When considering the design of houses, the heat conductivity through walls, roofs, and the floor pla — Physical Chemistry — Thermodynamics Chemistry Question
Heat Conductivity
When considering the design of houses, the heat conductivity through walls, roofs, and the floor plays an important role. The heat conductivities (λ) of some building materials are described in Table 1.
Table 1: Heat conductivity of different materials
Material: λ (W m–1 K–1)
Concrete: 1.10
Building brick: 0.81
Polystyrene insulation foam: 0.040
Linoleum (floor covering): 0.17
Gypsum: 0.35
Formula:
Heat flow through a wall: Pw = λ · A · (T2 – T1) · d–1
Area A, heat conductivity λ, temperature T, thickness d
Calculate the heat flow through a wall of 150 m2 (typical of a single–family house in Central Europe) that consists of a brick layer with a thickness of d = 24 cm and through the same wall that consists, however, of a brick layer with a thickness of d = 36 cm. There is a temperature of 25 °C inside and 10 °C outside.
Model Answer
The heat flows are:
PW = 150 m2 × (0.24 m)–1 × 0.81 W m–1 K–1 × (25 °C – 10 °C) = 7.59 kW and
PW = 150 m2 × (0.36 m)–1 × 0.81 W m–1 K–1 × (25 °C – 10 °C) = 5.06 kW
The heat loss can be minimized by using a layer of polystyrene foam. Calculate the heat loss through a 10 cm polystyrene insulation foam. The wall area again is 150 m2.
Model Answer
PW = 150 m2 × (0.1 m)–1 × 0.040 W m–1 K–1 × (25°C – 10 °C) = 0.90 kW
Although the wall is much thinner, the energy loss is much lower due to the much lower heat conductivity.
It is advantageous to use the heat resistance Λ–1 for the calculation of the heat conductivity through a wall consisting of different layers:
Λ–1 = d1/λ1 + d2/λ2 + d3/λ3 + ...
For the different parts of the house (window, wall) the diathermal coefficient can be calculated as:
ktot = (Λ1×A1 + Λ2×A2 + ...)/Atot
Energy–saving actions are of vital importance to decrease the energy requirements of the world. Good insulation is not only positive for the environment (reduction of CO2 emissions) but also good for the economy. Presently, an energy–saving house has a maximum diathermal coefficient of 0.50 W m–2·K–1
Calculate the thickness of a wall that only consists of brick to achieve this requirement.
Model Answer
k = λ · d–1 d = λ k–1 = 0.81 W m–1 K–1 × (0.5 W m–2 K–1 )–1
d = 1.62 m
The wall thickness can be minimized by insulation layers. A typical wall consists of a brick layer that has a thickness of d1 = 15 cm at the outside, a concrete layer with a thickness of d2 = 10 cm, an insulation layer (polystyrene foam) of thickness d3 and a gypsum layer with a thickness of d4 = 5 cm on the inside of the wall. Calculate the thickness of the insulation layer and the total thickness of the wall to fulfil the requirements of an energy–saving house.
Model Answer
Λ–1 = k–1 = (0.50 W m–2 K–1 )–1 = d1 × (λ1)–1 + d2 × (λ2)–1 + d3 × (λ3)–1 + d4 × (λ4)–1
= 0.15 m × (0.81 W m–1 K–1 )–1 + 0.10 m × (1.1 W m–1 K–1 )–1 + d3 × (0.040 W m–1 K–1 )–1 + 0.05 m × (0.35 W m–1 K–1 )–1
The thickness of the insulation foam layer is d3 = 6.3 cm
The total thickness is: 15 cm + 10 cm + 6.3 cm + 5 cm = 36.3 cm
Windows increase the mean value of the energy loss. Assume a wall of 15 m2 constructed as in 4.4 including a window of 4 m2 with a mean diathermal coefficient of 0.70 W m–2 K–1.
By what percentage has the thickness of the foam layer of 4.4 to be increased in order to achieve the same average k–value?
Model Answer
k = Λ1 × A1 × (Atot)–1 + Λ2 × A2 × (Atot)–1
0.50 W m–2 K–1 = 0.70 W m–2 K–1 × 4 m2 × (15 m2 )–1 + Λ2 × 11 m2 · (15 m2 )–1
Λ2 = 0.427 W m–2 K–1.
The calculation is similar to that of 4.4:
The thickness of the insulation foam layer is d3 = 7.7 cm
The total thickness is: 5 cm + 10 cm + 7.7 cm + 5 cm = 37.7 cm due to the much higher heat conductivity of the window.
The thickness of the foam layer has to be increased by 22 %.