## Information

- Post Date 2018-11-07T10:28:21+00:00
- Post Category Essays & Coursework

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## Order Details

### 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.

**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**

**R****i+****R****1+****R****2+****R****3 etc +****R****o**

**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.