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A conical cavity of base diameter 15 cm and height 20 cm has inside surface temperature 650 K. If emissivity of each surface is 0.85, determine the net radiative heat transfer from the cavity
- a)168.3 W
- b)158.3 W
- c)148.3 W
- d)138.3 W
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A conical cavity of base diameter 15 cm and height 20 cm has inside su...
Q = E 1 A 1 σ b T 14 [1 – F 11/1 – (1 – E 1) F 11]. Here, F 11 = 0.649 and A 1 = 0.0503 m2.
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A conical cavity of base diameter 15 cm and height 20 cm has inside su...
Given data:
- Base diameter of conical cavity = 15 cm
- Height of conical cavity = 20 cm
- Inside surface temperature = 650 K
- Emissivity of each surface = 0.85
Calculating the net radiative heat transfer:
1. Surface Area of the Conical Cavity:
- The slant height of the conical cavity can be calculated using the Pythagorean theorem:
Slant height (l) = √(r^2 + h^2), where r is the base radius and h is the height.
Given r = 15/2 = 7.5 cm and h = 20 cm, we get l = √(7.5^2 + 20^2) ≈ 21.54 cm
- The surface area of the conical cavity can be calculated as:
A = πrl, where r is the base radius and l is the slant height.
A = π * 7.5 * 21.54 ≈ 404.68 cm^2
2. Radiative Heat Transfer:
- The radiative heat transfer can be calculated using the formula:
Q = ε * σ * A * (T^4 - T_s^4), where ε is the emissivity, σ is the Stefan-Boltzmann constant,
A is the surface area, T is the temperature of the cavity, and T_s is the surroundings temperature.
Given ε = 0.85, σ = 5.67 x 10^-8 W/m^2K^4, T = 650 K, and T_s = 0 K (assuming surroundings is a perfect blackbody), we can calculate Q.
3. Calculating the Net Radiative Heat Transfer:
- Substituting the values in the formula, we get:
Q = 0.85 * 5.67 x 10^-8 * 404.68 * (650^4 - 0^4)
Q ≈ 168.3 W
Therefore, the net radiative heat transfer from the conical cavity is approximately 168.3 W. Hence, the correct answer is option 'A'.
- Base diameter of conical cavity = 15 cm
- Height of conical cavity = 20 cm
- Inside surface temperature = 650 K
- Emissivity of each surface = 0.85
Calculating the net radiative heat transfer:
1. Surface Area of the Conical Cavity:
- The slant height of the conical cavity can be calculated using the Pythagorean theorem:
Slant height (l) = √(r^2 + h^2), where r is the base radius and h is the height.
Given r = 15/2 = 7.5 cm and h = 20 cm, we get l = √(7.5^2 + 20^2) ≈ 21.54 cm
- The surface area of the conical cavity can be calculated as:
A = πrl, where r is the base radius and l is the slant height.
A = π * 7.5 * 21.54 ≈ 404.68 cm^2
2. Radiative Heat Transfer:
- The radiative heat transfer can be calculated using the formula:
Q = ε * σ * A * (T^4 - T_s^4), where ε is the emissivity, σ is the Stefan-Boltzmann constant,
A is the surface area, T is the temperature of the cavity, and T_s is the surroundings temperature.
Given ε = 0.85, σ = 5.67 x 10^-8 W/m^2K^4, T = 650 K, and T_s = 0 K (assuming surroundings is a perfect blackbody), we can calculate Q.
3. Calculating the Net Radiative Heat Transfer:
- Substituting the values in the formula, we get:
Q = 0.85 * 5.67 x 10^-8 * 404.68 * (650^4 - 0^4)
Q ≈ 168.3 W
Therefore, the net radiative heat transfer from the conical cavity is approximately 168.3 W. Hence, the correct answer is option 'A'.
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A conical cavity of base diameter 15 cm and height 20 cm has inside surface temperature 650 K. If emissivity of each surface is 0.85, determine the net radiative heat transfer from the cavitya)168.3 Wb)158.3 Wc)148.3 Wd)138.3 WCorrect answer is option 'A'. Can you explain this answer?
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