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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The passage contains a detailed problem statement followed by a series of calculations and answers for parts (a) and (b) of the problem.
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Given a text passage, determine:
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First, describe the content structure you found in 1-2 sentences.

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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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The passage contains a detailed problem statement followed by a series of calculations and answers for parts (a) and (b) of the problem.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS