Calculation of Heating and Cooling Loads - Term Paper Example

Name : xxxxxx Tutor : xxxxxxx Title : Calculation of Heating and Cooling Loads Institution : xxxxxxx @2015 ABSTRACT The paper covers the heating and cooling loads experienced in a building or a surface. The three modes of heat transfer and understanding of the solar geometry is of significant understanding. Heating and cooling loads are important in selection of air ventilation and conditioning system. The paper covers the heat gain into a building, heat that is transmitted through the elements of the building, and the heat that is required to warm the air from the outdoor entering the building or space. The solar system comprising the sun, the planets and the stars, where the sun is the only source of light and heating as far as heating and cooling loads is concerned is also covered. Lambert’s cosine law describes the amount of solar radiation received on the surface of the earth at any one point, and from the above equation/ law, a point will receive maximum solar radiation when the sun is at its overhead. Windows are important building elements. The location and design of a window is of critical significance as pertains to cooling or heating in a building. Their design is not mostly for beauty purposes and particularly in polar and temperate regions where there is intense low and high day temperatures respectively. INTRODUCTION Heat transfer is a process through which heat is transferred from one system to another. In heat transfer, the net heat transfer is from the hotter body to the less hot body and an equilibrium is reached when the two adjacent bodies are at the same temperature. Knowledge on heat transfer is very important in determining the type of material, sizes and shapes to use in construction of houses or thermal systems. In this case, passive thermal systems will require materials of poor thermal capacity or conductivity while active systems will demand material fabric that is of high thermal capacity. Cooling load refers to the rate at which heat is removed from a space so as to attain some desired temperatures or relative humidity within that space. Heat gain into a building should be analyzed first before the cooling load is determined because cooling loads is usually calculated when determining the ventilation, HVAC air conditioning systems and components, and when sizing up heating into a building. Heating load on the other hand involves calculations that are made to determine the heating load of a given building. Loss of heat from a building are of two kinds as listed below; i. Heat that is transmitted through the elements of the building like the ceiling, walls, floor, windows and other exposed surfaces. ii. The heat that is required to warm the air from the outdoor entering the building or space. The purpose of any building is to ensure safety and comfort to its occupants for a healthy living. This function is achieved through proper design and building material selection to enable it respond as require to climate changes. Thermal exchange takes place when the different elements of the building are exposed to external varying conditions. The exchange can be through conduction, convection or radiation as determined by how the structure interacts with its environment. The material used for building determines the thermal nature of the building as thermal conductivity is an intrinsic property of any material. Various forms of heat transfer processes take place in a building, mostly heat transfer occurs in areas like, roofs, through the walls, floor, and ceiling. MODES OF HEAT TRANSMISSION THROUGH BUILDING ELEMENTS There are three broad mechanisms through which heat is transferred from one system to another, they include; conduction, convection, and radiation. Conduction is the transfer of heat through excitation or bombardment of atoms between bodies that are in physical contact with each other. Knowledge on conduction is very important in buildings as it determines the temperature difference between the outside of the building and its inside particularly in the polar region where there is a vast range of temperature difference. Heat transfer though the above surfaces in a building is mostly by conduction mathematically, the amount of heat transferred to a body (W/m-2) or surface is computed using the formulae below; Where; Is the heat transfer coefficient or the constant of proportionality heat flux and the thermal driving force with units as W/m-2K-1. Is the area (m2) Is the temperature difference between the surfaces of the body between which heat is transferred in kelvins (K). Is the thickness (m) Heat transfer by convection involves heat movement within a fluid through diffusion of particle molecules or advection. The knowledge on convection is very important in building design as air movement or ventilation within the premise can conveniently be determined. Convection process can be natural or forced, for example, by use of fans, pumps, etc. Ventilation will be very useful in controlling the internal temperatures of a building, controlling the buildup of moisture and odours, and at the same time enhancing the comfort of the premise by the occupants. Radiation is the heat transfer by electromagnetic waves as they traverse through space or vacuum. As the electromagnetic waves hit the surface of an object, they transfer the heat in them to the particular object. All bodies with temperatures more than 10K emits thermal radiation but as discussed above, the net heat transfer is from the hotter body. The difference between the amount of radiation that is emitted by a body, and the amount of radiation absorbed by the body represents the net heat transfer at that instance in time. SOLAR ELEMENTS AND SOLAR GEOMETRY In the solar system, sun is the only source of heat and light. The amount of radiation that reaches the surface of the earth is determined by several factors, 1. The geometry of the sun and earth at any one instance in time. The position of the surface of the earth in relation to that of the sun determines the amount of solar radiation reaching that point. The geometry is most of the time determined by the position on the latitude, what time of the year it is as well as the time of the day. 2. Earth’s atmospheric conditions. Little or no solar insolation would reach the surface of the earth if it was all covered with clouds. Thus, the nature of the cloud cover on a point on the earth is a factor that principally determine the amount of solar radiation hitting that point. Low levels of cloud cover would result in high solar radiation received on a point on the earth surface and vice versa is also true. The earth-sun geometry comprises of the sun and the nine planets as its main elements. The earth takes a year to revolve about the sun and one day to rotate about its own axis. The axis of the earth is tilted at some angle I relation to its orbital plane. This explains we have a day, a night, and the four seasons. The radiation produced by the sun may be transmitted down to the earth’s surface, reflected by the air particles in the atmosphere, or diffused by the same particles. The radiation reaching or received by a point on the surface of the earth hence is by any of the three means. Direct radiation. Direct radiation represents the heat one feels on a sunny day. This radiation comprises the most significant portion of the total suns radiation. The radiation can hence be concentrated by use of optical devise to tap the energy for other uses. The devices include the lenses, parabolic reflectors, and the mirrors and in particular concave mirrors. Diffuse radiation. As the rays from the sun traverse through the atmosphere, water droplets and other small solid objects scatter away the radiation. Unlike the direct radiation, diffuse radiation cannot be concentrated, and is what causes shady sections in some areas during the day. Reflected radiation. The radiation from the sun may be reflected by the atmospheric environment, and particularly by solar devices. The effectiveness of the reflection of the radiation depends on what is the geometry as well as the reflective ability of the surrounding reflective objects. The amount of solar energy that is received by a point on the surface of the earth is given by the Lambert’s cosine law. The energy is determined by the angle at which the sun’s ray beam hits that point. The law is mathematically stated as below; is the solar radiation received on the earth’s surface is the angle of incidence of the suns radiation is the direct radiation received on the earth’ s surface. Fig 2; Radiation received at a point From the above equation/ law, it is evident that a point will receive maximum solar radiation when the sun is at its overhead, that is, when the sun hits the point at 0 degrees incidence angle since is maximum and is equal to 1. The solar radiation is zero at sunrise and at sunset when the angle of incidence of the radiation 90 degrees where is zero. Heat transferred to a building from the suns radiation is received through radiation. Radiation as described above is the transfer of heat or thermal energy through electromagnetic waves, when the rays hits the surface of the building, the energy is transferred. The radiant heat transferred or transmitted by a body depends on its absorptivity, transmissivity, and on its reflectivity. A body will have higher temperatures because its surface is black and rough while another body will be colder because most of the radiant heat is reflected from its surface. A body will transmit more heat than the other because its conductivity is higher as compared to the other body. Radiation involves the transfer of heat from the surface of a body by virtue of its body temperature. The transfer will be high as its temperature rises. Radiation must not require a media for heat propagation. When considering a building, its surfaces or elements like the walls, roofs, and the floor will always be exposed to the external atmosphere, and therefore, the heat exchange between these elements and the atmosphere is of significant importance when determining heating loads. The amount of heat transferred is determined by the use of the below equation. Where; is the exposed area of the building is the emmisivity of the exposed surface of the building is the Stefan-Boltzmann constant is the temperature of the exposed building surface is the temperature of the sun or the sky The temperature of the sky is equivalent to the temperature of the atmosphere. It takes into considerations that it’s hard for the atmosphere to be at a uniform temperature, and that it’s only a certain wavelengths, that the atmosphere radiates. Also, when considering heat transfer to buildings, the convective and the radiative heat transfer coefficients are usually combined so as to dive the overall heat transfer coefficient for the given surface. HOW SOLAR LOADS VARY WITH DIURNAL CYCLE Diurnal cycle is a pattern that repeats itself as a result of earth’s rotation. Diurnal cycle occurs ones in 24 hours which is the time the earth takes to complete one full rotation. Diurnal patterns are usually sinusoidal. The effect of solar radiation is significant at certain times of the day and some seasons in the countries of and southern Africa. Therefore, the design, orientation, and selection of materials used in the construction of a building will greatly influence the amount of solar radiation that falls on to those buildings. The roofs and walls of building in areas with high diurnal range of temperatures should be made of materials with high mass so as to increase their thermal capacity. The thermal capacity limits the day and night temperatures from going extremely high and extremely low respectively as well as reducing solar radiation effect to its minimum. HEAT TRANSMISSION THROUGH WINDOWS Windows design is very important particularly in Polar Regions where energy in the house ought to be retained from the outside frozen conditions. Likewise in the temperate houses are mostly oriented in the north south direction to prevent the direct heating by the sun’s radiation. Windows design determine their potentiality in conservation of energy. In order for one to determine the energy performance of any building, it is worth first knowing the window’s heat transfer properties that are affected by the parameters of the external environment such as temperature change or wind speed. Mostly windows are made transparent so as to offer; I. Daylight conditions inside the house during the day II. Provide room for visual communication with the outside world III. Beauty or aesthetics As a result of their transparency, windows will transmit radiations into the house or building and the heat transfer condition is different from that transmitted through opaque objects or surfaces. The reason behind this is because, any radiation incident on an opaque surface will be reflected by the surface, and a portion of it will be absorbed but not radiation is transmitted. Only a very small portion of the radiation absorbed by the surface is transferred to the inside of the house. The above is contrary to transparent surface because most of the incident radiation will be transmitted to the interior of the building as the other portion is absorbed and reflected by the surface. However, the heat transfer rate through these surfaces will depend on a number of factors like the orientation of the surfaces, climatic conditions, and the condition of the solar radiation falling on them. Fig 3; Heat transfer through a transparent window The amount of radiation transferred through a transparent window surface can be computed using the formulae below; Where; =Area of the exposed surface = Transmittivity of glass or material making the window = Total incident radiation =Fraction of the radiation that is transferred to the interior of the house by both conduction and convection processes. =Absorptivity of the material making the window Since the total incident radiation comprises of the direct, diffuse, and the reflected rays, then the values of absorptivity, and transmittivity are the same for the three different radiations. Under steady state condition, N can be given by Where; =overall heat transfer coefficient =external heat transfer coefficient EXAMPLES ON HEAT LOSS AND TRANSFER RATE Example 1 Calculat the rate of heat transfer on a cold day through a rectangular window that is 1.2 m wide and 1.8 m high, has a thickness of 6.2 mm, a thermal conductivity value of 0.27 W/m/°C. The temperature inside the home is 21°C and the temperature outside the home is -4°C. The heat transfer question is on conduction. The heat is transferred through the window through conduction. Heat transfer Where: q = heat conduction rate (W) k = thermal conductivity of material (W/m2.°C) A = cross-sectional area normal to the direction of heat flow (m2) ΔT = temperature gradient (°C) ΔL = thickness of the material conducting heat (m). The surface area of the window can be obtained by width × height. Area = (1.2 m) × (1.8 m) = 2.16 m2. d = 6.2 mm = 0.0062 m Now we can calculate the rate of heat transfer as; Rate = (0.27 W/m/°C) × (2.16 m2) × (21°C - -4°C)/(0.0062 m) Rate = 2400 W (rounded from 2352 W) Example 2 The homeowner wants to use very dark shingles with an emissivity of 0.90. The roof is made of 6 inch-thick fiberglass insulation and the attic temperature is always maintained at 26.7°C. if thermal conduction through the roof to the attic and radiative loss are the main sources of heat transfer determine the heat loss by radiation through the shingles. The solar power may be taken to be 1.0 kW/m². Solution: Energy absorption rate by the shingles =solar power × emissivity =1.0 kW/m² × 0.90 = 9.0 kW/m². Heat loss through conduction = (0.040 W/m·K) / (6.0×0.0254 m) T                    = 0.26 (T- 300) W/m² Heat loss through radiation = T4 = 0.90 × 5.67×10-8 W/(m²·K4) × T4                    = 5.6×10-8 T4 W/(m²·K4) Example 3 Considering two houses in Namibia, Africa, one of the houses is constructed with adobe block walls and a thatch roof, while the other is made of hollow core concrete blocks with a sheet metal roof. Each house measures 5 metres square, 2 metres high at the eaves, 3 metres at the ridge, has 1 m² of window and 1.5 m² of timber door. Determine the heat lost from each house if the outside temperature is 0 °C while the inside temperature of both the houses is 15 °C. From tables of overall heat transfer coefficients, the U value for a sheet metal roof is taken to be 3.85 W/(m².K); for a thatch roof, 0.26 W/(m².K); for an adobe wall, 2.5 W/(m².K); for a concrete block wall, 2.9 W/(m².K), and for single glass, 6 W/(m².K). The calculated U value for a 25 mm timber door is 2.4 W/(m².K). With heat loss equation as Q = A × U × ΔT Where: Q = total heat transfer rate through an element (W) A = area of the building element (m²) U = coefficient of heat transfer for the element (W/m2.K) ΔT = temperature differential across the element (K). For the thatched roof; Total area of the roof 5.4 × 5 = 27.0 m² Total heat loss from the roof=27.0 × 15 × 0.26 = 105 W Total area of the other exposed sections is obtained as follows; Walls 5 × 2 × 4 = 40.0 m² Gable ends + 5.0 m² Door and window - 2.5 m² The sign is negative because the area of the door and windows ought to be subtracted so as to obtain the area of the wall alone as the heat loss is calculated separately. Total area of the wall= 42.5 m² Total heat lost through the wall= 42.5 × 15 × 2.5 = 1 594 W Heat lost through the door =1.5 × 15 × 2.4 = 54 W Heat lost through the window= 1.0 × 15 × 6.0 = 90 W Total or overall heat loss from the thatched house =1 843 W Considering the house with a Metallic Roof; The areas of the house are equal to the above calculated, and therefore the heat lost from the respective areas is obtained as below; Heat lost through the roof = 27 × 15 × 3.85 = 1 559 W Heat lost through the wall =42.5 × 15 × 2.9 = 1 849 W Door 1.5 × 15 × 2.4 = 54 W Window 1.0 × 15 × 6.0 = 90 W Overall heat loss from the metallic house= 3 552 W. The above example indicates that a metallic house will lose more heat than the thatched house. The reason behind this is because metal are good conductors of heat and thus they will give out more heat than the thatched one. The compromise to this would be to use a thicker ceiling made of either rockwool or glasswool to reduce the heat loss through the roof. CONCLUSION In conclusion heat transfer is a very important area of study as it helps in the design of houses to ensure a comfortable working, and living conditions by its occupants. Heat transfer is influenced by the temperature difference between two bodies, and the bodies attain an equilibrium condition after some time, Zeroth law of heat transfer. Modes and factors that affect the amount of solar insolation are well looked at in the above discussion. Building material selection is of most important when constructing or designing any premise, location and orientation of its elements like the windows and the roof is also of great significance. References DOWN, P. G. (1969). Heating and cooling load calculations. Oxford, Pergamon Press. https://ezproxy.saskpolytech.ca/login?url=http://site.ebrary.com/lib/siastlibraries/Doc?id=10999804. BURDICK, A. (2011). Strategy guideline accurate heating and cooling load calculations. [Washington, D.C.], U.S. Dept. of Energy, Energy Efficiency & Renewable Energy, Building Technologies Program. http://purl.fdlp.gov/GPO/gpo9723. AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, RUDOY, W., & CUBA, J. F. (1979). Cooling and heating load calculation manual. New York, ASHRAE. Read More