2. Heat energy transfer mechanisms oConduction oConvection oRadiation

3. Conduction This is the mechanism by which heat is transferred from one part of an object to another part through molecular collisions. If one part of an object is hotter than its neighboring part, the molecules in the hotter part have more energy and vibrate more vigorously than their neighbors. When they collide with their neighboring molecules which vibrate less vigorously, energy is transferred to the latter and the temperature of the colder part increases. Free electrons (electrons that has become detached from their parent molecules) also play a important part in the thermal conduction as they provide an effective mechanism for carrying heat energy from one part of an object to another part. Metals are good electrical conductors because they have a lot of free electrons. Hence they are also good conductors of heat.

4. Thermal conductivity The rate of heat transfer through thermal conduction is proportional to the cross-sectional area and the temperature difference, and inversely proportional to the thickness: For a slab of infinitestimal thickness dx and temperature difference dT, the thermal conduction equation can be written as: The proportional constant k is called the thermal conductivity. The negative sign is introduced to adopt the convention that the direction of heat flow is opposite to the direction of increasing temperature.

5. Thermal conductivities of some substances Substancek (W/m o C) Metals (at 25 o C) Silver 427 Copper 390 Aluminum 238 Iron 79.5 Nonmetals (approximate values) Concrete 0.8 Glass 0.8 Water 0.6 Rubber 0.2 Wood 0.08 Asbestos 0.08 Air (at 20 o C) 0.0234

6. Steady state thermal conduction In steady state, the rate of heat flow through an insulated uniform rod can be determined using the equation: where T 2 is the temperature at the hotter end of the rod, T 1 the temperature at the colder end of the rod, A the cross- sectional area of the rod, and L its length.

7. Heat transfer through compound slabs Often compound slabs consisting several layers of different materials are used for insulation. The rate of heat transfer through such a slab can be calculated from the thermal conductivities of the materials that make up of the slab. Consider a compound slab consisting of two materials of thicknesses L 1, L 2 and thermal conductivities k 1, k 2 respectively. The temperatures of the outer surfaces are T 1 and T 2, where T 2 > T 1. Let the temperature at the interface be T. Since the heat transfer rates through the two layers must be the same, we have k 1 A(T – T 1 ) k 2 A(T 2 – T) L 1 L 2 =

8. Solving the equation for T, we have k 1 L 2 T 1 + k 2 L 1 T 2 k 1 L 2 + k 2 L 1 T = Substituting this expression for T in the heat transfer rate equation for either layer, we obtain A ( T 2 – T 1 ) L 1 / k 1 + L 2 / k 2 = Q tt

9. The rate of heat transfer through a compound slab consisting of n materials can be generalized from the 2- layer equation: A ( T 2 – T 1 )  L i / k i = Q tt i=1 n L/k for a particular substance is often referred to as the R value of the material, and the above equation can be written in terms of the R values: A ( T 2 – T 1 ) = Q tt  Ri Ri i=1 n

10. R values of some common building materials Material R value (ft 2 · o F · h/Btu) Hardwood siding (1”) 0.91 Brick (4”) 4.00 Fiberglass board (1”) 4.35 Flat glass (0.125”) 0.89 Air space (3.5”) 1.01 Drywall (0.5”) 0.45 Stagnant air layer* 0.17 *At any vertical surface open to the air, a very thin stagnant layer of air adheres to the surface and the R value of this stagnant air layer on an outside wall depends on wind speed. When determining the R value of a wall, one must consider the stagnant air layers on both sides of the wall.

11. Example: What is the R value of a wall consisting of a 4” brick layer, a 2” fiberglass board and a 0.5” dry wall? R 1 (outside stagnant air layer) 0.17 ft 2 · o F · h/Btu R 2 (brick) 4.00 R 3 (fiberglass board) 8.70 R 4 (dry wall) 0.45 R 5 (inside stagnant air layer) 0.17 RiRi 13.49 ft 2 · o F · h/Btu

12. Radial heat flow Consider a steam pipe of radius a, surrounded by a layer of insulating material of outer radius b, length L and thermal conductivity k. If the temperature of the steam pipe is T 2 and that of the air outside the insulating material is T 1 (T 1 < T 2 ), What is the rate of energy transfer through the insulating material and what is the temperature at r (a < r < b) when a steady state has been reached? Law of thermal conduction: Q tt is constant at steady state Representing by H, the law can be written as: Q tt Steam pipe Insulating material T1T1 T2T2 2a2a 2b2b L k

13. where c is an integration constant.  (equ. 1) At r = a, T = T 2 (equ. 2) From the above equation:At r = b, T = T 1 (equ. 3)

14. Solving equ. 2 and equ 3 for H and c: (This is the rate of energy transfer) Substituting the expressions for H and c in equ 1, we have

15. Convection This is the mechanism by which heat is transferred by actual motion of material. Natural convection: The material flows due to differences in density (caused by thermal expansion). Examples: Air flow at a beach. Water mixing in a lake when its surface is cooled. Forced convection: The material if forced to move by a blower of pump. Examples: Hot-water heating system. Flow of blood in the body.

16. Heat transfer by convection There is no simple equation for calculating the amount of heat transferred by convection. The heat lost or gained by a surface at one temperature in contact with a fluid at another temperature depends on many factors, such as the shape and orientation of the surface, the mechanical and thermal properties of the fluid and the nature of fluid flow (laminar or turbulent).

17. For practical calculations, the following equation is often used: Q tt = h A  T where, = rate of energy transfer or heat current Q tt A = surface area  T = temperature difference between the surface and the main body of the fluid h = convection coefficient

18. The convection coefficient defined in the above equation is temperature dependent and needs to be determined experimentally. Convection coefficient h, (cal·s -1 ·cm -2 · o C -1 ) Horizontal plate, facing upward0.0595 x 10 -4 (  T) 1/4 Horizontal plate, facing downward 0.0314 x 10 -4 (  T) 1/4 Vertical plate 0.0424 x 10 -4 (  T) 1/4 Horizontal or vertical pipe (dia. D) 1.00 x10 -4 (  T/D) 1/4 A situation of practical important is that of natural convection from a wall or a pipe that is at constant temperature and is surrounded by air at atmospheric pressure whose temperature is less than that of the wall or pipe by  T. The convection coefficients applicable in this situations are shown below:

19. Example: The air in a room is at a temperature of 25 o C, and the outside air is at –15 o C. What is the rate of heat transfer per unit area through a 2mm-thick glass window plane of thermal conductivity k=2.5x10 -3 cal·s -1 ·cm -1 ·(C o ) -1 ? In the steady state, the rates of heat transfer by (1) convection in the room, (2) by conduction through the glass and (3) by convection in the outside air are all the same. Let  T 2 = 25 - T 2 and  T 1 = T 1 + 15 Rate of heat transfer by convection in the room = h 2 A  T 2 Rate of heat transfer by convection outside = h 1 A  T 1 Rate of heat transfer by conduction thru glass = k A (T 2 - T 1 ) = k A (40 -  T 2 -  T 1 ) 25 o C-15 o C Inside Outside T2T2 T1T1

20. This “3-way” equality in the steady state gives 2 independent equations, with which one can solve for the two unknown  T 1 and  T 2 in principle. Once  T 1 and  T 2 have been determined, the rate of energy transfer can be calculated. However, h 1 and h 2 are proportional to (  T 1 ) 1/4 and (  T 2 ) 1/4, respectively, and therefore the solutions can only be determined numerically but not analytically. In the steady state : h 2 A  T 2 = h 1 A  T 1 = k A (40 -  T 2 -  T 1 )

21. As a first approximation in the solution of the problem, let us assume that  T 1 =  T 2 = 20 o C (or T 2 = T 1 = 5 o C). In this case, the convection coefficients inside and outside the window plane are the same: h  0.0424 x 10 -4 (20) 1/4 cal·s -1 ·cm -2 · o C -1  0.897 x 10 -4 cal·s -1 ·cm -2 · o C -1 The rate of heat transfer per unit area is: Q AtAt  h  T 1 = 0.897 x 10 -4 x 20 cal·s -1 ·cm -2  17.9 x 10 -4 cal·s -1 ·cm -2

22. T 2 and T 1 are actually not the same. Their difference can be computed using the approximated value of rate of heat transfer obtained: (T 2 - T 1 )  Q AtAt L k ·  17.9 x 10 -4 x 0.2 2.5 x 10 -3  0.14 C o Hence, T 1  5 – 0.07 = 4.93 C o T 2  5 + 0.07 = 5.07 C o

23. Radiation This is the mechanism by which heat is transferred by continual emission of energy from the surface of a body. This energy is called radiant energy and is in the form of electromagnetic waves. These waves travel with the speed of light and are transmitted through vacuum as well as through. When they fall on a body that is not transparent to them, such as the surface of one’s hand or the walls of a room, they are absorbed, resulting in a transfer of heat to the absorbing material. The radiant energy emitted by a surface depends on the nature of the surface and on its temperature. At any temperature, the radiant energy emitted is a mixture of waves of different wavelengths, and the mixture is a function of temperature.

24. The amount of radiant energy reaching the top of the Earth’s atmosphere from the Sun is approximately 1340 J per second. Some of this energy is reflected into the space and some absorbed by the atmosphere. The amount of the Sun’s radiant energy that reaches the surface of the Earth is hundreds of times of all the energy needed on this planet. The Sun’s radiant energy is primary visible and infrared light accompanied by a significant amount of ultraviolet radiation. The Sun influences the Earth’s average temperature, ocean currents, agriculture, rain patterns, etc. Radiant energy from the Sun

25. Stefan’s law The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. This is know as Stefan’s law: Q tt =  A e T 4 where  = 5.6696 x 10 -8 W·m -2 ·K -4 (Stefan-Boltzmann constant) A = surface area of the object in m 2 e = emissivity of the object T = the surface temperature of the object in kelvins. This relation was deduced by Josef Stefan (1835-1893) on the basis of experimental measurements made by John Tyndall (1820-1893) and was later derived from theoretical considerations by Ludwig Boltzmann (1844-1906). Hence, it is also called Stefen-Boltzmann law.

26. It is the fraction of the incoming radiation the surface absorbs. It has a value if the range 0 – 1. It depends on the properties of the surface of the object, generally larger for dark and rough surfaces than for light and smooth ones. A good radiant energy emitter has a large emissivity, hence it is also a good absorber and a poor reflector. A poor radiant energy emitter is also a poor absorber and a good reflector. An “ideal” absorber (e = 1) reflects no radiant energy, and hence would appear black in color. Such an ideal absorber is called blackbody. Emissivity

27. As an object radiates electromagnetic energy, it also absorbs electromagnetic radiation, otherwise its temperature would reduce to absolute zero. The amount of electromagnetic energy absorbed by an object depends on the temperature of its surroundings. If an object at a temperature T and its surroundings are at temperature T o, then the rate of net energy gained (or lost) by the object due to emission and absorption of electromagnetic radiation is: Q tt =  A e (T 4 - T o 4 ) Hence, if T > T o, The object radiates more energy than it absorbs, and its temperature decreases until equilibrium with its surroundings is reached.

28. Example: A metal block of dimensions 10 cm x 10 cm x 0.2 cm is heated by an electric heater to 800 o C in vacuum. If the emissivity of the metal block is 0.9, how much electric power input to the heater would be required to maintain the metal block at this temperature? To maintain the metal block at a fixed temperature, the electric power input to the heater must be equal to the rate of energy loss by radiation. where  = 5.6696 x 10 -8 W·m -2 ·K -4 A = 2 x (0.1 x 0.1 + 0.1 x 0.002 + 0.1 x 0.002) = 0.0208 e = 0.9 T = 800 +273.15 = 1073 K Q tt =  A e T 4, Rate of radiant energy loss : Q tt = 5.6696 x 0.0208 x 0.9 x (1073) 4 x 10 -8 = 1407 W

29. The Dewar flask (invented by James Dewar) is a container designed for storage of cold or hot liquids for long period of time. It is a Pyrex glass vessel and has a design as shown in the diagram below. The Dewar flask It has two walls with evacuated space in between to minimize energy transfer by conduction and convection. The surfaces of the wall are coated with silver (a very good reflector with low emissivity) to minimize energy transfer by radiation.