Chapter 11 Thermal Transport

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1 Chapter 11 Thermal Transport GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define the following terms, and use them in an operational definition: conduction radiation convection Energy Transfer Problems Solve numerical problems that involve a transfer of energy, temperature gradients, and conduction, convection, or radiation. Living System Thermal Properties Explain basic thermal effects in living systems. PREREQUISITES Before beginning this chapter, you should have achieved the goals of Chapter 9, Transport Phenomena, and Chapter 10, Temperature and Heat. 100

2 Chapter 11 Thermal Transport OVERVIEW From Chapter Nine you should recall that three Transport Phenomena were highlighted. In this chapter, the flow or transport of heat is considered in greater detail. Please note that the chapter deals with three major forms of Thermal Transport: Conduction (Section 11.2), Convection (Section 11.3), and Radiation (Section 11.4). SUGGESTED STUDY PROCEDURE To begin your study of this chapter, please read all the Chapter Goals: Definitions, Energy Transfer Problems, and Living System Thermal Properties. For an expanded treatment of each of the terms listed under Definitions, turn to the next section of this chapter. Next, read text sections Even though the algebra becomes involved in example 2 in section 11.2, try to go through the problem as you may find the result surprising. As you read the text sections, consider the questions asked. The answers to these questions are answered in this Study Guide chapter. Now read the Chapter Summary and complete Summary Exercises 1-9 and do Algorithmic Problems 1-4. Next, complete Exercises and Problems 1, 4, 6, 7, and 10. Additional examples of the major concepts presented in this chapter are found in the Examples section of this Study Guide chapter. Now you should be prepared to attempt the Practice Test in the Study Guide at the end of this chapter. Check your answers. If you have difficulties, please refer to the specific section of this chapter again. This study procedure is outlined below Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems Definitions 11.1, 11.2, , 6, Transfer 11.2, 11.3, , 7 Problems Living System Thermal Properties 101

3 DEFINITIONS CONDUCTION Transfer of thermal energy by the interaction of neighboring portions of a substance at rest. Conduction is a most effective transfer mechanism in solids, particularly in metals. CONVECTION Transfer of thermal energy by the motion of matter as in liquids and gases. Convection arises because of the different densities of warm and cool fluids which cause buoyant forces in the earth's gravitational field. In the weightless environment of space, convection ceases to be an energy transfer mechanism. RADIATION Transfer of energy by electromagnetic fields. Microwave cooking is a most recent example of using radiation to heat objects. ANSWERS TO QUESTIONS FOUND IN THE TEXT SECTION 11.6 Microwave Cooking The explanation of microwave cooking taken from the cookbook of the Litton Systems may seem more vivid but it certainly carries some wrong implications. Food does not "attract" microwaves anymore than you would say that sunglasses attract light. Sunglasses absorb light that would otherwise strike your eyes. Foods, especially fats and sugars, absorb the microwaves that pass through other materials. The absorption process leads to the transformation of microwave, electromagnetic energy into thermal energy. The Consumer Reports description is more in accord with our present day scientific model to explain these phenomena. EXAMPLES ENERGY TRANSFER PROBLEMS 1. The wooden handle of a frying pan has an area of 10 cm 2 and is 15 cm long. If the pan is at 120 ø C and the free end of the handle is at 30 ø C, how much heat will flow through the handle in 10 minutes? (K wood = 0.15 Wm -1 deg -1 ) The area of the handle, the high and low temperatures, the distance between those two temperatures, the time, and the thermal conductivity of the material. It is assumed that no heat escapes from the handle out the sides, but that heat only flows along the handle. The problem can be treated as a one-dimensional problem. The concepts of thermal conduction are needed. We can make use of the one dimensional thermal conductivity equation 11.2 H = -KA (ΔT/ΔX) (11.2) 102

4 Algebraic Solution H = -KA (ΔT/ΔX) where H is in joules/second Heat flow in 10 minutes = 600 H Numerical Solution Heat Flow in 10 minutes = 600 sec. x (-0.15/m deg)(j/sec) x (10 x 10-4 m 2 ) x [(30-120) ø C/0.15 m] = 600 (-.15)(10-3 ) (-90/.15) joules = 54 J Thinking About the Answer Notice that the units for the answer, joules, are correct. 2. In a room of temperature 20 ø C, two identical open pans, one containing ice water, the other boiling water, are placed in front of an air-circulating fan. If the pans contain equal amounts of water, compute the ratio of the initial rates of their temperature changes. Is the ratio a positive or negative number? The temperature of the room in a situation of forced air convection. The two pans are assumed to be in identical situations with respect to the forced air circulation. The only difference between the two pans is their initial starting temperatures. One pan has a temperature 0 ø C. The other pan has a temperature of 100 ø C. This problem can be solved using Newton's law of cooling; i.e., the rate of change of temperature of an object is proportional to the difference between the temperature of the object and its surroundings. The heat capacity equation, Q = mcδt, (Equation 10.9). The forced convection equation, rate of thermal energy flow = H = -K 1 (T s - T a ) (11.6) Algebraic Solution Let T c be initial temperature of the cold water Let T h be initial temperature of the hot water for cold water H c = -K 1 (T c - T a ) for hot water H h = -K 1 (T h - T a ). The rate of heat flow of an object is related to the temperature change of an object by Equation 10.9 Δ0 = mc ΔT, but H =ΔQ/Δtime. so: H = ΔQ/Δt = mcδt/δt Since the masses and specific heat of the two systems of water are equal, then ΔT c /Δt = -K 1 (T c - T a ) andδt h /Δt = -K 1 (T h - T a ) The ratio of the initial rates of temperature change are given by (ΔT c /Δt) / (ΔT h /Δt) = (T c - T h ) / (T h - T a ) Numerical Solution (ΔT c /Δt) / (ΔT h /Δt) = (0 ø - 20 ø )C / (100 ø - 20 ø )C = -20/80 = -1/4 The ratio is a negative number because one container is warming up so its ΔT/Δt is positive and the other container is cooling down so its ΔT/Δt is negative. Hence the ratio is negative. 103

5 Thinking About the Answer Notice that the answer for this question is a pure number with no units because it is a ratio of similar quantities. It is always a negative number. For the case given the hot water is cooling down four times as fast as the cold water is heating up. 3. In a cryogenics (low temperature) experiment the temperature of the sample is decreased from liquid helium temperatures (4 ø K) to 1 ø K. How much is the rate of heat loss by radiation decreased? The absolute temperature of the sample is given. Any heat radiated to the samples from the environment is neglected. The Stefan-Boltzmann equation for radiation. Rate of energy radiated = P = σat 4 (11.8) Algebraic Solution T i = initial temperature; T f = final temperature P f /P i = T f4 /T i4 = (T f /T i ) 4 (2) Numerical Solution P f /P i = (1 ø K/4 ø K) 4 = 1/256 = 3.9 x 10-3 Thinking About the Answer You can notice that even though the temperature is only changed 3 ø, there is a large change in the amount of energy lost by radiation. 4. Suppose you double the absolute temperature of an object. How much will the three types of heat transfer change? (Hint: Assume the ambient room temperature is one-half the original temperature of the object.) The initial temperature of the object is Tc. The initial ambient temperature is T 0 /2. The final temperature of the object is 2T 0. The situation is such that all of the first order heat transfer equations are valid, Equations (11.2), (11.6), and (11.8). The basic concepts of conduction, convection, and radiation are needed. Conduction H = -KA (ΔT/ΔX) (11.2) Convection H = -K 1 (T s - T a ) (11.6) Radiation P = σat 4 (11.8) Algebraic Solution Let T i = initial temperature, T f = final conduction temperature, and T a = ambient temp. Conduction: H i (T i - T a ) /Δx and H f (T f - T a )/Δx so ratio H f /H i = (T f - T a )/(T i - T a ) (3) Convection: H i (T i - T u ) and H f (T f - T a ) 104

6 ratio H f /H i = (T f - T a ) / (T i - T a ) (4) Radiation: P i T i4 - T a4 ; i.e., heat lost by the object goes at T 4, but heat received from the environment goes at T a4, so the net rate of radiated energy is the difference in the fourth power of the temperatures. P f T f 4 - T a 4 ratio P f /P i = (T f 4 - T a 4 ) / (T i 4 - T a 4 ) (5) Numerical Solutions Conduction H f /H i = (2T 0 - T 0 /2) / (T 0 - T 0 /2) = (3/2)/(1/2) = 3 Convection H f /H i = (2T 0 - T 0 /2) / (T 0 - T 0 /2) = (3/2)/(1/2) = 3 Radiation P f /P i = [(2T 0 ) 4 - (T 0 /2) 4 ] / [T (T 0 /2) 4 ] = (255/16)/(15/16) = 15.9 Thinking About the Answer Once again you see the strong temperature dependence of radiation as a form of energy transfer. 105

7 PRACTICE TEST 1. A hot cup of black coffee sits in a cool room in a black coffee cup. a. Name and describe the thermal processes which are actively involved in the transfer of heat from the cup to the room. b. Explain how the coffee, the cup, or the room environment could be altered to reduce the thermal transport by each of the processes named in Part A. 2. A modern home is insulated with 15.2 cm (6 inches) of glass wool (K =.042). How many cm thick must a brick wall be constructed (K =.147) to achieve the same insulating factor? 3. The human body has an intricate set of regulating systems which act together to keep the core of the body at a constant temperature of 37 ø C. a. What is the typical temperature of the body at the skin's surface? Does the temperature vary over the skin's surface? (If it does, give an example showing the magnitude of the variation.) b. The major core regulating process is by "forced convection". Explain briefly how this process works. ANSWERS: 1. Convection (to room air), conduction (to room air and to table), radiation (to each object in the room which is lower in temperature and to the walls, ceiling, and floor). To change; convection (reduce air currents), conduction (insulate cup), radiation (change color of cup and/or coffee or room walls as increase temperature of room walls cm ø C, Yes, about 3 ø C less at exposed extremities: vasodilation regulates the flow of blood and therefore the direction of heat flow (see section 11.5). 106

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