Given data-

Length = l

Inner radius = R₁

Outer radius = R₂

Thermal conductivity = k

The corresponding diagram is shown in the fig.

a. When the flat ends are maintained at temperature T₁ and T₂(T₂> T₁)

Now, the area of cross- section through which heat is flowing is given by –

Area,

A = π (R₂² – R₁ ²) (1)

Let q be the heat, then

Rate of flow of heat (H)–

= (2)

H=

Where

∆T = is change in temperature between the two walls of the tube.

A= Area of cross section of the tube

K = thermal conductivity of the tube

L = length of the tube

Hence

From (1) and (2),

Rate of flow of heat -

H= k × π(R₂² – R₁ ²) × (T₂ - T₁)

b. When the inside of the tube is maintained at temperature T₁ and the outside is maintained at T₂.

Let’s consider a small imaginary

cylinder of radius ”r” of

differential radius “dr” as shown in fig.

Rate of flow of heat (H) –

H = (1)

Where

K = thermal conductivity of the tube

A = area of cross section

L = length of the tube

= change in temperature between the two walls of the tube.

Negative sign since “r” increases, heat decreases.

Since, the cross-section of the tube is in cylindrical form

Hence Curved Surface Area of the Cylinder,

A =2πrl

Where

r = radius of the base

l = length of the tube

From (1), substituting the value of A,

H=

⇒

=

Integrating both sides –

⇒ =

⇒

=

⇒ H =