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Heat Transfer and Its Applications Introduction 1 Heat Exchange Insulation 2 Nature of heat flow Heat flows from the object at the higher temperature to that at the lower temperature. How it happens? 3 Three mechanisms : Conduction (热传导) Convection (对流传热) Radiation (热辐射) 4 Heat Transfer by Conduction 5 Conduction If a temperature gradient exists in a continuous substance, heat can flow unaccompanied by any observable motion of matter. Qualitative(定性的) Quantitative(定量的) 6 Heat transfer by conduction Assumption: considering heat flow in homogeneous ( 均一的 ) isotropic( 孤立的 ) solids because in these there is no convection and the effect of radiation is negligible. 7 Math. Model by Conduction According to Fourier’s law(傅里叶导热定 律), the heat flux is proportional to the temperature gradient and opposite to x it in sign. For one-dimensional heat flow q dq dT k dA dx (4.2-1) Where: q: heat transfer rate, W=J/s A: heat transfer area, m2 dq/dA: heat transfer flux, W/m2 T: temperature, K or oC x: distance, m dT/dx: temperature gradient, K/m k: thermal conductivity, W/(m∙K) 8 Thermal conductivity k The proportionality constant k is a physical property of the substance. The unit of k are W/(m∙℃) or W/(m∙K). For small ranges of temperature, k is constant For larger temperature ranges, k is approximated by k a bT (4.2-5) k vary over a wide range. They are highest for metals and lowest for finely powdered materials from which air has been evacuated (排空). 9 conductor k of metal cover a wide range of values, from about 17W/m∙ºC for stainless steel (不锈钢) and 45W/m∙ºC for mild steel (低碳钢) , to 380W/m∙ºC for copper and 415W/m∙ºC for silver For glass and most nonporous materials, the thermal conductivities are much lower, from about 0.35 to 3.5. For most liquid k is lower than that for solids, with typical values of about 0.17. k decreases by 3 ~ 4 %t for a 10 ºC rise in temperature, except water. insulator Gases have the smallest thermal conductivities, with values as low as 0.007. For air at 0ºC, k is 0.024 W/m·ºC. 10 metals: thermal conduction results from the motion of free electrons. In conductor solids : poor conductor of electricity, produced by momentum transport In In most liquids: thermal conduction results from momentum transfer between adjacent vibrating(振动) molecules or atoms. In gases: conduction occurs by the random motion of molecules. insulator 11 Steady-state conduction Limiting Condition Conduction under the condition of constant temperature distribution is called steady-state conduction. In the steady state, T is a function of position only, and the rate of heat flow at any point is a constant. Heat flow into the tank Heat flow from the tank 12 Since in steady state there can be neither accumulation nor depletion of heat within the slab, q is constant along the path of heat flow. If x is distance from the hot side. T For steady one-dimensional flow.Eq.(4.2-1) may be written q dT dx kA q (4.2-7) x 13 Since the only variables in this Eq. are x and T, direct integration gives T1 T2 q T k k A x2 x1 B (4.2-8) Where x2-x1=B= thickness of layer of insulation T1-T2=ΔT=temperature drop across layer When k varies linearly with temperature, Eq.(4.2-5), still can be used rigorously by taking an average value k. 14 Equation (4.2-8) can be written in the form q T A R (4.2-9) Where R=B/k is thermal resistance between points 1 and 2. m2∙℃/W 15 Compound resistance in series 16 Compound resistance in series Consider a flat wall constructed of a series of layers, as shown in Fig.4.2 Assume that the layers are in excellent thermal contact, so that no temperature difference exists across the interfaces between the layers. Then, if ΔT is the total temperature drop across the entire wall ΔT= ΔTA+ ΔTB+ΔTC 17 According to Eq.(4.2-8) BA TA q A kA A TA TB TC BC BB TC qC TB qB kC A kB A q A BA qB BB qC BC T Ak A Ak B AkC q A qB qC q T TA TB TC R RA RB RC (4.2-12) q T A BA k A BB k B BC kC T T RA RB RC R (4.2-11) Where RA, RB and RC=resistance of individual layers R=overall resistance 18 Example 4.2 A flat furnace wall is constructed of 120-mm layer of sil-ocel brick, with a thermal conductivity 0.08 W/(m·℃), backed by a 150-mm of common brick, of conductivity 0.8 W/(m·℃), the temperature of inner face of the wall is 1400℃, and that of the outer face is 200℃ BA=0.12m BB=0.15m To=200℃ Ti=1400℃ T2 kB=0. 8 W/(m∙℃) kA=0.08 W/(m∙℃) 19 Example 4.2 (a) What is the heat loss through the wall, in W per square meter T T0 T3 1400 200 1200℃ BA=0.12m BB=0.15m T3=200℃ T0=1400℃ T2 RA BA 0.12 1.5 m2 ℃/W k A 0.08 RB BB 0.15 0.1875 m2 ℃/W 0.8 kB According to Eq.(4.2-9) kB=0. 8 W/(m∙℃) kA=0.08 W/(m∙℃) 1200 q T 711.1W m 2 1.5 0.1875 A RA RB 20 Example 4.2 b) To reduce the heat loss to 600 W/m2 by adding a layer of cork (软木) with k= 0.2 W/(m·℃) on the outside of common brick, how many meters of cork are required? BA=0.12m BB=0.15m BC=? m q 600 W m 2 A According to Eq.(4.2-9) q T 1200 600 W m 2 A RA RB RC 1.5 0.1875 RC To=200℃ Ti=1400℃ T2 T3 BC BC RC 0.3125 m2 ℃/W k C 0.2 kC=0. 2 W/(m∙℃) B 0.0625m 62.5mm C kB=0. 8 W/(m∙℃) kA=0.08 W/(m∙℃) 21 Heat flow through a cylinder Consider the hollow cylinder represented by Fig.4.3 The inside radius of the cylinder is ri, the outside radius is ro, and the length of the cylinder is L. ri, Ti The temperature of outside surface is To, and the that of inside surface is Ti. The thermal conductivity of the material of which the cylinder is made is k. L ro, T0 22 It is desired to calculate the rate of heat flow outward for this case. At radius r from the center, the heat flow rate is q and the area through which it flows is A. At steady state the heat flow rate is constant. Since heat flows only in the r direction, Eq. (4.2-7) becomes ri, Ti r L q q dT k A 2rL dr (4.2-13) ro, T0 23 Rearranging Eq(4.2-13) and integrating between limits gives ro ri dr 2Lk To 2Lk dT T i r q q 2Lk ln ro ln ri q Ti To dT r Ti To 2Lk Ti To q ln ro ri ri, Ti L (4.2-14) the logarithmic (对数) mean rL ro ri ln ro ri q (4.2-15) k AL Ti To ro ri AL 2Lro ri ln ro ri (4.2-16) ro, T0 (4.2-17) 24 The logarithmic mean is less convenient than the arithmetic (算术) mean, and the latter can be used without appreciable error for thin-walled tubes, where ro/ri is nearly 1. 96% rL The ratio of the logarithmic mean to the arithmetic mean is a function of ro/ri as shown in Fig4.3 ro ri ln ro ri (4.2-15) 25 Formula Derivation r1 r0 k 2 2 1 k1 r2 26 Formula Derivation r1 r0 k 2 2 1 k1 r2 According to Eq.(4.2-14) q1 2Lk1 T1 T0 ln r0 r1 T1 T0 T1 q2 2Lk 2 T2 T1 ln r1 r2 T2 T1 T2 Steady-state conduction q q1 q2 ln ri 1 ri q1 ln r0 r1 q2 ln r1 r2 q T T1 T2 2 Lk 2Lk1 2Lk 2 i i Heat loss per meter for multi-layer cylinder q L T ln ri 1 ri i 2k i 27 Exercise A steel tube with dimension Ф60 ×3mm is backed by a 30-mm thick cork, with a thermal conductivity of 0.043 W/(m·℃) , and is followed by a 40-mm layer of some insulating material, of 0.07 W/(m·℃) . The temperature of the outer face of steel tube is -110℃, and that of the outer face of insulating material is 10℃. What is the heat loss per meter tube through the wall? 28 Summary Conduction—If a temperature gradient exists in a continuous substance, heat can flow unaccompanied by any observable motion of matter. For steady one-dimensional conduction q T A BA k A BB k B BC kC (4.2-11) Additional formula q L T ln ri 1 ri i 2k i Uniform units Draw schematic diagram 29 Problem 4.1 A layer of pulverized cork 152-mm thick is used as a layer of thermal insulation in a flat wall. The temperature of the cold side of the cork is 4.4 ℃, and that of the warm side is 82.2 ℃. The thermal conductivity of the cork at 0 ℃ is 0.036 W/(m ·℃), and that at 93.3℃ is 0.055 W/(m ·℃). The area of the wall is 2.32m2. What is the rate of heat flow through the wall in watts? 30 Steps & traps 1、draw a schematic diagram 2、write out the known quantity and unify the units 3、analysis the relationship between the known’s and unknown’s 4、be careful with some details such as diameter, temperature dependent physical properties, average temperature 31 Problem 4.3 A tube of 60-mm outer diameter (OD) is insulated with a 50mm layer of silica foam, for which the conductivity is 0.055 W/(m · ℃ ), followed with a 40-mm layer of cork with a conductivity of 0.05 W/(m ·℃). If the temperature of the outer surface of the pipe is 150℃ and the temperature of the outer surface of the cork is 30℃, calculate the heat loss in watts per meter of pipe. 32 Problem 4.5 The outer diameter of a steel tube is 150mm. The tube wall is backed by two insulating layers to reduce the heat loss. The ratio of thermal conductivity of two insulating materials is k2/k1=2 and both insulating materials have the same thickness 30mm. If the temperature difference between pipe wall and outer surface of insulating material is constant, which insulating material should be backed inside to enable less heat loss? 33