Chemical Engineering Exam > Chemical Engineering Questions > A sheet of Fe 1.0 mm thick is exposed to a ox...
Start Learning for Free
A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.
- a)9.87*10-12
- b)9.87*10-13
- c)9.87*10-11
- d)9.87*10-10
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side a...
Most Upvoted Answer
A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side a...
°C. The oxidizing gas reacts with the iron on the surface, forming a layer of iron oxide (FeO). The deoxidizing gas reduces the FeO back to iron (Fe) at the interface between the Fe and FeO layers. This process is known as the reduction of iron oxide.
The thickness of the FeO layer grows with time, as more Fe is oxidized and more FeO is formed. The rate of growth of the FeO layer depends on the concentration of the oxidizing gas, the temperature, and the rate of diffusion of the oxidizing gas through the FeO layer.
The reduction of FeO back to Fe depends on the concentration of the deoxidizing gas, the temperature, and the rate of diffusion of the deoxidizing gas through the FeO layer. The reduction reaction releases heat, which raises the temperature at the interface between the Fe and FeO layers.
The overall process of oxidation and reduction can lead to the formation of complex oxide layers with different chemical compositions and microstructures. These oxide layers can affect the mechanical, electrical, and corrosion properties of the Fe sheet. Therefore, it is important to control the oxidation and reduction processes in order to optimize the performance of the Fe sheet in various applications.
The thickness of the FeO layer grows with time, as more Fe is oxidized and more FeO is formed. The rate of growth of the FeO layer depends on the concentration of the oxidizing gas, the temperature, and the rate of diffusion of the oxidizing gas through the FeO layer.
The reduction of FeO back to Fe depends on the concentration of the deoxidizing gas, the temperature, and the rate of diffusion of the deoxidizing gas through the FeO layer. The reduction reaction releases heat, which raises the temperature at the interface between the Fe and FeO layers.
The overall process of oxidation and reduction can lead to the formation of complex oxide layers with different chemical compositions and microstructures. These oxide layers can affect the mechanical, electrical, and corrosion properties of the Fe sheet. Therefore, it is important to control the oxidation and reduction processes in order to optimize the performance of the Fe sheet in various applications.
|
Explore Courses for Chemical Engineering exam
|
|
Similar Chemical Engineering Doubts
A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. Can you explain this answer?
Question Description
A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. 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 sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. 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 sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. Can you explain this answer?.
A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. 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 sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. 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 sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. Can you explain this answer?.
Solutions for A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. 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 sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A sheet of Fe 1.0 mm thick is exposed to a oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.a)9.87*10-12b)9.87*10-13c)9.87*10-11d)9.87*10-10Correct answer is option 'A'. 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