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Consider one-dimensional steady state heat conducting along x-axis (0 < x < L), through a plane wall with the boundary surfaces (x = 0 and x = L) maintained at temperature of 0°C and 100°C. Heat is generated uniformly through out the wall.Choose the CORRECT statement
[2013]
- a)The direction of heat transfer will be from the surface at 100°C to the surface of 0°C.
- b)The maximum temperature inside the wall must be greater than 100°C
- c)The temperature distribution is linear within the wall
- d)The temperature distribution is symmetric about the mid-plane of the wall
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Consider one-dimensional steady state heat conducting along x-axis (0 ...
at x = 0, T1 = 0
x = l , T2 = 100
⇒
∴
For maximum temp:
Most Upvoted Answer
Consider one-dimensional steady state heat conducting along x-axis (0 ...
To solve the problem, we can use the one-dimensional heat conduction equation:
q = -kA(dT/dx)
where q is the heat flow rate, k is the thermal conductivity, A is the cross-sectional area, and dT/dx is the temperature gradient.
Since the system is in steady state, the heat flow rate is constant throughout the material. Therefore, we can write:
q = (T1 - T2)/R
where T1 and T2 are the temperatures at the two ends of the material, and R is the thermal resistance given by:
R = L/(kA)
where L is the length of the material.
Setting these two equations equal to each other, we get:
(T1 - T2)/R = -kA(dT/dx)
Rearranging and integrating both sides, we obtain:
(T1 - T2) = -kA/L * ∫dx
(T1 - T2) = -kA/L * (x2 - x1)
where x1 and x2 are the positions of the two ends of the material.
Plugging in the given values, we get:
(T1 - T2) = -40 * (0.01 * 0.01)/(0.1 * 0.001) * (0.02 - 0)
(T1 - T2) = -800 K
Therefore, the temperature difference between the two ends of the material is 800 K, and the temperature at x = 0.01 m is:
T(x=0.01) = T2 + (T1 - T2) * (x/L)
T(x=0.01) = 20 + (-800) * (0.01/0.1)
T(x=0.01) = 12°C
Therefore, the temperature at x = 0.01 m is 12°C.
q = -kA(dT/dx)
where q is the heat flow rate, k is the thermal conductivity, A is the cross-sectional area, and dT/dx is the temperature gradient.
Since the system is in steady state, the heat flow rate is constant throughout the material. Therefore, we can write:
q = (T1 - T2)/R
where T1 and T2 are the temperatures at the two ends of the material, and R is the thermal resistance given by:
R = L/(kA)
where L is the length of the material.
Setting these two equations equal to each other, we get:
(T1 - T2)/R = -kA(dT/dx)
Rearranging and integrating both sides, we obtain:
(T1 - T2) = -kA/L * ∫dx
(T1 - T2) = -kA/L * (x2 - x1)
where x1 and x2 are the positions of the two ends of the material.
Plugging in the given values, we get:
(T1 - T2) = -40 * (0.01 * 0.01)/(0.1 * 0.001) * (0.02 - 0)
(T1 - T2) = -800 K
Therefore, the temperature difference between the two ends of the material is 800 K, and the temperature at x = 0.01 m is:
T(x=0.01) = T2 + (T1 - T2) * (x/L)
T(x=0.01) = 20 + (-800) * (0.01/0.1)
T(x=0.01) = 12°C
Therefore, the temperature at x = 0.01 m is 12°C.
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Consider one-dimensional steady state heat conducting along x-axis (0 < x < L), through a plane wall with the boundary surfaces (x = 0 and x = L) maintained at temperature of 0°C and 100°C. Heat is generated uniformly through out the wall.Choose the CORRECT statement[2013]a)The direction of heat transfer will be from the surface at 100°C to the surface of 0°C.b)The maximum temperature inside the wall must be greater than 100°Cc)The temperature distribution is linear within the walld)The temperature distribution is symmetric about the mid-plane of the wallCorrect answer is option 'B'. Can you explain this answer?
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Consider one-dimensional steady state heat conducting along x-axis (0 < x < L), through a plane wall with the boundary surfaces (x = 0 and x = L) maintained at temperature of 0°C and 100°C. Heat is generated uniformly through out the wall.Choose the CORRECT statement[2013]a)The direction of heat transfer will be from the surface at 100°C to the surface of 0°C.b)The maximum temperature inside the wall must be greater than 100°Cc)The temperature distribution is linear within the walld)The temperature distribution is symmetric about the mid-plane of the wallCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Consider one-dimensional steady state heat conducting along x-axis (0 < x < L), through a plane wall with the boundary surfaces (x = 0 and x = L) maintained at temperature of 0°C and 100°C. Heat is generated uniformly through out the wall.Choose the CORRECT statement[2013]a)The direction of heat transfer will be from the surface at 100°C to the surface of 0°C.b)The maximum temperature inside the wall must be greater than 100°Cc)The temperature distribution is linear within the walld)The temperature distribution is symmetric about the mid-plane of the wallCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider one-dimensional steady state heat conducting along x-axis (0 < x < L), through a plane wall with the boundary surfaces (x = 0 and x = L) maintained at temperature of 0°C and 100°C. Heat is generated uniformly through out the wall.Choose the CORRECT statement[2013]a)The direction of heat transfer will be from the surface at 100°C to the surface of 0°C.b)The maximum temperature inside the wall must be greater than 100°Cc)The temperature distribution is linear within the walld)The temperature distribution is symmetric about the mid-plane of the wallCorrect answer is option 'B'. Can you explain this answer?.
Consider one-dimensional steady state heat conducting along x-axis (0 < x < L), through a plane wall with the boundary surfaces (x = 0 and x = L) maintained at temperature of 0°C and 100°C. Heat is generated uniformly through out the wall.Choose the CORRECT statement[2013]a)The direction of heat transfer will be from the surface at 100°C to the surface of 0°C.b)The maximum temperature inside the wall must be greater than 100°Cc)The temperature distribution is linear within the walld)The temperature distribution is symmetric about the mid-plane of the wallCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Consider one-dimensional steady state heat conducting along x-axis (0 < x < L), through a plane wall with the boundary surfaces (x = 0 and x = L) maintained at temperature of 0°C and 100°C. Heat is generated uniformly through out the wall.Choose the CORRECT statement[2013]a)The direction of heat transfer will be from the surface at 100°C to the surface of 0°C.b)The maximum temperature inside the wall must be greater than 100°Cc)The temperature distribution is linear within the walld)The temperature distribution is symmetric about the mid-plane of the wallCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider one-dimensional steady state heat conducting along x-axis (0 < x < L), through a plane wall with the boundary surfaces (x = 0 and x = L) maintained at temperature of 0°C and 100°C. Heat is generated uniformly through out the wall.Choose the CORRECT statement[2013]a)The direction of heat transfer will be from the surface at 100°C to the surface of 0°C.b)The maximum temperature inside the wall must be greater than 100°Cc)The temperature distribution is linear within the walld)The temperature distribution is symmetric about the mid-plane of the wallCorrect answer is option 'B'. Can you explain this answer?.
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