Sol-air temperature

In the first instance, heat transfer through a wall depends on the rate at which heat enters its outer surface. The concept of ‘Sol-air temperature’ has been in use for some time as an aid to the determination of the initial rate of entry of heat. It is defined as the value of the outside air temperature which would, in the absence of all radiation exchanges, give the same rate of heat flow into the outer surface of the wall as the actual combination of temperature differences and radiation exchanges really does.

The sol-air temperature, teo, otherwise termed the outside environmental temperature, is used in the following equation

Q’= hso(teo-tso) (7.18)

Where Q’ = rate of heat entry into the outer surface, in W m-2

Hso = outside surface heat transfer coefficient, in W m-2 K_1 teo = sol-air temperature, in °C tso = outside surface temperature, in °C

Q’ can be expressed in another way which does not involve the use of the sol-air temperature:

Q’ = a/8 + a7s + hs0(t0 — fso) + R (7.19)

In this basic heat entry equation, a and a’ are the absorption coefficients (usually about the same value) for direct, /5, and scattered, 7S, radiation which is normally incident on the wall surface. R is a remainder term which covers the complicated long wavelength heat exchanges by radiation between the wall and nearby surfaces. The value of R is difficult to assess; in all probability it is quite small and, if neglected, results in little error. Equations (7.18) and (7.19) can be combined to yield an expression for /eo in useful form:

A/8 + a7s + R

U o = *o + ————————————————————————— <7-20)

‘*so

If a’ is made equal to a, and R is ignored, this expression becomes

1 (7.2!)

•ho

The CIBSE Guide A5 (1999) quotes an equation virtually the same as this.

EXAMPLE 7.12

Calculate the sol-air temperature for a vertical, south-west wall at 13.00 h sun-time and latitude 40°N, given the following information:

Iv = 343 W m~2 = /5

(see example 7.11)

I& = 118 W m-2

(see example 7.11)

Os

O

II

S

II

A

Hs0 = 22.7 W nT2 K-1

T0 = 32°C

Answer

By equation (7.21)

0.9(343 + 118) reo — ^ + 22.7

= 50.3°C

Table 7.8 gives values of sol-air temperatures for average walls at different times and orientations at 51°33’N. From such a table the sol-air temperature for the above example is 42.8°C. So a difference of 11V2 degrees more southerly latitude and about 5 degrees more air temperature produces an increase of about 7V2 degrees in the value of fc, in this particular comparison.

Posted in Air Conditioning Engineering


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