Water vapor diffusion from attic heat store to cold air heater

David Delaney
Ottawa, December 7, 2004

I received a challenging comment on my proposal of a thermal scheme for a solar house.

I answered the comment (about the amount of energy that would be lost when the air heater cooled down) by calculating the energy lost by the cooling air contained in the air heater at the end of the day and by the condensation of the water vapor in the air heater at the end of the day . The commenter responded to this calculation with a further comment, writing:

Start quote:

Here's where I see a problem.  You have done meticulous calculations, but I don't believe the assumptions.  Here you assume that the only water vapor will be from the air heater [at the end of the day].  What about the rest of the structure?  When the vapor condenses, vapor pressure will push more water vapor into the air heater (probably bringing warmth with it).
....
According to my calculations:
30% RH at 70ºF = 0.74" of Hg
11.5% RH at 100ºF = 1.93" of Hg
95% RH at 14ºF = 0.09" of Hg

So when all three spaces are at equal vapor pressure, house space will be 12%RH, heat storage 5%RH.  All that water ends up on your glazing in the air heater.

End quote.

This note calculates an upper bound for the total diffusion of water vapor down through the ceiling ducts of the air heater during the 18 hours of an Ottawa December night.

First we construct a model of the air heater and its ducts. The total cross sectional area of the ducts of the air heater is 40 ft2. This area is made up of a 20 ft2 (40 ft x 0.5 ft) cool air duct through which cool air falls into the air heater, and a 20 ft2 (14 x 1.5 ft2)  hot duct through which hot air rises.  The duct system is modelled by a insulated (adiabatic walls) vertical duct 40 ft2 in horizontal cross sectional area, and 1.5 ft in length from top to bottom.

The top of the duct projects up into a large volume of air at 100 F (37.78C) and relative humidty (RH) 11.45%.  These are the conditions of the attic heat store air when its absolute humidity equals the absolute humidity of the air in the living space of the house, in which condition the vapor pressure of the water vapor in the attic heat store is equal to the vapor pressure of the water vapor in the living space.

The bottom of the duct is closed at night by a conducting plate which is maintained at exactly 14 F (-10C) by unspecified means while in its place closing the bottom of the duct.. Water condensing from air in the duct is removed from the plate by unspecified means.

At 3:00 pm, at the beginning of the 18 hour night, when there is a uniform column of hot air in the duct,  the bottom of the duct is suddenly closed by the cold plate.  There will be a downward flux of heat by conduction and diffusion of moisture. The vertical gradients of temperature and pressure start out at zero. (Constant temperature and density from top to bottom of the duct, with a discontinuity at the cold bottom plate.) As the low air in the duct is cooled by losing energy to the cold plate at the bottom of the duct, a stable monotonic density gradient forms, with denser air below, and less dense air above, at every point in the duct.  The adiabatic walls of the duct prevent either heating or cooling of the air by the walls, and hence do not support condensation.  At all times after 3:00 PM, the air in the duct has only one-dimensional gradients of  density and temperature.  In other words, the density (temperature) of the air at any point on a horizontal planar cross section of the duct will equal the density (temperature) of the air at any other point on that plane.  The density and temperature of the air at a particular elevation in the duct may vary with time, but all points at that elevation will have the same temperature and the same density at any point in time.

The equation reference numbers below, e.g. (3), and page references, e.g. F5.1, are to the 1997 ASHRAE Handbook of Fundamentals, SI Edition.

The appropriate diffusion mass flow equation for this problem is the one-dimensional Fick's law for diffusion of water vapor through stagnant air,

                                                             mBdot =  − Ddρ ⁄dy                     (5) F5.1

in which  mBdot is the rate of diffusion (kg/m2.s) of component B, the water vapor,  diffusing through air.   ρB is the density of  the water vapor.   y is the  displacement along the dimension of diffusion, in this case the vertical distance of any point in the duct from the bottom of the duct.  The bottom of the duct is at y = 0, the top of the duct is at y = L = 1.5 ft = 0.46 m.  dρB ⁄ dy is the vertical density gradient. Diffission is in the direction opposite to the direction of increasing density, hence the negative sign.

Eventually, by the inherent stability of the air in the duct, the temperature and density will cease varying with time throughout the volume of the duct.  At times after  this equilibrium has been reached, there will be a constant one dimensional downward flow of heat, and a constant one dimensional downward diffusion of water vapor. It is easy to see that these constant flows are larger than the flow at the top of the duct in the transient state, since the temperature gradient and the density gradient  at the top of the duct reach their maximum value at the beginning of the steady state.  In the steady state,  the rate of diffusion of water mass across a horizontal  cross section of the duct at one elevation  must be the same as at any other elevation, since there can be no continually increasing densification or rarefaction of the water vapor at any elevation. This means that  mBdot is a constant function of elevation.  As we can see from (5),  this means Dv and dρ ⁄dy are inversely proportional to each other.  The variation of Dv with temperature is shown in the following graph of relation  (11) F5.3. 
Graph of diffusivity of air


The diffusivity at the mean temperature of the air in the duct  is

                         Dv((-10+38)/2) = Dv(14C) =  24.048 mm2/s


From the following psychrometric data was displayed by PsyCalc for the three air masses of interest:

PsyCalc image    PsyCalc image    PsyCalc image

           70 F                                     100 F                                 14 F
   Living space air               Air at the top of the duct    Air at the bottom of the duct
         
The density of water vapor  at the bottom of the duct (14 F, -10 C) is

                           ρB (0 m) = 1 m3 / 0.7474 m3.kg *  1.606 g/kg =  2.149 g/m3

The density of water vapor at the top of the duct (100 F, 38 C) is

                          ρB (L) = ρB (0.46 m) = 1 m3 / 0.8875 m3.kg *  4.66  g/kg =  5.250  g/m3
                        

Calculating the total diffusion of water vapor

Method 1:     

We assume that  dρ ⁄dy is a constant throughout the height of the duct, equal to its mean value,   and that  Dv is a constant equal to the value of Dv at the mean temperature of the air in the duct,

                                          dρ ⁄dy = B (L)  −  ρB (0)) ⁄  L    and

                                           Dv((-10+38)/2) = (Dv(14 C)  = 24.048 mm2/s

(5) F5.1 becomes

                                           mBdot =  24.048  mm2/s  *  1e-6 m2/mm2 *  (5.250 - 2.149) g/m3 / 0.46 m  =  1.62e-4 g/m2

For the whole 40 ft2 of the duct cross section the mass of  water vapor diffused downward over 18  hours is

                                           40 ft2 * 0.30482 m2/ft2 * 18 hr * 3600 s/hr *  1.62e-4 g/m2.s  =  39 g

Method 2:


In the region from -10C (Dv =  20.245) to 38C (Dv = 28.127) Dv(T) is well approximated by the straight line 
                           
                                          Dv0.169T + 21.8 * mm2/s              T in Celsius, the underlined expression is treated as dimensionless

Therefore, from (5) F5.1,

                                         dρ ⁄dy   =  −  mBdot ⁄ ( (0.169T +  21.8) * 10^-6 m2/s)          (The underlined expression is treated as dimensionless)

We assume  T  varies linearly from y = 0 to y = L because the thermal conductivity of moist air is almost constant from -10C (0.024 W/m.C)  to 38C (0.27 W/m.C), so   T = − 10 + 48y, yielding

                                           dρ ⁄dy   =  −  mBdot * 10^6 s/m2  * 1 ⁄  (8.1y + 20.1)      (The underlined expression is treated as dimensionless)

Integrating with respect to y,

                                            ρB(y) = a constant  −  mBdot * 10^6 s/m2 *  ln (8.1y + 20.1 )  ⁄  8.1 * 1 m          (the 1 m is for the integration)

                                            ρB(L) − ρB(0) =   −  mBdot * 10^6 s/m(ln (8.1L + 20.1)  − ln (8.1 * 0 + 20.1) ) / 8.1
                      
                                           mBdot =  − 8.1 * 10^−6 m/s * (ρB(L) − ρB(0)) / ln((8.1/20.1)L + 1)

                                           mBdot =  − 8.1 * 10^−6 m/s * (5.251 -  2.149) * 10^-3 kg/m3  / 0.17
                        
                                            mBdot  =  − 1.48e-7 kg/m2.s =  − 1.48e-4 g/m2.s

For the whole 40 ft2 of the duct cross section the mass of  water vapor diffused downward over 18  hours is

                                           40 ft2 * 0.30482 m2/ft2 * 18 hr * 3600 s/hr *  1.48e-4 g/m2.s  =  36 g

Discussion


 I have calculated the downward diffusion of water vapor in a duct that models the ducts connecting the air in the air heater to the air in the attic heat store.  Two methods of calculation yielded 36 g and 39 g of water vapor diffused downward during the 18 hours of an Ottawa December night.  I believe that 40 g is a reasonable upper bound on the total diffusion, because the density gradient in the actual ducts would have a smaller magnitude. The diffusion path to the cold glazing would have to include the air in the air heater, increasing the water vapor density at the bottom of the actual ducts relative to the density at the bottom of the model duct.

 To put 40 g of water vapor lost  in 18 hours in  perspective, the ASHRAE Handbook of fundamentals says "Tenwolde (1988, 1994) reports production rates between 135 and 330 g/h [of water vapor] for one to two adults, with an average of 230 g/h".  ( See F23.5, section Indoor Humidity Control.)  Not all of the 40 g would be lost, since some of the frost on the glazing would sublime back into the air in the air heater in the morning. Only vapor that leaked out of the air heater to the outside would be lost from the house system.

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