Question

A cylindrical rod of cross-sectional area 5.0mm2 is made by joining a 0.30m rod of silver to a 0.12m rod of nickel. The silver end is maintained at 290K and the nickel end at 440K. given the thermal conductivities of silver and nickel are 0.42kWm-1K-1 and 91Wm-1K-1 respectively, calculate:

the temperature of the join under steady conditions, and
b the rate of conduction of heat down the rod
State any assumptions you need to make.

Answer 1

A=5•10⁻⁶ mm²,
Silver: k₁= 420 W/m•K, L₁= 0.3 m,
Nickel: k₂=91 W/m•K, L₂=0.12 m,
T=? q=?

Q/t=q=k•A•ΔT/L

Resistance of the silver rod:
R₁=L₁/k₁A₁ =0.3/420•5•10⁻⁶=142.9 Ω.
Resistance of the nickel rod:
R₂=L₂/k₂A₂ = 0.12/91•5•10⁻⁶= 263.7 Ω.
Two rods are the conductotrs in series:
R=R1+R2=406.6 Ω.
q= ΔT/R= (440-290)/406.6=0.369 W.

ΔT₁=q•R₁=0.369•142.9=52.7 K
ΔT₂=q•R₂=0.369•263.7=97.3 K
T=ΔT₁+T₁=52.7+290=342.7 K
or
T= T₂-ΔT₂=440-97.3=342.7 K
= >temperature of the join is
T=342.7 K ~343 K,
the rate of conduction of heat is
q= 0.369 W ~ 0.37 W