Question

A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K₁ and that of the outer cylinder is K₂. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is:
(A) (K₁ + K₂)/2
(B) (K₁ + K₂)
(C) (2K₁ + 3K₂)/5
(D) (K₁ + 3K₂)/4

Answer 1

Explanation:

A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K₁ and that of the outer cylinder is K₂. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is:

(A) (K₁ + K₂)/2

(B) (K₁ + K₂)

(C) (2K₁ + 3K₂)/5

(D) (K₁ + 3K₂)/4v

Let R₁ = Initial Radius = R  

Let R₂ = Final Radius = 2R  

The two conductors are in parallel. Therefore, equivalent thermal resistance R is :-

1/R = 1/R₁ + 1/R₂ ---------- (1)

R= KA/l by Q = KA (T₁ - T₂)/ l.

R₁= K₁A₁/l₁

R₂= K₂A₂/l₂

Substituting above in (1) we get,

KA/l = K₁A₁/l₁ + K₂A₂/l₂ ---------- (2)

Now, l₁ = l₂ = l

A₁ = ΠR²

A₂ = Π (2R)² - ΠR₂² = 3ΠR₂

A = Π (2R)² = 4ΠR²

Substituting in (2) we get,

K4ΠR²/ l = K₁ΠR²/l + K₂3ΠR₂/l

∴ K = (K₁ + 3K₂) /4