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For a furnace wall of thickness the thermal conductivity varies with temperature according to the relation
k = (a+bt2)W/m-°C where t is in °C Proceed to calculate the rate of heat transfer through the wall if ? = 0.25 m, t1 = 250°C, t2 = 25°C, a = 0.3 & b = 5 X 10-6 in W/m2
- a)373
- b)375
Correct answer is between '373,375'. Can you explain this answer?
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Verified Answer
For a furnace wall of thickness the thermal conductivity varies with ...
Invoking Fourier’s law of heat conduction
View all questions of this test
Separating the variables and integrating within the prescribed boundary conditions,
we get
which is the required expression Substituting the given data, the rate of heat transfer through the wall works out as
= 374.06 W/m2
Most Upvoted Answer
For a furnace wall of thickness the thermal conductivity varies with ...
To calculate the rate of heat transfer through the wall, we need to use the formula for heat transfer:
Q = (kAΔt)/d
Where:
Q is the rate of heat transfer
k is the thermal conductivity
A is the cross-sectional area of the wall
Δt is the temperature difference across the wall
d is the thickness of the wall
Given:
Δt = t1 - t2 = 250°C - 25°C = 225°C
d = 0.25 m
a = 0.3
To find k, we substitute the given values into the equation for k:
k = a bt^2 = 0.3 * b * t^2
Now we need to integrate k with respect to t to find the constant b:
∫k dt = ∫(a bt^2) dt
∫k dt = ∫(0.3 * b * t^2) dt
kt = (0.3 * b * t^3)/3 + C
Substituting the temperatures t1 and t2 into the equation, we can solve for C:
kt1 = (0.3 * b * t1^3)/3 + C
kt2 = (0.3 * b * t2^3)/3 + C
Since the wall is assumed to be uniformly heated, we can assume that the temperatures t1 and t2 are constant throughout the wall. Therefore, kt1 and kt2 are also constant. Subtracting the two equations:
kt1 - kt2 = (0.3 * b * t1^3)/3 + C - (0.3 * b * t2^3)/3 - C
kt1 - kt2 = (0.3 * b * t1^3)/3 - (0.3 * b * t2^3)/3
Simplifying:
k(t1 - t2) = 0.3 * b * (t1^3 - t2^3)
k = (0.3 * b * (t1^3 - t2^3))/(t1 - t2)
Now we can substitute the given values into the equation for k:
k = (0.3 * b * (250^3 - 25^3))/(250 - 25)
To calculate the rate of heat transfer, we substitute all the values into the formula:
Q = (kAΔt)/d
Q = ((0.3 * b * (250^3 - 25^3))/(250 - 25)) * A * Δt / d
Since we don't have the values for A or b, we cannot calculate the exact rate of heat transfer through the wall. However, we can use this formula once we have the values for A and b.
Q = (kAΔt)/d
Where:
Q is the rate of heat transfer
k is the thermal conductivity
A is the cross-sectional area of the wall
Δt is the temperature difference across the wall
d is the thickness of the wall
Given:
Δt = t1 - t2 = 250°C - 25°C = 225°C
d = 0.25 m
a = 0.3
To find k, we substitute the given values into the equation for k:
k = a bt^2 = 0.3 * b * t^2
Now we need to integrate k with respect to t to find the constant b:
∫k dt = ∫(a bt^2) dt
∫k dt = ∫(0.3 * b * t^2) dt
kt = (0.3 * b * t^3)/3 + C
Substituting the temperatures t1 and t2 into the equation, we can solve for C:
kt1 = (0.3 * b * t1^3)/3 + C
kt2 = (0.3 * b * t2^3)/3 + C
Since the wall is assumed to be uniformly heated, we can assume that the temperatures t1 and t2 are constant throughout the wall. Therefore, kt1 and kt2 are also constant. Subtracting the two equations:
kt1 - kt2 = (0.3 * b * t1^3)/3 + C - (0.3 * b * t2^3)/3 - C
kt1 - kt2 = (0.3 * b * t1^3)/3 - (0.3 * b * t2^3)/3
Simplifying:
k(t1 - t2) = 0.3 * b * (t1^3 - t2^3)
k = (0.3 * b * (t1^3 - t2^3))/(t1 - t2)
Now we can substitute the given values into the equation for k:
k = (0.3 * b * (250^3 - 25^3))/(250 - 25)
To calculate the rate of heat transfer, we substitute all the values into the formula:
Q = (kAΔt)/d
Q = ((0.3 * b * (250^3 - 25^3))/(250 - 25)) * A * Δt / d
Since we don't have the values for A or b, we cannot calculate the exact rate of heat transfer through the wall. However, we can use this formula once we have the values for A and b.
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For a furnace wall of thickness the thermal conductivity varies with temperature according to the relationk = (a+bt2)W/m-°C where t is in °C Proceed to calculate the rate of heat transfer through the wall if ? = 0.25 m, t1 = 250°C, t2 = 25°C, a = 0.3 & b = 5 X 10-6 in W/m2a)373b)375Correct answer is between '373,375'. Can you explain this answer?
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For a furnace wall of thickness the thermal conductivity varies with temperature according to the relationk = (a+bt2)W/m-°C where t is in °C Proceed to calculate the rate of heat transfer through the wall if ? = 0.25 m, t1 = 250°C, t2 = 25°C, a = 0.3 & b = 5 X 10-6 in W/m2a)373b)375Correct answer is between '373,375'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about For a furnace wall of thickness the thermal conductivity varies with temperature according to the relationk = (a+bt2)W/m-°C where t is in °C Proceed to calculate the rate of heat transfer through the wall if ? = 0.25 m, t1 = 250°C, t2 = 25°C, a = 0.3 & b = 5 X 10-6 in W/m2a)373b)375Correct answer is between '373,375'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a furnace wall of thickness the thermal conductivity varies with temperature according to the relationk = (a+bt2)W/m-°C where t is in °C Proceed to calculate the rate of heat transfer through the wall if ? = 0.25 m, t1 = 250°C, t2 = 25°C, a = 0.3 & b = 5 X 10-6 in W/m2a)373b)375Correct answer is between '373,375'. Can you explain this answer?.
For a furnace wall of thickness the thermal conductivity varies with temperature according to the relationk = (a+bt2)W/m-°C where t is in °C Proceed to calculate the rate of heat transfer through the wall if ? = 0.25 m, t1 = 250°C, t2 = 25°C, a = 0.3 & b = 5 X 10-6 in W/m2a)373b)375Correct answer is between '373,375'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about For a furnace wall of thickness the thermal conductivity varies with temperature according to the relationk = (a+bt2)W/m-°C where t is in °C Proceed to calculate the rate of heat transfer through the wall if ? = 0.25 m, t1 = 250°C, t2 = 25°C, a = 0.3 & b = 5 X 10-6 in W/m2a)373b)375Correct answer is between '373,375'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a furnace wall of thickness the thermal conductivity varies with temperature according to the relationk = (a+bt2)W/m-°C where t is in °C Proceed to calculate the rate of heat transfer through the wall if ? = 0.25 m, t1 = 250°C, t2 = 25°C, a = 0.3 & b = 5 X 10-6 in W/m2a)373b)375Correct answer is between '373,375'. Can you explain this answer?.
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