Project ForgeSign up free
Heat Transfer
Conduction
GATE 2009
Hard

For a composite wall (Case 1: layers P then Q, thickness L for P, hot side 500°C, interface 180°C, cold side 20°C), the steady state interface temperature is 180°C. If the thickness of layer P is doubled (Case 2, thickness 2L), then the rate of heat transfer (assuming 1-D conduction) is reduced by

  • A20%
  • 40%
  • C50%
  • D70%

Explanation

In Case 1, ΔTP=500−180=320°C and ΔTQ=180−20=160°C flow through the same Q1, so RP=320/Q1, RQ=160/Q1, giving RP=2RQ. Total R1=3RQ, Q1=480/(3RQ)=160/RQ. Doubling P's thickness doubles RP to 4RQ; total R2=5RQ, Q2=480/(5RQ)=96/RQ. Reduction=(160−96)/160=40%.

Source: GATE CH 2009 (2009)

This question is part of a 15-day revision cycle — sign in and Project Forge will bring it back the day before you'd forget it.

Start revising free