Q. 215.0( 1 Vote )

# A hollow tube has a length ℓ, inner radius R1 and outer radius R2. The material has a thermal conductivity K. Find the heat flowing through the walls of the tube(a) the flat ends are maintained at temperature T1 and T2(T2> T1)(b) the inside of the tube is maintained at temperature T1 and the outside is maintained at T2.

Given data-

Length = l

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

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 =

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Work done in Thermodynamic ProcessFREE Class
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses