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A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7 W/mK. For the same heat loss and temperature drop,this can be replaced by a layer of diatomite earth of conductivity 0.14W/mK and thickness
- a)0.05 m
- b)0.1 m
- c)0.2 m
- d)0.5 m
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A furnace is made of a red brick wall of thickness 0.5 m and conductiv...
k1/x1 = k2/x2,
0.7/0.5 = 0.14/x2,
= 0.07/0.7 = 0.1.
Most Upvoted Answer
A furnace is made of a red brick wall of thickness 0.5 m and conductiv...
Temperature drop or heat loss is dependent on thermal resistance offered by the brick material.
We have ,
Length = 0.5 meter
conductivity =. 0.7w/m k
Area = 1 metre^2
Thermal Resistance = L/KA = 0.5/0.7 = 0.71429
Now for thermal conductivity of 0.14 w/m K
Thermal resistance = L/ KA
0.71429 = L/0.14
L = 0.714290*0.14
L = 0.10 meters.
Thank you.
We have ,
Length = 0.5 meter
conductivity =. 0.7w/m k
Area = 1 metre^2
Thermal Resistance = L/KA = 0.5/0.7 = 0.71429
Now for thermal conductivity of 0.14 w/m K
Thermal resistance = L/ KA
0.71429 = L/0.14
L = 0.714290*0.14
L = 0.10 meters.
Thank you.
Hence Option (B) is correct
For Notes on Modes of Heat Transfer click on the link given below
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A furnace is made of a red brick wall of thickness 0.5 m and conductiv...
The Problem
We have a furnace made of a red brick wall, with a thickness of 0.5 m and a thermal conductivity of 0.7 W/mK. We want to replace this brick wall with a layer of diatomite earth, which has a thermal conductivity of 0.14 W/mK. We want to determine the thickness of the diatomite layer that will result in the same heat loss and temperature drop as the original brick wall.
Concepts to Understand
1. Heat conduction: Heat conduction is the transfer of heat through a material by means of atomic or molecular interactions. It occurs when there is a temperature gradient within a solid or between two solids in contact.
2. Fourier's Law of Heat Conduction: Fourier's law states that the rate of heat transfer through a material is directly proportional to the surface area, temperature difference, and inversely proportional to the thickness of the material. It can be expressed as: Q = (k * A * ∆T) / d, where Q is the heat transfer rate, k is the thermal conductivity, A is the surface area, ∆T is the temperature difference, and d is the thickness of the material.
Solution
To determine the thickness of the diatomite layer, we need to equate the heat transfer rate through the brick wall and the diatomite layer.
Step 1: Equate the heat transfer rates
Let Q_brick be the heat transfer rate through the brick wall and Q_diatomite be the heat transfer rate through the diatomite layer.
Q_brick = Q_diatomite
Step 2: Apply Fourier's law of heat conduction
Q_brick = (k_brick * A * ∆T) / d_brick
Q_diatomite = (k_diatomite * A * ∆T) / d_diatomite
where k_brick and k_diatomite are the thermal conductivities of the brick and diatomite, A is the surface area, ∆T is the temperature difference, d_brick is the thickness of the brick wall, and d_diatomite is the thickness of the diatomite layer.
Step 3: Substitute the given values
Q_brick = (0.7 * A * ∆T) / 0.5
Q_diatomite = (0.14 * A * ∆T) / d_diatomite
Step 4: Equate the heat transfer rates
(0.7 * A * ∆T) / 0.5 = (0.14 * A * ∆T) / d_diatomite
Step 5: Solve for d_diatomite
Cross-multiplying and simplifying the equation:
0.7 * d_diatomite = 0.14 * 0.5
d_diatomite = (0.14 * 0.5) / 0.7
d_diatomite ≈ 0.1 m
Therefore, the thickness of the diatomite layer that will result in the same heat loss and temperature drop as the original brick wall is approximately 0.1 m.
We have a furnace made of a red brick wall, with a thickness of 0.5 m and a thermal conductivity of 0.7 W/mK. We want to replace this brick wall with a layer of diatomite earth, which has a thermal conductivity of 0.14 W/mK. We want to determine the thickness of the diatomite layer that will result in the same heat loss and temperature drop as the original brick wall.
Concepts to Understand
1. Heat conduction: Heat conduction is the transfer of heat through a material by means of atomic or molecular interactions. It occurs when there is a temperature gradient within a solid or between two solids in contact.
2. Fourier's Law of Heat Conduction: Fourier's law states that the rate of heat transfer through a material is directly proportional to the surface area, temperature difference, and inversely proportional to the thickness of the material. It can be expressed as: Q = (k * A * ∆T) / d, where Q is the heat transfer rate, k is the thermal conductivity, A is the surface area, ∆T is the temperature difference, and d is the thickness of the material.
Solution
To determine the thickness of the diatomite layer, we need to equate the heat transfer rate through the brick wall and the diatomite layer.
Step 1: Equate the heat transfer rates
Let Q_brick be the heat transfer rate through the brick wall and Q_diatomite be the heat transfer rate through the diatomite layer.
Q_brick = Q_diatomite
Step 2: Apply Fourier's law of heat conduction
Q_brick = (k_brick * A * ∆T) / d_brick
Q_diatomite = (k_diatomite * A * ∆T) / d_diatomite
where k_brick and k_diatomite are the thermal conductivities of the brick and diatomite, A is the surface area, ∆T is the temperature difference, d_brick is the thickness of the brick wall, and d_diatomite is the thickness of the diatomite layer.
Step 3: Substitute the given values
Q_brick = (0.7 * A * ∆T) / 0.5
Q_diatomite = (0.14 * A * ∆T) / d_diatomite
Step 4: Equate the heat transfer rates
(0.7 * A * ∆T) / 0.5 = (0.14 * A * ∆T) / d_diatomite
Step 5: Solve for d_diatomite
Cross-multiplying and simplifying the equation:
0.7 * d_diatomite = 0.14 * 0.5
d_diatomite = (0.14 * 0.5) / 0.7
d_diatomite ≈ 0.1 m
Therefore, the thickness of the diatomite layer that will result in the same heat loss and temperature drop as the original brick wall is approximately 0.1 m.
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A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7 W/mK. For the same heat loss and temperature drop,this can be replaced by a layer of diatomite earth of conductivity 0.14W/mK and thickness a)0.05 mb)0.1 mc)0.2 md)0.5 mCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7 W/mK. For the same heat loss and temperature drop,this can be replaced by a layer of diatomite earth of conductivity 0.14W/mK and thickness a)0.05 mb)0.1 mc)0.2 md)0.5 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7 W/mK. For the same heat loss and temperature drop,this can be replaced by a layer of diatomite earth of conductivity 0.14W/mK and thickness a)0.05 mb)0.1 mc)0.2 md)0.5 mCorrect answer is option 'B'. Can you explain this answer?.
A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7 W/mK. For the same heat loss and temperature drop,this can be replaced by a layer of diatomite earth of conductivity 0.14W/mK and thickness a)0.05 mb)0.1 mc)0.2 md)0.5 mCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7 W/mK. For the same heat loss and temperature drop,this can be replaced by a layer of diatomite earth of conductivity 0.14W/mK and thickness a)0.05 mb)0.1 mc)0.2 md)0.5 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7 W/mK. For the same heat loss and temperature drop,this can be replaced by a layer of diatomite earth of conductivity 0.14W/mK and thickness a)0.05 mb)0.1 mc)0.2 md)0.5 mCorrect answer is option 'B'. Can you explain this answer?.
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