A cavity wall is made of bricks 0.1m thick with an air space 0.1m thick between them.
a) Assuming the thermal conductivity of brick is 20 times that of air; calculate the thickness of the brick which conducts he same quantity of heat per second per unit area as 0.1m of air.
b) If the thermal conductivity of the brick is 0.5 W m-¹ K-¹, calculate the rate of heat conducted per unit area through the cavity wall when the outside surfaces of the brick walls are respectively 19°C and 4°C
a) The thickness of the brick which conducts the same quantity of heat per second per unit area as 0.1m of air is 0.005m.
b) The rate of heat conducted per unit area through the cavity wall is 0.45 W m-² K-¹.
a) In order to find the thickness of the brick which conducts the same quantity of heat per second per unit area as 0.1m of air, we need to equate the rate of heat conduction through the brick to the rate of heat conduction through the air.
Rate of heat conduction through the brick = Rate of heat conduction through the air
Using Fourier's Law of heat conduction:
(Q/ A)brick = (Q/ A)air
(𝑘𝑏𝑟𝑖𝑐𝑘 × Δ𝑇𝑏𝑟𝑖𝑐𝑘/𝑡𝑏𝑟𝑖𝑐𝑘 × A) = (𝑘𝑎𝑖𝑟 × Δ𝑇𝑎𝑖𝑟/𝑡𝑎𝑖𝑟 × A)
kbrick × Δ𝑇brick/𝑡brick = kair × Δ𝑇air/𝑡air
Given:
𝑑𝑎𝑖𝑟 = 0.1m (thickness of air)
𝑘𝑎𝑖𝑟 = ?
𝑑𝑏𝑟𝑖𝑐𝑘 = ?
𝑘𝑏𝑟𝑖𝑐𝑘 = 20 × 𝑘𝑎𝑖𝑟
From the given data, we can deduce the following relationship:
𝑑𝑎𝑖𝑟 × 𝑘𝑎𝑖𝑟 = 𝑑𝑏𝑟𝑖𝑐𝑘 × 𝑘𝑏𝑟𝑖𝑐𝑘
0.1 × 𝑘𝑎𝑖𝑟 = 𝑑𝑏𝑟𝑖𝑐𝑘 × (20 × 𝑘𝑎𝑖𝑟)
Canceling out 𝑘𝑎𝑖𝑟 on both sides, we get:
0.1 = 20 × 𝑑𝑏𝑟𝑖𝑐𝑘
Dividing both sides by 20, we find:
𝑑𝑏𝑟𝑖𝑐𝑘 = 0.1 / 20
𝑑𝑏𝑟𝑖𝑐𝑘 = 0.005m (thickness of the brick)
Therefore, the thickness of the brick that conducts the same quantity of heat per second per unit area as 0.1m of air is 0.005m.
b) To calculate the rate of heat conducted per unit area through the cavity wall, we can again use Fourier's Law of heat conduction:
Rate of heat conduction (Q/ A) = (𝑘 × Δ𝑇)/𝑡
Given:
𝑑𝑡 = 0.1m (thickness of the cavity wall)
𝑘 = 0.5 W m⁻¹ K⁻¹ (thermal conductivity of the brick)
Δ𝑇 = (19°C - 4°C) = 15°C (temperature difference)
Using the formula:
Rate of heat conduction (Q/ A) = (𝑘 × Δ𝑇)/𝑡
= (0.5 × 15°C)/0.1m
= 7.5 W/m²
Therefore, the rate of heat conducted per unit area through the cavity wall is 7.5 W/m².
a) To find the thickness of the brick that conducts the same quantity of heat per second per unit area as 0.1m of air, we need to compare their respective thermal conductivities.
Let's assume the thickness of the brick is x meters.
The thermal conductivity of air is denoted by k_air, and since the thickness of the air gap is 0.1m, the total resistance of the air gap is:
R_air = thickness_air / thermal_conductivity_air
R_air = 0.1 / k_air
The thermal conductivity of the brick is 20 times that of air, so we can write:
k_brick = 20 * k_air
The resistance of the brick is:
R_brick = thickness_brick / thermal_conductivity_brick
R_brick = x / k_brick
R_brick = x / (20 * k_air)
Now, since the total resistance of the cavity wall is equal to the sum of the individual resistances of the air gap and the brick, we have:
R_total = R_air + R_brick
Given that the thickness of the air gap and its thermal conductivity is the same as the brick (0.1m and 20 * k_air, respectively), we can substitute these values into the equation:
R_total = 0.1 / k_air + x / (20 * k_air)
Now, we want the thermal conductivities to be the same for both the brick and the air gap:
R_total = 0.1 / k_brick + x / (20 * k_brick)
To make the heat conduction the same for both materials, we can equate the two expressions for R_total:
0.1 / k_air + x / (20 * k_air) = 0.1 / k_brick + x / (20 * k_brick)
Multiplying through by 20 * k_air * k_brick:
2 * k_brick + x / k_air = 2 * k_air + x / k_brick
Simplifying:
2 * k_brick - x / k_brick = 2 * k_air - x / k_air
Multiplying through by k_air * k_brick:
2 * k_brick^2 - x^2 = 2 * k_air^2 - x^2
Simplifying:
2 * k_brick^2 = 2 * k_air^2
Dividing by 2:
k_brick^2 = k_air^2
Taking the square root of both sides:
k_brick = k_air
Since the thermal conductivity of the brick is 20 times that of air, this means:
20 * k_air = k_air
Substituting back into the equation for R_total:
R_total = 0.1 / k_brick + x / (20 * k_brick)
R_total = 0.1 / (20 * k_air) + x / (20 * k_air)
Now, we can simplify further:
R_total = 0.005 / k_air + 0.05 / k_air
R_total = (0.005 + 0.05) / k_air
R_total = 0.055 / k_air
Since we're looking for the thickness of the brick that conducts the same amount of heat as the air gap (0.1m), we can set R_total equal to the thickness of the air gap and solve for x:
0.1 = 0.055 / k_air
Solving for k_air:
k_air = 0.055 / 0.1
Now, substituting this value back into the equation for R_total:
R_total = 0.055 / k_air
R_total = 0.055 / (0.055 / 0.1)
R_total = 0.1
Therefore, the thickness of the brick that conducts the same quantity of heat per second per unit area as 0.1m of air is 0.1 meters.
b) To calculate the rate of heat conducted per unit area through the cavity wall, we can use Fourier's Law of Heat Conduction:
Q = k * A * (T2 - T1) / L
Where:
Q = rate of heat conducted per unit area
k = thermal conductivity of the material (0.5 W m-¹ K-¹ for the brick)
A = area through which heat is conducted
T2 = temperature of one side of the wall (19°C)
T1 = temperature of the other side of the wall (4°C)
L = thickness of the wall (0.1m)
Substituting the values into the equation:
Q = 0.5 * A * (19 - 4) / 0.1
As we don't have the value of the area through which heat is conducted, we cannot calculate the exact rate of heat conducted per unit area.
To solve this problem, we need to use the concept of thermal conductivity and heat transfer through materials. Let's go step by step to find the answers to the given questions.
a) To find the brick thickness that conducts the same amount of heat per unit area as 0.1m of air, we can use the formula for heat transfer through materials:
Q = (k * A * ΔT) / d
Where:
Q is the heat conducted per second per unit area,
k is the thermal conductivity of the material,
A is the area of the material through which heat is conducted,
ΔT is the temperature difference across the material,
d is the thickness of the material.
First, we need to find the heat conducted through 0.1m of air per second per unit area. The thermal conductivity of air is generally low, approximately 0.024 W m-¹ K-¹. Therefore:
Q_air = (k_air * A * ΔT_air) / d_air
= (0.024 * 1 * (T1 - T2)) / 0.1
Next, we need to find the equivalent brick thickness that conducts the same amount of heat. The thermal conductivity of brick is given as 20 times that of air, so k_brick = 20 * k_air.
Q_brick = (k_brick * A * ΔT_brick) / d_brick
= (20 * k_air * 1 * (T1 - T2)) / d_brick
To find the equivalent thickness of the brick that carries the same amount of heat, we can equate Q_air and Q_brick, and solve for d_brick:
(0.024 * 1 * (T1 - T2)) / 0.1 = (20 * 0.024 * 1 * (T1 - T2)) / d_brick
By simplifying the equation, the thickness of the brick (d_brick) would be:
d_brick = 0.1 / 20
= 0.005m = 5 mm
Therefore, the thickness of the brick required to conduct the same quantity of heat per second per unit area as 0.1m of air is 5 mm.
b) To calculate the rate of heat conducted through the cavity wall, we can use a similar approach. The formula remains the same:
Q = (k * A * ΔT) / d
Given the thermal conductivity of the brick as 0.5 W m-¹ K-¹, the temperature difference across the wall as (19 - 4) = 15°C, and assuming the area (A) to be 1 m²:
Q_wall = (k_wall * A * ΔT_wall) / d_wall
= (0.5 * 1 * 15) / d_wall
= 7.5 / d_wall
Here, we need the thickness of the cavity wall (d_wall) to calculate the rate of heat conducted through it. However, the thickness of the air space is not provided in the question.
If we consider the thickness of the air space as x meters:
Total thickness of the cavity wall = thickness of brick + thickness of air space
= 0.1 + x meters
Using this information, we can rewrite the equation:
Q_wall = 7.5 / (0.1 + x)
To obtain the exact value for Q_wall, you need to know the thickness of the air space (x).