Download Aptitude Crack
img not found img not found

Chemical Engineering Basics - Engineering

Q1:

One face of a furnace wall is at 1030°C and the other face is exposed to room temperature (30°C). If the thermal conductivity of furnace wall is 3 W . m-1 . k-1 and the wall thickness is 0.3 m, the maximum heat loss (in W/m) is

A 100

B 900

C 9000

D 10000

ANS:D - 10000

To calculate the maximum heat loss through the furnace wall, we can use Fourier's law of heat conduction: Q=dk⋅A⋅ΔT​ Where:

  • Q = Heat transfer rate (W)
  • k = Thermal conductivity of the material (W/m·K)
  • A = Surface area perpendicular to heat flow (m²)
  • ΔT = Temperature difference across the wall (K)
  • d = Thickness of the wall (m)
Given:
  • Thermal conductivity (k) = 3 W/m·K
  • Thickness of the wall (d) = 0.3 m
  • Temperature difference (ΔT) = 1030°−30°=1000°=10001030°C−30°C=1000°C=1000K (since 1°C = 1K)
  • Surface area (A) = Area of one face of the wall (since heat is transferred through one face to the other, we consider only one face) =1×1=1 m2=1×1=1m2
Now, let's calculate the heat transfer rate (Q): Q=0.3m3W/m⋅K×1m2×1000K​ Q=0.3m3000W​=10000W/m So, the maximum heat loss through the furnace wall is 10000 W/m10000W/m. Therefore, the correct answer is 10000 W/m.