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Two flat steel sheets, each of 2.5 mm thickness, are being resistance spot welded using a current of 6000 A and weld time of 0.2 s. The contact resistance at the interface between the two sheets is 200 μ? and the specific energy to melt steel is 10×109 J/m3. A spherical melt pool of diameter 4 mm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy. The ratio of the heat used for melting to the total resistive heat generated is ______
(Important - Enter only the numerical value in the answer)
Correct answer is '0.2327'. Can you explain this answer?
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Two flat steel sheets, each of 2.5 mm thickness, are being resistance ...
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Two flat steel sheets, each of 2.5 mm thickness, are being resistance ...
The contact resistance at the interface between the two sheets is given as 200.
To calculate the resistance of the sheets, we can use the formula:
Resistance = Resistivity x Length / Area
Since the sheets are flat, we can assume they have the same length and width. Let's assume the length and width of the sheets are L and W, respectively.
The area of each sheet can be calculated as:
Area = Length x Width
The thickness of each sheet is given as 2.5 mm, so the length of the sheets is:
Length = Thickness x Number of Sheets = 2.5 mm x 2 = 5 mm = 0.005 m
The width of the sheets is not given, so we'll assume it to be 1 meter for simplicity.
Therefore, the area of each sheet is:
Area = 0.005 m x 1 m = 0.005 m^2
Now we can calculate the resistance of each sheet:
Resistance = Resistivity x Length / Area
The resistivity of steel is typically around 6.9 x 10^-7 ohm-m.
Resistance = (6.9 x 10^-7 ohm-m) x (0.005 m) / (0.005 m^2)
Resistance = 6.9 x 10^-7 ohm
Since there are two sheets being resistance spot welded, the total resistance at the interface between the two sheets is double that:
Total Resistance = 2 x 6.9 x 10^-7 ohm = 1.38 x 10^-6 ohm
Therefore, the total resistance at the interface between the two sheets is 1.38 x 10^-6 ohm.
To calculate the resistance of the sheets, we can use the formula:
Resistance = Resistivity x Length / Area
Since the sheets are flat, we can assume they have the same length and width. Let's assume the length and width of the sheets are L and W, respectively.
The area of each sheet can be calculated as:
Area = Length x Width
The thickness of each sheet is given as 2.5 mm, so the length of the sheets is:
Length = Thickness x Number of Sheets = 2.5 mm x 2 = 5 mm = 0.005 m
The width of the sheets is not given, so we'll assume it to be 1 meter for simplicity.
Therefore, the area of each sheet is:
Area = 0.005 m x 1 m = 0.005 m^2
Now we can calculate the resistance of each sheet:
Resistance = Resistivity x Length / Area
The resistivity of steel is typically around 6.9 x 10^-7 ohm-m.
Resistance = (6.9 x 10^-7 ohm-m) x (0.005 m) / (0.005 m^2)
Resistance = 6.9 x 10^-7 ohm
Since there are two sheets being resistance spot welded, the total resistance at the interface between the two sheets is double that:
Total Resistance = 2 x 6.9 x 10^-7 ohm = 1.38 x 10^-6 ohm
Therefore, the total resistance at the interface between the two sheets is 1.38 x 10^-6 ohm.
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Two flat steel sheets, each of 2.5 mm thickness, are being resistance spot welded using a currentof 6000 A and weld time of 0.2 s. The contact resistance at the interface between the two sheetsis 200 μ?and the specific energy to melt steel is 10×109 J/m3. A spherical melt pool of diameter4 mm is formed at the interface due to the current flow. Consider that electrical energy iscompletely converted to thermal energy. The ratio of the heat used for melting to the totalresistive heat generated is ______(Important - Enter only the numerical value in the answer)Correct answer is '0.2327'. Can you explain this answer?
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Two flat steel sheets, each of 2.5 mm thickness, are being resistance spot welded using a currentof 6000 A and weld time of 0.2 s. The contact resistance at the interface between the two sheetsis 200 μ?and the specific energy to melt steel is 10×109 J/m3. A spherical melt pool of diameter4 mm is formed at the interface due to the current flow. Consider that electrical energy iscompletely converted to thermal energy. The ratio of the heat used for melting to the totalresistive heat generated is ______(Important - Enter only the numerical value in the answer)Correct answer is '0.2327'. 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 Two flat steel sheets, each of 2.5 mm thickness, are being resistance spot welded using a currentof 6000 A and weld time of 0.2 s. The contact resistance at the interface between the two sheetsis 200 μ?and the specific energy to melt steel is 10×109 J/m3. A spherical melt pool of diameter4 mm is formed at the interface due to the current flow. Consider that electrical energy iscompletely converted to thermal energy. The ratio of the heat used for melting to the totalresistive heat generated is ______(Important - Enter only the numerical value in the answer)Correct answer is '0.2327'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two flat steel sheets, each of 2.5 mm thickness, are being resistance spot welded using a currentof 6000 A and weld time of 0.2 s. The contact resistance at the interface between the two sheetsis 200 μ?and the specific energy to melt steel is 10×109 J/m3. A spherical melt pool of diameter4 mm is formed at the interface due to the current flow. Consider that electrical energy iscompletely converted to thermal energy. The ratio of the heat used for melting to the totalresistive heat generated is ______(Important - Enter only the numerical value in the answer)Correct answer is '0.2327'. Can you explain this answer?.
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