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Resistance spot welding is performed on two plates of 1.5 mm thickness with 6 mm diameter electrode, using 15000 A current for a time duration of 0.25 seconds. Assuming the interface resistance to be 0.0001 Ω , the heat generated to form the weld is
- a)5625 W-sec
- b)8437 W-sec
- c)22500 W-sec
- d)33750 W-sec
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Resistance spot welding is performed on two plates of 1.5 mm thickness...
H = I2Rt = 150002 × 0.0001× 0.25J = 5625 W − s
Most Upvoted Answer
Resistance spot welding is performed on two plates of 1.5 mm thickness...
Given:
- Thickness of the plates (t) = 1.5 mm
- Diameter of electrode (d) = 6 mm
- Current (I) = 15000 A
- Time (t) = 0.25 seconds
- Interface resistance (R) = 0.0001 ohms
Assumptions:
- The resistance of the plates and the electrode is negligible compared to the interface resistance.
- The heat generated is given by the formula: Heat (Q) = I^2 * R * t
Calculation:
The first step is to calculate the resistance of the weld interface using the formula: Resistance (R) = (ρ * l) / (A)
where:
- ρ is the resistivity of the material
- l is the length of the interface (which can be approximated as the perimeter of the electrode)
- A is the cross-sectional area of the interface (which can be approximated as the area of the electrode)
The resistivity of the material is not given, but assuming it to be constant, we can substitute the values into the formula to get the resistance.
Step 1: Calculating the resistance of the weld interface
- Diameter of the electrode (d) = 6 mm
- Radius of the electrode (r) = d/2 = 6/2 = 3 mm = 0.003 m
- Length of the interface (l) = Perimeter of the electrode = 2πr = 2 * 3.14 * 0.003 = 0.01884 m
- Cross-sectional area of the interface (A) = Area of the electrode = πr^2 = 3.14 * (0.003)^2 = 0.00002826 m^2
- Resistance (R) = (ρ * l) / (A) = 0.0001 ohms (given)
Step 2: Calculating the heat generated
- Heat (Q) = I^2 * R * t = 15000^2 * 0.0001 * 0.25 = 5625 W-sec
Answer:
The heat generated to form the weld is 5625 W-sec (option A).
- Thickness of the plates (t) = 1.5 mm
- Diameter of electrode (d) = 6 mm
- Current (I) = 15000 A
- Time (t) = 0.25 seconds
- Interface resistance (R) = 0.0001 ohms
Assumptions:
- The resistance of the plates and the electrode is negligible compared to the interface resistance.
- The heat generated is given by the formula: Heat (Q) = I^2 * R * t
Calculation:
The first step is to calculate the resistance of the weld interface using the formula: Resistance (R) = (ρ * l) / (A)
where:
- ρ is the resistivity of the material
- l is the length of the interface (which can be approximated as the perimeter of the electrode)
- A is the cross-sectional area of the interface (which can be approximated as the area of the electrode)
The resistivity of the material is not given, but assuming it to be constant, we can substitute the values into the formula to get the resistance.
Step 1: Calculating the resistance of the weld interface
- Diameter of the electrode (d) = 6 mm
- Radius of the electrode (r) = d/2 = 6/2 = 3 mm = 0.003 m
- Length of the interface (l) = Perimeter of the electrode = 2πr = 2 * 3.14 * 0.003 = 0.01884 m
- Cross-sectional area of the interface (A) = Area of the electrode = πr^2 = 3.14 * (0.003)^2 = 0.00002826 m^2
- Resistance (R) = (ρ * l) / (A) = 0.0001 ohms (given)
Step 2: Calculating the heat generated
- Heat (Q) = I^2 * R * t = 15000^2 * 0.0001 * 0.25 = 5625 W-sec
Answer:
The heat generated to form the weld is 5625 W-sec (option A).
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Resistance spot welding is performed on two plates of 1.5 mm thickness with 6 mm diameter electrode, using 15000 A current for a time duration of 0.25 seconds. Assuming the interface resistance to be 0.0001 Ω , the heat generated to form the weld is a)5625 W-sec b)8437 W-sec c)22500 W-sec d)33750 W-secCorrect answer is option 'A'. Can you explain this answer?
Question Description
Resistance spot welding is performed on two plates of 1.5 mm thickness with 6 mm diameter electrode, using 15000 A current for a time duration of 0.25 seconds. Assuming the interface resistance to be 0.0001 Ω , the heat generated to form the weld is a)5625 W-sec b)8437 W-sec c)22500 W-sec d)33750 W-secCorrect answer is option 'A'. 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 Resistance spot welding is performed on two plates of 1.5 mm thickness with 6 mm diameter electrode, using 15000 A current for a time duration of 0.25 seconds. Assuming the interface resistance to be 0.0001 Ω , the heat generated to form the weld is a)5625 W-sec b)8437 W-sec c)22500 W-sec d)33750 W-secCorrect answer is option 'A'. 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 Resistance spot welding is performed on two plates of 1.5 mm thickness with 6 mm diameter electrode, using 15000 A current for a time duration of 0.25 seconds. Assuming the interface resistance to be 0.0001 Ω , the heat generated to form the weld is a)5625 W-sec b)8437 W-sec c)22500 W-sec d)33750 W-secCorrect answer is option 'A'. Can you explain this answer?.
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