a quantitative analysis
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a quantitative analysis
the status quo vs. our whitewash mixture on housing materials
With our solution in place, we know that it would be affordable, durable, and improve the housing materials’ ability to withstand the severe temperatures that come with droughts. But, anyone can assume that it’s better than the status quo.
How do we know this is really the case?
This post is a breakdown of the quantitative analysis to understand how temperature affects houses in Kenya and how our solution provides a 5x increase in thermal mass, making it a better material to store heat and take longer to release it.
Currently, Kenya has shown that residents think “it is very uncomfortable to stay in [their] houses that are made of metal sheet” and “Houses are made of metal sheet that doesn’t keep the heat”. The reason is iron has a low thermal mass since it absorbs and loses heat quickly (it gets hot in the house fast, making it inconvenient).
To understand the status quo with temperature, we calculated at which temperature a house in rural Kenya starts cracking and collapsing:
The initial slope in both the graphs being constant show that the material first absorbs the heat as the temperature increases until it reaches a threshold. This threshold being 29.97°C marks the point where the strength of the house deteriorates exponentially (as shown in left diagram) or the cracks increase exponentially (as shown in the right diagram) as the temperature increases. The thing is, Kenya’s average temperature is 29°C, so half the time the temperature is above that, making the house more vulnerable.
To come up with this graph, there were a few assumptions made from a paper [source] on the strength, toughness, and thermal shock resistance of ancient ceramics. These ceramics referred to clay pottery vessels, but since they are both made from clay (which makes up the mud that majority of houses are build of), this assumption was rational. The first equation was calculated to find the stress on the material from a certain temperature difference:
The variable S can be set to 1 because the shape of the house isn’t globular-shaped, which tends to have a higher thermal shock resistance than those with a sharp angle at the junction between base and wall.
Young’s modulus is the ability of the material to resist elastic deformation under load and retain it’s shape even with the heat. It is calculated as the ratio of tensile stress (σ) to tensile strain (ε). When you have a lower modulus material, strain stresses it and creates the risk of breaking. In this case, values of Young’s modulus for clay vary at around 8 GPa, up to 800 °C (which is obviously the correct temperature in Kenya).
The linear thermal expansion is a material property that shows how much a material expands when heated. This value is 7.2 × 10−6/°C [source].
The Poisson ratio is the ratio of the change in the width per unit width of a material, to the change in its length per unit length because of strength. Another way to think of it is, ‘when you stretch a material in one way, what is the ratio of its deformation to it’s stretch in the other way?’
Think of the material as a rubber band. When you stretch the rubber band the longitudinal strain pulls it outwards and lateral strain pulls it thinner → you find ratio of that, but for clay.
The type of clay used in Kenya is somewhere between medium and stiff since it is used for building, though the info isn’t available because it is locally sources → good assumption is 0.3 [source 1, source 2].
Substituting all these variables into the equation:
After finding the stress, you need to find the ΔT, which is the temperature difference between the outside temperature and the temperature of the material → used to find the threshold at which the material starts cracking.
Substituting the previous variables into the equation with σ being the stress:
You get a temperature difference of 9.97°C. This means (as stated before) that when the temperature is above 29.97°C, the housing in Kenya starts cracking.
Comparing the thermal mass of status quo vs. our solution
status quo
Thermal mass is a property of a material that refers to its ability to absorb, store, and release heat energy. Many homes are made up of iron (82% of roofs and 8% of walls) - the problem with iron is that it has low thermal mass. To quantify this, we used: Thermal mass = Specific heat capacity × Mass.
To calculate the mass of iron needed to cover 125 square meters with a 0.375 mm thickness, we will need to know the density of iron and use Mass = Volume × Density. The density of iron is approximately 7,874 kg/m³. So:
Coverage area: 125 square meters
Thickness: 0.375 mm, which is equal to 0.000375 meters
Volume: Coverage area × Thickness = 125 m² × 0.000375 m = 0.046875 m³
Density of iron: 7,874 kg/m³
Now, using the formula, Mass = 0.046875 m³ × 7,874 kg/m³ ≈ 369.11 kg
Using this and the specific heat capacity, thermal mass of iron = 369.11 kg × 449 J/(kg·K) ≈ 165,667.39 J/K
Thermal mass of iron with a mass of 369.11 kg is approximately 165,667.39 J/K
our solution - the whitewash mixture
To find the heat capacities of our solution:
One part hydrated lime and one part water (total 2 parts) for the whitewash:
Whitewash specific heat capacity = (750 J/(kg·K) + 4,186 J/(kg·K)) / 2 ≈ 2,468 J/(kg·K)
One part sand and four parts whitewash (total 5 parts) for the sand mixture.
The sand mixture has 1/5 part sand and 4/5 part whitewash:
Sand mixture specific heat capacity = (1/5 * 830 J/(kg·K)) + (4/5 * 2,468 J/(kg·K)) ≈ 2,240.4 J/(kg·K)
One part crushed recycled glass and four parts whitewash (total 5 parts) for the glass mixture:
Glass mixture specific heat capacity = (1/5 * 750 J/(kg·K)) + (4/5 * 2,468 J/(kg·K)) ≈ 2,280 J/(kg·K)
To determine the overall specific heat capacity of the final mixture, combine the sand mixture and glass mixture → the final specific heat capacity will lie between the specific heat capacities of the sand mixture (2,240.4 J/K) and the glass mixture (2,280 J/K). Taking the average of this, you get 2260.2 J/K as the specific heat capacity of the whitewash mixture.
Again, to find the thermal mass: Thermal mass = mass × specific heat capacity
To find the mass: Mass = Volume × Density
Coverage area: Let's take the average of 100 and 150 square meters, which is 125 square meters. + Thickness: 2mm = 0.002 meters
Volume = Coverage area × Thickness Volume = 125 m² × 0.002 m ≈ 0.25 m³
To find the density, these are the approximate densities of the individual components:
Hydrated lime (calcium hydroxide): Approximately 2,200 kg/m³
Water: Approximately 1,000 kg/m³
Fine-grained sand (silicon dioxide): Approximately 2,600 kg/m³
Crushed recycled glass (primarily silicon dioxide): Approximately 2,500 kg/m³
Since the whitewash is made of equal parts hydrated lime and water, we can calculate the average density of the whitewash mixture:
Whitewash density ≈ (2,200 kg/m³ + 1,000 kg/m³) / 2 ≈ 1,600 kg/m³
For the sand mixture, we have one part sand and four parts whitewash (total 5 parts). The density of the sand mixture can be approximated by a weighted average based on these proportions:
Sand mixture density ≈ (1/5 * 2,600 kg/m³) + (4/5 * 1,600 kg/m³) ≈ 1,840 kg/m³
For the glass mixture, we have one part crushed recycled glass and four parts whitewash (total 5 parts). The density of the glass mixture can also be approximated by a weighted average based on these proportions:
Glass mixture density ≈ (1/5 * 2,500 kg/m³) + (4/5 * 1,600 kg/m³) ≈ 1,780 kg/m³
The final density will lie between the densities of the sand mixture (1,840 kg/m³) and the glass mixture (1,780 kg/m³). Taking the average of these two densities:
Approximate final mixture density ≈ (1,840 kg/m³ + 1,780 kg/m³) / 2 ≈ 1,810 kg/m³
SO:
Mass = 0.25 m³ × 1,810 kg/m³ ≈ 452.5 kg
Specific Heat Capacity = 2260.2 J/K
Thermal mass = mass × specific heat capacity ≈ 1,022,740.5 J/K
Average thermal mass with our solution ≈ 1,022,740.5 J/K
(note: kelvin equivalent to Celsius is a direct conversation- temperature difference in kelvin and degrees Celsius is the same. This means that 1 K is equal to 1 °C in terms of temperature difference)
OVERALL
If you have the thermal mass of iron as ≈ 165,667.39 J/°C and the thermal mass with solution ≈ 1,022,740.5 J/°C, then:
Percentage growth = ((final value - initial value) / initial value) * 100
Percentage growth = ((1,022,740.5 - 165,667.39) / 165,667.39) * 100
Percentage growth = (857,073.11 / 165,667.39) * 100 ≈ 517.16%
So, the percentage growth from the thermal mass of iron (165,667.39 J/°C) to the thermal mass with the whitewash solution (1,022,740.5 J/°C) is approximately 517.16%.
percentage growth from 165,667.39 to 1,022,940.25 ≈ 517.16% or about 5x increase