Heat energy transfer mechanisms

1. Heat energy transfer mechanisms • Conduction • Convection • Radiation

2. 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.

3. 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: Thermal conductivity 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.

4. Thermal conductivities of some substances Substance k (W/moC) Metals (at 25oC) 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 20oC) 0.0234

5. Steady state thermal conduction In steady state, the rate of heat flow through an insulated uniform rod can be determined using the equation: where T2 is the temperature at the hotter end of the rod, T1 the temperature at the colder end of the rod, A the cross-sectional area of the rod, and L its length.

6. Heat transfer through compound slabs Consider a compound slab consisting of two materials of thicknesses L1, L2 and thermal conductivities k1, k2 respectively. The temperatures of the outer surfaces are T1 and T2, where T2 > T1. Let the temperature at the interface be T. Since the heat transfer rates through the two layers must be the same, we have k1A(T – T1) k2A(T2 – T) = L1L2 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.

7. Solving the equation for T, we have k1L2T1 + k2L1T2 T = k1L2 + k2L1 Substituting this expression for T in the heat transfer rate equation for either layer, we obtain A ( T2 – T1 ) Q = L1 / k1 + L2 / k2 t

8. The rate of heat transfer through a compound slab consisting of n materials can be generalized from the 2-layer equation: L/k for a particularsubstance 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 ( T2 – T1 ) Q = n t Ri i=1 A ( T2 – T1 ) Q = n t Li / ki i=1

9. Material R value (ft2·oF · 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 R values of some common building materials *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.

10. 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? R1 (outside stagnant air layer) 0.17 ft2·oF · h/Btu R2 (brick) 4.00 R3 (fiberglass board) 8.70 R4 (dry wall) 0.45 R5 (inside stagnant air layer) 0.17 Ri 13.49 ft2·oF · h/Btu

11. Law of thermal conduction: T2 Steam pipe Insulating material k L Q Q is constant at steady state T1 t t Representing by H, the law can be written as: 2a 2b 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 T2 and that of the air outside the insulating material is T1 (T1 < T2), 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?

12.  From the above equation: (equ. 1) where c is an integration constant. At r = a, T = T2 (equ. 2) At r = b, T = T1 (equ. 3)

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

14. 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.

15. 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).

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

17. 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: h, (cal·s-1·cm-2 ·oC-1) Horizontal plate, facing upward 0.0595x10-4(T)1/4 Horizontal plate, facing downward 0.0314x10-4(T)1/4 Vertical plate 0.0424x10-4(T)1/4 Horizontal or vertical pipe (dia. D) 1.00x10-4(T/D)1/4 Convection coefficient The convection coefficient defined in the above equation is temperature dependent and needs to be determined experimentally.

18. Inside Outside 25oC -15oC T2 T1 Example: The air in a room is at a temperature of 25oC, and the outside air is at –15oC. 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·(Co)-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 T2 = 25 - T2 and T1 = T1 + 15 Rate of heat transfer by convection in the room = h2 A T2 Rate of heat transfer by convection outside = h1 A T1 Rate of heat transfer by conduction thru glass = kA (T2 - T1) = kA (40 -  T2 -  T1)

19. In the steady state : h2 A T2 = h1 A T1 = kA (40 -  T2 -  T1) This “3-way” equality in the steady state gives 2 independent equations, with which one can solve for the two unknown T1 and T2 in principle. Once T1 and T2 have been determined, the rate of energy transfer can be calculated. However, h1 and h2 are proportional to (T1 )1/4and (T2)1/4, respectively, and therefore the solutions can only be determined numerically but not analytically.

20. As a first approximation in the solution of the problem, let us assume that T1 = T2 = 20 oC (or T2 = T1 = 5 oC). 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 ·oC-1  0.897 x 10-4cal·s-1·cm-2 ·oC-1 The rate of heat transfer per unit area is: Q hT1 = 0.897 x 10-4x 20 cal·s-1·cm-2 At 17.9 x 10-4 cal·s-1·cm-2

21. Q L (T2 - T1)  · At k 17.9 x 10-4 x 0.2  2.5x 10-3  0.14 Co T2 and T1 are actually not the same. Their difference can be computed using the approximated value of rate of heat transfer obtained: Hence, T1 5 – 0.07 = 4.93 Co T2  5 + 0.07 = 5.07 Co

22. 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.

23. Radiant energy from the Sun • 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.

24. Q =  A e T4 t 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. 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: where  = 5.6696x10-8 W·m-2·K-4 (Stefan-Boltzmann constant) A = surface area of the object in m2 e = emissivity of the object T = the surface temperature of the object in kelvins.

25. Emissivity • 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.

26. Q =  A e (T 4 -To4) t 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 To, then the rate of net energy gained (or lost) by the object due to emission and absorption of electromagnetic radiation is: Hence, if T > To , The object radiates more energy than it absorbs, and its temperature decreases until equilibrium with its surroundings is reached.

27. Q Rate of radiant energy loss: =  A e T 4, t where  = 5.6696x10-8 W·m-2·K-4 A = 2 x (0.1x0.1 + 0.1x0.002 + 0.1x0.002) = 0.0208 e = 0.9 T = 800 +273.15 = 1073 K Q = 5.6696 x 0.0208 x 0.9 x (1073)4 x 10-8 = 1407 W t Example: A metal block of dimensions 10 cm x 10 cm x 0.2 cm is heated by an electric heater to 800 oC 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.

28. The Dewar flask 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. • 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.