HEAT TRANSFER

Heat can be transferred from one place to the other by any of three possible ways: conduction, convection and radiation. In the conduction, convection processes, a medium is necessary for the heat transfer. Radiation, however, does no have this restriction. This is also the fastest mode of heat transfer, in which heat is transferred from one place to the other in the form of electromagnetic radiation.

(i) Conduction

Figure shows a rod whose ends are in thermal contact with a hot reservoir at temperature T₁ and a cold reservoir at temperature T₂. The sides of the rod are covered with insulating medium, so the transport of heat is along the rod, not through the sides. The molecules at the hot reservoir have greater vibrational energy. This energy is transferred by collisions to the atoms at the end face of the rod. These atoms in turn transfer of heat through a substance in which heat is transported without direct mass transport is called conduction.

Most metals use another, more effective mechanism to conduct heat. The free electrons, which move throughout the metal can rapidly carry energy from the hotter to cooler regions, so metals are generally good conductors of heat. The presence of 'free' electrons also causes most metals to be good electrical conductors. A metal rod at 5°C feels colder than a piece of wood at 5°C because heat can flow more easily from your hand into the metal.

Heat transfer occurs only between regions that are at different temperature, and the rate of heat flow is $\frac{dQ}{dt}$. This rate is also called the heat current, denoted by H. Experiments show that the heat current is proportional to the cross-section area A of the rod and to the temperature gradient $\frac{dT}{dx}$, which is the rate of change of temperature with distance along the bar. In general

H = $\frac{dQ}{dt}=-kA\frac{dT}{dx}$ … (1)

The negative sign is used to make $\frac{dQ}{dt}$ a positive quantity since $\frac{dT}{dx}$ is negative. The constant k, called the thermal conductivity is measure of the ability of a material to conduct heat.

A substance with a large thermal conductivity k is a good heat conductor. The value of k depends on the temperature, increasing slightly with increasing temperature, but k can be taken to be practically constant throughout a substance if the temperature difference between its ends is not too great.

Let us apply Eq. (1) to a rod of length L and constant cross sectional area A in which a steady state has been reached. In a steady state the temperature at each point is constant in time. Hence,

$-\frac{dT}{dx}=T_1-T_2$

Therefore, the heat $\Delta$Q transferred in time $\Delta$t is

$\Delta Q=kA\left(\frac{T_1-T_2}{L}\right)\Delta t$ …(2)

Thermal Resistance (R)

Eq. (2) in differential form can be written as

$\frac{dQ}{dt}=H=\frac{\Delta T}{l/kA}=\frac{\Delta T}{R}$ …(3)

Here, $\Delta$T = temperature difference (T.D) and

$R=\frac{l}{kA}$ = thermal resistance of the rod.

(ii) Convection

Although conduction does occur in liquids and gases also, heat is transported in these media mostly by convection. In this process, the actual motion of the material is responsible for the heat transfer. Familiar examples include hot-air and hot-water home heating systems, the cooling system of an automobile engine and the flow of blood in the body.

You probably have warmed your hands by holding them over an open flame. In this situation, the air directly above the flame is heated and expands. As a result, the density of this air decreases and then air rises. When the movement results from differences in density, as with air around free, it is referred to as natural convection. Air flow at a beach is an Illustration of natural convection. When the heated substance is forced to move by a fan or pump, the process in called forced convection. If it were not for convection currents, it would be very difficult to boil water. As water is heated in a kettle, the heated water expands and rises to the top because its density is lowered. As the same time, the denser, cool water at the surface sinks to the bottom of the kettle and is heated. Heating a room by a radiator is an Illustration of forced convection.

Ingen Hausz Experiment:

Ingen Hausz provided a method to compare the thermal conductivities of different materials. He took wax coated rods of different materials but of the same area. One end of the rods is kept in a hot water bath and the other end is kept at the temperature of surroundings. If $\ell$₁, $\ell$₂, $\ell$₃ . . . represent the lengths upto which the wax has melted and K₁, K₂, K₃ . . . are their thermal conductivities, then

$\frac{K_1}{{\ell }_1^2}=\frac{K_2}{{\ell }_2^2}$ = . . . = constant

or $\frac{K_1}{K_2}=\frac{{\ell }_1^2}{{\ell }_2^2}$

Illustration 1. A wall is made of two equally thick layers A and B of different materials. The thermal conductivity of A is twice that of B. In the steady state, the temperature difference across the wall is 36°C. The temperature difference across the layer A will be

Solution: $K\Delta \theta =cons\tan t$

\begin{align}  \therefore \,\,\,\dfrac{\Delta {{\theta }_{A}}}{\Delta {{\theta }_{B}}}=\dfrac{K}{2K}=\dfrac{1}{2} \\  \therefore \,\,\,\Delta {{\theta }_{A}}=\left( \dfrac{1}{2+1} \right)36=12{}^\circ C \\ \end{align}

(iii) Radiation

The third means of energy transfer is radiation which does not require a medium. The best known Illustration of this process is the radiation from sun. All objects radiate energy continuously in the form of electromagnetic waves.

$\left(Energy=\frac{1}{2}{\epsilon }_0{E}^2\right)$ The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. This is known as the Stefan's law and is expressed in equation form as

P = $\sigma$AeT⁴

Here P is the power in watts (J/s) radiated by the object, A is the surface area in m², e is a fraction between 0 and 1 called the emissivity of the object and $\sigma$ is a universal constant called Stefan's constant, which has the value

$\sigma$ = 5.67 x 10^–8 W/m²-K⁴

BLACK BODY RADIATION

(i) Perfectly black body

A body that absorbs all the radiation incident upon it and has an emissivity equal to 1 is called a perfectly black body. A black body is also an ideal radiator. It implies that if a black body and an identical another body are kept at the same temperature, then the black body will radiate maximum power as is obvious from equation p = east⁴ also. Because e = 1 for a perfectly black body while for any other body e < 1.

Materials like black velvet or lamp black come close to being ideal black bodies, but the best practical realization of an ideal black body is a small hole leading into a cavity, as this

Absorbs 98% of the radiation incident on them.

(ii) Absorptive power 'a'

"It is defined as the ratio of the radiant energy absorbed by it in a given time to the total radiant energy incident on it in the same interval of time."

$a=\frac{energyabsorbed}{energyincident}$

As a perfectly black body absorbs all radiations incident on it, the absorptive power of a perfectly black body is maximum and unity.

(iii) Spectral absorptive power 'e'

The absorptive power 'a' refers to radiations of all wavelengths (or the total energy) while the spectral absorptive power is the ratio of radiant energy absorbed by a surface to the radiant energy incident on it for a particular wavelength $\lambda$. It may have different values for different wavelengths for a given surface. Let us take an example, suppose a = 0.6, a_$\lambda$ = 0.4 for 1000 Å and a_$\lambda$ = 0.7 for 2000 Å for a given surface. Then it means that this surface will absorbs only 60% of the total radiant energy incident on it. Similarly it absorbs 40% of the energy incident on it corresponding to 1000 Å and 70% corresponding to 2000 Å. The spectral absorptive power a_$\lambda$ is related to absorptive power a through the relation

$a={\int }_0^{\infty }{a}_{\lambda }d\lambda$

(iv) Emissive power 'e'

(Don't confuse it with the emissivity e which is different from it, although both have the same symbols e).

"For a given surface it is defined as the radiant energy emitted per second per unit area of the surface." It has the units of W/m² or J/s–m². For a black body e=$\sigma$T⁴.

(v) Spectral emissive power 'e_$\lambda$'

"It is emissive power for a particular wavelength $\lambda$." Thus,

$e={\int }_0^{\infty }{e}_{\lambda }d\lambda$

Kirchhoff's law: "According to this law the ratio of emissive power to absorptive power is same for all surfaces at the same temperature."

Fig.

Hence, $\frac{e_1}{a_1}=\frac{e_2}{a_2}=\left(\frac{e}{a}\right)$ $Perfectlyblackbody$

but (a)_{black body} = 1

and (e)_{black body} = E (say)

Then, ${\left(\frac{e}{a}\right)}_{foranysurface}=cons\tan t=E$

Similarly for a particular wavelength $\lambda$,

${\left(\frac{e_{\lambda }}{a_{\lambda }}\right)}_{foranybody}=E_{\lambda }$

Here E = emissive power of black body at temperature T

= $\sigma$T⁴

From the above expression, we can see that

e$\lambda$ $\propto$ a$\lambda$

i.e., good absorbers for a particular wavelength are also good emitters of the same wavelength.

COOLING BY RADIATION

Consider a hot body at temperature T placed in an environment at a lower temperature T₀. The body emits more radiation than it absorbs and cools down while the surrounding absorb radiation from the body and warm up. The body is losing energy by emitting radiations at a rate.

P₁ = eA$\sigma$T⁴

and is receiving energy by absorbing radiations at a rate

P₂ = aA$\sigma$T₀⁴

Here 'a' is a pure number between 0 and 1 indicating the relative ability of the surface to absorb radiation from its surroundings. Note that this 'a' is different from the absorptive power 'a'. In thermal equilibrium, both the body and the surrounding have the same temperature (say T_c) and,

P₁ = P₂

or eA$\sigma$T_c⁴ = aA$\sigma$T_c⁴

or e = a

Thus, when T > T₀, the net rate of heat from the body to the surroundings is,

$\frac{dQ}{dt}=$eA$\sigma$ (T₄ – T₀⁴)

or $mc\left(-\frac{dT}{dt}\right)=$eA$\sigma$ (T⁴ – T₀⁴)

$\Rightarrow$ Rate of cooling

$\left(-\frac{dT}{dt}\right)=\frac{eA\sigma }{mc}(T^4-{T_0}^4)$

or $-\frac{dT}{dt}$ $\propto$ (T⁴ – T₀⁴)