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Chemical Engineering Basics - Engineering

Q1:

A furnace is made of a refractory brick wall of thickness 0.5 metre and thermal conductivity 0.7 W/m.°K For the same temperature drop and heat loss, this refractory wall can be replaced by a layer of diatomaceous earth of thermal conductivity 0.14 W/m.K and thickness __________ metre.

A 0.01

B 0.1

C 0.25

D 0.5

ANS:B - 0.1

To find the thickness of the diatomaceous earth layer that would provide the same heat loss as the refractory brick wall, we can use Fourier's Law of Heat Conduction: Q=tk⋅A⋅ΔT​ where:

  • Q = heat loss (constant)
  • k = thermal conductivity of the material
  • A = surface area perpendicular to the direction of heat flow
  • ΔT = temperature difference across the material
  • t = thickness of the material
Since the heat loss (Q), temperature difference (ΔT), and surface area (A) are constant, we can set up a proportionality between the thicknesses of the refractory brick wall (t1​) and the diatomaceous earth layer (t2​): ​k1​​=t2​k2​​ Given:
  • Thermal conductivity of refractory brick wall (k1​) = 0.7 W/m·K
  • Thermal conductivity of diatomaceous earth (k2​) = 0.14 W/m·K
  • Thickness of refractory brick wall (t1​) = 0.5 m
We need to find t2​. t2​=k1​k2​⋅t1​​ t2​=0.7W/m\cdotpK0.14W/m\cdotpK×0.5m​ t2​=0.7W/m\cdotpK0.07m2/s​ t2​=0.1m So, the thickness of the diatomaceous earth layer that would provide the same heat loss as the refractory brick wall is 0.1 m0.1m.