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# Basic factors that affect human comfort

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### Basic factors that affect human comfort

1. 1. BASIC FACTORS THAT AFFECT HUMAN COMFORT IN THE INTERNAL ENVIRONMENT UNIT 4 LEVEL 3
2. 2. Thermal and air quality  What affects the surroundings you live in?  Air quality is affected by how hot it is outside or inside your environment  What is humidity and what affects humidity?  The amount of moisture that is present within the air will have an effect on humidity, which is linked to the amount of ventilation entering  What is the normal temperature of a human being?  Human temperature maintain an average core temperature of 37º depending on the metabolic rate
3. 3. Nature of heat • What is the measure of temperature • Temperature is measured in degrees celsius • The lower is 0 fixed at a melting point of ice at a stand at atmospheric pressure of 101.32kN/m2 • The upper point is 100 degrees – temperature of steam above the boiling point • What is the acceptable value of temperature taken at normal design? • Normal design temperature are taken at 21 degrees inside and -1 degrees outside on average
4. 4. Thermodynamic temperature scale • This is another measure of temperature in degrees Kelvin • 0 degree celsius= 273.16 Kelvin (K) • 100 degree celsius = 317.16 Kelvin • The unit of thermodynamic temperature is the fraction of the thermodynamic temperature at the triple point water • (equilibrium point of the temperature and pressure at which three known phases of substance can exist i.e. liquid, water vapour and pure ice)
5. 5. Quantity of heat  How do we measure the quantity of heat?  Heat is measured in joules (J) which is a measure of work done  The rate of expenditure of energy or doing work or of heat loss is measured in watts (W)  1 watt is = 1 Joule per second  1 W =1 J/s
6. 6. Heat transfer  Name three ways heat is transferred from one mass to another, for instance a person sitting next to a radiator.  Conduction  Convection  Radiation
7. 7. Thermal comfort  In high activity the temperature rises and the more heat you will give off. Several factors influences the level heat is generated (metabolic rate) including:  Your surface area  Age  Gender  Level of activity  e.g.  Sleeping heat output 70W. Lifting 440W.
8. 8. Typical heat output of an adult male Activity Example Heat output Immobile Sleeping 70W Seated Watching TV 115W Light work Office 140W Medium work Factory Work 265W Heavy work Lifting 440W
9. 9. Clothing  The amount of clothing that we wear generally depends on the season and affects our thermal comfort  Clothing is measured in a scale called clo value  1 clo= 0.155m2 K/W of insulation to the body  Typical values vary from 1-4 clo
10. 10. Typical clothing values Clo value Clothing Typical comfort temperature when sitting 0 clo Swimwear 29ºC 0.5 clo Light clothing 25ºC 1 clo Suit , jumper 22ºC 2 clo Coat, gloves, hat 14ºC
11. 11. Heat losses from buildings  Comfortable temperature for humans is provided by balancing the heat lost through conduction and ventilation through the fabric with similar heat  Optimum temperature will depend on material used , type of construction, orientation of the building and degree of exposure to the rain and wind
12. 12. Room temperatures  What would you consider in design to maintain temperature in buildings?  The resistance of a material to the passage of heat and the thermal conductivity of the material in passing the heat along are the basics of understanding of maintaining a steady temperature and a comfortable thermal indoor environment  In order to maintain a comfortable room temperature the building must be provided with as much heat as is lost through ventilation
13. 13. What will the loss of heat in buildings depend on?  Materials used  Type of construction  Orientation of the building in relation to the sun  Degree of exposure to rain and wind
14. 14. Thermal conductivity (k)  The amount of heat loss in one second through 1m2 of material, whose thickness is 1 metre  The units are W/mK (watts per metre Kelvin)
15. 15. K-Values Material K Value (W/mK) Brickwork (internal/exposed) (1700kg/m3) 0.84 Concrete, dense (2100kg/m3) 1.40 Concrete, lightweight (1200kg/m3) 0.38 Plaster, dense 0.50 Rendering 0.50 Concrete block, medium, weight (1400kg/m3) 0.51 Concrete block, lightweight (600kg/m3) 0.19
16. 16. Thermal resistivity (r)  Thermal resistivity is the reciprocal of thermal conductivity:  R=1/K
17. 17. Air movement  Properties are tested for airtightness  Draught seals are fitted to all openings to restrict thermal losses  If warmer air enter a room is not mixed with cooler air the room becomes hotter near the ceiling and colder at floor level
18. 18. Humidity & Ventilation Humidity- the amount of water or moisture in the air measured in grams per cubic metre(g/m3)  Relative Humidity or percentage saturation  This the percentage saturation  Actual amount of water vapour/maximum amount of water vapour that can be held X 100% of the temperature
19. 19. RELATIVE HUMIDITY  Humans are used to a relative humidity of between 40 and 60%. Greater than this we start to describe air as being ‘Humid’.
20. 20. HEAT LOSS DUE TO VENTILATION  Natural ventilation leads to the complete volume of air in a room changing a certain number of times in one hour  Type of room Air changes in hr  Halls 1.0  Bedrooms /lounges 1.5  WCs and bathrooms 2.0
21. 21. HEAT LOSS DUE TO VENTILATION  The fresh air entering the room will need to be heated to the internal temperature of the room. This is calculated with the formula:  Volume of room x air change rate x volumetric specific heat for air x temperature difference  The volumetric specific heat for air is approximately 1300j/m3K and is considered a constant in this formula which will give an answer in joules per hour.  This then has to be converted into watts in order to find the rate of heat loss which is achieved by dividing the number of joules by the number of seconds in one hour
22. 22. Heat loss to ventilation  This then has to be converted into watts in order to find the rate of heat loss which is achieved by dividing the number of joules by the number of seconds in one hour  Volume of room/building x air changes hr x 1300J x Temperature difference / 3600s = Watts  It is convenient when carrying out heat loss calculations to assume an average internal temperature of 19°C minus average of -1°C in winter which gives 20°C difference between inside and outside temperatures
23. 23. U-Values  A measurement of the rate of heat loss through a structure