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.

Answer :


Given data-


21.PNG


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.


work1.PNG


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 =


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