Question

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 mmsuperscript{-2 (rounded off to 2 decimal places).}

 

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.

Answer 1

Correct Answer: 9.92

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}.