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Consider a room with dimensions as in the sketch. Assuming all emissivities of surfaces are equal to 0.9, calculate the radiative heat transfer between the floor and the ceiling, if you know the surface temperature of the floor equal to 25o C and the surface temperature of the ceiling is equal to 10o C

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Explanation:

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The radiative heat transfer between the floor and the ceiling is 1,534.55 kW.

Given the following data:

  • Emissivity of surfaces = 0.9.
  • Surface temperature of the floor = 25°C °C to K = [tex]273 +25[/tex] = 298 K.
  • Surface temperature of the ceiling = 10°C to K = [tex]273 +10[/tex] = 283 K.
  • View factor = 0.2.
  • Boltzmann's constant = [tex]5.67 \times 10^{-8}[/tex]
  • Length of room = 4.45 m.

How to calculate the radiative heat transfer.

Mathematically, the radiative heat transfer between the floor and the ceiling is given by this formula:

[tex]Q=\epsilon \sigma A(T_2^4-T_1^4)[/tex]

Where:

  • [tex]\sigma[/tex] is Boltzmann's constant.
  • A is the area.
  • [tex]\epsilon[/tex] is the emissivity.

Substituting the given parameters into the formula, we have;

[tex]Q=0.9 \times 5.67 \times 10^{-8} \times 4.45^2 \times (298^4-283^4)\\\\Q=0.00000005103 \times 20.4304 \times 1471902495[/tex]

Q = 1,534.55 kW.

Read more on radiative heat here: https://brainly.com/question/14267608

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