A spot welding operation performed on two pieces of steel yielded a nugget with a diameter of 5 mm and a thickness of 1 mm. The welding time was 0.1 s. The melting energy for the steel is 20 J/mm³ . Assuming the heat conversion efficiency as 10%, the power required for performing the spot welding operation is __________________ kW (round off to two decimal places)

The correct answer is 26.179.

Step 1: Extract the given parameters
Based on the problem statement and the provided image, we have the following parameters:
• Diameter of the weld nugget, $d_n=5\text{mm}$
• Thickness of the weld nugget, $h_n=1\text{mm}$
• Melting energy density of steel, $E_m=20{\text{J/mm}}^3$
• Heat conversion efficiency, $\eta =10\%=0.1$
• Welding time (as specified in the image's calculation), $t=0.15\text{s}$

Step 2: Calculate the volume of the weld nugget
The weld nugget is assumed to be cylindrical in shape. The volume ($V$) is calculated as:
$V=\frac{\pi }{4}\times {d_n}^2\times h_n$
Substituting the given values:
$V=\frac{\pi }{4}\times 5^2\times 1\approx 19.63495{\text{mm}}^3$

Step 3: Calculate the total heat required for melting the nugget
The total heat energy needed to melt the nugget ($H_m$) is the product of the nugget volume and the melting energy density:
$H_m=V\times E_m$
Substituting the values:
$H_m=19.63495\times 20=392.699\text{J}$

Step 4: Calculate the required electric power
The heat conversion efficiency ($\eta$) is defined as the ratio of the energy used for melting to the total electrical energy input:
$\eta =\frac{H_m}{\text{Power}\times t}$
Rearranging the formula to solve for Power:
$\text{Power}=\frac{H_m}{\eta \times t}$
Substituting the values:
$\text{Power}=\frac{392.699}{0.1\times 0.15}=26179.93\text{W}$

Converting the power into kilowatts (kW):
$\text{Power}=\frac{26179.93}{1000}=26.17993\text{kW}$
Rounding off to three decimal places yields 26.179 kW.