Application of scientific principles to the design and use of buildings; solving scientific problems in construction and the built environment.
Task 3 (LO 3.3)
A) Determine U-value for the cavity wall shown below if thermal conductives for brick work, block work and plaster are respectively 0.84, 0.65 and 0.50 W/mk. Take outside and inside surface resistances as 0.055 and 0.123 m2K/W respectively and air gap (cavity) resistance as 0.18 m2 K/W.
To find the U value for the cavity wall we need to find the thermal resistivity of each material in the wall using the following equation.
L = thickness of material in meters
K= Thermal conductivity = K value
0.100 ÷ 0.84 = 0.119 m2 K/W
0.18 m2 K/W
0.100 ÷ 0.65=0.153
0.020 ÷ 0.50 = 0.04
Inside surface resistances
From this we can now work out the U value of the cavity wall from this equation.
Ri+R1+R2+R3 etc +Ro
0.67 = 1.49 w/m2 K
If insulation was added was added with a thermal conductivity of 0,03
First of all work out the thermal resistivity for the thickness cavity of 80mm with the following equation
0.080 ÷ 0.03 = 2.66 M2K/L
So changing the above equations and removing the value for air gap in cavity and replacing it with the insulation thermal resistivity 2.66 m2K/L.
0.005+0.119+2.66+0.153+0.04+0.123 = 0.322 w/m2 K
As the workings show by adding in insulation gives it a higher thermal resistivity which intern gives it a lower U value which shows a better heat insulator.
Determine the total heat losses due to fabric and ventilation for the building shown below.
Rectangular building dimensions 6.0 meters long x 3.0 meters wide x 2.5 meters high. The air change rate due to natural ventilation is 2 air changes per hour.