text: Question
  
  # The wall of an industrial furnace is constructed from a fireclay brick layer which is covered...
  
  The wall of an industrial furnace is constructed from a fireclay brick layer which is covered with another layer of ordinary brick. The outer surface wall temperature has a steady-state temperature of 110 °C when the ambient air has a temperature of 20°C. Heat is lost from the outer surface wall via convection and radiation. The internal furnace wall has a surface temperature of 500 DegreeC.
  
  Convection heat transfer coefficient between outer wall and air: 5 W/m2K
  
  Stefan-Boltzmann constant σ: 5.67 × 10-8 W/m2K4
  
  Fireclay wall thickness: 0.1m
  
  Fireclay wall thermal conductivity: 1.7 W/mK
  
  Ordinary brick thermal conductivity: 0.7 W/mK
  
  (a) (i) Assuming that the outside wall is a grey body with emissivity ε= 0.7, calculate the steady state heat flux to the surroundings by radiation, and use this to determine the radiation heat transfer coefficient.
  
  (ii) Taking into account the effects of convection and radiation, determine the overall heat flux from the out side surface to the surroundings.
  
  (b) Calculate the specific resistance of the ordinary brick layer and use this to obtain the temperature at the interface between the ordinary brick and the fireclay brick.
  
  Part a (i)
  
  Q/A = ? x ? x (T4out - T4surr)
  
  = 5.67 × 10^-8 W/m2-K4 x 0.7 x (3834 - 2934) K4
  
  = 561.52 W/m2
  
  Now calculate heat transfer coefficient of radiation
  
  hrad = 561.52/(383-293) = 6.24 W/m2-K
  
  Part a (ii)
  
  overall heat flux from the out side surface to the surroundings by convection and radiation
  
  = h x (T0 - T surr) + 561.52
  
  = 5 x (383 - 293) + 561.52
  
  = 450 + 561.52
  
  = 1011.52 W/m2
  
  Part b
  
  Q = (Ti - T0) / (R1 + R2)
  
  R1 = specific resistance of the fireclay brick layer
  
  = L/k = 0.1/1.7
  
  = 0.0588 m2-K/W
  
  R2 = specific resistance of the ordinary brick layer
  
  1011.52 = (500-110)/(0.0588+R2)
  
  (0.0588+R2) = 0.3855
  
  R2 = 0.3267 m2-K/W
  
  Q = (Tinterface - T0) / R2
  
  Q x R2 = (Tinterface - T0)
  
  T interface = 1011.52 x 0.3267 + 110 = 440.5 °C
  
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explanation: The document contains a detailed math problem related to heat transfer in an industrial furnace wall. It includes multiple parts with questions and corresponding answers. The questions involve calculating heat flux, heat transfer coefficients, and temperatures at different points in the wall.
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  # The wall of an industrial furnace is constructed from a fireclay brick layer which is covered...
  
  The wall of an industrial furnace is constructed from a fireclay brick layer which is covered with another layer of ordinary brick. The outer surface wall temperature has a steady-state temperature of 110 °C when the ambient air has a temperature of 20°C. Heat is lost from the outer surface wall via convection and radiation. The internal furnace wall has a surface temperature of 500 DegreeC.
  
  Convection heat transfer coefficient between outer wall and air: 5 W/m2K
  
  Stefan-Boltzmann constant σ: 5.67 × 10-8 W/m2K4
  
  Fireclay wall thickness: 0.1m
  
  Fireclay wall thermal conductivity: 1.7 W/mK
  
  Ordinary brick thermal conductivity: 0.7 W/mK
  
  (a) (i) Assuming that the outside wall is a grey body with emissivity ε= 0.7, calculate the steady state heat flux to the surroundings by radiation, and use this to determine the radiation heat transfer coefficient.
  
  (ii) Taking into account the effects of convection and radiation, determine the overall heat flux from the out side surface to the surroundings.
  
  (b) Calculate the specific resistance of the ordinary brick layer and use this to obtain the temperature at the interface between the ordinary brick and the fireclay brick.
  
  Part a (i)
  
  Q/A = ? x ? x (T4out - T4surr)
  
  = 5.67 × 10^-8 W/m2-K4 x 0.7 x (3834 - 2934) K4
  
  = 561.52 W/m2
  
  Now calculate heat transfer coefficient of radiation
  
  hrad = 561.52/(383-293) = 6.24 W/m2-K
  
  Part a (ii)
  
  overall heat flux from the out side surface to the surroundings by convection and radiation
  
  = h x (T0 - T surr) + 561.52
  
  = 5 x (383 - 293) + 561.52
  
  = 450 + 561.52
  
  = 1011.52 W/m2
  
  Part b
  
  Q = (Ti - T0) / (R1 + R2)
  
  R1 = specific resistance of the fireclay brick layer
  
  = L/k = 0.1/1.7
  
  = 0.0588 m2-K/W
  
  R2 = specific resistance of the ordinary brick layer
  
  1011.52 = (500-110)/(0.0588+R2)
  
  (0.0588+R2) = 0.3855
  
  R2 = 0.3267 m2-K/W
  
  Q = (Tinterface - T0) / R2
  
  Q x R2 = (Tinterface - T0)
  
  T interface = 1011.52 x 0.3267 + 110 = 440.5 °C
  
  #### Earn Coins
  
  Coins can be redeemed for fabulous gifts.
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distilabel_metadata.raw_output_q_a_detector_0: The document contains a detailed math problem related to heat transfer in an industrial furnace wall. It includes multiple parts with questions and corresponding answers. The questions involve calculating heat flux, heat transfer coefficients, and temperatures at different points in the wall.
  
  Label: multiple-question-and-answer