# How much power is required to weld 280 by 20 MM cold rolled steel? How is power related to the cross section of the material escpecially thickness?

This cross section of steel is large and would probably not be a candidate for spot welding. Butt or flash welding would be a better choice. I am not aware of any chart available in the marketplace to offer power requirements or suggested schedules by cross section for this type of material.

If your facility has equipment in house or you can rent time somewhere, trial and error would be my suggestion. Always start trials at low power settings and work up to higher power levels for safety sake. When you begin to get some sort of welding hone in slowly to good results and record all welding conditions.

Another approach might be to calculate the heat required to melt the steel.

HOW MUCH HEAT IS REQUIRED TO BUTT WELD
280 X 20 MM COLD ROLLED STEEL

1. SPECIFIC HEAT
CP =0.49 kJ/ kg K

2. ASSUME 1MM MELTING ON EACH SIDE OF THE JOINT
VOLUME OF MELTING IS 28 CM X 2 CM X 0.2 CM = 11.2 CM3

3. DENSITY OF COLD ROLLED STEEL = 7.85 g/CM3

4. MELTED MASS IS: 7.85 g/CM3 X 11.2 CM3 =87.92 g of melted steel

5. ASSUME THE CHANGE IN TEMPERATURE WAS = 1500 °C

6. CHANGE IN HEAT ΔQ = M X CP X ΔT
87.92 X ((0.49)(KJ)/(kg X K)) X 1500K X Kg/1000G
Q = 64.62 KJ

7. TO THIS WE ADD THE LATENT HEAT OF FUSHION TO CHANGE THE MATERIAL FROM SOLID TO LIQUID.

ASSUME 250 KJ/Kg = L
L = 250 KJ/Kg (87.92 g) (1Kg/1000g)
L = 22 KJ

8. TOTAL HEAT REQUIRED TO MELT CR STEEL Q + L = 64.62 + 22 = 86.62 KJ

9. Q - HEAT IS GENERATED BY

JOULES LAW Q= I2RT = 86.62 KJ

10. R = ρL/A ρ = resistivity of cold rolled steel (material being welded) [Ω cm]

L = distance between centers of die clamps [cm]
A = 28 x 2 (cross section) [cm2]

11. ASSUMMING A RESISTIVITY OF 10-5 Ω cm FOR STEEL AND A 20 cm SPACING OF THE CENTER OF THE DIE CLAMPS, WE HAVE

R = 10-5 Ω cm X 20 cm / 28 CM X 2 CM = 3.57 µ Ω

12. THIS VALUE OF R CAN BE APPLIED TO THE EQUATION IN 9 ABOVE

86.62 kJ= I2RT
86.62 kJ= I2 X 3.57 µ Ω X T
86.62 kJ / 3.57 µ Ω = I2 X T

24.26kJ/µ Ω = I2 X T

13. VALUES OF CURRENT AND TIME CAN NOW BE TRIED TO PRODUCE THE REQUIRED HEAT VALUE.

In the formulas the mass of material being melted do affect the heat being generated. Therefor thicker sections will require more heat.

Reference:  RWMA Resistance Welding Manual 4th Edition

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