A resistance spot-welding operation performed on two pieces of 2.5 mm thick sheet steel uses 12,000 amps for a 0.20 s duration. The electrodes are 6 mm in diameter at the contacting surfaces. Resistance is assumed to be 0.0001  and the resulting weld nugget is 6 mm in diameter and averages 3 mm in thickness. The unit melting energy for the metal Um = 12.0 J/mm3. What portion of the heat generated was used to form the weld nugget, and what portion was dissipated into the work metal, electrodes, and surrounding air?

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A resistance spot-welding operation performed on two pieces of 2.5 mm thick sheet steel uses 12,000 amps for a 0.20 s duration. The electrodes are 6 mm in diameter at the contacting surfaces. Resistance is assumed to be 0.0001  and the resulting weld nugget is 6 mm in diameter and averages 3 mm in thickness. The unit melting energy for the metal Um = 12.0 J/mm3. What portion of the heat generated was used to form the weld nugget, and what portion was dissipated into the work metal, electrodes, and surrounding air?

EXPERT ANSWER

Given data:

Thickness of the work piece is, Z = 2.5 mm

current, I = 12000 A

Time, t = 0.2 s

Diameter of the electrodes, d = 6 mm

Resistance, R = 0.0001 Ohm

diameter of the nugget, dn = 6 mm

thickness of the nugget, Zn = 3 mm

The unit melting energy, q = 12.0 J/mm3

Solution:

Volume of the nugget,

V_{n} = \frac{\pi }{4}\cdot d^{2}_{n}\cdot Z_{n}

Putting the values,

V_{n} = \frac{\pi }{4}* 6^{2}*3

Vn = 84.82 mm3

So, the total melting energy = unit melting energy * Volume of the nugget

Q = q * Vn

Q = 12 * 84.82 = 1017.876 Joules Answer

This is the portion of the consumed energy wivh is used to form the weld nugget.

We know that,

The total consumed heat is,

H = i^{2}\cdot R\cdot t

putting the values,

H = 12000^{2}*0.0001*0.2

H = 2880 Joules

So, the heat dissipated to the work metal, electrodes and surrounding air is,

H_{d} = H-Q

Putting the values,

Hd = 2880 – 1017.876

Hd = 1862.124 Joules Answer

So, the dissipated heat is 1862.124 Joules.