The resistance spot welding of two 1.55 mm thick metal sheets is performed using welding current of 10000 A for 0.25 s. The contact resistance at the interface of the metal sheets is 0.0001 Ω. The volume of weld nugget formed after welding is 70 mm³ . Considering the heat required to melt unit volume of metal is 12 J/mm³ , the thermal efficiency of the welding process is ________ % (round off to one decimal place).

The correct answer is 33.6.

To find the thermal efficiency of the resistance spot welding process, we need to compare the heat actually utilized to melt the weld nugget with the total electrical heat supplied to the interface.

Step 1: Calculate the total heat supplied (H_s)
The heat generated due to the flow of welding current through the contact resistance is given by Joule's law of heating:

$$H_s=I^2\cdot R\cdot t$$

Where:
- Current (I) = 10000 A
- Contact resistance (R) = 0.0001 Ω
- Time (t) = 0.25 s

Substituting these values:

$$H_s={\left(10000\right)}^2\cdot 0.0001\cdot 0.25$$

$$H_s=100000000\cdot 0.0001\cdot 0.25$$

$$H_s=10000\cdot 0.25=2500\text{J}$$

Step 2: Calculate the heat required for melting (H_m)
The heat needed to melt the volume of the weld nugget is:

$$H_m=\text{Melting energy per unit volume}\cdot \text{Volume of weld nugget}$$

Given:
- Melting energy per unit volume = 12 J/mm³
- Volume of weld nugget = 70 mm³

Substituting these values:

$$H_m=12\cdot 70=840\text{J}$$

Step 3: Calculate the thermal efficiency (η)
The thermal efficiency is the ratio of the heat required for melting to the total heat supplied:

$$\eta =\frac{H_m}{H_s}$$

$$\eta =\frac{840}{2500}$$

$$\eta =0.336$$

Expressed as a percentage:

$$\eta =0.336\cdot 100=33.6\%$$