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THERMAL LOADING OF BUILDINGS
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U-value
Thermal conductivity
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Cavity walls and other air gaps
Factors affecting surface resistances
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Temperature dependence
Effects of Pressure and Density
Effect of moisture content
Roofs
Ground floors and basements
Windows
Effects of blinds and curtains
Energy flows associated with ventilation and infiltration
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Natural ventilation
Mechanical ventilation
Heat transfer associated with ventilation
U-VALUE
When selecting materials for construction, we choose good insulators to reduce heat energy losses. In
practice, it is usual to use a figure for a particular structure, e.g. a brick wall, rather than take the values
of the conductivities of its constituent materials. The figure used is called the U-value. It refers to
Unit Heat Loss Rate , and its unit is the watt per metre squared per kelvin, W m-2 K-1.
The U-value is defined as the rate at which thermal energy is conducted through unit area, per kelvin
temperature difference between its two sides.
U = rate of loss of energy/(surface area)(temperature difference)
A good insulating structure therefore has a low U-value, if a structure has a U-value of 1, this means 1
J per second will pass through each square metre for each kelvin (degree Celsius) difference in
temperature between the two sides of the structure.
U-value is also referred to as an ‘overall heat transfer co-efficient’ and measures how well parts of a
building transfer heat. This means that the higher the U-value the worse the thermal performance of
the building envelope. A low U-value usually indicates high levels of insulation. They are useful as it is
a way of predicting the composite behaviour of an entire building element rather than relying on the
properties of individual materials.
IMPORTANCE OF U-VALUE
U-values are important because they form the basis of any energy or carbon
reduction standard. In practice, nearly every external building element has to
comply with thermal standards that are expressed as a maximum U-value.
Knowledge of how to simply calculate U-values at an early stage in the design
process, avoids expensive re-working later on in a project. It allows the designer
to test the feasibility of their project at an early stage to ensure it is fit for purpose
and will comply with regulatory frameworks.
EFFECT OF INSULATION ON U-VALUES
U-VALUES OF WINDOWS
CALCULATION OF U-VALUE
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To calculate the U value of a building element such as a wall, floor or roof, you need to know the
build up of that element. Each building material should be positioned properly in sequence. The
thickness of each building material also is required.
The other key property you need to obtain is the conductivity of each building material. This is a
measure of its inherent ability to facilitate the passage of heat. It is normally referred to as a ‘k
value’ and values for materials can be found in publications such as the New Metric Handbook
and the Architects’ Pocket Guide
The properties of the internal and external faces of the constructional element under scrutiny
need to be allowed for. These are called external resistances and are fixed values.
The U-value is defined as being reciprocal of all the resistances of the materials found in the
building element.
The resistance of a building material is derived by the following formula:
R = (x/k)
where k is the conductivity of the building material and x is the material thickness.
The formula for the calculation of a U value is
U(element) = 1 / (Rso + Rsi + R1 + R2 ...)
U-VALUE
The reciprocal of the total thermal resistance of a body is regarded as the overall
thermal transmittance or U-value.
R = 1/U
Q = kA(T1-T2)/x
Q = A(T1-T2)/R
Where
R = thermal resistance = x/k or L/k
Q =UA(T1-T2)
U-VALUE OF WINDOWS
U-VALUE OF WINDOWS
ONLINE U-VALUE CALCULATOR
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http://www.uvalue.co.uk/U-Value-Calculator
Celotex insulation
http://www.celotex.co.uk/Other-Resources/U-value-Calculator
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EXAMPLE
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Determine the U-value for an external brick cavity wall consisting of 100mm
brick outer leaf of thermal conductivity of 0.84W/mK; a 50mm unvented air
gap; and a 100mm block inner leaf of thermal conductivity 0.17W/mK. The
internal surface, external surface and air resistances are respectively 0.06, 0.12
and 0.18m2K/W
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1/U = Ros+x1/k1+Ra+x2/k2+Ris
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1/U = 0.12+0.1/0.84+0.18+0.1/0.17+0.06 = 1.06m2K/W
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U = 1/1.06 = 0.94W/m2K
EXAMPLE
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A cold-storage room has walls constructed of 0.23 m brick on the outside, 0.08 m of insulation
foam, and finally 0.015 m of wood on the inside. The total wall area is 90 m2. The outdoor air
temperature is constant at 22°C, and the air temperature inside is constant at -2°C. The
convection coefficient from wood to air is 29 W/m2 K, and the convection coefficient from air to
brick is 12 W/m2 K, and the thermal conductivities of wood, foam and brick are 0.17,0.024, and
0.98 W/m K respectively. It can be assumed that the heat transfer by radiation may be neglected.
Determine:
the overall heat transfer coefficient
the energy to be removed by the refrigeration equipment in twenty four hours
the temperature of the inner surface of the brick.
Solution
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The overall heat transfer coefficient is given by
1/U=R=1/ho+x1/k1+x2/k2+x3/k3+1/h1
1/U=1/12 +0.23/0.98+0.08/0.024+0.015/0.17+1/29 = 3.77m2K/W
U=0.265W/m2K
Q = UA∆T = 0.265 x 90 x [22-(-2)] = 572W
E =24 x 572=13.7kWh
Q = Ak3/x3 ∆T
572 = 90 x 0.98/0.23 (22-T)
T=20.5C