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.

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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.

R= L÷K

L = thickness of material in meters

K= Thermal conductivity = K value

Brick work:

0.100 ÷ 0.84 = 0.119 m2 K/W

Cavity

0.18 m2 K/W

Block work

0.100 ÷ 0.65=0.153

Plaster

0.020 ÷ 0.50 = 0.04

Inside surface resistances

0.055

0.123

From this we can now work out the U value of the cavity wall from this equation. 

1

Ri+R1+R2+R3 etc +Ro

 

 1

 

                                                           1                                                                                            

                                                        0.67                     = 1.49 w/m2 K

B)

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

R= L÷K

Insulation

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.

1

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.

C)

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.

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