Notes for Heat Transfer chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.
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Modes of Transference of Heat
Heat can be transferred from one place to another by three different methods, namely, conduction, convection and radiation. Conduction usually takes place in solids,Â convection in liquids and gases and no medium is required for radiation.
Heat Conduction
This transfer takes place due to molecular collisions. The molecules at one end of the rod gain heat from the heat source and their average kinetic energy increases. As these molecules collide with neighboring molecules having less kinetic energy, the energy is shared between these two groups. The kinetic energy of these neighboring molecules increases. As they collide with their neighbors on the colder side, they transfer energy to them. This way, heat is passed along the rod from molecule to molecule. The average position of a molecule does not change and hence, there is no mass movement of matter.
The transfer of energy arising form the temperature difference between adjacent parts of a body is called heat conduction. Consider a slab of material of cross-sectional area A and thickness , whose faces are kept at different temperatures. We measure the heat Â that flows perpendicular to the faces during time . Experiment shows that Â is proportional to Â and to the cross-sectional area A for a given temperature difference , and that Â is proportional to /Â for a given Â and A, providing both Â and Â are small. That is,
In the limit of a slab of infinitesimal thickness dx, across which there is a temperature difference dT, we obtain the fundamental law of heat conduction
Hereis the time rate of heat transfer across the areais called the temperature gradient, and k is a constant proportionality called the thermal conductivity. We choose the direction of heat flow to be the direction in which x increases; since heat flows in the direction of decreasing T, we introduce a minus sign in equation, (i.e., we wish to the positive when is negative).
The phenomenon of heat conduction also shows that the concepts of heat and temperature are distinctly different. Different rods, having the same temperature difference between their ends, may transfer entirely different quantities of heat in the same time.
Example 1
A rod of length l with thermally insulated lateral surface consists of a material whose heat conductivity coefficient varies with temperature as K = a/T, where a is a constant. The ends of the rod are kept at temperatures T_{1}, and T_{2 }(T_{1} > T_{2}). Find the function T(x) where x is the distance from the end whose temperature is T, and the heat flow density.
Solution:
Use fundamental law of heat conduction.
Heat flow density,
Hence
Once again,
Hence
Alternative method.
The rate of heat flow through a conductor is given by
where K is the average conductivity.
or heat flow density Â Â Â Â Â Â (1)
Let T be the temperature at a distance x from the left end as shown in the figure.
ThenÂ and heat flow density is
Equating (1) and (2) we get
T =Â
Thermal and Electrical Conductivity
There exists an useful analog between thermal conductivity and electrical conductivity
Thermal Conduction
Heat flows from higher temperature to lower temperature The rate of heat flow is called the heat current. Â Â Â Â Â Â Â Â I =Â Â Thermal resistance is defined as Â Â Â Â R_{T }Â = Ohmâ€™s law for heat conduction may be stated as Â Â Â Â Â Â I = |
Electrical Conduction
Charges flow from higher potential to lower potential The rate of charge flow is called the electric current Â Â Â Â Â I = Â Electrical resistance is defined as Â Â Â R_{E} = Â Ohmâ€™s law for electric conduction may be stated as Â Â Â Â Â Â Â I = |
Using the above analogy, a problem of heat conduction may be transformed into a problem of electrical conduction and can be easily using the formulae of electric circuits. Such as, for a series and parallel combination the equivalent resistance is defined a
(series)
(parallel)
Example 2
Three rods of same length l and cross-sectional area A are joined in series between two heat reservoirs as shown in the figure. Their conductivity are 2K, K and K/2 respectively. Assuming that the conductors are logged from the surroundings find the temperatures T_{1} and T_{2} at the junction in the steady state condition.
Solution:
The thermal resistance of the three conductors are
Let R_{1}Â = R then R_{2} = 2RÂ and R_{3} = 4R
Thus, the electric analogy of the heat conduction system is shown in the figure. The equivalent resistance of the system is
R_{eq} = R + 2R + 4R = 7R
The heat current is
The temperature T_{1} and T_{2} of the junction are
Example 3
A double - pane window consists of two glass sheets each of area 1m^{2}Â and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state,Â the room glass interface and the glass outdoor interface are at constant temperatures of 27Â°C and 0Â°C , respectively.
(a) Calculate the rate of heat flow through the window pane.
(b) Find the temperatures of other interfaces.
Take, thermal conductivitiesÂ as K_{glass} = 0.8 Wm^{-1}K^{-1} K_{air} = 0.08 Wm^{-1}K^{-1Â
}
Solution:
(a) Total thermal resistance is
Here,Â A_{1} = A_{2} = 1m^{2} , t_{1} = 0.01 m,
t_{2} = 0.05 m;
K_{1} = 0.8 Wm^{-1}K^{-1},
K_{2} = 0.08Wm^{-1}K^{-1}.
Heat current =
(b)
In convection, heat is transferred from one place to the other by the actual motion of heated material. For example, in a hot air blower, air is heated by a heating element and is blown by a fan. The air carries the heat wherever it goes. When water is kept in a vessel and heated on a stove, the water at the bottom gets heat due to conduction through the vessel's bottom. Its density decreases and consequently it rises. Thus, the heat is carried from the bottom to the top by the actual movement of the parts of the water. If the heated material is forced to move, say by a blower or a pump, the process of heat transfer is called forced convection. If the material moves due to difference in density, it is called natural or free convection.
The process in which heat is transferred from one place to the other without any intervening medium is called radiation. All bodies at all temperatures and at all times radiate energy in the form of electromagnetic waves, called radiant energy. Emission rate is faster at higher temperatures. In this radiant energy, electromagnetic waves of wavelength ranging from 10^{-4} mm to 1 mm are called infrared radiation or heat radiation. According to prevostâ€™s theory of exchange besides radiating heat radiation or thermal radiation all bodies also absorb part of the thermal radiation falling on them emitted byÂ the surrounding bodies. If a body radiates more, what it absorbs, its temperature falls and vice versa.
Basic Definitions
(i) Perfectly Black Body
A body which absorbs all the radiations incident on it is called a perfectly black body.
(ii) Absorptive Power of a surface (a)
The ratio of the radiant energy absorbed by it in a given time to the total radiant energy incident on it in the same time is called the absorptive power (a) of the surface. Since it is a pure ratio it has no units and dimensions. The absorptive power of a perfectly black body is maximum and its value is unity.
(iii) Spectral Absorptive Power ()
Absorptive power refers to all wavelengths (total radiant energy). However, any surface will have different values of absorptive powers for different wavelengths. A surface may be a good absorber for a wavelengthÂ and bad absorber for wavelength , it means its absorptive power for wavelength is greater than the absorptive power forÂ .
Thus, spectral absorptive powerÂ is defined as the ratio of the radiant energy of a given wavelength absorbed by a given surface in a given time to the total radiant energy of that wavelength incident in the same time on the same surface. Or we can say that it is absorptive power for that particular wavelength. It is now obvious that wavelength incident in the same time on the same surface. Or we can say that it is absorptive power for that particular wavelength. It is now obvious that
For perfectly black bodyÂ
Example 4
100 units of energy is incident on a surface. In this 20 units are of wavelength , 30 units are of wavelengthÂ and rest 50 units are other wavelengths. The total 60 units of energy is absorbed by the surface. In this 60 units 5 units are ofÂ and 25 units are of . Find a, Â and.
Solution:
Total absorptive power
and spectral absorptive power for
Spectral absorptive power for
From this example it is clear that total absorption power of the surface is only 0.6 whereas = is 0.83 (> 0.6) i.e. the surface is good absorber of wavelength .
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