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Assume that the two 2 mm thick steel sheets are being sport welded at a current of 5500 A and current-flow time t= 0.15s. Using electrodes 6 mm in diameter, estimate the amount of heat generated in resistance spot welding.
(Take Rc = 250 μΩ)
(Take Rc = 250 μΩ)
- a)1032 J ’
- b)1120 J
- c)995 J.
- d)1134 J
Correct answer is option 'D'. Can you explain this answer?
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The amount of heat generated in resistance spot welding can be estimated using the formula:
Q = I^2 R t
where Q is the heat generated (in Joules), I is the current (in Amperes), R is the resistance (in Ohms), and t is the time (in seconds).
First, we need to calculate the resistance of the joint. The resistance of a spot weld can be estimated using the formula:
Rc = ρ (t1 + t2) / (A1 + A2)
where Rc is the resistance of the joint (in Ohms), ρ is the resistivity of the material (in Ohm-m), t1 and t2 are the thicknesses of the two sheets (in meters), and A1 and A2 are the cross-sectional areas of the two sheets (in square meters).
Assuming that the resistivity of steel is 2.5 x 10^-7 Ohm-m, we can calculate the cross-sectional area of a 2 mm thick steel sheet with a diameter of 6 mm:
A = π (d/2)^2 = π (6/2)^2 = 28.27 x 10^-6 m^2
Therefore, the total cross-sectional area of the joint is:
A1 + A2 = 2A = 56.54 x 10^-6 m^2
Using the values t1 = t2 = 2 mm = 2 x 10^-3 m and Rc = 250 Ohms, we can solve for the resistivity:
ρ = Rc (A1 + A2) / (t1 + t2) = 250 (56.54 x 10^-6) / (4 x 10^-3) = 3.54 x 10^-6 Ohm-m
Now we can calculate the heat generated:
Q = I^2 R t = (5500)^2 (250) (0.15) = 7.69 x 10^7 Joules
Therefore, the amount of heat generated in resistance spot welding is approximately 76.9 MJ.
Q = I^2 R t
where Q is the heat generated (in Joules), I is the current (in Amperes), R is the resistance (in Ohms), and t is the time (in seconds).
First, we need to calculate the resistance of the joint. The resistance of a spot weld can be estimated using the formula:
Rc = ρ (t1 + t2) / (A1 + A2)
where Rc is the resistance of the joint (in Ohms), ρ is the resistivity of the material (in Ohm-m), t1 and t2 are the thicknesses of the two sheets (in meters), and A1 and A2 are the cross-sectional areas of the two sheets (in square meters).
Assuming that the resistivity of steel is 2.5 x 10^-7 Ohm-m, we can calculate the cross-sectional area of a 2 mm thick steel sheet with a diameter of 6 mm:
A = π (d/2)^2 = π (6/2)^2 = 28.27 x 10^-6 m^2
Therefore, the total cross-sectional area of the joint is:
A1 + A2 = 2A = 56.54 x 10^-6 m^2
Using the values t1 = t2 = 2 mm = 2 x 10^-3 m and Rc = 250 Ohms, we can solve for the resistivity:
ρ = Rc (A1 + A2) / (t1 + t2) = 250 (56.54 x 10^-6) / (4 x 10^-3) = 3.54 x 10^-6 Ohm-m
Now we can calculate the heat generated:
Q = I^2 R t = (5500)^2 (250) (0.15) = 7.69 x 10^7 Joules
Therefore, the amount of heat generated in resistance spot welding is approximately 76.9 MJ.
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Assume that the two 2 mm thick steel sheets are being sport welded at a current of 5500 A and current-flow time t= 0.15s. Using electrodes 6 mm in diameter, estimate the amount of heat generated in resistance spot welding.(Take Rc = 250 μΩ)a)1032 J ’b)1120 Jc)995 J.d)1134 JCorrect answer is option 'D'. Can you explain this answer?
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Assume that the two 2 mm thick steel sheets are being sport welded at a current of 5500 A and current-flow time t= 0.15s. Using electrodes 6 mm in diameter, estimate the amount of heat generated in resistance spot welding.(Take Rc = 250 μΩ)a)1032 J ’b)1120 Jc)995 J.d)1134 JCorrect answer is option 'D'. 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 Assume that the two 2 mm thick steel sheets are being sport welded at a current of 5500 A and current-flow time t= 0.15s. Using electrodes 6 mm in diameter, estimate the amount of heat generated in resistance spot welding.(Take Rc = 250 μΩ)a)1032 J ’b)1120 Jc)995 J.d)1134 JCorrect answer is option 'D'. 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 Assume that the two 2 mm thick steel sheets are being sport welded at a current of 5500 A and current-flow time t= 0.15s. Using electrodes 6 mm in diameter, estimate the amount of heat generated in resistance spot welding.(Take Rc = 250 μΩ)a)1032 J ’b)1120 Jc)995 J.d)1134 JCorrect answer is option 'D'. Can you explain this answer?.
Assume that the two 2 mm thick steel sheets are being sport welded at a current of 5500 A and current-flow time t= 0.15s. Using electrodes 6 mm in diameter, estimate the amount of heat generated in resistance spot welding.(Take Rc = 250 μΩ)a)1032 J ’b)1120 Jc)995 J.d)1134 JCorrect answer is option 'D'. 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 Assume that the two 2 mm thick steel sheets are being sport welded at a current of 5500 A and current-flow time t= 0.15s. Using electrodes 6 mm in diameter, estimate the amount of heat generated in resistance spot welding.(Take Rc = 250 μΩ)a)1032 J ’b)1120 Jc)995 J.d)1134 JCorrect answer is option 'D'. 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 Assume that the two 2 mm thick steel sheets are being sport welded at a current of 5500 A and current-flow time t= 0.15s. Using electrodes 6 mm in diameter, estimate the amount of heat generated in resistance spot welding.(Take Rc = 250 μΩ)a)1032 J ’b)1120 Jc)995 J.d)1134 JCorrect answer is option 'D'. Can you explain this answer?.
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