Introduction in to heat transfer

1. Introduction in to heat transfer
Heat is defined as energy transferred by virtue of a temperature difference. It flows from regions of higher temperature to regions of lower temperature. It is customary to refer to different types of heat transfer mechanisms as modes. The basic modes of heat transfer are conduction, radiation, and convection.
Heat transfer by conduction
Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. Conduction can take place in solids, liquids, or gases. In gases and liquids, conduction is due to the collisions and diffusion of the molecules during their random motion. In solids, it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons. A cold canned drink in a warm room, for example, eventually warms up to the room temperature as a result of heat transfer from the room to the drink through the aluminum can by conduction.
The rate of heat conduction through a medium depends on the geometry of the medium, its thickness, and the material of the medium, as well as the temperature difference across the medium. We know that wrapping a hot water tank with glass wool (an insulating material) reduces the rate of heat loss from the tank. The thicker the insulation, the smaller the heat loss. We also know that a hot water tank will lose heat at a higher rate when the temperature of the room housing the tank is lowered. Further, the larger the tank, the larger the surface area and thus the rate of heat loss. Consider steady heat conduction through a large plane wall of thickness ?x = L and area A, as shown in Fig. 1.

FIGURE 1 Heat conduction through a large plane wall of thickness ? x and area A.
The temperature difference across the wall is ?T = T2 _ T1. Experiments have shown that the rate of heat transfer Q • through the wall is doubled when the temperature difference ?T across the wall or the area A normal to the direction of heat transfer is doubled, but is halved when the wall thickness L is doubled. Thus we conclude that the rate of heat conduction through a plane layer is proportional to the temperature difference across the layer and the heat transfer area, but is inversely proportional to the thickness of the layer. That is,

(1)
where the constant of proportionality k is the thermal conductivity of the material, which is a measure of the ability of a material to conduct heat (Fig. 2).


FIGURE 2 The rate of heat conduction through a solid is directly proportional to its thermal conductivity.
In the limiting case of ?x ?0, the equation above reduces to the differential form
(2)
which is called Fourier’s law of heat conduction after J. Fourier, who expressed it first in his heat transfer text in 1822. Here dT/dx is the temperature gradient, which is the slope of the temperature curve on a T-x diagram (the rate of change of T with x), at location x. The relation above indicates that the rate of heat conduction in a direction is proportional to the temperature gradient in that direction. Heat is conducted in the direction of decreasing temperature, and the temperature gradient becomes negative when temperature decreases with increasing x. The negative sign in Eq. 2 ensures that heat transfer in the positive x direction is a positive quantity.
NOTE : Heat is conducted in the direction of decreasing temperature, and the temperature gradient becomes negative when temperature decreases with increasing x. The negative sign in Eq. 2 ensures that heat transfer in the positive x direction is a positive quantity.
Example 1 One face of a copper plate 3 cm thick is maintained at 400?, and the other face is maintained at 100?.Calculate the heat transfer per unit area ?
Solution. From Appendix A the thermal conductivity for copper is 370 W/M•? at 250?. From Fourier’s law
= -k
Integration gives

= -k = = 3.7 MW/m2 ? [1.173 106 Btu/h•ft2]
Example 2
The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and is made of a flat layer of concrete whose thermal conductivity is k = 0.8 W/m • °C .(Fig. blow) . The temperatures of the inner and the outer surfaces of the roof one night are measured to be 15°C and 4°C, respectively, for a period of 10 hours. Determine (a) the rate of heat loss through the roof that night and (b) the cost of that heat loss to the homeowner if the cost of electricity is $0.08/kWh.

Solution (a) Noting that heat transfer through the roof is by conduction and the area of the roof is A = 6 m * 8 m = 48 m2, the steady rate of heat transfer through the roof is determined to be

(b) The amount of heat lost through the roof during a 10-hour period and its cost are determined from

Example 3 the conductivity of the wall having 0.2 m thickness is given by eq.
K = 0.1 + 0.001T+0.002 T2 . Calculate the heat transfer per area of the wall, when the wall exposed to temperature at 200 and 100 °C
Solution :
= -k
q / A = - {-[0.1 T + 0.001 T2 / 2 + 0.002 T3 / 3) ]?(200@100) / 0.2}
q / A = 3.408 KW / m2
Thermal Conductivity
We have seen that different materials store heat differently, and we have defined the property specific heat Cp as a measure of a material’s ability to store thermal energy. For example, Cp = 4.18 kJ/kg • °C for water and Cp = 0.45 kJ/kg • °C for iron at room temperature, which indicates that water can store almost 10 times the energy that iron can per unit mass. Likewise, the thermal
Conductivity k is a measure of a material’s ability to conduct heat. For example, k = 0.608 W/m • °C for water and k = 80.2 W/m • °C for iron at room temperature, which indicates that iron conducts heat more than 100 times faster than water can. Thus we say that water is a poor heat conductor relative to iron, although water is an excellent medium to store thermal energy . The thermal conductivity of a material can be defined as the rate of heat transfer through a unit thickness of the material per unit area per unit temperature difference. The thermal conductivity of a material is a measure of the ability of the material to conduct heat. A high value for thermal conductivity indicates that the material is a good heat conductor, and a low value indicates that the material is a poor heat conductor or insulator.
Thermal Diffusivity
The product ?Cp, which is frequently encountered in heat transfer analysis, is called the heat capacity of a material. Both the specific heat Cp and the heat capacity ?Cp represent the heat storage capability of a material. But Cp expresses it per unit mass whereas ?Cp expresses it per unit volume, as can be noticed from their units J/kg • °C and J/m3 • °C, respectively. Another material property that appears in the transient heat conduction analysis is the thermal diffusivity, which represents how fast heat diffuses through a material and is defined as


Note: that the thermal conductivity k represents how well a material conducts heat, and the heat capacity ?Cp represents how much energy a material stores per unit volume. Therefore, the thermal diffusivity of a material can be viewed as the ratio of the heat conducted through the material to the heat stored per unit volume. A material that has a high thermal conductivity or a low heat capacity will obviously have a large thermal diffusivity. The larger the thermal diffusivity, the faster the propagation of heat into the medium. A small value of thermal diffusivity means that the material mostly absorbs heat and a small amount of heat will be conducted further.
Example:
A common way of measuring the thermal conductivity of a material is to sandwich an electric thermo foil heater between two identical samples of the material, as shown in Fig. blow. The thickness of the resistance heater, including its cover, which is made of thin silicon rubber, is usually less than 0.5 mm. A circulating fluid such as tap water keeps the exposed ends of the samples
at constant temperature. The lateral surfaces of the samples are well insulated to ensure that heat transfer through the samples is one-dimensional. Two thermocouples are embedded into each sample some distance L apart, and a differential thermometer reads the temperature drop ?T across this distance along each sample. When steady operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage. In a certain experiment, cylindrical samples of diameter 5 cm and length 10 cm are used. The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.4 A at 110 V,
and both differential thermometers read a temperature difference of 15°C. Determine the thermal conductivity of the sample.

Solution :
The electrical power consumed by the resistance heater and converted to heat is
, the rate of heat flow for each sample

The heat transfer area is the area normal to the direction of heat flow, which is
the cross-sectional area of the cylinder in this case:

Noting that the temperature drops by 15°C within 3 cm in the direction of heat
flow, the thermal conductivity of the sample is determined to be