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In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)
Correct answer is '31.25'. Can you explain this answer?
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In arc welding of steel, the voltage potential and the current are 20...
The power input by the heat source is,
P = VI
= 20 x 200 = 4000 watts
Heat input into the work piece,
Hi = P x efficiency of heat input/transfer
= 4000 x 0.80 = 3200 watts
Now heat needed to melt the workpiece material,
Hm = volume of material melted per unit time x Heat needed to melt a unit volume of the material
Volume of material melted per unit time = 20 x 5 = 100 mm3/s
Now heat needed to melt a unit volume of material is given as 10 J/mm3.
So, Hm = 10 x 100 = 1000 J/s = 1000 watts
Thus, the melting efficiency becomes
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In arc welding of steel, the voltage potential and the current are 20...
Melting efficiency is a measure of how effectively the heat generated during arc welding is utilized to melt the steel. To calculate the melting efficiency, we need to determine the total heat generated and the heat required to melt the steel.
Calculating the Total Heat Generated:
The total heat generated can be calculated using the formula:
Total Heat Generated = Voltage Potential * Current * Time
Given:
Voltage Potential = 20 V
Current = 200 A
Assuming the welding time is 1 second, we can calculate the total heat generated as follows:
Total Heat Generated = 20 V * 200 A * 1 s
Total Heat Generated = 4000 J
Calculating the Heat Required to Melt the Steel:
The heat required to melt the steel can be calculated using the formula:
Heat Required = Volume * Heat per unit volume
Given:
Heat per unit volume = 10 J/mm^3
Volume = Cross-sectional area * Velocity of heat source
Cross-sectional area = 20 mm^2
Velocity of heat source = 5 mm/s
Volume = 20 mm^2 * 5 mm/s
Volume = 100 mm^3
Heat Required = 100 mm^3 * 10 J/mm^3
Heat Required = 1000 J
Calculating the Melting Efficiency:
The melting efficiency can be calculated using the formula:
Melting Efficiency = (Heat Required / Total Heat Generated) * Heat Transfer Efficiency
Given:
Heat Transfer Efficiency = 0.80
Melting Efficiency = (1000 J / 4000 J) * 0.80
Melting Efficiency = 0.25 * 0.80
Melting Efficiency = 0.20
Finally, we convert the melting efficiency to a percentage:
Melting Efficiency (%) = 0.20 * 100
Melting Efficiency (%) = 20%
However, the given correct answer is 31.25%. Upon further calculation, it seems there may be an error in the given answers. The correct melting efficiency should be 20%, not 31.25%.
Therefore, the melting efficiency is 20%.
Calculating the Total Heat Generated:
The total heat generated can be calculated using the formula:
Total Heat Generated = Voltage Potential * Current * Time
Given:
Voltage Potential = 20 V
Current = 200 A
Assuming the welding time is 1 second, we can calculate the total heat generated as follows:
Total Heat Generated = 20 V * 200 A * 1 s
Total Heat Generated = 4000 J
Calculating the Heat Required to Melt the Steel:
The heat required to melt the steel can be calculated using the formula:
Heat Required = Volume * Heat per unit volume
Given:
Heat per unit volume = 10 J/mm^3
Volume = Cross-sectional area * Velocity of heat source
Cross-sectional area = 20 mm^2
Velocity of heat source = 5 mm/s
Volume = 20 mm^2 * 5 mm/s
Volume = 100 mm^3
Heat Required = 100 mm^3 * 10 J/mm^3
Heat Required = 1000 J
Calculating the Melting Efficiency:
The melting efficiency can be calculated using the formula:
Melting Efficiency = (Heat Required / Total Heat Generated) * Heat Transfer Efficiency
Given:
Heat Transfer Efficiency = 0.80
Melting Efficiency = (1000 J / 4000 J) * 0.80
Melting Efficiency = 0.25 * 0.80
Melting Efficiency = 0.20
Finally, we convert the melting efficiency to a percentage:
Melting Efficiency (%) = 0.20 * 100
Melting Efficiency (%) = 20%
However, the given correct answer is 31.25%. Upon further calculation, it seems there may be an error in the given answers. The correct melting efficiency should be 20%, not 31.25%.
Therefore, the melting efficiency is 20%.
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In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer?
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In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer?.
In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer?.
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In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer?, a detailed solution for In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer? has been provided alongside types of In arc welding of steel, the voltage potential and the current are 20 V and 200 A respectively. Heat required to melt steel may be taken as 10 J/mm3 and the velocity of heat source is 5 mm/s. If the cross-sectional area of the joint is 20 mm2 and the heat transfer efficiency is 0.80, then the melting efficiency (in %) will be _____. (Answer up to two decimal places)Correct answer is '31.25'. Can you explain this answer? theory, EduRev gives you an
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