Question : Condensation and Freezing of Moisture in Walls The condensat...

Condensation and Freezing of Moisture in Walls

The condensation and even freezing of moisture in walls without effective vapor retarders is a real concern in cold climates, and it undermines the effectiveness of insulations. Consider a wood frame wall that is built around 38  mm \times 90  mm ( 2 \times 4 nominal) wood studs. The 90-mm-wide cavity between the studs is filled with glass fiber insulation. The inside is finished with 13-mm gypsum wallboard and the outside with 13-mm wood fiberboard and 13-mm \times 200-mm wood bevel lapped siding. Using manufacturer’s data, the thermal and vapor resistances of various components for a unit wall area are determined to be as:

 

Construction R-Value,  m^{2} \cdot ^{\circ}C/W R_{\nu }-Value,  s \cdot m^{2} \cdot Pa/ng
1. Outside surface, 24 km/h wind 0.030
2. Painted wood bevel lapped siding 0.14 0.019
3. Wood fiberboard sheeting, 13 mm 0.23 0.0138
4. Glass fiber insulation, 90 mm 2.45 0.0004
5. Painted gypsum wallboard, 13 mm 0.079 0.012
6. Inside surface, still air 0.12
TOTAL 3.05 0.0452

 

The indoor conditions are 20^{\circ}C and 60 percent relative humidity while the outside conditions are - 16^{\circ}C and 70 percent relative humidity. Determine if condensation or freezing of moisture will occur in the insulation.

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SOLUTION  The thermal and vapor resistances of different layers of a wall are given. The possibility of condensation or freezing of moisture in the wall is to be investigated.

Assumptions  1 Steady operating conditions exist. 2 Heat transfer through the wall is one-dimensional. 3 Thermal and vapor resistances of different layers of the wall and the heat transfer coefficients are constant.

Properties  The thermal and vapor resistances are as given in the problem statement. The saturation pressures of water at 20^{\circ}C and – 16^{\circ}C are 2339 Pa and 151 Pa, respectively (Table 14–9).

Analysis  The schematic of the wall as well as the different elements used in its construction are shown in Figure 14–25. Condensation is most likely to occur at the coldest part of insulation, which is the part adjacent to the exterior sheathing. Noting that the total thermal resistance of the wall is 3.05  m^{2} \cdot  ^{\circ}C/W , the rate of heat transfer through a unit area A = 1  m^{2} of the wall is

\dot{Q}_{wall} = A \frac{T_{i}  –  T_{o}}{R_{total}}

 

              = \left ( 1  m^{2} \right ) \frac{\left [ 20  –  \left ( –  16 \right )^{\circ}C \right ]}{3.05  m^{2} \cdot  ^{\circ}C/W} = 11.8  W

 

The thermal resistance of the exterior part of the wall beyond the insulation is 0.03 + 0.14 + 0.23 = 0.40  m^{2} \cdot ^{\circ}C/W . Then the temperature of the insulation– outer sheathing interface is

T_{I} = T_{o} + \dot{Q}_{wall} R_{ext}

 

            = –  16^{\circ}C + \left ( 11.8  W \right )\left ( 0.40^{\circ}C/W \right ) = –  11.3^{\circ}C

 

The saturation pressure of water at – 11.3°C is 234 Pa, as shown in Table 14–9, and if there is condensation or freezing, the vapor pressure at the insulation– outer sheathing interface will have to be this value. The vapor pressure at the indoors and the outdoors is

P_{\nu, 1} = \Phi _{1} P_{sat, 1} = 0.60 \times \left ( 2340  Pa \right ) = 1404  Pa

 

P_{\nu, 2} = \Phi _{2} P_{sat, 2} = 0.70 \times \left (151  Pa \right ) = 106  Pa

 

Then the rate of moisture flow through the interior and exterior parts of the wall becomes

\dot{m}_{\nu , interior} = A \left ( \frac{\Delta P}{R_{\nu }} \right )_{interior} = A \frac{P_{\nu , 1} – P_{\nu , 1}}{R_{\nu , interior}}

 

      = \left ( 1  m^{2} \right ) \frac{\left ( 1404  –  234 \right ) Pa}{\left ( 0.012 + 0.0004 \right )  Pa \cdot m^{2} \cdot s/ng} = 94,355  ng/s = 94.4  \mu g/s

 

\dot{m}_{\nu , exterior} = A \left ( \frac{\Delta P}{R_{\nu }} \right )_{exterior} = A \frac{P_{\nu , 1}  –  P_{\nu , 1}}{R_{\nu , exterior}}

 

      = \left ( 1  m^{2} \right ) \frac{\left ( 234  –  106 \right ) Pa}{\left ( 0.019 + 0.0138 \right )  Pa \cdot m^{2} \cdot s/ng} = 3902  ng/s = 3.9  \mu g/s

 

That is, moisture is flowing toward the interface at a rate of 94.4  \mu g/s but flowing from the interface to the outdoors at a rate of only 3.9  \mu g/s . Noting that the interface pressure cannot exceed 234 Pa, these results indicate that moisture is freezing in the insulation at a rate of

\dot{m}_{\nu , freezing} = \dot{m}_{\nu , interior}  –  \dot{m}_{\nu , exterior} = 94.4  –  3.9 = 90.5 \ mu g/s

 

This corresponds to 7.82 g during a 24-h period, which can be absorbed by the insulation or sheathing, and then flows out when the conditions improve. However, excessive condensation (or frosting at temperatures below 0°C) of moisture in the walls during long cold spells can cause serious problems. This problem can be avoided or minimized by installing vapor barriers on the interior side of the wall, which will limit the moisture flow rate to 3.9  \mu g/s . Note that if there were no condensation or freezing, the flow rate of moisture through a 1  m^{2} section of the wall would be 28.7  \mu g/s (can you verify this?).

 

TABLE 14–9

Saturation pressure of water at
various temperatures

Temperature, °C Saturation Pressure, Pa
-40 13
-36 20
-32 31
-28 47
-24 70
-20 104
-16 151
-12 218
-8 310
-4 438
0 611
5 872
10 1,228
15 1,705
20 2,339
25 3,169
30 4,246
35 5,628
40 7,384
50 12,349
100 101,330
200 1.55 \times 10^{6}
300 8.58 \times 10^{6}
14.15