Basic factors that affect human comfort

11,732 views 23 slides Oct 02, 2013
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BASIC FACTORS THAT
AFFECT HUMAN
COMFORT IN THE
INTERNAL ENVIRONMENT
UNIT 4
LEVEL 3

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

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

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)

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

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

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.

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

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

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

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

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

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

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)

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

Thermal resistivity (r)
Thermal resistivity is the reciprocal of thermal conductivity:
R=1/K

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

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

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’.

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

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

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

U-Values
A measurement of the rate of heat loss through a structure
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