Chemical Engineering Exam > Chemical Engineering Questions > A steel tube of length 20 cm with internal a...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
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
View all questions of this test
= 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.
|
Explore Courses for Chemical Engineering exam
|
|
Similar Chemical Engineering Doubts
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for Chemical Engineering.
Download more important topics, notes, lectures and mock test series for Chemical Engineering Exam by signing up for free.
Here you can find the meaning 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? defined & explained in the simplest way possible. Besides giving the explanation 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?, 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.
|
|
Explore Courses for Chemical Engineering exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup