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In BOF steelmaking, 5 metric ton of lime containing 90 wt.% CaO is used to refine 100 metric ton of hot metal containing 93.2 wt.% Fe. The slag produced during refining contains 40 wt.% CaO and 22 wt.% FeO. Neglecting material losses, the yield of Fe (in %) is _________________
Correct answer is between '97.5,98.3'. Can you explain this answer?
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In BOF steelmaking, 5 metric ton of lime containing 90 wt.% CaO is use...
Solution:
Given:
- Lime used = 5 metric ton
- Lime contains 90 wt.% CaO
- Hot metal used = 100 metric ton
- Hot metal contains 93.2 wt.% Fe
- Slag contains 40 wt.% CaO and 22 wt.% FeO
Calculation:
Step 1: Calculation of the amount of CaO required
The amount of CaO required can be calculated by using the following formula:
CaO required = (Amount of hot metal × % of CaO in slag) / % of CaO in lime
= (100 × 40) / 90
= 44.44 metric ton
Therefore, the amount of CaO required is 44.44 metric ton.
Step 2: Calculation of the amount of FeO produced in the slag
The amount of FeO produced in the slag can be calculated as follows:
Amount of FeO produced = (Amount of CaO used × % of FeO in slag) / % of CaO in slag
= (5 × 22) / 40
= 2.75 metric ton
Therefore, the amount of FeO produced in the slag is 2.75 metric ton.
Step 3: Calculation of the yield of Fe
The yield of Fe can be calculated by using the following formula:
Yield of Fe = (Amount of Fe produced × 100) / Amount of hot metal used
= [(100 – % of FeO in slag) × Amount of hot metal used – Amount of FeO produced] / Amount of hot metal used
= [(100 – 22) × 100 – 2.75] / 100
= 97.25 %
Therefore, the yield of Fe is 97.25%.
Final Answer: The yield of Fe (in %) is 97.25%.
Given:
- Lime used = 5 metric ton
- Lime contains 90 wt.% CaO
- Hot metal used = 100 metric ton
- Hot metal contains 93.2 wt.% Fe
- Slag contains 40 wt.% CaO and 22 wt.% FeO
Calculation:
Step 1: Calculation of the amount of CaO required
The amount of CaO required can be calculated by using the following formula:
CaO required = (Amount of hot metal × % of CaO in slag) / % of CaO in lime
= (100 × 40) / 90
= 44.44 metric ton
Therefore, the amount of CaO required is 44.44 metric ton.
Step 2: Calculation of the amount of FeO produced in the slag
The amount of FeO produced in the slag can be calculated as follows:
Amount of FeO produced = (Amount of CaO used × % of FeO in slag) / % of CaO in slag
= (5 × 22) / 40
= 2.75 metric ton
Therefore, the amount of FeO produced in the slag is 2.75 metric ton.
Step 3: Calculation of the yield of Fe
The yield of Fe can be calculated by using the following formula:
Yield of Fe = (Amount of Fe produced × 100) / Amount of hot metal used
= [(100 – % of FeO in slag) × Amount of hot metal used – Amount of FeO produced] / Amount of hot metal used
= [(100 – 22) × 100 – 2.75] / 100
= 97.25 %
Therefore, the yield of Fe is 97.25%.
Final Answer: The yield of Fe (in %) is 97.25%.
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In BOF steelmaking, 5 metric ton of lime containing 90 wt.% CaO is used to refine 100 metric ton of hot metal containing 93.2 wt.% Fe. The slag produced during refining contains 40 wt.% CaO and 22 wt.% FeO. Neglecting material losses, the yield of Fe (in %) is _________________Correct answer is between '97.5,98.3'. Can you explain this answer?
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In BOF steelmaking, 5 metric ton of lime containing 90 wt.% CaO is used to refine 100 metric ton of hot metal containing 93.2 wt.% Fe. The slag produced during refining contains 40 wt.% CaO and 22 wt.% FeO. Neglecting material losses, the yield of Fe (in %) is _________________Correct answer is between '97.5,98.3'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about In BOF steelmaking, 5 metric ton of lime containing 90 wt.% CaO is used to refine 100 metric ton of hot metal containing 93.2 wt.% Fe. The slag produced during refining contains 40 wt.% CaO and 22 wt.% FeO. Neglecting material losses, the yield of Fe (in %) is _________________Correct answer is between '97.5,98.3'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In BOF steelmaking, 5 metric ton of lime containing 90 wt.% CaO is used to refine 100 metric ton of hot metal containing 93.2 wt.% Fe. The slag produced during refining contains 40 wt.% CaO and 22 wt.% FeO. Neglecting material losses, the yield of Fe (in %) is _________________Correct answer is between '97.5,98.3'. Can you explain this answer?.
In BOF steelmaking, 5 metric ton of lime containing 90 wt.% CaO is used to refine 100 metric ton of hot metal containing 93.2 wt.% Fe. The slag produced during refining contains 40 wt.% CaO and 22 wt.% FeO. Neglecting material losses, the yield of Fe (in %) is _________________Correct answer is between '97.5,98.3'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about In BOF steelmaking, 5 metric ton of lime containing 90 wt.% CaO is used to refine 100 metric ton of hot metal containing 93.2 wt.% Fe. The slag produced during refining contains 40 wt.% CaO and 22 wt.% FeO. Neglecting material losses, the yield of Fe (in %) is _________________Correct answer is between '97.5,98.3'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In BOF steelmaking, 5 metric ton of lime containing 90 wt.% CaO is used to refine 100 metric ton of hot metal containing 93.2 wt.% Fe. The slag produced during refining contains 40 wt.% CaO and 22 wt.% FeO. Neglecting material losses, the yield of Fe (in %) is _________________Correct answer is between '97.5,98.3'. Can you explain this answer?.
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