Mist and fog
For condensation to occur in the atmosphere the presence is required of small, solid particles termed condensation nuclei. Any small solid particle will not do; it is desirable that the particles should have some affinity for water. Hygroscopic materials such as salt and sulphur dioxide then, play some part in the formation of condensation. The present opinion appears to be that the products of combustion play an important part in the provision of condensation nuclei and that the size and number of these nuclei vary tremendously. Over industrial areas there may be several million per cubic centimetre of air whereas, over sea, the density may be as low as a few hundred per cubic centimetre.
These nuclei play an important part in the formation of rain as well as fog, but for fog to form the cooling of moist air must also take place. There are two common sorts of fog: advection fog, formed when a moist sea breeze blows inland over a cooler land surface, and radiation fog. Radiation fog forms when moist air is cooled by contact with ground which has chilled as the result of heat loss by radiation to an open sky. Cloud cover
discourages such heat loss and, inhibiting the fall of surface temperature, makes fog formation less likely. Still air is also essential; any degree of wind usually dissipates fog fairly rapidly. Fog has a tendency to occur in the vicinity of industrial areas owing to the local atmosphere being rich in condensation nuclei. Under these circumstances, the absence of wind is helpful in keeping up the concentration of such nuclei. The dispersion of the nuclei is further impeded by the presence of a temperature inversion, that is, by a rise in air temperature with increase of height, instead of the reverse. This discourages warm air from rising and encourages the products of combustion and fog to persist, other factors being helpful.
Long-wave thermal radiation, /Lw directed to the sky at night can be considered as radiation to a black body at a temperature of absolute zero. This is modified by a correction factor to account for variations in the absorption by water vapour in the lower reaches of the atmosphere. This changes as the amount of cloud cover and the area of the sky seen by the surface varies. According to Brunt (1932) the absorption effect can be accounted for by a vapour correction factor, K, expressed by
K = 0.56 — 0.08-VaT (5.1)
Where ps is the vapour pressure of the air in millibars. The amount of sky seen by a surface is given by an angle factor, B,
B = 0.5(1 + cos 8) (5.2)
Where 8 is the acute angle between the surface and the horizontal. Thus for a flat roof 8 =
0 And B = 1, whereas for a wall 8 = 90° and B = 0.5. Cloud cover can play a part as well and if we assume it is wholly effective in suppressing loss from the surface to outer space, a cloud cover factor, C, can be introduced. Table 5.1 gives typical, approximate, average cloud cover factors for Kew, based on data recorded for the amount of bright sunshine received between sunrise and sunset in relation to the maximum amount of sunshine that could be received during the same period of the day.
Table 5.1 Typical, approximate cloud cover factors for Kew
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
C 0.2 0.2 0.3 0.4 0.4 0.4 0.4 0.4 0.4 0.3 0.2 0.2
273 + ts 100
/LW = 5.77B(1 — C)K
The surface temperature of a building could be calculated if steady-state heat flow is assumed but if the building has appreciable thermal inertia, as is usually the case, it becomes complicated.
Calculate the emission from a flat roof to the night sky in January, at Kew, under steady — state conditions, when the temperature of the room beneath the roof is 20°C and the outside
Hence the long-wave thermal radiation from a surface to the sky can be expressed by
Air is at -1°C, assuming (a) 20 per cent cloud and (b) no cloud cover. Assume the U — value of the roof is 0.25 W m~2 KT1 and the outside surface film resistance is 0.0455 m2 K W-1.
(a) lfU = 0.25 W m-2 K-1 then the overall thermal resistance is 1/0.25 = 4.0 m2 K W_1 and the roof surface temperature is
Ts = -1 + (0.0455/4) x (20 + 1) = -0.76°C 8 = 0 therefore, by equation (5.2), B — 1.0
For an outside temperature as low as -0.76°C the air is likely to be saturated in the UK and hence, from CIBSE psychrometric tables:
Ps = 0.5297 kPa (5.297 mbar) and, by equation (5.1),
K = 0.56 — 0.08V5.297 = 0.376
And hence, by equation (5.3),
/LW = 5.77 x 1 x (1 — 0.2) x 0.376[(273 — 0.76)/100]4 = 95.3 W m“2
(b) C = 0, B = ,K = 0.376
/LW = 5.77 x 1 x (1 — 0) x 0.376[(273 — 0.76)/100]4 = 119.2 W nT2
Posted in Engineering Fifth Edition