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Two flat steel sheets, each of 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.?
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Two flat steel sheets, each of 2.52.5 mmmm thickness, are being resist...
The resistance spot welding process involves joining two metal sheets by passing an electrical current through them. In this question, we are given the dimensions of the metal sheets, the current, the weld time, the contact resistance, and the specific energy to melt steel. We are also told that the electrical energy is completely converted to thermal energy.
1. Calculation of electrical power:
- The electrical power can be calculated using the formula P = IV, where P is power, I is current, and V is voltage.
- Since we are given the current (6000 A), we can assume a voltage of 1 V (as the electrical energy is completely converted to thermal energy).
- Therefore, the electrical power is 6000 W.
2. Calculation of energy input:
- The energy input can be calculated using the formula E = Pt, where E is energy, P is power, and t is time.
- Substituting the values, we get E = 6000 W * 0.2 s = 1200 J.
3. Calculation of contact resistance:
- The contact resistance can be calculated using the formula R = ρ * (L/A), where R is resistance, ρ is resistivity, L is length, and A is area.
- Substituting the values, we get R = 200 μΩ * (2.5 mm / (2.5 mm * 2.5 mm)) = 200 μΩ.
4. Calculation of heat generated:
- The heat generated can be calculated using the formula Q = I^2 * R * t, where Q is heat, I is current, R is resistance, and t is time.
- Substituting the values, we get Q = (6000 A)^2 * 200 μΩ * 0.2 s = 1.44 J.
5. Calculation of energy required to melt steel:
- The energy required to melt steel can be calculated using the formula E = V * A * h * ρ, where E is energy, V is volume, A is area, h is height, and ρ is density.
- Substituting the values, we get E = (π * (22 mm)^3 / 6) * 10^9 J/m^3 = 12.1 J.
6. Comparison of heat generated and energy required to melt steel:
- The heat generated (1.44 J) is less than the energy required to melt steel (12.1 J), indicating that the steel will not melt completely.
- However, a spherical melt pool of diameter 4 mm is formed at the interface due to the current flow.
In summary, in resistance spot welding, the electrical energy is converted to thermal energy. The heat generated at the interface between the metal sheets is not sufficient to melt the steel completely, but it forms a spherical melt pool. This process facilitates the joining of the two metal sheets.
1. Calculation of electrical power:
- The electrical power can be calculated using the formula P = IV, where P is power, I is current, and V is voltage.
- Since we are given the current (6000 A), we can assume a voltage of 1 V (as the electrical energy is completely converted to thermal energy).
- Therefore, the electrical power is 6000 W.
2. Calculation of energy input:
- The energy input can be calculated using the formula E = Pt, where E is energy, P is power, and t is time.
- Substituting the values, we get E = 6000 W * 0.2 s = 1200 J.
3. Calculation of contact resistance:
- The contact resistance can be calculated using the formula R = ρ * (L/A), where R is resistance, ρ is resistivity, L is length, and A is area.
- Substituting the values, we get R = 200 μΩ * (2.5 mm / (2.5 mm * 2.5 mm)) = 200 μΩ.
4. Calculation of heat generated:
- The heat generated can be calculated using the formula Q = I^2 * R * t, where Q is heat, I is current, R is resistance, and t is time.
- Substituting the values, we get Q = (6000 A)^2 * 200 μΩ * 0.2 s = 1.44 J.
5. Calculation of energy required to melt steel:
- The energy required to melt steel can be calculated using the formula E = V * A * h * ρ, where E is energy, V is volume, A is area, h is height, and ρ is density.
- Substituting the values, we get E = (π * (22 mm)^3 / 6) * 10^9 J/m^3 = 12.1 J.
6. Comparison of heat generated and energy required to melt steel:
- The heat generated (1.44 J) is less than the energy required to melt steel (12.1 J), indicating that the steel will not melt completely.
- However, a spherical melt pool of diameter 4 mm is formed at the interface due to the current flow.
In summary, in resistance spot welding, the electrical energy is converted to thermal energy. The heat generated at the interface between the metal sheets is not sufficient to melt the steel completely, but it forms a spherical melt pool. This process facilitates the joining of the two metal sheets.
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Two flat steel sheets, each of 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.?
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Two flat steel sheets, each of 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.? 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 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.? 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 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.?.
Two flat steel sheets, each of 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.? 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 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.? 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 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.?.
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Two flat steel sheets, each of 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.?, a detailed solution for Two flat steel sheets, each of 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.? has been provided alongside types of Two flat steel sheets, each of 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.? theory, EduRev gives you an
ample number of questions to practice Two flat steel sheets, each of 2.52.5 mmmm thickness, are being resistance spot welded using a current of 60006000 AA and weld time of 0.20.2 s.s. The contact resistance at the interface between the two sheets is 200μΩ200μΩ and the specific energy to melt steel is 10×109J/m3.10×109J/m3. A spherical melt pool of diameter 44 mmmm is formed at the interface due to the current flow. Consider that electrical energy is completely converted to thermal energy.? tests, examples and also practice GATE tests.
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