m3] the air density ρv [kg/m3] the vapor concentration Dvd [m2/s] the binary diffusion coefficient of water vapor in dry air, given by

81.1101069.8 TDvd−⋅= (according to Schirmer)

41


• let’s simplify this formula

a

vvdav Dg

ρρρ ∇−=

(assume that ρa is constant) vvdv Dg ρ∇−=

(vapor behaves as an ideal gas)

TRpDgv

vvdv ∇−=

(assume that T is constant) v

v

vdv p

TRDg ∇−=

42


• and some further simplifications

(introduce the variable δa = vapor permeability of air)

vav pg ∇−= δ

(definition of a gradient)

vv

vdv p

TRDg ∇−=

va

v pd

g ∆=δ

s 102 10−⋅≈aδ at 20°C

43

Large vapor flux if: • Large vapor pressure difference • Large vapor permeability of air • Small thickness of (air) layer d

𝛥𝛥𝑝𝑝𝑣𝑣 𝛿𝛿𝑠𝑠


• the mass flow rate by diffusion in air is thus

with 𝐾𝐾𝑣𝑣𝑠𝑠 [s/m] the (vapor) diffusion permeance of air 𝛿𝛿𝑠𝑠 [s] the vapor permeability of air 𝑑𝑑 [m] the thickness of the air layer 𝐴𝐴 [m2] the frontal area of the air layer

diffusion is an extremely slow process, but since buildings have a long life-time …

44

( ),1 2 ,1 ,2aa

v v v v vG A p AK p pdδ

− = ∆ = − a avK

dδ

=


• … diffusion can lead to severe problems

© Marsden Property Services

45


• the diffusion constant 𝑁𝑁 [s-1] is the inverse of the vapor permeability of air

• 𝑁𝑁 was introduced because small quantities give the (wrong) impression that the associated phenomena are unimportant

a

Nδ1

= -19 s 105 ⋅≈N at 20°C and

46

vapor diffusion in materials

• just like (advective) air transport, also diffusive transport can take place through materials with open porosity

accessible

non-accessible

solid material

since there is only a limited number of pathways through the material, diffusion will be slower than in plain air

47


• the shortest path through a material is at least as long as the material thickness, but probably much longer

48

accessible

non-accessible

solid material

this is due to the «tortuosity» of the pore system d

L

L ≥ d


• the vapor resistance factor μ [-] quantifies how much slower diffusive vapor transport in a material is, compared to in air

v

a

δδµ =

with δa [s] the vapor permeability of air δv [s] the vapor permeability of the material

by definition, 1 ≤ μ < +∞

or N

av µµ

δδ 1==

49

vapor diffusion in materials material vapor resistance factor [-]

Brick 10-20

Concrete 100

Glass +∞

Gypsum board 12

Metal +∞

Mineral wool 1.5

Natural stone (sandstone) 15

Polymer >10000

Plastic foam >30

Wood (pine) 50

porous materials are more vapor open

50


• the (vapor) diffusion resistance 𝑍𝑍 [m/s] was defined as

• 𝑍𝑍 is –literally– the resistance against diffusion • the higher 𝑍𝑍, the smaller the vapor flux

through the material for a given difference in partial vapor pressure

dNddZav

µδµ

δ===

51

wall ii

Z Z= ∑


• material documentation – for most materials, documentation specifies the

vapor permeability δv [s] or the vapor resistance factor μ [-]

– for some materials, especially foils, it is hard to measure the thickness. For those materials, the diffusion thickness μd [m] is specified.

from any of these, the diffusion resistance Z [m/s] can be calculated

52


• mass flow rate 𝐺𝐺𝑣𝑣,1

𝜙𝜙→2

[kg/s] through a layer of

open porous material from location 1 to 2:

with 𝐾𝐾𝑣𝑣𝜙𝜙 = 1

𝑍𝑍⁄ [s/m] the vapor permeance, and 𝐴𝐴 [m2] the frontal area of the open porous layer

53

𝑝𝑝𝑣𝑣,1 𝑝𝑝𝑣𝑣,2 𝐺𝐺𝑣𝑣,1

𝜙𝜙→2

𝐺𝐺𝑣𝑣,1

𝜙𝜙→2

= 𝐾𝐾𝑣𝑣𝜙𝜙𝐴𝐴 𝑝𝑝𝑣𝑣,1 − 𝑝𝑝𝑣𝑣,2

Always use same convention: Draw flow rate from point 1 -> 2 Write flow rate as 𝑝𝑝𝑣𝑣,1 − 𝑝𝑝𝑣𝑣,2

( ) vvv pAKG ∆= φ

To remember

• in air layer

• in porous layer

• in joints

• in holes

( ) vavv pAKG ∆= with 𝐾𝐾𝑣𝑣𝑠𝑠 [s/m] the vapor permeance

and 𝐴𝐴 [m2] the area of the layer

( ) vvv pLKG ∆= ψ

( ) vvv pnKG ∆= χ

with 𝐾𝐾𝑣𝑣𝜓𝜓 [s] the vapor permeance

and 𝐿𝐿 [m] the length of the joint

with 𝐾𝐾𝑣𝑣𝜒𝜒 [m.s] the vapor permeance

and 𝑛𝑛 [-] the number of holes

54

with 𝐾𝐾𝑣𝑣𝜙𝜙 [s/m] the vapor permeance

and 𝐴𝐴 [m2] the area of the layer

• Flow rate

• Vapour resistance – permeance

( ) ( ),1 ,2,1 2 ,1 ,2

v vv v v v

A p pG K A p p

Zφ

−

−= − =

To remember

55

1

v v a

d dZ N dKϕ

µ µδ δ

= = = =

Always use same convention Draw flow rate Gv,1-2 from point 1→2 Write pressure difference as (pv,1-pv,2)

9 -11 5 10 sa

Nδ

= ≈ ⋅v

a

δδµ =

• Mass balance example – Draw all flow rates towards point of interest

( ) ( ) ( )1 ,1 ,i 2 ,2 ,i 3 ,3 ,i

1 2 3

0v v v v v vA p p A p p A p pZ Z Z− − −

+ + =

To remember

56

,1vp ,2vp

,3vp

,ivp

,1 ,2 ,3 0v i v i v iG G G− − −+ + =

57

Exercises next week

• Bring slides / book / calculator or laptop

Additional exercises Vapour - part 1

58

Exercise 2: Vapour resistance • A wall consists of two material layers (from inside to outside):

– layer 1 has thickness d = 0.115 m and vapor resistance factor µ = 4 – layer 2 has thickness d = 0.50 m and vapor resistance factor µ = 2

• What is the vapour permeance Kv

φ of the wall (in s/m) (N = 5.0x109 s-1)

• What is the total vapour flux through this wall (in kg/(m2s)) for following vapor pressures at the interior and exterior: – pv,i = 2100 Pa and pv,e = 600 Pa

• What is the vapor pressure (in Pa) at the interface between layer 1 and

2?

(Hint: gv is constant throughout wall so gv,i->e =gv,i->interface

59

101.37 10vsKm

φ −= ×

722.05 10v

kggm s

−= ×

,interface 1627vp Pa=

Exercise 3: Retrofitting basement

• A basement is transformed into a disco. Above the basement, there is a heated space. Due to the new room purpose, the air temperature and RH in the basement will increase. To avoid surface condensation, the basement wall was retrofitted with insulation.

– Air temperature 25°C – Soil temperature 12°C – Relative humidity 80% – Heat transfer coefficient 8 W/m2K

60 Original basement wall (at soil temperature)

Insulation d = 80 mm λ = 0.2 W/mK

Mortar d = 35 mm λ = 0.7 W/mK

Exercise 3: Retrofitting basement

• Calculate the surface temperature on the inner surface with and without insulation.

• Is retrofitting necessary and sufficient to avoid surface condensation (calculate!)?

61 Original basement wall (at soil temperature)

Insulation d = 80 mm λ = 0.2 W/mK

Mortar d = 35 mm λ = 0.7 W/mK

θs,i,without-insul = 15.7 °C θs,i,with-insul = 22.2 °C

Yes, retrofitting is necessary

Documents

transport of water vapour...– Winterday -10°C 90% – Summerday 25°C 40% • On which day is the air more humid (absolute humidity)? 32 . vv v v. mp V RT. ρ= = To remember •