# Calculating Insulation Thickness for a Building (Engineering)

1. Jul 15, 2014

### btec_noob

1. The problem statement, all variables and given/known data

A single roomed building measureing 12.0m x 12.0m x 2.5m is maintained throughout the day at a temperature of between 18-23°C. Each wall is to be constructed from 2 layers of common brick, an internal particle board and a layer of insulation. The building is heated using standard space heaters that will run on natural gas. The gas currently costs 7p/kWh and is burnt at an efficiency of 70%. During heating calculations it is assumed that the same thickness of insulation is applied to floor, wall and roof and that mineral wool at a current price of £55/m3 is utilised. Calculate the the optimum thickness for insulation over a 12 months period.

2. Relevant equations

q = ΔT/∑Rth
therefore: q with insulation/q without insulation = 0.3 = ∑Rth (with)/∑Rth (without)
Thermal resistance Rth = Δx/k

3. The attempt at a solution

So far I've assumed that due to it being over 12 months, there will be a change in outside temperature during the different seasons. Also assumed that there is no air gap between the wall. The k values I have found so far are: brick = 0.7, particle board = 0.15 and k for wool is 0.04, so I can calculate the Thermal resistance and heat loss.
R wool = L/kA = x/0.4*1 = x/0.4 °C/W
R brick = L/kA = 0.1/0.7*1 = 1.429*2 = 0.2858 °C/W
R P.board = L/kA = 0.01/0.15*1 = 0.067 °C/W

I just need help in finding the equation for calculating Energy cost.

Last edited: Jul 15, 2014
2. Jul 16, 2014

### CWatters

In order to maintain a constant temperature the energy supplied by the heating must equal the heat loss through the structure. So calculate the energy loss over a year in Joules. The same amount of energy must be delivered by the gas heating.

Because the heating is not 100% efficient more energy must be supplied from the gas main than calculated above so do that calculation.

Once you have a figure for the total energy needed (in Joules) remember that:

1 Watt = 1 Joule per second
or
1 Joule = 1 Watt for 1 second

You have been provided a cost figure of 7p/kWh so a bit more work is needed to arrive at the total cost in pence.

3. Jul 19, 2014

### RTW69

You will also need to know something about the outdoor temperatures where the building is located. An average winter outdoor temperature is a start but not very accurate, average temperature by month would be better, heating degree days are also used, bin data is also used.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted