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Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will be
- a)600 W/m²
- b)1000 W/m²
- c)1200 W/m²
- d)1800 W/m²
Correct answer is option 'C'. Can you explain this answer?
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Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque s...
The radiosity of a surface is the total amount of radiation leaving the surface per unit area, including both emitted and reflected radiation. It is given by the equation:
Radiosity = Emissive Power + Reflective Power
1. Emissive Power:
Emissive power is the amount of radiation emitted by the surface. In this case, the emissive power is given as 600 W/m².
2. Reflective Power:
The reflective power is the amount of radiation reflected by the surface. Since the surface is opaque, it does not transmit any radiation, so all the incident radiation is either absorbed or reflected. The surface is gray, which means it reflects a portion of the incident radiation. The reflectivity of a gray surface is equal to 1 - emissivity. In this case, the emissivity is given as 0.5, so the reflectivity is 1 - 0.5 = 0.5.
The reflective power can be calculated as:
Reflective Power = Reflectivity * Incident Radiation
3. Incident Radiation:
The incident radiation is the solar radiation falling perpendicularly on the surface. It is given as 1200 W/m².
Calculating the Reflective Power:
Reflective Power = 0.5 * 1200 = 600 W/m²
Calculating the Radiosity:
Radiosity = Emissive Power + Reflective Power
Radiosity = 600 W/m² + 600 W/m²
Radiosity = 1200 W/m²
Therefore, the radiosity of the surface is 1200 W/m², which is option C.
Radiosity = Emissive Power + Reflective Power
1. Emissive Power:
Emissive power is the amount of radiation emitted by the surface. In this case, the emissive power is given as 600 W/m².
2. Reflective Power:
The reflective power is the amount of radiation reflected by the surface. Since the surface is opaque, it does not transmit any radiation, so all the incident radiation is either absorbed or reflected. The surface is gray, which means it reflects a portion of the incident radiation. The reflectivity of a gray surface is equal to 1 - emissivity. In this case, the emissivity is given as 0.5, so the reflectivity is 1 - 0.5 = 0.5.
The reflective power can be calculated as:
Reflective Power = Reflectivity * Incident Radiation
3. Incident Radiation:
The incident radiation is the solar radiation falling perpendicularly on the surface. It is given as 1200 W/m².
Calculating the Reflective Power:
Reflective Power = 0.5 * 1200 = 600 W/m²
Calculating the Radiosity:
Radiosity = Emissive Power + Reflective Power
Radiosity = 600 W/m² + 600 W/m²
Radiosity = 1200 W/m²
Therefore, the radiosity of the surface is 1200 W/m², which is option C.
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Solar radiation of 1200 W/m² falls perpendicularly on a gray opaque surface of emissivity 0.5. If the surface temperature is 50°C and surface emissive power is 600 W/m², the radiosity of that surface will bea)600 W/m²b)1000 W/m²c)1200 W/m²d)1800 W/m²Correct answer is option 'C'. Can you explain this answer?
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