- Thermal Mass
- Thermal Resistance
- Time Lag and Decrement Factor
- Thermal Insulation
- Transmittance or U-Value
- Thermal Conductivity

In real life situations in almost all the building components, heat transfer takes place through all the three modes. For example, in the diagram below, the wall receives heat through convection from the ambient air, through conduction the heat gets transmitted upto the inner surface where again through convective heat transfer due to inside room air, and heat goes into the building. Using the concept of thermal resistance, the overall heat transfer can be found using:

Qc = (T outside – T inside) / {(1/h_{o}A) + (l/ kA) + (1/h_{i}A)}

The above equation can be written in a simplified form as:

Qc = U * A * (T outside – T inside)

Where,

U = 1/ {(1/h_{o}) + (l/ k) + (1/hi)}

The term ‘U’ represents overall thermal conductance from the outside to inside covering all modes of heat transfer. From the above equation, ‘U- value’ can be defined as the rate of heat flow over unit area of any building component through unit overall temperature difference between both sides of the component.

The U-Value is an important concept in building design. It represents the air-to-air transmittance of an element. This refers to how well an element conducts heat from one side to the other, which makes it the reciprocal of its thermal resistance. Thus, if we calculate the thermal resistance of an element, we can simply invert it to obtain the U-Value

U = 1 / R_{t}

U = 1/ (R_{so} + ∑R_{n} + R_{si})

The U-Value is a property of a material. Thus, its units are Watts per meter squared Kelvin (W/m² K). This means that, if a wall material had a U-Value of 1 W/m² K, for every degree of temperature difference between the inside and outside surface, 1 Watt of heat energy would flow through each meter squared of its surface.

Calculation of U – value

Calculate the U- value for the 220 m brick wll with 16 mm plaster on the inside face. Assume normal exposure.

For brick,k=0.84W m^{-1}K^{-1}; for plaster k= 0.5 W m^{-1}K^{-1}

Internal resistance, Rsi = 0.123 m^{2} K W ^{-1}

Plaster resistance = l/k = 0.016/0.5 = 0.032 m^{2} K W ^{-1}

Brick resistance, R_{so}=0.055m^{2}K W^{-1}

Hence,

Total resistance = 0.123 + 0.032 + 0.262.+ 0.055 = 0.472 m^{2}K W^{-1}

Therefore,

*U* = 1/∑*R* = 2.12 W m ^{-2}K^{-1}