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A furnace with an aperture of 20-mm diameter and emis- sive power of 3. 72 x 105 W/m, is used to calibrate a heat flux gage having a sensitive area of 1. 6 × 10-5 m2. (a) At what distance, measured along a normal from the aperture, should the gage be positioned to receive irradiation of 1000 W/m2? (b) If the gage is tilted off normal by 20°, what will be its irradiation?

User Nikhil
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To calculate the distance at which the gage should be positioned to receive irradiation of 1000 W/m², we can use the inverse square law for radiation. The formula is:

Irradiation = Emissive Power / (4 * π * distance^2)

where Irradiation is in W/m², Emissive Power is in W, and distance is in meters.

(a) Let's solve for the distance:

1000 W/m² = 3.72 × 10^5 W/m / (4 * π * distance^2)

Rearranging the equation to solve for distance:

distance^2 = (3.72 × 10^5 W/m) / (4 * π * 1000 W/m²)

distance^2 = 296.94 m²

distance ≈ √(296.94) ≈ 17.23 meters

Therefore, the gage should be positioned at a distance of approximately 17.23 meters from the furnace aperture to receive irradiation of 1000 W/m².

(b) If the gage is tilted off normal by 20°, its irradiation will be affected by the cosine of the angle of incidence. The formula is:

Irradiation = Irradiation_normal * cos(angle)

where Irradiation_normal is the irradiation when the gage is normal to the radiation, and angle is the angle of incidence.

Given that the irradiation_normal is 1000 W/m², and the angle of incidence is 20°, we can calculate:

Irradiation = 1000 W/m² * cos(20°)

Irradiation ≈ 1000 W/m² * 0.9397 ≈ 939.7 W/m²

Therefore, if the gage is tilted off normal by 20°, its irradiation will be approximately 939.7 W/m².

User Chinatsu
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