Steady State of Heat Flow: Thermal Conductivity

STEADY STATE OF HEAT FLOW

Let us take a bar of uniform cross-section and heat it at one end. Each cross-section of the bar receives heat by conduction from the adjacent cross-section towards the heated end.

This heat is spent in three ways; a part is absorbed by the cross-section to increase its temperature, another part escapes out from the sides of the cross-section, and the third is conducted to the next cross-section. The same happens at the next cross-section, and so on.

Thus the temperature of each part of the bar rises, and the bar is said to be in a "variable state".

After some time, however, a state is reached when temperature at each point of the bar becomes stationary. This is called the 'steady state'.

In this state no heat is absorbed by the bar. The heat that reaches any section is transferred to the next, except that some heat may escape out from the sides. (The steady state is not the same as thermal equilibrium, in which all parts of the bar must have the same temperature).

The steady state is tried to be achieved in an experiment in which the heat supplied to a body in a given time is to be measured, such as Searle's experiment for determining the thermal conductivity of a rod.

In this experiment the heat supplied at one end of the rod is determined by measuring the rise in temperature of a known amount of water flowing around the other end.

The measurement is done after the steady state is reached because it is only then that all the heat supplied at one end reaches the other end where it is taken in by water.

Before the steady state, a part of the heat supplied is used up in increasing the temperature of the rod and hence the heat taken in by water is not equal to the heat supplied to the rod.

 

THERMAL CONDUCTIVITY

The ability of a substance to conduct heat energy is measured by its thermal conductivity.

Expression for Thermal Conductivity

Consider a slab of material of length x metre (thickness) and area of cross-section A as shown in fig. 1.5.

One end of the slab is maintained at a higher temperature e1 (hot end) and the other end at a lower temperature 02 (cold end). Now, heat flows from hot end to cold end.

It is found that the amount of heat (Q) conducted from one end to the other end is

• directly proportional to area of cross-section (A). n (A). I

• directly proportional to temperature difference between the ends (θ₁ – θ₂).

• directly proportional to time of conduction (t).

• inversely proportional to length (x) between the faces (Shaded in fig. 1.5)

Combining all these factors, we have

where K is a proportionality constant. It is known as coefficient of thermal conductivity or simply thermal conductivity. Its value depends on the nature of material.

K = Q _x  / A (θ₁-θ₂) t        θ₁ - θ₂  = 1kelvin

x = 1 metre                      t = 1 second

 Then,

K = Q

This condition defines othe coefficient of thermal conductivity.

Definition

It is defined as the amount of heat conducted per second normally across unit area of cross-section of the material per unit temperature difference per unit length.

The  quantity  θ₁ – θ₂ / x denotes the rate of fall temperature with respect to distance. It is known as  temperature gradient.

For the smaller values, θ₁ – θ₂ / x is written as dθ / dx

Rewriting the expression (2), we have is

Q = - KA dθ / dx  t                   ...(3)

 

To indicate dθ / dx  is negative, a negative sign is included in the R.H.S of the equation, since it signifies that the temperature decreases with distance.

Unit:

We know that K = 

Substituting the corresponding units, we have

= joule x metre /  metre² x kelvin x second

= joule / second x metre x kelvin

= watt / metre x kelvin  W /m / K 

= Wm^-1 K^-1

Therefore, the unit of thermal conductivity is Wm^-1 K^-1

 Note

Thermal conductivity denotes the heat heat conducting characteristics of the substances. Generally metals are good conductors of heat (e.g., silver, copper) and non-metals are bad conductors of heat (Air, glass, wood).

Thermal conductivities of some common materials are given in the table. 1.1.

The knowledge about thermal conductivity of the material is very much essential in selecting the materials for suitable engineering applications (design and construction).

 

ANNA UNIVERSITY SOLVED PROBLEMS

Problem 1.1

A rod 0.25 m long and 0.892 x 10^-4 m² area of cross section is heated at one end through 393 K while the other end is kept at 323 K. The quantity of heat which will flow in 15 minutes along the rod is 8.811 x 10 joule. Calculate thermal conductivity of the rod. (A.U Dec 2020)

 

Given data

Area of cross section of the rod (A) = 0.892 × 10^-4 m²

Distance between two ends of the rod (x) = 0.25 m

Temperature difference ( θ₁ – θ₂ ) = 393 K - 323 K = 70 K

Quantity of heat conducted (Q) = 8.811 × 10³ joule

Time of flow of heat (t) = 15 minutes = 15 x 60 = 900 second

 

Solution

We know that K =  Q _x  / A (θ₁ – θ₂) t

Substituting the given values, we have

 K = 8.811 × 10³ × 0.25 / 0.892 x 10^-4 x 70 x 900

Thermal conductivity  K = 392 Wm ^-1K^-1

Problem 1.2

How much heat will be conducted through a slab of area 90 x 10^-4 m² and thickness 1.2 × 10^-3 m in one one second when its opposite faces are maintained at difference in temperature of 20 K. The The coefficient of thermal conductivity of that material is 0.04 Wm^-1 K^-1

 

Given data

Area of the slab A = 90 x 10^-4 m²

Thickness of the slab x = 1.2 x 10^-3 m

Temperature difference (θ₁ – θ₂) = 20 K

Thermal conductivity K = 0.04 Wm ^-1 K-¹

Time taken t = 1 second

 

Solution:

Amount of heat conducted

Q = KA (θ₁ – θ₂) t / x

Substituting the given values, we have

Q = 0.04 × 90 × 10^-4  ×  20 × 1 / 1.2 x 10^-3

Q = 6 joule

Amount of heat conducted in one second = 6 joule.

 

Problem 1.3

The total area of the glass window is 0.5 m². Calculate how much heat is conducted per hour through the glass window if thickness of the glass is 7 × 10 me the temperature of the inside surface is 25°C and of the outside surface is 40°C. Thermal conductivity of glass is 1.0 Wm^{- 1}K^-1    [A.U jan 2020]

 

Given data

Area of glass window A = 0.5 m²

Thickness of the glass x = 7 x 10^-3 m

 θ₁ = 40°C = 273 + 40 = 313 K

θ₂ = 25°C 273 + 25 = 298 K

Thermal conductivity K = 1 Wm^-1 K^-1

t = 1 hour = 60 x 60 second

= 3600 second

 

Solution

Amount of heat conducted

Q = KA (θ₁ – θ₂) t / x

Q =  1 x 0.5 × (313 - 298) × 3600 / 7 x 10^-3

Q  = 27000 / 7 × 10^-3

Q = 3857 × 10³ J

Q = 3.857 × 10⁶ J