Conduction of Heat Through Compound Media

CONDUCTION OF HEAT THROUGH COMPOUND MEDIA

Material bars in series

Let us consider a compound media of two different materials A and B with thermal conductivities K₁ and K₂ and thicknesses x₁ and x₂ (fig. 1.6).

After the steady state is reached,

Amount of heat flowing through the material (A) per second

Q = KA (θ₁ – θ) t / x ₁ ………………….(1)

Amount of heat flowing through the material (B) per second

Q = KA (θ – θ₂) t / x₂ …………………….(2)

The amount of heat flowing through the materials A and B is equal in steady state conditions.

Hence, the eqns (1) & (2) are equal

K₁A (θ₁ – θ) t / x ₁ = K₂A (θ – θ₂) t / x₂ …………….(3)

Rearranging the eqn (3), we have

Rearranging,

This is the expression for interface temperature of two composite slabs in series.

Substituting for θ in equation (1), We get

The amount of heat flowing per second through compound wall of two materials.

This method can also be extended to composite slab with more than two slabs.

In general for any number of slabs, the amount of heat conducted per sec is given by

Materials in parallel

Consider a composite media of two different materials A and B with thermal conductivities K₁ and K₂ and thicknesses x₁ and x₂. They are arranged in parallel as shown in fig. 1.7.

The faces of the material A and B are at temperature e, and the other end faces of A and B are at temperature θ₂. A1  and  A2 are the are  of cross-section of the materials.

Amount of heat flowing through the first material (A) in one second.

Fig. 1.7 Heat conduction through the bodies in parallel

Similarly

Amount of heat flowing through the second material (B) in one second.

Q₂ = K₂ A₂ (θ₁-θ₂) / x_{2………………..(2)}

The total heat flowing through these materials per second and is equal to the sum of Q₁ and Q₂

 Q = Q₁ + Q₂

∴ Amount of heat flowing per second

In general, the net amount of heat flowing per second parallel to the composite slabs is given by

Q  =  (θ₁ - θ₂) Σ KA/ x

Note:

This expression is valid where the heat is conducted through various materials simultaneously under temperature condition.

Problem 1.4

A cooper rod of length 50cm and cross-sectional area 6 x 10^-2 cm ² is connected in sereis with an iron rod of same area of cross-section and length 25 cm. One end of copper is immersed in boiling water. The far end of the iron is in an ice bath of 0°C. Find the rate of transfer of heat from boiling water to ice bath. (Thermal conductivity of copper and iron are 401 Wm^-1 K^-1 and 80 Wm^-1 K^-1 respectively.)

Given

Length of copper rod  x₁ = 50cm

= 50 x10^-2 cm

Length  of the iron rod x₂ = 25cm

= 25 x10^-2 cm

Temperature of ice bath θ₂ = 0^oC

Temperature of boiling water θ1 = 100^oC

Thermal conductivity of copper

K₁ = 401 Wm ^-1K^-1

Thermal conductivity of iron

K₂ = 80 Wm ^-1K^-1

Area of cross section of Copper and Iron

= 6 × 10^-2 cm²

= 6 x 10^-2 x (10^-2 m)2

= 6 x 10^-2 x 10-⁴ m²

Solution:

The rods are connected in series. Therefore, the rate of heat flow from boiling end to ice

Methods to determine thermal Conductivity

The thermal conductivity of a material is determined by various methods

1. Searle's method for good conductors like metallic rods

2. Forbe's method for determining the absolute conductivity of metals.

3. Lee's disc method for bad conductors

4. Radial flow method for bad conductors.

Radial flow method

This is interesting because there is no loss of heat as in the case of other methods. In this method heat will flow from the inner side towards the other side along the radius of the spherical shell and cylindrical shell.

Spherical method is suitable for poor conductors available in powder form, like asbestos, cork, charcoal, clay, sand, etc. In the cylindrical shell method, the thermal conductivity of a thick walled glass tube or rubber tube can be determined by allowing heat to flow radially through the walls of the tube.

Spherical shell method

The material to which, thermal conductivity K has to be found is placed between two thin spherical shells A and B of r₁ radii and r₂ as shown in figure 1.8.

A source of constant heat supply is fixed at the centre of the shells. Heat flows radially across the wall of the shell from inner to outer side. The heat is passed through the material and subsequently lost by emission from the surface of the outer shell.

After a steady state is reached let the temperature of the inner and the outer shells be θ₁ and θ₂.

Imagine spherical surfaces at a distance r and r + dr from O at temperature θ and θ +d θ.

The rate of flow of heat across this shell per second

This quantity of heat is equal to the energy supplied by the source at the centre per second.

Rearranging

We have dr / r2 = -4 Πk/Q dθ…………………..(2)

Q is constant for the values of r as a steady state has reached.

Integrating (2) between the limits r₁ and r₂ and the corresponding temperatures θ₁ and θ₂.

Cylindrical shell method

This is similar to the spherical shell method.

Consider a cylindrical tube of length l, inner radius r₁ and outer radius r₂ as shown in fig. 1.9. The tube carries steam or some hot liquid.

Heat is conducted radially across the walls of the tube. After the steady state is reached, the temperature on the inner surface is θ₁ and on the outer surface is θ ₂.

This thick pipe is imagined to consist of a large number of thin co-axial cylinders of increasing radius. Any such thin imaginary cylinder of the material of thickness 'dr' at a distance r from the axis of the pipe is taken.

Amount of heat flowing per second through this elementary cylinder

Q = - KA dθ / dr  …………….. (1)

Now, surface area of the imaginary cylinder

A = 2π r x l

∴ Q = -2 π r Lk dθ / dr ………….(2)

 After steady state is reached, the amount of heat flowing (Q) through all the imaginary cylinders is same.

Re-arranging the equation (2), we get

Integrating both sides between their proper limits, we have