A furnace wall is to be constructed of bricks having a standard thickness of 7.6

Question

A furnace wall is to be constructed of bricks having a standard thickness of 7.6 cm (3 in). Two types of brick material are available. Type 1 has a maximum service temperature of 1100C, thermal conductivity of 1.6 W/m-K, and emissivity of 0.75. Type 2 has a maximum service temperature of 870C, thermal conductivity of 0.7 W/m-K, and emissivity of 0.85. The bricks have the same cost. The inside surface temperature of the furnace wall is 1000C and the outside surface temperature must be no higher than 150C.

a. If the ambient air and surroundings (large) outside the furnace are at 25C and the convection heat transfer coefficient is 8 W/m2 -K, determine the most economical arrangement of the available bricks that meets the requirements.

b. Determine the heat flux (W/m2 ) through the furnace wall and outside surface temperature (C) for your design.

Hint: estimate the heat flux using the maximum outer surface temperature and the known convection and radiation conditions, then determine the required bricks. Adjust if needed

Explanation / Answer

Type 1 bricks Maximum useful temperature (T1, max) = 1100°C

Thermal conductivity (k1) = 1.6 W/(m K)

Type 2 bricks Maximum useful temperature (T2, max) = 870°C

Thermal conductivity (k2) = 0.7 W/(m K) Bricks cost the same

Wall hot side temperature (THOT) = 1000°C and wall cold side temperature (TCOLD) = 150°C

Maximum permissible heat transfer (qmax/A) = h x T = 8 x(25 + 273)

                                                                        = 2384 W/m2

The most economical arrangement for the bricks

One-dimensional, steady state heat transfer conditions Constant thermal conductivities The contact resistance between the bricks is negligible

SKETCH Type 2 Bricks

Type 1 Bricks

Since the type 1 bricks have a higher thermal conductivity at the same cost as the type 2 bricks, the most economical wall would use as few type 1 bricks as possible. However, there should be thick enough layer of type 1 bricks to keep the type 2 bricks at 870°C or less.

For one-dimensional conduction through type 1 bricks (from Equation 1.2)

(Thot – T12) where L1 is the minimum thickness of the type 1 bricks. Solving for L1

L1 = 1 maxk

(THOT – T12)

q = kA T / L

qmax / A = K / L

Therfore L1 =

L1 =

   = 0.606 m

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