Primary objective of low energy design of buildings
Design to maintain building function through the use of
  • low embodied energy materials
  • low energy consumption in use
Maintenance of building function generally requires maintenance of a comfortable climate for the occupants.
This effectively adds a third requirement
  • Maintaining a thermally comfortable environment
The three requirements will not be entirely compatible
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The requirements of low energy in use and thermal comfort may be assessed through the study of heat flows in buildings.
Thermal Comfort assessment requires the examination of building dynamic performance.
Energy Consumption is best evaluated by dynamic methods but may be assessed by simpler techniques involving steady state methods..
However, potential problems of achieving thermal comfort may be diagnosed and minimised using a combination of these methods
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Dynamic Performance
This involves studying the manner in which the internal environment changes during the day due to changes in
  • climate
  • the use of active components
  • building services inputs
  • occupant behaviour
  • use of machinery and lighting.
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Steady State
Steady State assumes that all impacts on the building are fixed.ie
  • internal and external temperatures
  • solar gains
  • internal loads
These are derived by taking average values, for example
external temperatures
mean temp = (Max + Min)/2
solar radiation
mean irradiance = daily insolation/24
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Balance Point Temperature-Solar Gains only
Even with no activity a building will be warmer inside as the solar gain becomes a heat source following transmission through glazing it is absorbed by building surfaces. Balance is achieved when
solar heat gain=thermal losses
Solar Gains are:
  • absorbed by opaque fabric
  • transmitted and absorbed by translucent elements
Thermal losses are lost
  • through fabric
  • through mass transport
  • through thermal radiation
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Effect of Temperature Gradients
Heat is lost through fabric due to temperature differences.
double the temperature difference=>double the rate of heat loss
There is a linear relationship between MOST heat losses and the difference between internal and external temperatures
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Deriving the balance point temperature
Balance Point Temperature = External Temperature which gives rise to minimum comfort temperature inside.
if
Min Comfort Temp=20oC
Balance Temperature Difference=5oC
 Then
Balance point Temperature = 15oC
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Effect of changes on balance Point Temperature
An increase in heat losses due to higher U value will increase the balance point temperature
An increase in solar gains will decrease the balance point temperature
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Internal Comfort Temperature
Comfort Temperature* will also affect the balance point temperature
* Humphreys, 1978 Outdoor temperatures and comfort indoors BRE Current Paper 53/78
Based on data for Kew
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Thermal Losses
Thermal Losses are generally categorised as
  • fabric
  • ventilation
Thermal radiation losses are generally incorporated in fabric losses and are incorporated in U value calculations.
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Fabric Losses
Fabric Losses are calculated by summing the product
Uvalue x external surface Area
for each external surface
and multiplying the result by the temperature difference
For heating requirements the temperature difference is relative to the Balance Point Temperature
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Ventilation Losses
Heat is Lost in ventilation and infiltration because the air is a store of heat. The rate at which heat is lost may be found by multiplying
  • number of air changes per hour N
  • Volume of air in room/building V
  • Specific Heat Capacity of air Cp
  • Density of air r
  • temperature difference dT
 By substituting in known values this is simplified

from

N xV x1000 x1.2 xdT/(60x60)

to

NVdT/3

Again, for heating requirements the temperature difference is relative to the Balance Point Temperature
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Degree Days
The basis of many steady state methods for calculating heating and cooling energy requirements is the Degree Day
One heating degree day is experienced when the mean external temperature is one degree below the balance point temperature for one day.

Example uses a Balance Point temperature of 15.5oC
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Annual Heating Energy Requirement
The annual heating requirement may be derived by multiplying
  • Thermal Losses
by
  • Total Annual Degree Days
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Calculating U values
Thermal Transmittance (U value) represents the ability of a construction to transfer heat between itself and its surroundings.
All materials (which includes air) offer a resistance to the passage of heat.

In calculating U values we evaluate the resistance of the construction and the resistance of the surroundings to the flow of heat.

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Resistances
When heat passes through a solid construction due to a temperature difference either side, the heat has to pass through 3 layers
  • internal air Ri
  • construction Rc
  • external air Re
Rtotal = Ri + Rc + Re U = 1/Rtotal
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Conduction
Heat passes through the solid elements by conduction
The rate of heat flow depends upon
  • Construction thickness t
  • Surface Area A
  • Temperature Difference dT
  • Thermal Conductivity k
 Effect
  • Double - Halve heat flow
  • Double - Double heat flow
  • Double - Double heat flow
  • Double - Double heat flow
Rate of Heat Flow = kAdT/t Rc = t/k
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Convection
Heat passes through air layers by convection
The rate of heat flow depends upon
  • Surface Area A
  • Temperature Difference dT
  • direction of flow
  • type of flow
  • containment
Relationships are complex with main influences grouped and represented by Heat Transfer Coefficient hconv
Rate of Heat Flow = hconvAdT Rconv = 1/hconv
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Radiation
Heat is emitted from exposed surfaces due to temperature differences between surfaces.
The rate of heat flow depends upon
  • Surface Area A
  • Temperature Difference dT
  • Surface Geometry
  • Surface Temperatures
  • Surface emissivity
Relationships are complex with main influences grouped and represented by Heat Transfer Coefficient hrad
Rate of Heat Flow = hradAdT Rrad = 1/hrad
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Emissivity
Surface emissivity is the only material property that affects radiation and it has a linear effect
Most building materials (including glazings) have high emissivities of the order of 90% (0.9)
Highly reflective materials such as polished metals have emissivities of the order of 10% (0.1).
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Total Surface Resistance
Total Surface Resistance can be derived from Convection and Radiation Resistances.1/R = 1/Rconv+1/Rrad
A simpler relationship connects the Total Surface Heat Transfer Coefficients, since 1/R = hh = hconv + hrad
Convective and Radiative Resistances tend to be combined in U value calculations with 'typical' values used rather than complex calculations
Typical Values* for walls in m2K/W
Exposure High emissivity Low emissivity
Internal 0.12 0.30
External Sheltered 0.08 0.11
Normal 0.06 0.07
Severe 0.03 0.03
* CIBSE Guide A
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Solar Gain
Solar Gain through transparent elements may be derived by multiplying
  • Unshaded Mean Irradiance
  • Glazed Area
  • Total solar transmittance
  • geometrical shading factor
Using data from sources such as
  • Page and Lebens
  • -
  • Solar shading of Buildings
  • solar geometry analysis
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Sol-Air Temperature
External absorptance of solar gains on opaque surfaces depends upon
  • Solar absorptance of surface a
  • External Surface Resistance Re
  • Mean irradiance reaching the surface I
The concept of Sol-Air Temperature is used to represent this: it is the effective increase in external temperature due to the absorbed radiation Sol-Air Temperature = External Temperature + aIRe
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Derivation of mean daily irradiance from daily insolation: the kilo-watt hour
Insolation is the total energy received from the sun on a given surface over a specified period of time.
Internationally the units used for daily insolation are kWh/m2. Btu/m2 are used in the USA.

1 kWh (kiloWatt-hour) is the equivalent amount of energy received over 1 hour if the rate of energy arriving is 1kiloWatt (ie 1 Joule every second).

A 3kiloWatt electric fire will consume
  • 1kWh after 20 minutes
  • 3kWh over 1 hour
  • 72kWh over 1 day
 		1kWh is equivalent to
  • 1kW for 1 hour
  • 2kW for 1/2 hour (twice the power half the time)
  • 500W for 2 hours (half the power over twice the time)
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suppose the following average irradiances are received on each square metre of a south facing wall:
  1. 50W 1hour (0.05kWh)
  2. 150W 1hour (0.15kWh)
  3. 350W 2hour (0.70kWh)
  4. 250W 1hour (0.25kWh)
  5. 100W 1hour (0.10kWh)
  6. 25W 1/2hour (12.5Wh)
    7.
  1. The total radiation over the day will be the sum of these: 1.26kWh
  2. If radiation was received every hour at the same level then this would be 1.26kWh/24 = 0.0525kWh per hour (52.5Wh per hour)
  3. This represents 52.5Watts for one hour. Thus the average irradiance is 52.5W
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Thermal Conductivities of common building materials
These are typical values. Use product data where available.
MaterialConductivity (W/m2oC)Resistance of 1cm thickness(moC/W)
Aluminium 211 0.00005
Brick 1.0 0.01
Concrete 1.5 0.007
Gypsum plaster 0.25 0.04
Softwood 0.12 0.08
Copper 385 0.00003
Glass 1.0 0.01
Polystyrene 0.04 0.25
External surface high emissivity - 0.06
Internal surface high emissivity - 0.12
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Calculating U values
One fairly straightforward way is to use the table on the previous page (or generate one yourself).

Multiply the number of centimetres thickness of each layer by its 1cm thick resistance

add the resistances obtained for each layer

add external and internal resistances

invert the answer

10 x 0.01, 5 x 0.25, 0.04

0.1 + 1.25 + 0.04 = 1.39

1.39 + 0.12 + 0.06 = 1.57

1/1.57 = 0.64W/m2oC
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Resistance of cavities
Cavities need special care.
They are a special case of two surface-air resistances
The resistance depends upon

* the width of the cavity

* the surface properties (as radiation will be emitted here), and

* whether or not it is ventilated (the height of the cavity will also be a factor).

Derivation of the resistance is difficult.

It is usual to read data from a table or graph which fits the

width, surface emissivity and
ventilated/unventilated status.
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Calculating U values for constructions with cavities
  • Multiply the number of centimetres thickness of each glazing by its 1cm thick resistance
  • add the resistances to the appropriate cavity resistance
  • add external and internal resistances
  • invert the answer
    •
  • 0.6 x 0.01, 0.6 x 0.01
  • 2 x 0.006 + 0.20 = 0.212
  • 0.212 + 0.12 + 0.06 = 0.392
  • 1/0.392 = 2.5W/m2oC
    •
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    Balance Point Temperature calculation: calculating overall heat losses
    Assume construction and climate as previously used. To determine the Balance Point temperature:
    Determine the heat loss from the external surfaces for 1degree difference between inside and outside (assuming no ventilation heat loss)
    Uwall x Areawall + Uglazing x Areaglazing
    0.64 x 27 + 2.5 x 3 = 25W/oC
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    Balance Point Temperature calculation: calculating solar gains
    Calculate Mean Solar Gains by multiplying
    • Solar Irradiance
      (52.5W/m2)
    • Window Area
      (5m2)
    • Glazing Solar Transmittance
      (typically 0.7 for clear double glazing)
    52.5 x 3 x 0.7 = 110W
    Thus on average 110W enter the appartment throughout the day. At the Balance Point Temperature these will balance out the losses of the building. Thus no additional heating will be required if the outside temperature no less than 110/25oC below the inside temperature. For an inside temperature of 20oC this will thus occur at approximately 15.5oC.
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    Start Again?
    Dynamic and Steady State
    Dynamic
    Steady State
    Balance Point Temperature
    Effect of Temperature Gradients
    Deriving the balance point temperature
    Effect of changes on balance Point Temperature
    Internal Comfort Temperature
    Thermal Losses
    Fabric Losses
    Ventilation Losses
    Degree Days
    Annual Heating Energy Requirement
    Calculating U values
    Resistances
    Conduction
    		 Convection
    Radiation
    Emissivity
    Total Surface Resistance
    Solar Gain
    Sol-Air Temperature
    The kilo-watt hour
    Mean Daily Irradiance calculation
    Thermal Conductivities
    U values - Solid Constructions
    Cavity Resistance
    U values - Cavity constructions
    BPT - Overall Heat Losses
    BPT - Solar Gains


    Other Tutorials