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The literature on heat transfer generally recognizes three distinct modes of heat transmission; conduction, convection and radiation. These three modes are similar in that a temperature differential must exist and the heat exchange is in the direction of decreasing temperature. Each method has, however, different physical picture and different controlling laws.

**Mode # 1. Conduction: **

Thermal conduction is a mechanism of heat propagation from a region of higher temperature to a region of low temperature within a medium (solid, liquid or gaseous) or between different mediums in direct physical contact. Conduction does not involve any movement of macroscopic portions of matter relative to one another.

The thermal energy may be transferred by means of electrons which are free to move through the lattice structure of the material. In addition, or alternatively, it may be transferred as vibrational energy in the lattice structure. Irrespective of the exact mechanism, the observable effect of conduction is an equalization of temperature.

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Consider the flow of heat along a metal rod, one end of which is placed adjacent to a flame. The elementary particles (molecules, atoms, electrons) composing the rod, and which are in immediate vicinity of the flame, get heated.

Because of the resulting temperature growth, their kinetic energy increases and this puts them in a violent state of agitation, and they start vibrating about their mean positions. Consequently, these more active particles collide with less active molecules lying next to them. During collision, the less active particles also get excited, i.e., thermal energy is imparted to them.

The process is repeated for layer after layer of molecules until the other end of the rod is reached. Each layer of molecules is at a slightly higher temperature than the preceding one, i.e., a temperature gradient exists along the length of the rod. The rate of heat flow between the two ends depends upon the length of the rod, temperature difference between the two ends, and the physical and chemical composition of the bar material.

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**The rate equation for one-dimensional steady flow of heat by conduction is prescribed by the Fourier Law: **

Q = – kA dt/dx …. (1.1)

Where, Q is the heat transfer rate, A is the area of heat transfer surface, dt is the temperature difference for a short perpendicular distance dx, and the thermal conductivity k is a characteristic of the surface material. Since the temperature gradient is negative in the positive x-direction, the minus sign in the equation gives positive heat flow.

If δ is the path length in the direction of heat flow and (t_{1} – t_{2}) is the temperature difference, then-

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Q = kA (t_{1} – t_{2})/δ … (1.2)

The heat flux q is the heat conducted per unit time per unit area and is given by

q = Q/A = k (t_{1} – t_{2})/δ …. (1.3)

Heat transfer in metal rods, in heat treatment of steel forgings and through the walls of heat exchange equipment are some practical examples of heat conduction.

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**Example 1:**

A 7.5 cm thick side wall of an oven is primarily made of insulation with a thermal conductivity of 0.04 W/mK. Conditions on the inside of wall fix the temperature on that side at 420 K. The electric coils within the oven dissipate 36.5 watts of electrical energy to make up for the heat loss through the wall. Calculate the wall surface area, perpendicular to heat flow, so that temperature on the other side of the wall does not exceed 310 K.

**Solution: **

Under the stipulations of one-dimensional steady state heat conduction, the electrical energy dissipation rate within the oven must equal the conduction heat transfer rate across the wall.

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that is-

Q = kA (t_{1} – t_{2})/δ

36.5 = 0.04 A (420 – 310)/0.75 = 5.87 A

Hence, the required wall surface area,

A = 36.5/5.87 = 6.218 m^{2}

**Example 2: **

A plane wall of 10 cm thickness and 3 m^{2} area is made of a material whose conductivity is 8.5 W/mK. The temperatures of the wall surfaces are steady at 100°C and 30°C respectively. Find the temperature gradient and heat flow across the wall.

**Solution: **

The temperature gradient in the direction of heat flow is-

dt/dx = t_{2} – t_{1}/δ = 30 – 100/0.1 = – 700°C/m

(b) Heat flow across the wall is given by Fourier’s heat conduction equation-

Q = -kA dt/dx = – 8.5 × 3 × (-700) = 17850 W or 17.85 kW

**Mode # ****2. Convection: **

Thermal convection is a process of energy transport affected by the circulation or mixing of a fluid medium (gas, liquid or a powdery substance). Convection is possible only in a fluid medium and is directly linked with the transport of medium itself. The effectiveness of heat transfer by convection depends largely upon the mixing motion of the fluid.

With respect to origin, two types of convection are distinguished; forced and natural or free convection.

In natural or free convection, the circulation of the fluid medium is caused by buoyancy effects, i.e., by the difference in the densities of the cold and heated particles. Consider heat flow from a hot plate to atmosphere. The stagnant layer of air in the immediate vicinity of the plate gets thermal energy by conduction.

The energy thus transferred serves to increase the temperature and internal energy of the air particles. Because of temperature rise these particles become less dense (and therefore lighter) than the surrounding air. The lighter air particles move upwards to a region of low temperature where they mix with and transfer a part of their energy to the cold particles.

Simultaneously the cold air particles descend downwards to fill the space vacated by the hot air particles. The circulation pattern, upward movement of the warm air and the downward movement of the cold air, is called the convection currents.

**Some examples of free convection are: **

i. Chilling effect of a cold wind on a warm body,

ii. Heat flow from a hot pavement to surrounding atmosphere and heating of air in a room by a stove,

iii. Cooling of billets in the atmosphere,

iv. Heat exchange on the outside of cold and warm pipes,

v. Hot-water heating system.

In forced convection, the flow of fluid is caused by a pump, fan or by atmospheric winds. These mechanical devices provide a definite circuit for the circulating currents and that speeds up the heat transfer rate.

**Example of forced convection are: **

i. Flow of water in condenser tubes,

ii. Fluid passing through the tubes of a heat exchanger,

iii. Cooling of internal combustion engine,

iv. Air conditioning installation and nuclear reactors.

Regardless of the particular nature, the appropriate rate equation for the convective heat transfer between a surface and an adjacent fluid is prescribed by Newton’s law of cooling.

Q = hA (t_{s} – t_{f}) … (1.4)

where, Q is the convective heat flow rate, A is the area exposed to heat transfer, t_{s} and t_{f} are the surface and fluid temperatures respectively. The heat transfer co-efficient h depends upon the thermodynamic and transport properties (e.g. density, viscosity, specific heat and thermal conductivity of the fluid), the geometry of the surface, the nature of fluid flow, and the prevailing thermal conditions.

Convection mechanism involving phase changes leads to the important fields of boiling (evaporation) and condensation.

**Example 3: **

An oil cooler in a high performance engine has an outside surface area 0.12 m^{2} and a surface temperature of 65°C. The air rushes over the surface of the cooler at a temperature of 30°C and gives rise to a surface coefficient of heat transfer equal to 45.4 W/m^{2}K. Calculate the heat transfer rate from the cooler.

**Solution: **

The conditions described imply a convective process, that is, heat transfer from a solid surface (the oil cooler) to an adjacent moving fluid (the air passing over the cooler). The rate of heat transfer by convection from oil cooler to the air is then,

Q = hA (t_{s} – t_{f}) = 45.4 × 0.12 (65 – 30) = 190.68 W

**Example 4: **

A wire 10 cm long and 1 mm in diameter is held between two conducting supports in a water tank and is submerged. A controlled amount of current is made to pass through the wire until the temperature of water becomes 100°C and it starts boiling. Make calculations for the steady temperature of wire if 23.5 watts of electric power is consumed. Take convective heat transfer coefficient to be 5000 W/m^{2}-deg.

**Solution: **

When steady state is reached, the power supplied to the wire equals the convective heat loss which is given by-

Q = h A Δl = h (πdl) × (t_{s} – t_{w})

Or 23.5 = 5000 × (π × 0.001 × 0.1) × (t_{s} – 100) = 1.57 (t_{s} – 100)

.** ^{.}**. Surface temperature of wire is-

t_{s} = 23.5/1.57 + 100 @ 115°C

**Mode # ****3. Radiation: **

Thermal radiation is the transmission of heat in the form of radiant energy or wave motion from one body to another across an intervening space. Unlike heat transfer by conduction and convection, transport of thermal radiation does not necessarily affect the material medium between the heat source and the receiver.

An intervening medium is not even necessary and the radiation can be affected through vacuum or a space devoid of any matter. Radiation exchange, in fact, occurs most effectively in vacuum. A material present between the heat source and the receiver would either reduce or eliminate entirely the propagation of radiation energy.

**The mechanism of the heat flow by radiation consists of three distinct phases: **

**(i) Conversion of Thermal Energy of the Hot Source into Electromagnetic Waves: **

All bodies above absolute zero temperature are capable of emitting radiant energy. Energy released by a radiating surface is not continuous but is in the form of successive and separate (discrete) packets or quanta of energy called photons. The photons are propagated through the space as rays; the movement of swarm of photons is described as the electromagnetic waves.

**(ii) Passage of Wave Motion through Intervening Space: **

The photons, as carriers of energy, travel with unchanged frequency in straight paths with speed equal to that of light.

**(iii) Transformation of Waves into Heat: **

When the photons approach the cold receiving surface, there occurs reconversion of wave motion into thermal energy which is partly absorbed, reflected or transmitted through the receiving surface.

The most vivid evidence of radiation heat transfer is that represented by solar energy which passes through inter-stellar space (conditions close to that for perfect vacuum) on its way to the earth surface. Solar radiation plays an important part in the design of heating and ventilating systems.

Heat transfer by radiation is encountered in boiler furnaces, billet reheating furnaces and other types of heat exchange apparatus. The design and construction of engines, gas turbines, nuclear reactors and solar collectors is also significantly influenced by the radiation heat transfer.

**The basic rate equations for radiation heat transfer are based on Stefan-Boltzmann Law: **

E_{b }= σ_{b }A T^{4} … (1.5)

where, E_{b} is the energy radiated per unit time, T is the absolute temperature of the surface, and σ_{b} is the Stefan-Boltzmann constant.

σ_{b} = 5.67 × 10^{-8} W/m^{2} K^{4 }

Equation 1.6 is essentially valid for an ideal radiator or a black body-suffix b designates a black surface. The radiant energy emitted by a real surface is less than that for an ideal emitter and is given by

E = ϵ σ_{b} A T^{4} … (1.6)

where, ϵ is a radiative property of the surface and is called emissivity, its value depends upon surface characteristics and temperature. It indicates how effectively the surface emits radiations compared to an ideal or black body radiator. Normally a body radiating heat is simultaneously receiving heat from other bodies as radiation. Consider that surface 1 at temperature T_{1} is completely enclosed by another black surface 2 at temperature T_{2}. The net radiant heat transfer is

Q = σ_{b }A_{1} (T_{1}^{4} – T_{2}^{4}) … (1.7)

Likewise, the net rate of heat transfer between the real surface (called gray surface) at temperature T_{1} to a surrounding black surface at temperature T_{2} is

Q = σ_{b} A_{1} ϵ_{1} (T_{1}^{4} – T_{2}^{4}) … (1.8)

The net exchange of heat between the two radiating surfaces is due to the fact that one at the higher temperature radiates more and receives less energy for its absorption. An isolated body which remains at constant temperature emits just as much energy by radiation as it receives.

**Example 5: **

A radiator in a domestic heating system operates at a surface temperature of 60°C. Calculate the heat flux at the surface of the radiator if it behaves as a black body.

**Solution: **

The heat flux at the surface is the rate at which radiant energy leaves the surface per unit area.

q = Q/A

= σ_{b} T^{4} = 5.67 × 10^{-8} (273 + 60)^{4} = 697.2 W/m^{2}

**Example 6: **

A cylindrical rod, 1.5 m long and 2 cm in diameter, is heated electrically and positioned in a vacuum furnace which has interior walls at 800 K temperature. A controlled amount of current is passed through the rod and its surface is maintained at 1000 K. Calculate the power supplied to the heating rod if its surface has an emissivity of 0.9.

**Solution: **

For steady state conditions, the electric power supplied to the rod equals the radiant heat loss from it. Further, since the walls of the furnace completely enclose the heating rod, all the radiant energy emitted by the surface of the rod is intercepted by the furnace walls. Thus-

Q = σ_{b} A ϵ (T_{1}^{4} – T_{2}^{4})

= σ_{b} (πdl) ϵ (T_{1}^{4} – T_{2}^{4})

= 5.67 × 10^{-8} (π × 0.02 × 1.5) (0.9) (1000^{4} – 800^{4})

= 5.67 (π × 0.02 × 1.5) (0.9) [(1000/100)^{4} – (800/100)^{4}]

= 5.67 (π × 0.02 × 1.5) (0.9) (10000 – 4096) = 2838 W.

Thus the rate of electrical input to the rod must equal 2838 W.