Heat Loss or Heat Gain
Just as the human body has heat exchange processes with the environment, the building can be similarly considered as a defined unit and its heat exchange processes with the outdoor environment can be examined. Heat energy tends to distribute itself evenly until a perfectly diffused uniform thermal field is achieved. Heat tends to flow from higher temperatures to lower temperature zones by conduction, convection and radiation. The rate of heat flow by any of these three forms is determined by the temperature difference between the two zones or areas considered. The greater the temperature difference, the faster the rate of heat flow.
The equations and the calculations methods given below are valid only when both the out-door and indoor temperature are constant. Such static conditions do not occur in the nature, and hence the assumption of the steady state conditions is a simplification.
Calculations based on steady state assumptions are useful to determine the maximum rate of heat loss or gain and also for establishing the cooling or heating load for mechanical installations.
Figure illustrates the following:
Qi + Qs +- Qc +- Qv +- Qm -Qe = 0
Thermal balance i.e. the existing thermal condition is maintained if the sum of the above equation is zero. If the sum of this equation is less than zero (negative), the building will be cooling and if more than zero, the temperature in the building will increase.
- Conduction of heat may occur through the walls either inwards or outwards, the rate of which is denoted by Qc (convective and radiant components in the transfer of the same heat at the surfaces are included in the term transmittance).
Qc = U* A * ΔT
U = transmittance (W/m2 K)
= 1 / Rt
= 1/ (Rso + ∑Rn + Rsi)
Rt = the total overall resistance of the element (m2K/W)
Rn = the resistance of the nth material within a composite element (m2K/W)
Rso & Rsi are the outside and inside surface resistances respectively (m2K/W)
A = the surface area through which the heat flows (m2)
ΔT = the temperature difference between the warm and
cold sides of the material (K).
- As an example, assume a wall with a U-Value of 4.5 W/m² K and a surface area of 10 m². If the outside temperature was 30°C and the inside was 25°C, we could calculate the total heat gain due to conduction through the wall as follows:
Q =U* A * ΔT
= 4.5 x 10.0 x (30-25)
= 225 Watts
Q = the resultant heat flow (Watts)
A = the surface area through which the heat flows (m²)
ΔT = the temperature difference between the warm and cold sides of the material (K), and
R = the thermal resistance per unit area of the piece of material (m²K/W).
- The effects of solar radiation on opaque surfaces can be included in the above by using the sol air temperature concept, but through transparent surfaces (windows) the solar heat gain must be denoted separately and denoted by Qs.
Qs = A * I * θ
A = the surface area through which the heat flows (m²)
I = radiation heat flow density (W/m²)
θ = solar gain factor of the window glass.
Heat exchange may take place in either direction with the movement of air, i.e. ventilation; and the rate is denoted as Qv.
Qv= 1300 * V * ΔT
1300 = volumetric specific heat of air (J/m3°C)
V= ventilation rate (m3/s)
θT = the temperature difference (°C)
If the number of air changes per hour (N) is given, ventilation rate
V= (N* room volume) /3600
(3600 is the number of seconds in an hour)
- An internal heat gain may result from the heat output of human bodies, lamps motors and appliances. This may be denoted as Qi.
Sitting, moderate movement = 130- 160 W
Walking, lifting, pushing = 290- 410 W
Sustained work = 500- 700 W.
In buildings such as office buildings, commercial stores, shopping centers, entertainment halls etc much of the overheating problem during the summer can be caused by heat produced by equipment or by a high level of artificial lighting. When there are a large number of occupants or clients their metabolic heat can also add to the problem. In residential buildings the cooling needs are significantly lower. The main reasons are the lack of large equipment and the relative ease of applying natural cooling strategies such as ventilation and shading.
The use of energy efficient lighting, domestic appliances and office equipment reduces the cooling load. The boilers and the hot water tanks should be carefully insulated. In this way not only is their energy efficiency improved, but heat flow to the internal spaces is minimized as well. Electrical appliances and equipment should be placed in positions in the building from where it will relatively easy to expel the heat produced through natural ventilation.
- There may be deliberate introduction or removal of heat (heating or cooling), using some form of outside energy supply. The heat flow rate of such mechanical controls is denoted as Qm. The heat flow rate of the mechanical systems is subject to the designer's intention and is deliberately controllable. It can thus be taken as a dependent variable in the equation, i.e. it can be adjusted according to the balance of the other factors.
- If evaporation takes place on the surface of he building (e.g. roof pool) or within the building (human sweat or water in a fountain) the vapors are removed, this will produce a cooling effect, the rate of which is denoted as Qe.
Qe = 666 * kg/hr
As the latent heat of evaporation of water at around 20 deg C ~ 2400 kJ/ kg
2400000 J/h = 2400000 J/ 3600 s = 666 W