Calculating Resistances of a Cavity Wall

In summary, the U value of 0.48 for a cavity wall with brickwork of 103mm, insulation of 50mm, lightweight concrete block of 100mm, and plaster board of 13mm is not enough information to solve for the resistances. The U value accounts for both resistance values and convection coefficients on the inside and outside faces of the wall, making it impossible to solve for the individual resistances without all the necessary information. Furthermore, the thermal conductivity values for each material are needed in order to calculate the resistances. Without these values, the problem cannot be solved.
  • #1
laheer
2
0
Hi,

I am trying to work out the resistances for the following cavity wall. I am having to work backwards. I have the U value, which is 0.48. The brickwork is 103mm, the insulation in cavity is 50mm, lightweight concrete block is 100mm and plaster board is 13mm.

I know

total thickness (0.103+ 0.050 +.100+.013) / total resistance = a

1/a = .048.

Thanks
 
Physics news on Phys.org
  • #2
Simply stated, you cannot find what you are looking for if this is all the information that you have.

The U value is the overall heat transfer coefficient, and it means just that. It accounts for not only the resistance values of the materials that apply for conductance and a given temperature difference, but also convection coefficents on the inside and outside faces of the wall. Generally speaking, the U value will be defined under typical conditions for a given application, but the convection is still accounted for.

If you chose to neglect this and simplify the problem to a temperature difference from one face to the other, you are still short of information needed to solve your problem. You would have one equation, but four unknowns. The U value is the inverse of the sum of the resistances in the series, i.e. convection inside surface, brickwork, insulation, concrete block, plaster board, and convection on the outside surface. That's six but since we will neglect convection, we'll say four. So:

U=1/sum(R)

or

1=U*(R1+R2+R3+R4)

where R=l/k

l being the thickness and k being the thermal conductivity

so:

1=U*[(l1/k1)+(l2/k2)+(l3/k3)+(l4/k4)]

You have the U value, the thickness for each, but no thermal conductivity values. Finding the resistance values you want given the thicknesses you have provided, requires the respective thermal conductivity values. You would need the other three to solve for one of them.

Taking the thicknesses and materials you have specified and using this page as a reference:
http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

Brick work: 0.69
Insulation: 0.035
Concrete, light: 0.42
Gypsum or plaster board: 0.17

0.48*[(0.103/0.69)+(0.05/0.035)+(0.1/0.42)+(0.013/0.17)=0.908

This value being lower than one is not surprising because the resistance for the inside and outside surface convection has been neglected. Also of course, the values from this reference source and another are likely to vary some (especially the insulation which in this case accounts for over 75% of the total resistance). Hopefully this helps.
 
  • #3
for reaching out with your question. Calculating resistances for a cavity wall can be a complex process, but I am happy to help guide you through it. First, let's define what we mean by resistance in this context. Resistance, also known as thermal resistance, is a measure of how well a material or structure can resist the flow of heat. In this case, we are looking at the resistance of the wall to heat flow from the inside to the outside.

To calculate the resistance of a cavity wall, we need to consider the individual resistances of each layer in the wall. These include the thermal resistance of the brickwork, insulation, lightweight concrete block, and plasterboard. Each of these materials has a specific thermal conductivity, which is a measure of how easily heat can pass through it. The thermal resistance is then calculated by dividing the thickness of the material by its thermal conductivity.

In your case, you already have the total thickness of the wall (0.266m) and the U-value (0.48). The U-value is a measure of the overall thermal resistance of the entire wall. To calculate the individual resistances, we can rearrange the formula for U-value to solve for the total resistance (R). This would be R = 1/U = 1/0.48 = 2.08 m2K/W.

Next, we need to calculate the individual resistances for each layer. Using the formula for thermal resistance (R = thickness/thermal conductivity), we can calculate the resistance for each layer as follows:

- Brickwork: 0.103m / 0.7 W/mK = 0.147 m2K/W
- Insulation: 0.050m / 0.035 W/mK = 1.43 m2K/W
- Lightweight concrete block: 0.100m / 0.15 W/mK = 0.667 m2K/W
- Plasterboard: 0.013m / 0.16 W/mK = 0.081 m2K/W

Finally, we can add up all the individual resistances to get the total resistance of the wall:

0.147 + 1.43 + 0.667 + 0.081 = 2.325 m2K/W

This value is slightly higher than the calculated total resistance of 2.08 m2K/W, but this could be due to rounding errors or slight
 

Related to Calculating Resistances of a Cavity Wall

1. How do you calculate the total resistance of a cavity wall?

To calculate the total resistance of a cavity wall, you will need to add up the resistance values of each layer of the wall. This includes the resistance of the inside and outside layer of bricks, insulation material, and any air gaps or thermal bridges. The formula for calculating resistance is R = L / k, where R is the resistance, L is the thickness of the material, and k is the thermal conductivity of the material. The total resistance of the wall will be the sum of all these individual resistances.

2. What is the purpose of calculating the resistance of a cavity wall?

Calculating the resistance of a cavity wall is important in determining the energy efficiency and thermal performance of a building. It helps in identifying areas of heat loss and understanding the overall insulation of the building. This information can then be used to improve the design and construction of the wall to reduce energy consumption and save on heating and cooling costs.

3. How does the type of insulation used affect the resistance of a cavity wall?

The type of insulation used in a cavity wall can greatly affect its resistance. Different materials have different thermal conductivity values, which will impact the overall resistance of the wall. For example, foam insulation has a higher thermal conductivity than mineral wool, which means it will have a lower resistance. It is important to choose the right type of insulation for the desired level of thermal performance.

4. Can the resistance of a cavity wall change over time?

The resistance of a cavity wall can change over time due to factors such as settling of insulation, moisture, and deterioration of building materials. It is important to regularly assess and monitor the condition of the wall to ensure its resistance is maintained. Any changes in resistance should be addressed to maintain the energy efficiency and performance of the building.

5. How accurate are the calculations for the resistance of a cavity wall?

The accuracy of the calculations for the resistance of a cavity wall will depend on the accuracy of the input values used. This includes the thickness and thermal conductivity of each layer of the wall, as well as any assumptions made about the building materials. It is important to use reliable and up-to-date data when performing these calculations to ensure the most accurate results.

Similar threads

Replies
2
Views
919
  • Introductory Physics Homework Help
Replies
9
Views
2K
Replies
46
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Thermodynamics
Replies
3
Views
808
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
Back
Top