During a welding operation, thermal power of 2500 W is incident normally on a metallic surface. As shown in the figure below (figure is NOT to scale), the heated area is circular. Out of the incident power, 85% of the power is absorbed within a circle of radius 5 mm while 65% is absorbed within an inner concentric circle of radius 3 mm. The power density in the shaded area is _________ W mm^-2 (rounded off to 2 decimal places).
Show Hint
To find power density, divide the power in the shaded area by the area of the shaded region. Make sure to use the correct units and convert as needed.
We are given the following:
Total thermal power incident: \( P_{{total}} = 2500 \, {W} \)
85% of the power is absorbed within a circle of radius 5 mm.
65% of the power is absorbed within an inner circle of radius 3 mm. Step 1: Calculate the power absorbed in the circle of radius 5 mm.
Out of the total power, 85% is absorbed within the circle of radius 5 mm:
\[
P_{5{mm}} = 0.85 \times 2500 = 2125 \, {W}
\]
Step 2: Calculate the power absorbed in the inner circle of radius 3 mm.
Out of the total power, 65% is absorbed within the inner circle of radius 3 mm:
\[
P_{3{mm}} = 0.65 \times 2500 = 1625 \, {W}
\]
Step 3: Calculate the power absorbed in the shaded area.
The power absorbed in the shaded area is the difference between the power absorbed in the outer circle and the inner circle:
\[
P_{{shaded}} = P_{5{mm}} - P_{3{mm}} = 2125 - 1625 = 500 \, {W}
\]
Step 4: Calculate the area of the shaded region.
The area of a circle is given by \( A = \pi r^2 \). Therefore:
Area of the outer circle (radius 5 mm):
\[
A_{{outer}} = \pi (5)^2 = 25 \pi \, {mm}^2
\]
Area of the inner circle (radius 3 mm):
\[
A_{{inner}} = \pi (3)^2 = 9 \pi \, {mm}^2
\]
Area of the shaded region:
\[
A_{{shaded}} = A_{{outer}} - A_{{inner}} = 25 \pi - 9 \pi = 16 \pi \, {mm}^2
\]
Approximating \( \pi \approx 3.1416 \), we get:
\[
A_{{shaded}} \approx 16 \times 3.1416 = 50.265 \, {mm}^2
\]
Step 5: Calculate the power density in the shaded area.
The power density is given by the formula:
\[
{Power Density} = \frac{P_{{shaded}}}{A_{{shaded}}} = \frac{500}{50.265} \approx 9.95 \, {W/mm}^2
\]
Thus, the power density in the shaded area is approximately 9.95 W/mmsuperscript{-2}.
