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A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.
- a)80
- b)81
Correct answer is between '80,81'. Can you explain this answer?
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Verified Answer
A steel tube of length 20 cm with internal and external diameters of ...
Di = 10 cm,do = 12 cm, ? = 20 cm, T∞ = 10°C
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= 691 cm3
Cooling from 500°C to 100°C
∴ t = 62.12 s
Cooling from 100°C to 30°C
∴ t = 18.35 s
Total time for quenching = 62.12 + 18.38 = 80.5 s
Most Upvoted Answer
A steel tube of length 20 cm with internal and external diameters of ...
And 100°C is 0.5 kW/m2K. Calculate the time required for the centre of the tube to cool from 500°C to 30°C.
To calculate the time required for the center of the tube to cool from 500°C to 30°C, we can use the lumped capacitance method. This method assumes that the temperature at any point within the tube is uniform and can be represented by a single value.
The equation for the lumped capacitance method is:
θ = θi * exp(-Bt)
Where:
θ = temperature difference (θf - θi)
θi = initial temperature difference (θi - θ∞)
B = Biot number
t = time
First, let's calculate the Biot number:
B = h * L / k
Where:
h = heat transfer coefficient
L = characteristic length (in this case, the length of the tube)
k = thermal conductivity of the tube material
For temperatures below 100°C:
B = 1.5 * (0.2) / k
For temperatures above 100°C:
B = 0.5 * (0.2) / k
To find the effective mean value of the heat transfer coefficient between 500°C and 100°C, we can use the logarithmic mean temperature difference (LMTD) method:
1/UA = (1/hi) + (1/ho)
Where:
UA = overall heat transfer coefficient
hi = heat transfer coefficient at the hot side (above 100°C)
ho = heat transfer coefficient at the cold side (below 100°C)
Since we know the values of UA and hi, we can solve for ho:
ho = 1/(1/UA - 1/hi)
Now, we can calculate the time required for the center of the tube to cool from 500°C to 30°C using the lumped capacitance method equation:
θ = (θf - θ∞) * exp(-Bt)
Rearranging the equation:
t = -(1/B) * ln((θ - θ∞) / (θi - θ∞))
Substituting the given values, we can calculate the time required for cooling.
To calculate the time required for the center of the tube to cool from 500°C to 30°C, we can use the lumped capacitance method. This method assumes that the temperature at any point within the tube is uniform and can be represented by a single value.
The equation for the lumped capacitance method is:
θ = θi * exp(-Bt)
Where:
θ = temperature difference (θf - θi)
θi = initial temperature difference (θi - θ∞)
B = Biot number
t = time
First, let's calculate the Biot number:
B = h * L / k
Where:
h = heat transfer coefficient
L = characteristic length (in this case, the length of the tube)
k = thermal conductivity of the tube material
For temperatures below 100°C:
B = 1.5 * (0.2) / k
For temperatures above 100°C:
B = 0.5 * (0.2) / k
To find the effective mean value of the heat transfer coefficient between 500°C and 100°C, we can use the logarithmic mean temperature difference (LMTD) method:
1/UA = (1/hi) + (1/ho)
Where:
UA = overall heat transfer coefficient
hi = heat transfer coefficient at the hot side (above 100°C)
ho = heat transfer coefficient at the cold side (below 100°C)
Since we know the values of UA and hi, we can solve for ho:
ho = 1/(1/UA - 1/hi)
Now, we can calculate the time required for the center of the tube to cool from 500°C to 30°C using the lumped capacitance method equation:
θ = (θf - θ∞) * exp(-Bt)
Rearranging the equation:
t = -(1/B) * ln((θ - θ∞) / (θi - θ∞))
Substituting the given values, we can calculate the time required for cooling.
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A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. Can you explain this answer?
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A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. 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 A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. 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 A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. Can you explain this answer?.
A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. 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 A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. 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 A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. Can you explain this answer?.
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A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. Can you explain this answer?, a detailed solution for A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. Can you explain this answer? has been provided alongside types of A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A steel tube of length 20 cm with internal and external diameters of 10 and 12 cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below 100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing to a film of vapour being produced at the surface, and an effective mean value between 500°C & 100°C is 0.5 kW/m2K.The density of steel is 7500 kg/m3 and the specific heat is 0.47 kJ/kg-K. Neglecting internal thermal resistance of the steel tube, determine the quenching time.a)80b)81Correct answer is between '80,81'. Can you explain this answer? tests, examples and also practice Chemical Engineering tests.
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