THERMAL LOADING OF BUILDINGS U-value Thermal conductivity Temperature dependence Effects of Pressure and Density Effect of moisture content Cavity walls and other air gaps Factors affecting surface resistances Roofs Ground floors and basements Windows Effects of blinds and curtains Energy flows associated with ventilation and infiltration 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 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 http://www.uvalue.co.uk/U-Value-Calculator Celotex insulation http://www.celotex.co.uk/Other-Resources/U-value-Calculator

EXAMPLE 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 1/U = Ros+x1/k1+Ra+x2/k2+Ris 1/U = 0.12+0.1/0.84+0.18+0.1/0.17+0.06 = 1.06m2K/W U = 1/1.06 = 0.94W/m2K

EXAMPLE 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 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