quiz

Chapter Two- Handout Four

Radiation Heat Transfer Reminder: In the first handout of radiation heat transfer, we calculated the radiation exchange between surfaces inside buildings. This class is the exchange of thermal radiation with outside surfaces. Outside surfaces can be opaque like walls, roof, and doors, or transparent surfaces like windows. This handout treats radiation between outside opaque surfaces

Solar Heat Gain Through Opaque Surfaces (walls and Roof) The sun is the main heat source of the earth, and without the sun, the environment temperature would not be much higher than the deep space temperature of -270

o C. The solar energy stored in the atmospheric air, the ground, and the

structures such as buildings during the day is slowly released at night, and thus the variation of the outdoor temperature is governed by the incident solar radiation and the thermal inertia of the earth. Heat gain from the sun is the primary reason for installing cooling systems, and thus solar radiation has a major effect on the peak or design cooling load of a building, which usually occurs early in the afternoon as a result of the solar radiation entering through the glazing directly and the radiation absorbed by the walls and the roof that is released later in the day. The effect of solar radiation for glazing such as windows is expressed in terms of the solar heat gain factor (SHGF). For opaque surfaces such as the walls and the roof, on the other hand, the effect of solar radiation is conveniently accounted for by considering the outside temperature to be higher by an amount equivalent to the effect of solar radiation. This is done by replacing the ambient temperature in the heat transfer relation through the walls and the roof by the sol-air temperature, which is defined as the equivalent outdoor air temperature that gives the same rate of heat transfer to a surface as would the combination of incident solar radiation, convection with the ambient air, and radiation exchange with the sky and the surrounding surfaces . Heat flow into an exterior surface of a building subjected to solar radiation can be expressed as

Figure: The sol-air temperature (Tsol-air) represents the equivalent outdoor air temperature that gives the same rate of heat flow to a surface as would the combination of incident solar radiation and convection/radiation with the environment

Where qsolar is the solar irradiance falling on the surface, a function of orientation and latitude. Then, the heat transfer through a wall (or roof) can be expressed as:

accounting for conduction, convection, and radiation Note the temperature rise due to solar radiation is

Then, the rate of additional heat gain from solar through the wall becomes:

accounting for radiation only

The total solar radiation incident on the entire wall is Qsolar=A× qsolar. Therefore, the fraction of the incident solar heat transferred to the interior of the house is

Example: The west masonry wall of a house located in Bahrain (latitude 26

o N) has an area of 19.5 m

2 and is made of 200-mm thick

concrete blocks with overall U-value of 1.65 W/(m 2 .K) including both outside and inside convection coefficients. The

reflectivity of the wall surface is 55%. If the interior of the house is to be maintained at 24 o C, answer the following

questions:

a. The sol-air temperature on July 21 which the design date for cooling load. The solar irradiance is 772 W/m 2 on

west.

αs=1-ρ =1-0.55 =0.45

Tsol-air = To + αs×q/ho = 39.2 +0.45×772/22.7 = 54.5 o C

b. The total heat gain through the wall (conduction, convection, and radiation).

Qwall = UA (Tsol-air – Ti) = 1.65×19.5 (54.5-24) = 981 W

c. What is the temperature rise and heat gain due to radiation only. Temperature rise due to radiation only (∆Tsol) = 54.5 – 39.2 =15.3

o C

The heat gain due to radiation only Qrad =UA ∆Tsol = 1.65×19.5×15.3 =492 W

d. Calculate the fraction of radiation heat gain from the total heat gain.

Fraction = (492/981)×100 =50%

e. Calculate the fraction of radiation transferred to that of total incident on the wall. Comment on the answer.

Solar fraction transferred =

=1.65×0.45/22.7 = 0.033 (3.3%)