BASIC FACTORS THAT AFFECT HUMAN COMFORT IN THE INTERNAL ENVIRONMENTUNIT 4 - Science and Materials in constructionLEVEL 3 – BTEC Construction & Built EnvironmentBy Kenneth Bowazi – MSc BE, BSc Hons QS, BSc CE

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 maleActivity 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 valuesClo 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-ValuesMaterial 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 & VentilationHumidity- 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

Theory into practice Calculate the rate of heat loss due to ventilation for the

building measuring 4.5m x 3.25 in plan and has a ceiling height of 2.6m. The number of air changes in one hour is 1.35. The outside temperature is 6°C and the inside temperature is 19°C.

Calculation {(4.5x3.25x2.6)m3 x 1.35 x 1300J x (19-6)°}/ 3600s

240.983 Watts

Theory into practice

A domestic semi-detached dwelling is subject to 1.5 changes per hour. Calculate the total heat loss due to ventilation. In this example we have removed the circulation space which is uninhabited.

Room Dimensions Lounge is 3.5m x 3.5m Kitchen/diner is 4.0m x 2.5m Bedroom 1 is 3.0m x 3.0m Bedroom 2 is 2.75m x 2.75m Bathroom 3 is 2.5m x 2m Storey height is 2.4m Air changes for all rooms 1.5 per hour Temperature difference -1°C outside, 19°C inside.

Calculation Lounge 3.5 x 3.5 x 2.4 =29.4 Kitchen 4.0 x 2.5 x 2.4 =24.0 Bedroom One 3.0 x 3.0 x 2.4 =21.6 Bedroom Two 2.75 x 2.75 x 2.4 =18.15 Bathroom Three 2.5 x 2.0 x 2.4 =12.0

Total volume = 105.91m3

Calculation

condensation This is formed when hot , humid air meets a cold surface, it

condenses onto this surface forming droplets of water vapour. What are the effects of condensation in the internal

environment? Cause timber rot Encourage mould growth Produce cold spots Produce high humidity Cause corrosion to steelwork Dampen insulation, reducung its effectiveness

Heat flow through a structure

Acceptable values The acceptable values of heat loss or U-values is a

complicated topic and you will need to refer to the Building regulations Part L Conservation of fuel and power for guidance on the acceptable U- values.

Ventilation is linked to the Building Regulation Part L that it restricts air tightness of modern structure. Forced ventilation has to be provided in form of fans in bathrooms and cooking areas

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)

P= kA (T1-T2)/ x

A= Area X= thickness in m² and m respectively T1-T2= temperature difference in °C or K Which can be written as follows

W=k x m² x °C/m ; k = W x m/(m² x °C) = W/m°C or W/mK

U-Values A measurement of the rate of heat loss through a structure Thermal resistivity is the reciprocal of thermal conductivity: R=1/K