COOLING LOAD ESTIMATION

The components of a cooling load are:

1. Sensible heat gain: This is the direct addition of heat to the space to be conditioned. It may occur due to any one or all of the following sources of heat transfer:
   1. The heat flowing into the building by conduction through exterior walls, floors, ceilings, doors and windows due to the temperature difference on their two sides.
   2. The heat received from solar radiation. It consists of the following:
      1. The heat transmitted directly through glass of windows, ventilators, and doors.
      2. The heat absorbed by walls and roofs exposed to solar radiation and later on transferred to the space to be conditioned by conduction.
   3. The heat conducted from adjoining un-conditioned rooms.
   4. The heat generated by fans, lights, machinery, cooking operations, industrial processes, etc.
   5. Occupancy load.
   6. Air infiltration load through doors, windows, and their frequent opening.
   7. Heat gain from walls of ducts, etc.
2. Latent heat gain: The latent heat load occurs due to the presence of water vapour in the air of conditioned space. This load occurs due to the following sources:
   1. Moisture in the air entering by infiltration
   2. Condensation of moisture from occupants, food cooking process.
   3. Moisture entering directly into the conditioned space through permeable walls, partitions, etc.
   Total cooling load = Sensible heat load + Latent heat load.

1 Heat Transfer Through Walls and Roofs

The heat transferred through a plane wall by conduction is given by Fourier law. For a single wall shown in Fig. 20.28(a), we have

where   k = thermal conductivity of wall, W/(m. °C)

A_c = area of wall cross-section, m²

Δt = temperature difference on two sides of wall, °C

Δx = wall thickness, m

The heat transfer by convection from a plane wall is given by Newton’s law

Q_conv = hA_s Δt

where   h = heat transfer coefficient, W/(m² °C)

A_s = surface area of wall, m²

Δt = temperature difference between wall and surroundings, °C

Heat gained through a wall by the combined effect of conduction and convention becomes,

Figure 20.28 Heat transfer by conduction through a plane wall: (a) Single wall, (b) Composite wall, (c) Composite wall with air space

where subscripts o, i refer to outside and inside wall conditions respectively

and   U = overall coefficient of heat transfer of wall

 for a single wall (Fig. 20.28a)

 for a composite wall (Fig. 20.28b)

 for a composite wall with air gap(Fig. 20.28c)

where  k_a = thermal conductance of air space

2 Heat Gain from Solar Radiation

The heat from solar radiation is received by building surfaces by direct radiation and diffuse radiation. The direct radiation is the impingement of the Sun’s rays upon the surface. The diffuse radiation is received from moisture and dust particles in atmosphere which absorbs part of the energy of the Sun’s rays. The diffuse radiation is received by surfaces which do not face the sun.

The heat gain through outside walls and roofs is given by,

where  Δt_e = equivalent temperature differential.

3 Sol Air Temperature

It is a hypothetical temperature used to calculate the heat received by the outside surface of a building wall by the combined effect of convection and radiation.

Total heat received by the outside surface of the wall per unit area.

 

q_os = q_conv + q_rad = h_o (t_o − t_os) + I α

where  = sol air temperature.

t _o = temperature of outside air

t_os = temperature of outside surface of wall

h_o = outside film coefficient

I = total radiation intensity

α = absorptivity of the surface

4 Solar Heat Gain Through Glass Areas

5 Heat Gain Due to Infiltration

There are two methods of estimating the infiltrated air

(i) Crack length method, and (ii) Air change method.

The crack length method is usually used where greater accuracy is required. In most cases the air change method is used for calculating the quantity of infiltrated air. According to this method, the quantity of infiltrated air through windows and walls is,

where L, W, H = length, width and height of room respectively, m

A_c = air changes per hour

Factor  is used because infiltration takes place on the windward side of building only.

6 Heat Gain from Products

This heat gain is very important in case of cold storages.

1. Chilling load above freezing, where m = mass of productc_pm = mean specific heat of productT₁, T₂ = initial and final temperature of product respectivelyt_ch = chilling time.
2. where h_fg = latent heat of freezingt_f = freezing time
3. Cooling load below freezing, where c’ _pm = mean specific heat of freezing productT_1′ T_2′ = actual storage and freezing temperatures of productt_c = cooling time
4. 

7 Heat Gain from Lights

Heat gain from lights, Q_t = total wattage of lights × use factor × allowance factor

= 0.5 for industrial workshops

Allowance factor = 1.25 for flourescent tubes.

8 Heat Gain from Power Equipments

where η_em = efficiency of electric motor.

9 Heat Gain Through Ducts

where U = overall heat transfer coefficient

A_D = surface area of duct

t_a, t_s = temperatures of ambient and supply airs respectively

10 Empirical Methods to Evaluate Heat Transfer Through Walls and Roofs

There are two approaches to calculate empirically heat transfer through walls and roofs.

They are:

1. The decrement factor and time lag method.
2. The equivalent temperature differential method.

Both the methods use analytical-experimental results for their formulations. The equivalent temperature differential method is more commonly used by the air-conditioning engineers as it is also applicable to sunlit walls and roofs.

1. Using Decrement Factor and Time Lag: If the thermal capacity of the wall is ignored, then the instantaneous rate of heat transfer through the wall at any time τ is given byQ_τ = UA(t_e − t_i)where t_e = sol-air temperature.and on an average basis, the mean heat flow is given byQm = UA(tem – ti)where t_em = mean sol-air temperature.For the sake of simplicity, the dot above Q has been dropped to indicate the rate.But most building materials have a finite thermal capacity which is expressed asmc = ρcV = ρc(AΔx)where m = Mass of wall, V = volume of wallρ, c = Density and specific heat of wall materialA = Cross-sectional area of wallΔx = Wall thickness.It has been seen that there is a two-fold effect of thermal capacity on heat transfer
   1. There is a time lag between the heat transfer at the outside surface q_o and the heat transfer at the inside surface q_i, defined by ϕ.
   2. There is a decrement in the heat transfer due to the absorption of heat by the wall and subsequent transfer of a part of this heat back to the outside air when its temperature is lower, defined by λ.
   The rigorous analytical method to determine the time lag ϕ and decrement factor λ is quite complicated. In the limiting case, when the wall thickness approaches zero λ → 1 and ϕ → 0Then Q_τ = UA(t_e – t_i)i.e. the heat transfer through the wall is equal to its instantaneous value.Considering the effect of thermal capacity, the actual heat transfer at any time τ is,Q_τ = UA(t_em − t_i) + UA λ(t_{e τ − ϕ} − t_em)where t_{e τ − ϕ} = sol-air temperature at time τ − ϕ. i.e., ϕ hours before the heat transfer is to be calculated.In a locality where the daily range of variation of the outside air temperature is small it is immaterial what thickness of wall is provided. But in a locality when the daily range of temperature is large, it is desirable to have thick walls so as to cut the cooling load in summer and the heating load in winter. Such walls also do not allow the inside temperature to rise very much during the day and drop at night.
2. Equivalent temperature differential (ETD) method: The actual heat transfer can be written in terms of an equivalent temperature differential Δt_E defined by the equation.Q = UAΔt_Ewhere Δt_E = (t_em − t_i) + λ(t_{e τ − ϕ} − t_em)Δt_E depends on the following:
   1. l and f, which in turn depend on the thermo-physical properties of construction.
   2. Outside air temperature t₀ and solar radiation intensity I.
   3. Room temperature t_i.
   Thus, the equivalent temperature differential approach takes care of the exposure of the wall or roof to the sun.