Principles of Chemical Engineering Chapter 4. Heat Transfer Yanwei Wang (王衍伟)* Nan Fu (傅楠)* Department of Polymer Science and Engineering College of Chemistry, Chemical Engineering and Materials science 苏州大学材化部高分子科学与工程系 *Email: [email protected]. Office: 907-1427 Chemical Engineering Innovation Laboratory College of Chemistry, Chemical Engineering and Materials science 苏州大学材化部化工创新实验室 *Email: [email protected]. Office: 701-1517 Soochow University 1 What is heat? Soochow University 2 Heat is energy. Heat is always associated with a process of some kind. Heat naturally flows from hotter to colder systems. Soochow University 3 Common Temperature Scales Hot Cold Soochow University 4 Conversion between temperature units [°C] = ( [°F] – 32 )(5/9) [°K] = ( [°C] + 273.15 ) Soochow University 5 Zeroth Law of Thermodynamics If bodies A and B are each in thermal equilibrium with a third body C, then they are in thermal equilibrium with each other. Translation: Every body has a property called temperature. When two bodies (one might be a thermometer) are found to be in thermal equilibrium, then their temperatures are the same. This may be used to determine the temperature of a third body (through calibration). Soochow University 6 DEFINITION OF HEAT Heat is energy that flows from a higher temperature object to a lower temperature object because of a difference in temperatures. SI Unit of Heat: joule (J) OTHER UNITS 1 kcal = 4186 joules 1 cal = 4.186 joules 1 BTU = 1055 J The heat that flows from hot to cold originates in the internal energy of the hot substance. It is NOT correct to say that a substance contains heat. Soochow University 7 Heat and Temperature Change: Specific Heat Capacity The heat that must be supplied or removed to change the temperature of a substance is Q mc p T specific heat capacity Common Unit for Specific Heat Capacity: J/(kg·K) Soochow University 8 Exercise: When you drink cold water, your body must expend metabolic energy to maintain normal body temperature of 37 oC by warming up the water in your stomach. Could drinking ice water substitute for exercise as a way to “burn calories?” Suppose you expend 430 kilocalories during a brisk one-hour walk. How many liters of ice water would you have to drink in order to use 430 kilocalories of metabolic energy? Note: the stomach can hold about one liter. Soochow University 9 The three modes of heat transfer Soochow University 10 Soochow University 11 Heat conduction • Heat conduction is a mode of transfer of energy due to a temperature gradient. • Heat flows from the region of higher temperature to that at a lower temperature unaccompanied by any observable macroscopic motion. • Conduction is a microscopic–level mechanism, resulting from the exchange of translational, rotational, and vibrational energy among the molecules comprising the medium. • In metals, heat conduction results from the motion of free electrons. Soochow University 12 Convection Convection is the process in which heat is carried from one place to another by the bulk movement of a fluid. Natural convection: Movement of a fluid is due to buoyant forces created by density differences, and the density differences are caused by temperature gradients in the fluid. Soochow University 13 Thermal Radiation RADIATION A process in which energetic particles or energetic waves travel through a vacuum, or through mattercontaining media that are not required for their propagation. Soochow University 14 Heat Transfer by Conduction Soochow University 15 Heat conduction • In solids that are poor conductors of electricity and in most liquids, heat conduction results from momentum transfer between adjacent vibrating molecules or atoms. • In gases, heat conduction occurs by the random motion of molecules, so heat is diffused from hotter regions to colder ones. • Check Appendices 9-12 for thermal conductivity data. Soochow University 16 Fourier’s law for heat conduction Soochow University 17 Thermal conductivity The magnitude of thermal conductivity spans five orders of magnitude. Soochow University 18 Steady state heat conduction • steady flow energy equation q0 all surfaces q : rate of heat flow in direction normal to surface. • For the 1D case shown in the figure, the heat transfer rate in at the left (at x) is q( x dx) q ( x) dT q( x) kA dx x • The heat transfer rate on the right is q( x dx) q( x) • Hence, q( x dx) q( x) 0 dq dx dx x (Taylor expansion) dq d dT kA dx dx dx Soochow University 0 A and k may be functions of x. 19 Thermal Resistance Circuits R = B/K is the thermal resistance ((m2·K)/W). 1/R – Thermal conductance (heat transfer coefficient h (W/(m2·K))). Rth = B/(kA) is the absolute thermal resistance (K/W). Thermal resistors in series TL TR TL TR q = Rtotal R1 R2 Thermal resistors in Parallel q q1 q2 q3 1 Rtotal R11 R21 R31 q q Soochow University 20 Example in the textbook RA BA k A A RB BB RC BC kB A kC A TA TB TC q RA RB RC T TA TB TC q RA RB RC q TA = BA kA A Requirement: Understand and be able to draw the temperature profile. Soochow University 21 Supplementary - 1 Fourier’s law in cylindrical and spherical coordinates Fourier's law: j k T T T T xˆ yˆ zˆ x y z where xˆ, yˆ , zˆ are the unit vectors in their respective directions. In Cartesian coordinates ( x, y, z ) : T T 1 T ˆ T In Cylindrical coordinates ( r , , z ) : T rˆ zˆ r r z where rˆ, ˆ, zˆ are the unit vectors in their respective directions. In Spherical coordinates ( r , , ) : T T 1 T ˆ 1 T ˆ rˆ r r r sin where rˆ, ˆ, ˆ are the unit vectors in their respective directions. Soochow University 22 Sample derivation (for the cylindrical case) Soochow University 23 T in cylindrical coordinates (r , , z ) xˆ cos rˆ sin ˆ & yˆ sin rˆ cos ˆ zˆ zˆ rˆ cos xˆ sin yˆ ˆ sin xˆ cos yˆ zˆ zˆ x r cos y r sin z z r x 2 y 2 arctan y x z z T T x r T T y r z z r T x r T y & T sin T cos x r r T cos T sin y r r & T T r x T T x z z x T y T T cos sin r y r x y x T y T T r sin r cos y x y T T T T sin T ˆ xˆ yˆ zˆ cos cos rˆ sin x y z r r T cos T ˆ T zˆ T rˆ 1 T ˆ T zˆ sin sin rˆ cos r r z r r z T Soochow University 24 Example-1: Heat conduction through a plane slab Steady state heat conduction d dT kA 0 dx dx For constant k and A, we have dT C1 T C1 x C2 dx where C1 and C2 are constants of integration. Boundary conditions: T T1 at x x1 T T2 at x x2 T T1 x x1 T2 T1 x2 x1 Temperature distribution through the thin slab. Rate of heat flow T T T1 T2 dT q kA kA 1 2 dx x2 x1 x2 x1 kA Soochow University Note: • the linear temperature profile • the direction of heat flow (from regions of high temperature to regions of low temperature.) • the analogy with Ohm’s law and Poiseuille’s law 25 Steady Quasi-One-Dimensional Heat Flow in Non-Planar Geometry Heat is lost from a hot-water pipe to the air outside in the radial direction. Heat transfer from a long pipe is one dimensional Soochow University 26 Steady heat flow through a cylindrical shell q dT k A dr q 2rL(k dT T T k AL (T1 T2 ) ) 2rLk 1 2 dr ln( r2 / r1 ) (r2 r1 ) r2 r1 where AL 2 r L L, and r L ln( r2 / r1 ) Heat transfer q k T B through a thin A slab kA(T1 T2 ) q Soochow University ( x2 x1 ) 27 Logarithmic mean radius The logarithmic mean is a function of two non-negative numbers which is equal to their difference divided by the logarithm of their quotient. The logarithmic mean of two numbers is smaller than the arithmetic mean but larger than the geometric mean (unless the numbers are the same, in which case all three means are equal to the numbers). rg rL ra rrms When r0/ri is close to 1, rL approximately equals ra. logarithmic mean: rL ro ri ln ro ri arithmetic mean: ra ro ri 2 geometric mean: rg ro ri root mean square: rrms 2 2 r r o i 2 Soochow University 28 Steady heat flow through a cylindrical shell d dT kA 0 , A 2 rL dr dr dT r C1 where C1 is a constant of integration. dr d r r1 dT C1 T C1 ln r r1 C2 r r1 Boundary conditions: T T1 at r r1 , T T2 at r r2 T T1 ln r r1 T2 T1 ln r2 r1 Temperature distribution through the cross-section of the cylindrical shell. Heat transfer through a thin T T1 x x1 T2 T1 x2 x1 slab Soochow University 29 Exercise: A tube of 60-mm outer diameter (OD) is insulated with a 50-mm layer of silica foam, for which the conductivity is 0.055 W/(m·°C), followed with a 40-mm layer of cork with a conductivity of 0.05 W/(m·°C). If the temperature of the outer surface of the pipe is 150°C and the temperature of the outer surface of the cork is 30oC, calculate the heat loss in watts per meter of pipe. Soochow University 30 Supplementary-2: The heat equation Equation of change for temperature, in terms of the heat flux vector q and the viscous (for unsteady heat conduction) momentum flux tensor τ ^ C p is the volumetric specific heat capacity, describing the ability to store internal energy without phase transition. Soochow University 31 Derivation of the heat equation (1/2) • The heat equation is derived based on the conservation of energy and Fourier’s law. • For simplicity, we assume – temperature is uniform over cross-sections (one-dimensional) – Heat transfer is only by conduction – No heat escapes from sides (perfect insulation) x Ax c p T q( x, t ) q( x x, t ) t T ( x, t ): temperature q ( x, t ): rate of heat flow c p : specific heat capacity q ( x, t ) T ( x, t ) T ( x x, t ) k A x : density q ( x x, t ) T ( x x, t ) T ( x, t ) k A x k: thermal conductivity Soochow University 32 Derivation of the heat equation (1/2) Ax c p T q( x, t ) q( x x, t ) t Ax c p T kA T ( x, t ) T ( x x, t ) T ( x x, t ) T ( x, t ) kA t x x T ( x x, t ) T ( x, t ) T ( x, t ) T ( x x, t ) T k x x t c p x T k 2T t c p x 2 (for constant k ) T k 2 T where = t cp Soochow University 33 Heat Transfer by Conduction Summary Soochow University 34 • The relationship between temperature and heat • Three concepts: – Specific heat capacity – Thermal conductivity – Thermal resistance (circuits) • Definition - heat conduction • Calculation - Fourier’s law • Steady state heat conduction in thin slab and cylindrical shell • Logarithmic mean radius • Supplementary – The heat equation (for unsteady heat conduction) – Transformation between Cartesian coordinates and Cylindrical coordinates Soochow University 35
© Copyright 2026 Paperzz