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- 1. A Course on HEATTRANSFER
- 2. Content of the Course UnitNo. 01 • Introduction to Heat Transfer • Steady State Heat Conduction Unit No. 02 • Overall Heat Transfer • Heat Source System • Extended Surfaces • Unsteady state heat Conduction Unit No. 03 • Principles of Convection • Free and Forced Convection Unit No. 04 • Heat Exchanger • Parallel and Counter flow HE • TEMA Standards Unit No. 05 • Radiation Heat Transfer
- 3. Prof. G. P.Badarkhe Course Outcomes At the end of the unit you will be able to : CO1 • Explain the laws of heat transfer and deduce the general heat conduction equation and to explain it for 1-D steady state heat transfer in regular shape bodies CO2 • Describe the critical radius of insulation, overall heat transfer coefficient, thermal conductivity and lumped heat transfer CO3 • Interpret the extended surfaces CO4 • Illustrate the boundary layer concept, dimensional analysis, forced and free convection under different conditions CO5 • Describe the Boiling heat transfer, Evaluate the heat exchanger and examine the LMTD and NTU methods applied to engineering problems CO6 • Explain the thermal radiation black body, emissivity and reflectivity and evaluation of view factor and radiation shields
- 4. Assessment Strategy Continuous Assessment (20Marks) • CAI- 10 Marks • CAII- 10 Marks Mid Sem Exam (20 Marks) End Sem Exam (60 Marks)
- 5. Books to Referfor Heat Transfer 1. S. P. Sukhatme, “A Textbook on Heat Transfer”, Tata McGraw Hill Publications, 3rd edition. 2. Y. A. Cengel, “Heat Transfer A Practical Approach”, Tata McGraw Hill Publication s,3rd edition, 2006. 3. F. P. Incoropera, D. P. Dewitt, “Fundamentals of Heat and Mass Transfer”, John- Wiley, 5th edition, 1990. 4. J. P. Holman, “Heat Transfer”, Tata McGraw Hill Publications, 9th edition, 2004. 5. R. K. Rajput, “Heat and Mass Transfer”
- 6. Pre-Requisites for Heattransfer Calculus Fluid Mechanics Thermodynamics
- 7. Approach towards theCourse Theoretical Experimental Computational
- 8. Macroscopic and Microscopic Pointof View Macro means Big or at Bulk • Considers certain quantity of matter(Overall behaviour) • Requires simple mathematical Formulae • Properties of the system are average values • Few properties are needed to describe the system Micro means Small or at Molecular level • Concerned directly with the structure of matter • Analysis requires statistical method as the number of molecules are very large • Properties can not be easily measured by instruments • Large number of variables are needed to describe the system
- 9. Difference Between Heatand Temperature HEAT Heat is energy in transit. Heat flows from hot body to cold body Temperature diff. is the driving force Usually denoted by Q Unit is joules or calorie TEMPERATURE Temperature is a measure of the amount of energy possessed by the molecules of a substance. Degree of Hotness can be used to predict the direction of heat transfer Usually denoted by T Unit is degree Celsius or Kelvin
- 10. Heat Transfer Heat,which is the form of energy that can be transferred from one system to another as a result of temperature difference. The science that deals with the determination of the rates of such energy transfers is heat transfer. Temperature difference is the driving force. Heat always flows in the direction of decreasing temperature. Heat transfer rate is expressed in terms of Watt or Kilowatt.
- 11. Difference Between Thermodynamicsand Heat Transfer Thermodynamics tells us: •How much heat is transferred How much work is done Final state of the system Heat transfer tells us: How (with what modes) Q is transferred At what rate Q is transferred Temperature distribution inside the body Example: Hot steel bar placed in a cold water bath
- 12. Applications of HeatTransfer Energy production and conversion Steam Power plant, Solar energy conversion etc. Refrigeration and air conditioning Domestic applications Ovens, Stoves, Toaster Cooling of electronic equipment Manufacturing / materials processing Welding, Casting, Soldering, Laser machining Automobiles / aircraft design Nature (weather, climate etc) Chemical Industries Human Body
- 13. Applications of HeatTransfer Contd..
- 14. Rating and Sizingtype of Problem Rating type of problem deal with the determination of heat transfer rate for an existing system at a specified temperature difference. Sizing type of problem deal with the determination of the size of a system in order to transfer heat at a specified rate for a specified temperature difference.
- 15. Outcome of UnitNo. 01 At the end of the unit you will be able to : • Explain the laws of heat transfer and deduce the general heat conduction equation and explain it for 1-D steady state heat transfer in regular shape bodies • Identify different modes of heat transfer in engineering applications
- 16. Modes of HeatTransfer Conduction Convection Radition
- 17. Conduction It is thetransfer of energy from the more energetic particles of substance to the adjacent less energetic ones as a result of interaction between the particles
- 18. Convection Mode of energytransfer between a solid surface and the adjacent fluid that is in motion and it involves the combined effect of conduction and fluid motion In the absence of any bulk fluid motion, heat transfer between the solid surface and adjacent fluid is by pure conduction
- 19. Types of Convection NaturalConvection Fluid motion occurs due to density variation caused by temperature difference Forced Convection Fluid motion caused by external agency
- 20. Practical Examples ofNatural and Forced Convection
- 21. • Liquids getsconverted into vapours Boiling • Vapours gets converted into liquids Condensation
- 22. Radiation Energy transfer inthe form of electromagnetic waves because of vibrational rotational motion of molecules and atoms which makes up the matter or substance • All physical matter emits thermal radiation at a particular level of temperature • It does not require medium for the energy transfer to occur
- 23. Examples of radiationmode of heat transfer
- 24. Combined example ofall modes of heat transfer
- 25. Laws of HeatTransfer Fundamental Laws • Law of conservation of mass Mass of a closed system is constant in the absence of nuclear reaction For a control volume : Mass entering –Mass leaving = increase of mass • Newton’s law of motion Rate of change of momentum is directly proportional to applied force 𝑭 = 𝒅 𝒅𝒕 (𝒎𝑽) • First law of thermodynamics Principle of conservation of energy dE = dQ - dW Subsidiary Laws • Fourier’s law of Heat Conduction • Newton’s law of Cooling • Laws of thermal radiation
- 26. Fourier’slaw of HeatConduction 𝑸 ∝ ∆𝑻 𝑸 ∝ 𝟏 ∆𝒙 𝑸 ∝ 𝑨
- 27. 𝑹𝒂𝒕𝒆 𝒐𝒇 𝑯𝒆𝒂𝒕𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒐𝒏 ∝ 𝑨𝒓𝒆𝒂)(𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝑻𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 Fourier’s law of heat conduction states that rate of heat flow through a simple homogeneous solids is directly proportional to the area which is normal to the direction of heat flow and the change in temperature with respect to the length of the path of the heat flow 𝑸 ∝ 𝑨 ∗ 𝒅𝑻 𝐝𝒙 𝑸 𝑨 = 𝒒 ∝ 𝐝𝑻 𝐝𝒙 𝒒 = − 𝒌 𝐝𝑻 𝐝𝒙 Where q is heat flux - Sign indicates heat flow in the direction of decreasing temp.
- 28. ThermalConductivity Ability of amaterial to conduct heat Rate of heat transfer through a unit thickness of the material per unit area per unit temperature difference 𝒌 = 𝑸 ∗ 𝒅𝒙 𝑨 ∗ 𝒅𝑻 Unit of thermal conductivity : 𝑾 𝒎𝒌 𝒐𝒓 𝑾 𝒎℃ For Example Copper has Thermal Conductivity of 401 𝐖 𝐦𝐤 Thermal Conductivity depend on Material Structure Density of Material Moisture Content Temperature
- 29.
- 30. Effect of Alloying: k(Copper)=401 W/m.K K(Nickel)=91 W/m.K K(constantan: 55% Cu & 45% Ni)= 23 w/m.K Effect of Temperature: K(Metals) α 1/T Aluminium and Uranium as exception (Mercury) K(Non Metals) α T K(Gases) α T K(Liquids) α 1/T Water as exception Determination of Thermal Conductivity
- 31. Heat Capacity andThermal Diffusivity Specific Heat Cp: J/kg.K Cp(water)=4.18 kJ/kg K Cp(iron)=0.45 kJ/kg K Cp is the capacity to store thermal energy k(water)=0.608 W/m K k(iron)=80.2 W/m K k is the capacity to conduct thermal energy Heat Capacity 𝝆𝑪𝒑 : J/cubic m. k Thermal Diffusivity ∝= 𝐻𝑒𝑎𝑡 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝐻𝑒𝑎𝑡 𝑆𝑡𝑜𝑟𝑎𝑔𝑒 = 𝑘 𝜌𝐶𝑝 Unit of ∝ : Sq. m /Sec
- 32. Newtons Law ofCooling Rate of heat transfer is directly proportional to the temperature difference between surface & surrounding and area which is in contact. 𝑄𝐶𝑜𝑛𝑣. = ℎ 𝐴𝑠 𝑇𝑠 − 𝑇∞ 𝑄𝐶𝑜𝑛𝑣. ∝ 𝐴𝑠 𝑇𝑠 − 𝑇∞ 𝑯𝒆𝒂𝒕 𝑻𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕: ℎ = 𝑄𝐶𝑜𝑛𝑣. 𝐴𝑠 𝑇𝑠− 𝑇∞ = 𝑊 𝑚2𝑘 𝑜𝑟 ℃ Whose value depends upon: Surface Geometry Nature of Fluid Motion Properties of Fluid Bulk Fluid Velocity
- 33. Stefan-Boltzmann's Law It satesthat emissive power of a black body is directly proportional to fourth power of its absolute temperature. 𝑄 𝐴 = 𝐸𝑏𝑙𝑎𝑐𝑘 ∝ 𝑇4 𝐸𝑏𝑙𝑎𝑐𝑘 = 𝜎 𝑇 4 or𝑄𝐵𝑙𝑎𝑐𝑘 𝑜𝑟 𝑚𝑎𝑥= 𝜎𝐴𝑠 𝑇𝑠 4 Where 𝜎 is Stefan-Boltzmann constant and has the value of 5.67 ∗ 10−8 𝑊 𝑚2𝑘4 For Real Surfaces: 𝑄𝑅𝑒𝑎𝑙 = 𝜀𝐴𝑠 𝜎 𝑇𝑠 4
- 35.
- 36. Generalised Heat Conductionequation in Cartesian Co-ordinate systems Consider a infinitesimal small elemental volume of side dx, dy & dz as shown in figure in a medium in which temperature is varying with location and time
- 37. Let kx kykz be the thermal conductivities in x, y, and z direction respectively Energy balance equation for volumetric element (control volume) is given by: 𝑁𝑒𝑡 ℎ𝑒𝑎𝑡 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑓𝑟𝑜𝑚 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒𝑑(𝐴) + ℎ𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡(𝐵) = 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓𝑒𝑛𝑒𝑟𝑦 𝑐𝑜𝑛𝑡𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡(C)………………………………..(1) A. Net heat accumulated in the element due to conduction of heat from all the direction considered Quantity of heat entering the element in x direction per unit time i.e. Heat influx 𝑄𝑥 = − 𝑘𝑥. 𝐴. 𝜕𝑇 𝜕𝑥 𝑄𝑥 = − 𝑘𝑥. 𝑑𝑦. 𝑑𝑧 . 𝜕𝑇 𝜕𝑥
- 38. Quantity of heatleaving the element in x direction per unit time i.e. Heat efflux 𝑄𝑥+𝑑𝑥 = 𝑄𝑥 + 𝜕 𝜕𝑥 𝑄𝑥 𝑑𝑥 Heat accumulated in the element due to heat flow in x direction per unit time 𝑑𝑄𝑥 = 𝑄𝑥 − 𝑄𝑥 + 𝜕 𝜕𝑥 𝑄𝑥 𝑑𝑥 𝑑𝑄𝑥 = − 𝜕 𝜕𝑥 𝑄𝑥 𝑑𝑥 𝑑𝑄𝑥 = − 𝜕 𝜕𝑥 − 𝑘𝑥 𝑑𝑦. 𝑑𝑧 𝜕𝑇 𝜕𝑥 𝑑𝑥 𝑑𝑄𝑥 = 𝜕 𝜕𝑥 𝑘𝑥 𝜕𝑇 𝜕𝑥 𝑑𝑥. 𝑑𝑦. 𝑑𝑧
- 39. Similarly heat accumulateddue to heat flow by conduction along y and z direction will be: 𝑑𝑄𝑦 = 𝜕 𝜕𝑦 𝑘𝑦 𝜕𝑇 𝜕𝑦 𝑑𝑥. 𝑑𝑦. 𝑑𝑧 𝑑𝑄𝑧 = 𝜕 𝜕𝑧 𝑘𝑧 𝜕𝑇 𝜕𝑧 𝑑𝑥. 𝑑𝑦. 𝑑𝑧 Net heat accumulated in the element due to conduction of heat from all the direction considered per unit time = 𝜕 𝜕𝑥 𝑘𝑥 𝜕𝑇 𝜕𝑥 + 𝜕 𝜕𝑦 𝑘𝑦 𝜕𝑇 𝜕𝑦 + 𝜕 𝜕𝑧 𝑘𝑧 𝜕𝑇 𝜕𝑧 𝑑𝑥. 𝑑𝑦. 𝑑𝑧 --------------------(i) B. Total Heat generated within the element per unit time 𝑄𝑔 = 𝑞𝑔 ′′′ ∗ 𝑑𝑥. 𝑑𝑦. 𝑑𝑧 --------------- (ii)
- 40. C. Rate ofchange of energy content of the element 𝑚. 𝐶𝑝. 𝜕𝑇 𝜕𝜏 𝜌. 𝑉. 𝐶𝑝 . 𝜕𝑇 𝜕𝜏 𝜌. 𝐶𝑝. 𝜕𝑇 𝜕𝜏 . 𝑑𝑥. 𝑑𝑦. 𝑑𝑧 ------------------(iii) Putting value of (i), (ii),(iii) in equation no. (1) the heat balance equation will be 𝝏 𝝏𝒙 𝒌𝒙 𝝏𝑻 𝝏𝒙 + 𝝏 𝝏𝒚 𝒌𝒚 𝝏𝑻 𝝏𝒚 + 𝝏 𝝏𝒛 𝒌𝒛 𝝏𝑻 𝝏𝒛 𝒅𝒙. 𝒅𝒚. 𝒅𝒛 + 𝒒𝒈 ′′′ ∗ 𝒅𝒙. 𝒅𝒚. 𝒅𝒛 = 𝛒. 𝑪𝒑. 𝝏𝑻 𝝏𝝉 . 𝐝𝐱. 𝐝𝐲. 𝐝𝐳
- 41. 𝝏 𝝏𝒙 𝒌𝒙 𝝏𝑻 𝝏𝒙 + 𝝏 𝝏𝒚 𝒌𝒚 𝝏𝑻 𝝏𝒚 + 𝝏 𝝏𝒛 𝒌𝒛 𝝏𝑻 𝝏𝒛 + 𝒒𝒈 ′′′ = 𝛒.𝑪𝒑. 𝝏𝑻 𝝏𝝉 This is known as generalised heat conduction equation in cartesian co-ordinate system If material or medium is Isotropic then kx=ky=kz=k then above equation becomes 𝑘 𝜕 𝜕𝑥 𝜕𝑇 𝜕𝑥 + 𝜕 𝜕𝑦 𝜕𝑇 𝜕𝑦 + 𝜕 𝜕𝑧 𝜕𝑇 𝜕𝑧 + 𝑞𝑔 ′′′ = 𝜌. 𝐶𝑝. 𝜕𝑇 𝜕𝜏 𝝏𝟐 𝑻 𝝏𝒙𝟐 + 𝝏𝟐 𝑻 𝝏𝒚𝟐 + 𝝏𝟐 𝑻 𝝏𝒛𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝝆. 𝑪𝒑 𝒌 . 𝝏𝑻 𝝏𝝉 𝝏𝟐𝑻 𝝏𝒙𝟐 + 𝝏𝟐𝑻 𝝏𝒚𝟐 + 𝝏𝟐𝑻 𝝏𝒛𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉 𝜶= 𝒌 𝝆.𝑪𝒑 Thermal diffusivity
- 42. Simplified forms ofheat conduction equation in cartesian coordinate system Unsteady state heat flow with no internal heat generation (Fourier’s Equation) 𝝏𝟐𝑻 𝝏𝒙𝟐 + 𝝏𝟐𝑻 𝝏𝒚𝟐 + 𝝏𝟐𝑻 𝝏𝒛𝟐 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉 Steady state heat flow with heat generation(Poisson’s equation) 𝝏𝟐𝑻 𝝏𝒙𝟐 + 𝝏𝟐𝑻 𝝏𝒚𝟐 + 𝝏𝟐𝑻 𝝏𝒛𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝟎 Steady state heat flow without internal heat generation(Laplace Equation) 𝝏𝟐𝑻 𝝏𝒙𝟐 + 𝝏𝟐𝑻 𝝏𝒚𝟐 + 𝝏𝟐𝑻 𝝏𝒛𝟐 = 𝟎
- 43. Steady state,one dimensional heat flow with heat generation 𝝏𝟐 𝑻 𝝏𝒙𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝟎 Steady state, one dimensional heat flow without heat generation 𝝏𝟐 𝑻 𝝏𝒙𝟐 = 𝟎 Unsteady state, one dimensional heat flow without heat Generation 𝝏𝟐𝑻 𝝏𝒙𝟐 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉
- 44. Generalised Heat Conductionequation in Polar/Cylindrical Co-ordinate systems Consider a infinitesimally small elemental volume having the coordinates r,𝝓 & z as shown in figure in a medium in which temperature is varying with location and time
- 45. Let us assumethat Thermal Conductivity k, Density 𝛒 and specific heat Cp do not change with space Volume of the element rd𝝓.dr.dz Energy balance equation for volumetric element (control volume) is given by: 𝐍𝐞𝐭 𝐡𝐞𝐚𝐭 𝐚𝐜𝐜𝐮𝐦𝐮𝐥𝐚𝐭𝐞𝐝 𝐢𝐧 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐝𝐮𝐞 𝐭𝐨 𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐡𝐞𝐚𝐭 𝐟𝐫𝐨𝐦 𝐚𝐥𝐥 𝐭𝐡𝐞 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐜𝐨𝐧𝐬𝐢𝐝𝐞𝐫𝐞𝐝(𝐀) + 𝐡𝐞𝐚𝐭 𝐠𝐞𝐧𝐞𝐫𝐚𝐭𝐞𝐝 𝐰𝐢𝐭𝐡𝐢𝐧 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭(𝐁) = 𝐑𝐚𝐭𝐞 𝐨𝐟 𝐜𝐡𝐚𝐧𝐠𝐞 𝐨𝐟𝐞𝐧𝐞𝐫𝐲 𝐜𝐨𝐧𝐭𝐞𝐧𝐭 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 (C)………………………………..(1 ) Net heat accumulated in the element due to conduction of heat from all the direction considered Heat Generated within the element Rate of change of energy content of the element + =
- 46. A. Net heataccumulated in the element due to conduction of heat from all the direction considered Heat Flow in Radial Direction Quantity of heat entering the element in r direction (z- 𝝓 plane) per unit time i.e. Heat influx in r direction 𝑸𝒓 = −𝒌. 𝑨. 𝝏𝑻 𝝏𝒓 𝑸𝒓 = −𝒌. 𝒓𝒅𝝓. 𝒅𝒛 . 𝝏𝑻 𝝏𝒓 Quantity of heat leaving the element in r direction per unit time i.e. Heat efflux 𝐐𝐫+𝐝𝒓 = 𝐐𝒓 + 𝛛 𝛛𝒓 𝐐𝒓 𝐝𝐫
- 47. Net Heat accumulatedin the element due to heat flow in radial direction per unit time 𝐝𝐐𝒓 = 𝐐𝒓 − 𝐐𝒓 + 𝛛 𝛛𝒓 𝐐𝒓 𝐝𝒓 𝐝𝐐𝒓 = − 𝛛 𝛛𝒓 𝐐𝒓 𝐝𝐫 𝒅𝑸𝒓 = − 𝝏 𝝏𝒓 − 𝒌 𝒓𝒅𝝓. 𝒅𝒛 𝝏𝑻 𝝏𝒓 𝒅𝒓 𝒅𝑸𝒓 = 𝒌 𝒅𝒓. 𝒅𝝓. 𝒅𝒛 𝝏 𝝏𝒓 𝒓 𝝏𝑻 𝝏𝒓 𝒅𝑸𝒓 = 𝒌(𝒅𝒓. 𝒅𝝓. 𝒅𝒛) 𝒓 𝝏𝟐𝑻 𝝏𝒓𝟐 + 𝝏𝑻 𝝏𝒓 𝒅𝑸𝒓 = 𝒌(𝒅𝒓. 𝒓𝒅𝝓. 𝒅𝒛) 𝝏𝟐𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓
- 48. Heat Flow inTangential i.e. 𝝓 (r-z plane) Direction Quantity of heat entering the element in 𝝓 direction (r-z plane) per unit time i.e. Heat influx in 𝝓 direction 𝑸𝝓 = −𝒌. 𝑨. 𝝏𝑻 𝒓. 𝝏𝝓 𝑸𝝓 = −𝒌. 𝒅𝒓. 𝒅𝒛 . 𝝏𝑻 𝒓. 𝝏𝝓 Quantity of heat leaving the element in 𝝓 direction per unit time i.e. Heat efflux 𝐐𝝓+𝐝𝝓 = 𝐐𝝓 + 𝛛 𝒓. 𝛛𝝓 𝐐𝝓 𝒓𝐝𝝓 Net Heat accumulated in the element due to heat flow in tangential direction per unit time 𝐝𝐐𝝓 = 𝐐𝝓 − 𝐐𝝓+𝒅𝝓
- 49. Net Heat accumulatedin the element due to heat flow in tangential direction per unit time 𝐝𝐐𝝓 = 𝐐𝝓 − 𝐐𝝓 + 𝛛 𝒓. 𝛛𝝓 𝐐𝝓 𝒓𝐝𝝓 𝐝𝐐𝝓 = − 𝛛 𝒓. 𝛛𝝓 𝐐𝝓 𝒓𝐝𝝓 𝒅𝑸𝝓 = − 𝝏 𝒓. 𝛛𝝓 − 𝒌 𝒅𝒓. 𝒅𝒛 𝝏𝑻 𝒓. 𝛛𝝓 𝒓𝐝𝝓 𝒅𝑸𝝓 = 𝒌 𝒅𝒓. 𝒅𝝓. 𝒅𝒛 𝝏 𝝏𝝓 𝟏 𝒓 𝝏𝑻 𝝏𝝓 𝒅𝑸𝝓 = 𝒌 𝒅𝒓. 𝒓𝒅𝝓. 𝒅𝒛 𝟏 𝒓𝟐 𝝏 𝝏𝝓 𝝏𝑻 𝝏𝝓 𝒅𝑸𝝓 = 𝒌 𝒅𝒓. 𝒓𝒅𝝓. 𝒅𝒛 𝟏 𝒓𝟐 𝝏𝟐 𝑻 𝝏𝝓𝟐
- 50. Heat Flow inAxial i.e. z (r- 𝝓 plane) Direction Quantity of heat entering the element in z direction (r- 𝛟 plane) per unit time i.e. Heat influx in 𝒛 direction 𝑸𝒛 = −𝒌. 𝑨. 𝝏𝑻 𝝏𝒛 𝑸𝒛 = −𝒌. 𝒓𝒅𝝓. 𝒅𝒓 . 𝝏𝑻 𝝏𝒛 Heat accumulated in the element due to heat flow in axial direction per unit time 𝐝𝐐𝒛 = 𝐐𝒛 − 𝐐𝒛 + 𝛛 𝛛𝒛 𝐐𝒛 𝐝𝒛 𝐝𝐐𝒛 = − 𝛛 𝛛𝒛 𝐐𝒛 𝐝𝐳 Quantity of heat leaving the element in r direction per unit time i.e. Heat efflux 𝐐𝐳 = 𝐐𝒛 + 𝛛 𝛛𝒛 𝐐𝒛 𝐝𝐳
- 51. 𝐝𝐐𝐳 = − 𝛛 𝛛𝐳 −𝐤 𝐫𝐝𝛟. 𝐝𝐫 𝛛𝐓 𝛛𝐳 𝐝𝐳 𝐝𝐐𝐳 = 𝐤(𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳) 𝛛𝟐 𝐓 𝛛𝐳𝟐 Net heat accumulated in the element due to conduction of heat from all the direction considered per unit time = 𝝏𝟐𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝝏𝟐𝑻 𝝏𝝓𝟐 + 𝛛𝟐𝐓 𝛛𝐳𝟐 𝐤(𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳) --------------------(i) B. Total Heat generated within the element per unit time 𝑸𝒈 = 𝒒𝒈 ′′′ ∗ 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳 --------------- (ii)
- 52. C. Rate ofchange of energy content of the element 𝐦. 𝐂𝐩. 𝛛𝐓 𝛛𝛕 𝛒. 𝐕. 𝐂𝐩 . 𝛛𝐓 𝛛𝛕 𝛒. 𝐂𝐩. 𝛛𝐓 𝛛𝛕 . 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳 ------------------(iii) Now A + B = C Net heat accumulated in the element due to conduction of heat from all the direction considered Heat Generated within the element Rate of change of energy content of the element
- 53. 𝝏𝟐 𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝝏𝟐 𝑻 𝝏𝝓𝟐 + 𝝏𝟐 𝑻 𝛛𝒛𝟐 𝐤(𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳) 𝒒𝒈 ′′′∗ 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳 𝛒. 𝑪𝒑. 𝝏𝑻 𝝏𝝉 . 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳 + = 𝝏𝟐𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝝏𝟐𝑻 𝝏𝝓𝟐 + 𝛛𝟐𝐓 𝛛𝐳𝟐 𝐤 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳 + 𝒒𝒈 ′′′ ∗ 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳 = 𝛒. 𝐂𝐩. 𝛛𝐓 𝛛𝛕 . 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳
- 54. Dividing both sideby 𝑑𝑟. 𝑟𝑑𝜙. 𝑑𝑧 𝝏𝟐 𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝝏𝟐 𝑻 𝝏𝝓𝟐 + 𝝏𝟐 𝑻 𝝏𝒛𝟐 𝒌 + 𝒒𝒈 ′′′ = 𝝆. 𝑪𝒑. 𝝏𝑻 𝝏𝝉 Or 𝝏𝟐𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝝏𝟐𝑻 𝝏𝝓𝟐 + 𝝏𝟐𝑻 𝝏𝒛𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝝆. 𝑪𝒑 𝒌 . 𝝏𝑻 𝝏𝝉 𝝏𝟐𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝝏𝟐𝑻 𝝏𝝓𝟐 + 𝝏𝟐𝑻 𝝏𝒛𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉 For Steady state, without internal heat generation and one dimensional flow 𝝏𝟐𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 = 𝟎
- 55. Generalised Heat Conductionequation in Spherical Co-ordinate systems Consider a infinitesimally small elemental volume having the coordinates r, 𝜙 & 𝜃 as shown in figure in a medium in which temperature is varying with location and time
- 56. Let us assumethat Thermal Conductivity k, Density 𝜌 and specific heat Cp do not change with space Volume of the element dr. rd𝜃. 𝑟𝑠𝑖𝑛 𝜃d𝜙 Energy balance equation for volumetric element (control volume) is given by: 𝑁𝑒𝑡 ℎ𝑒𝑎𝑡 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑓𝑟𝑜𝑚 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒𝑑(𝐴) + ℎ𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡(𝐵) = 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓𝑒𝑛𝑒𝑟𝑦 𝑐𝑜𝑛𝑡𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡(C)………………………………..(1) Net heat accumulated in the element due to conduction of heat from all the direction considered Heat Generated within the element Rate of change of energy content of the element + =
- 57. A. Net heataccumulated in the element due to conduction of heat from all the direction considered Heat Flow in Radial (r) Direction Quantity of heat entering the element in r direction (𝜃 - 𝜙 plane) per unit time i.e. Heat influx in r direction 𝑄𝑟 = −𝑘. 𝐴. 𝜕𝑇 𝜕𝑟 𝑄𝑟 = −𝑘. rd𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙 . 𝜕𝑇 𝜕𝑟 Quantity of heat leaving the element in r direction per unit time i.e. Heat efflux 𝑄r+𝑑𝑟 = 𝑄𝑟 + 𝜕 𝜕𝑟 𝑄𝑟 𝑑r
- 58. Net Heat accumulatedin the element due to heat flow in radial direction per unit time 𝑑𝑄𝑟 = 𝑄𝑟 − 𝑄𝑟 + 𝜕 𝜕𝑟 𝑄𝑟 𝑑𝑟 𝑑𝑄𝑟 = − 𝜕 𝜕𝑟 𝑄𝑟 𝑑r 𝑑𝑄𝑟 = − 𝜕 𝜕𝑟 −𝑘 𝑟𝑑𝜃. 𝑟 sin 𝜃𝑑𝜙 𝜕𝑇 𝜕𝑟 𝑑𝑟 𝑑𝑄𝑟 = 𝑘 𝑑𝑟. 𝑑𝜃. sin 𝜃𝑑𝜙 𝜕 𝜕𝑟 𝑟2 𝜕𝑇 𝜕𝑟 𝑑𝑄𝑟 = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 1 𝑟2 𝜕 𝜕𝑟 𝑟2 𝜕𝑇 𝜕𝑟
- 59. Heat Flow in𝜙 (r-𝜃 plane) Direction Quantity of heat entering the element in 𝜙 direction (r-𝜃 plane) per unit time i.e. Heat influx in 𝜙 direction 𝑄𝜙 = −𝑘. 𝐴. 𝜕𝑇 𝑟. sin 𝜃 𝜕𝜙 𝑄𝜙 = −𝑘. 𝑑𝑟. 𝑟𝑑𝜃 . 𝜕𝑇 𝑟. sin 𝜃 𝜕𝜙 Quantity of heat leaving the element in 𝜙 direction per unit time i.e. Heat efflux 𝑄𝜙+𝑑𝜙 = 𝑄𝜙 + 𝜕 𝑟. sin 𝜃 𝜕𝜙 𝑄𝜙 𝑟. sin 𝜃 𝑑𝜙 Net Heat accumulated in the element due to heat flow in 𝜙 per unit time 𝑑𝑄𝜙 = 𝑄𝜙 − 𝑄𝜙+𝑑𝜙
- 60. Net Heat accumulatedin the element due to heat flow in 𝜙 per unit time 𝑑𝑄𝜙 = 𝑄𝜙 − 𝑄𝜙 + 𝜕 𝑟 sin 𝜃 𝜕𝜙 𝑄𝜙 𝑟 sin 𝜃. 𝑑𝜙 𝑑𝑄𝜙 = − 𝜕 𝑟 sin 𝜃 𝜕𝜙 𝑄𝜙 𝑟 sin 𝜃. 𝑑𝜙 𝑑𝑄𝜙 = − 𝜕 𝑟 sin 𝜃 𝜕𝜙 − 𝑘 𝑑𝑟. 𝑟𝑑𝜃 𝜕𝑇 𝑟 sin 𝜃 𝜕𝜙 𝑟 sin 𝜃. 𝑑𝜙 𝑑𝑄𝜙 = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 1 𝑟2 sin2 𝜃 𝜕2 𝑇 𝜕𝜙2
- 61. Heat Flow in𝜃(r-𝜙plane) Direction Quantity of heat entering the element in 𝜃 direction (r-𝜙 plane) per unit time i.e. Heat influx in 𝜃 direction 𝑄𝜃 = −𝑘. 𝐴. 𝜕𝑇 𝑟𝜕𝜃 𝑄𝜃 = −𝑘. 𝑑𝑟. 𝑟sin 𝜃𝑑𝜙 . 𝜕𝑇 𝑟𝜕𝜃 Quantity of heat leaving the element in 𝜃 direction per unit time i.e. Heat efflux 𝑄𝜃+𝑑𝜃 = 𝑄𝜃 + 𝜕 𝑟. 𝜕𝜃 𝑄𝜃 𝑟𝑑𝜃 Net Heat accumulated in the element due to heat flow in 𝜃 direction per unit time 𝑑𝑄𝜃 = 𝑄𝜃 − 𝑄𝜃+𝑑𝜃
- 62. Net Heat accumulatedin the element due to heat flow in 𝜃direction per unit time 𝑑𝑄𝜃 = 𝑄𝜃 − 𝑄𝜃 + 𝜕 𝑟𝜕𝜃 𝑄𝜃 𝑟. 𝑑𝜃 𝑑𝑄𝜃 = − 𝜕 𝑟𝜕𝜃 𝑄𝜃 𝑟𝑑𝜃 𝑑𝑄𝜃 = − 𝜕 𝑟𝜕𝜃 − 𝑘 𝑑𝑟. 𝑟sin 𝜃𝑑𝜙 𝜕𝑇 𝑟𝜕𝜃 𝑟𝑑𝜃 𝑑𝑄𝜃 = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 𝜕 𝑟2 sin 𝜃 𝜕𝜃 sin 𝜃 𝜕𝑇 𝜕𝜃
- 63. Net heat accumulatedin the element due to conduction of heat from all the direction considered per unit time = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 1 𝑟2 𝜕 𝜕𝑟 𝑟2 𝜕𝑇 𝜕𝑟 + 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 1 𝑟2 sin2 𝜃 𝜕2𝑇 𝜕𝜙2 + 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 𝜕 𝑟2 sin 𝜃 𝜕𝜃 sin 𝜃 𝜕𝑇 𝜕𝜃 1 𝑟2 𝜕 𝜕𝑟 𝑟2 𝜕𝑇 𝜕𝑟 + 1 𝑟2 sin2 𝜃 𝜕2 𝑇 𝜕𝜙2 + 𝜕 𝑟2 sin 𝜃 𝜕𝜃 sin 𝜃 𝜕𝑇 𝜕𝜃 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 --------------------(i) B. Total Heat generated within the element per unit time 𝑄𝑔 = 𝑞𝑔 ′′′ ∗ 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 --------------- (ii)
- 64. C. Rate ofchange of energy content of the element 𝑚. 𝐶𝑝. 𝜕𝑇 𝜕𝜏 𝜌. 𝑉. 𝐶𝑝 . 𝜕𝑇 𝜕𝜏 𝜌. 𝐶𝑝. 𝜕𝑇 𝜕𝜏 . 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 ------------------(iii) Now A + B = C Net heat accumulated in the element due to conduction of heat from all the direction considered Heat Generated within the element Rate of change of energy content of the element
- 65. 𝟏 𝒓𝟐 𝝏 𝝏𝒓 𝒓𝟐 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝐬𝐢𝐧𝟐 𝜽 𝝏𝟐 𝑻 𝝏𝝓𝟐 + 𝝏 𝒓𝟐𝐬𝐢𝐧 𝜽 𝝏𝜽 𝐬𝐢𝐧 𝜽 𝝏𝑻 𝝏𝜽 𝒌 𝒅𝒓. 𝒓𝒅𝜽. 𝒓𝐬𝐢𝐧 𝜽𝒅𝝓 𝒒𝒈 ′′′ ∗ 𝒅𝒓. 𝒓𝒅𝜽. 𝒓𝒔𝒊𝒏 𝜽𝒅𝝓 𝝆. 𝑪𝒑. 𝝏𝑻 𝝏𝝉 . 𝒅𝒓. 𝒓𝒅𝜽. 𝒓𝒔𝒊𝒏 𝜽𝒅𝝓 + = 1 𝑟2 𝜕 𝜕𝑟 𝑟2 𝜕𝑇 𝜕𝑟 + 1 𝑟2 𝑠𝑖𝑛2 𝜃 𝜕2 𝑇 𝜕𝜙2 + 𝜕 𝑟2 𝑠𝑖𝑛 𝜃 𝜕𝜃 𝑠𝑖𝑛 𝜃 𝜕𝑇 𝜕𝜃 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙 + 𝑞𝑔 ′′′ ∗ 𝑑𝑟. 𝑟𝑑𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙 = 𝜌. 𝐶𝑝. 𝜕𝑇 𝜕𝜏 . 𝑑𝑟. 𝑟𝑑𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙
- 66. Dividing both sideby 𝑑𝑟. 𝑟𝑑𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙 𝟏 𝒓𝟐 𝝏 𝝏𝒓 𝒓𝟐 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝒔𝒊𝒏𝟐 𝜽 𝝏𝟐 𝑻 𝝏𝝓𝟐 + 𝝏 𝒓𝟐 𝒔𝒊𝒏 𝜽 𝝏𝜽 𝒔𝒊𝒏 𝜽 𝝏𝑻 𝝏𝜽 𝒌 + 𝒒𝒈 ′′′ = 𝝆. 𝑪𝒑. 𝝏𝑻 𝝏𝝉 Or 𝟏 𝒓𝟐 𝝏 𝝏𝒓 𝒓𝟐 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝒔𝒊𝒏𝟐 𝜽 𝝏𝟐𝑻 𝝏𝝓𝟐 + 𝝏 𝒓𝟐 𝒔𝒊𝒏 𝜽 𝝏𝜽 𝒔𝒊𝒏 𝜽 𝝏𝑻 𝝏𝜽 + 𝒒𝒈 ′′′ 𝒌 = 𝝆. 𝑪𝒑 𝒌 . 𝝏𝑻 𝝏𝝉 𝟏 𝒓𝟐 𝝏 𝝏𝒓 𝒓𝟐 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝒔𝒊𝒏𝟐 𝜽 𝝏𝟐𝑻 𝝏𝝓𝟐 + 𝝏 𝒓𝟐 𝒔𝒊𝒏 𝜽 𝝏𝜽 𝒔𝒊𝒏 𝜽 𝝏𝑻 𝝏𝜽 + 𝒒𝒈 ′′′ 𝒌 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉 For Steady state, without internal heat generation and one dimensional flow 𝟏 𝒓𝟐 𝝏 𝝏𝒓 𝒓𝟐 𝝏𝑻 𝝏𝒓 = 𝟎
- 67. For CartesianCoordinate system 𝝏𝟐 𝑻 𝝏𝒙𝟐 + 𝝏𝟐 𝑻 𝝏𝒚𝟐 + 𝝏𝟐 𝑻 𝝏𝒛𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉 For Polar/Cylindrical Coordinate system 𝝏𝟐 𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝝏𝟐 𝑻 𝝏𝝓𝟐 + 𝝏𝟐 𝑻 𝝏𝒛𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉 For Spherical Coordinate system 𝟏 𝒓𝟐 𝝏 𝝏𝒓 𝒓𝟐 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝒔𝒊𝒏𝟐 𝜽 𝝏𝟐𝑻 𝝏𝝓𝟐 + 𝝏 𝒓𝟐 𝒔𝒊𝒏 𝜽 𝝏𝜽 𝒔𝒊𝒏 𝜽 𝝏𝑻 𝝏𝜽 + 𝒒𝒈 ′′′ 𝒌 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉
- 68. Initial and BoundaryCondition Solving a differential equation is essentially a process of removing derivatives, or an integration process, and thus the solution of a differential equation typically involves arbitrary constants The mathematical expressions of the thermal conditions at the boundaries are called the boundary conditions Thermal Condition which is usually specified at time t=0 is called initial Condition As the generalized heat conduction equation is second order in space coordinate thus the boundary condition may involve first derivative at the boundaries as well as specified value of temperature Heat conduction equation is first order in time thus the initial condition can not involve any derivatives
- 69. 1. Specified TemperatureBoundary Condition i.e. Surface temperature is constant 𝑻 𝟎, 𝒕 = 𝑻𝟏 𝑻 𝑳, 𝒕 = 𝑻𝟐 2. Specified Heat Flux Boundary Condition i.e. Heat Flux(Heat rate per unit surface area) is constant 𝒒𝒔 = −𝒌 𝝏𝑻 𝝏𝒙 𝒙=𝟎,𝒕
- 70. 3. Adiabatic orinsulated wall Boundary Condition i.e. Heat Flux(Heat rate per unit surface area) is constant 𝒒𝒔 = 𝟎 = 𝒌 𝝏𝑻 𝝏𝒙 𝒙=𝟎,𝒕 𝝏𝑻 𝝏𝒙 𝒙=𝟎,𝒕 = 𝟎 4. Convection Boundary Condition i.e. Heat Flux(Heat rate per unit surface area) is constant −𝒌 𝝏𝑻 𝝏𝒙 𝟎,𝒕 = 𝒉𝟏 𝑻∞𝟏 − 𝑻 𝟎,𝒕) −𝒌 𝝏𝑻 𝝏𝒙 𝑳,𝒕 = 𝒉𝟐 𝑻 𝑳,𝒕) − 𝑻∞𝟐
- 71. Solution Procedure forsolving heat conduction problem Heat Transfer Problem Mathematical Formulation (Diff. Equation & Boundary condition) General Solution of Diff. Equation Application of Boundary Condition Solution of the Problem
- 72. One Dimensional SteadyState Heat Conduction Through Slab Consider a plane wall of homogeneous material through which heat is flowing in x direction Let L – Thickness of slab/wall k – Thermal Conductivity A – Area perpendicular to heat flow T1 & T2 are the temperatures at two faces The general heat conduction equation in cartesian coordinate system is given by 𝝏𝟐𝑻 𝝏𝒙𝟐 + 𝝏𝟐𝑻 𝝏𝒚𝟐 + 𝝏𝟐𝑻 𝝏𝒛𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉 = 0 Final equation in case of steady, one dimensional and without heat generation will be 𝝏𝟐𝑻 𝝏𝒙𝟐 = 𝟎 or 𝒅𝟐𝑻 𝒅𝒙𝟐 = 𝟎
- 73. On integrating twice 𝑑𝑇 𝑑𝑥 =𝐶2 and 𝑇 = 𝐶1𝑥 + 𝐶2 …………….(i) 𝐶𝟏 & 𝐶𝟐 are the arbitrary constant and can be calculated by applying boundary condition At x = 0, T=T1 At x = L, T=T2 Applying first boundary condition we get 𝑇1 = 𝐶2 Applying second boundary condition we get 𝑇2 = 𝐶1𝐿 + 𝐶2 Here 𝐶1 = 𝑻𝟐−𝑻𝟏 𝑳 Equation i reduced to 𝑻 = 𝑻𝟐 − 𝑻𝟏 𝑳 𝒙 + 𝑻𝟏
- 74. For Heat Flowthrough Slab/Wall We have from Fourier’s law of heat conduction 𝑸 = −𝒌𝑨 𝒅𝑻 𝒅𝒙 𝒅𝑻 𝒅𝒙 = 𝒅 𝒅𝒙 𝑻𝟐 − 𝑻𝟏 𝑳 𝒙 + 𝑻𝟏 𝒅𝑻 𝒅𝒙 = 𝑻𝟐 − 𝑻𝟏 𝑳 ∴ 𝑸 = −𝒌𝑨 𝑻𝟐 − 𝑻𝟏 𝑳 Or 𝑸 = 𝒌𝑨 𝑻𝟏 − 𝑻𝟐 𝑳 𝑸 = 𝑻𝟏 − 𝑻𝟐 𝑳 𝒌𝑨
- 75. Alternative Method We havefrom Fourier’s law of heat conduction 𝑄 = −𝑘𝐴 𝑑𝑇 𝑑𝑥 𝑄. 𝑑𝑥 = −𝑘𝐴. 𝑑𝑇 Integrating on both side 𝑄 𝑑𝑥 = −𝑘𝐴 𝑑𝑇 𝑇2 𝑇1 𝐿 0 𝑄 𝑑𝑥 = 𝑘𝐴 𝑑𝑇 𝑇1 𝑇2 𝐿 0 𝑄. 𝑥 0 𝐿 = 𝑘𝐴 𝑇 𝑇2 𝑇1 𝑄. 𝐿 = 𝑘𝐴 𝑇1 − 𝑇2 ……….(i) 𝑸 = 𝒌𝑨 𝑻𝟏 − 𝑻𝟐 𝑳 𝑄 𝑑𝑥 = −𝑘𝐴 𝑑𝑇 𝑇 𝑇1 𝑥 0 𝑄. 𝑥 = 𝑘𝐴 𝑇1 − 𝑇 ……….(ii) Dividing ii by I we get 𝑄. 𝑥 𝑄. 𝐿 = 𝑘𝐴 𝑇1 − 𝑇 𝑘𝐴 𝑇1 − 𝑇2 𝑻𝟏 − 𝑻 𝑻𝟏 − 𝑻𝟐 = 𝒙 𝑳
- 76. Variable Thermal Conductivity Thedependence of thermal conductivity on temperature can be expressed as- 𝐤 = 𝐤𝟎 𝟏 + 𝛃𝐓 𝐤𝟎 - thermal conductivity at reference 𝛃 – is constant for given material T – Temperature According to Fourier’s Law 𝑸 = −𝒌𝑨 𝒅𝑻 𝒅𝒙 𝑸 = −𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻 𝒅𝒙 𝑸. 𝒅𝒙 = −𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻 Integrating on both side and applying boundary condition as At x = 0, T=T1 At x = L, T=T2 We get 𝑸 𝒅𝒙 = − 𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻 𝑻𝟐 𝑻𝟏 𝑳 𝟎 𝑸 𝒅𝒙 = − 𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻 𝑻𝟐 𝑻𝟏 𝑳 𝟎 𝑸. 𝒙 𝟎 𝑳 = −𝒌𝟎𝑨 𝑻 + 𝜷 𝑻𝟐 𝟐 𝑻𝟏 𝑻𝟐 𝑸. 𝑳 = −𝒌𝟎𝑨 𝑻𝟐 − 𝑻𝟏 + 𝜷 𝟐 𝑻𝟐 𝟐 − 𝑻𝟏 𝟐 𝑸. 𝑳 = 𝒌𝟎𝑨 𝑻𝟏 − 𝑻𝟐 + 𝜷 𝟐 𝑻𝟏 − 𝑻𝟐 𝑻𝟏 + 𝑻𝟐 𝑸. 𝑳 = 𝒌𝟎𝑨 𝑻𝟏 − 𝑻𝟐 𝟏 + 𝜷 𝟐 𝑻𝟏 + 𝑻𝟐 𝒌𝒎 = 𝒌𝟎 𝟏 + 𝜷 𝟐 𝑻𝟏 + 𝑻𝟐 𝑸 = 𝒌𝒎𝑨 𝑻𝟏 − 𝑻𝟐 𝑳
- 77. Electrical Analogy As perFourier’s Law 𝑸 = −𝒌𝑨 𝒅𝑻 𝒅𝒙 𝑸 = 𝒌𝑨 𝑻𝟏 − 𝑻𝟐 𝑳 𝑸 = 𝑻𝟏 − 𝑻𝟐 𝑳 𝒌𝑨 𝑸 = 𝒅𝑻 𝑹𝒕𝒉 𝑹𝒕𝒉𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒐𝒏) = 𝑳 𝒌𝑨 𝑹𝒕𝒉𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒐𝒏) = 𝟏 𝒉𝑨 As per Ohm’s Law 𝑰 = 𝒅𝑽 𝑹 I – Current flow dV – Voltage difference R – Electrical Resistance I is analogous to Q dV is analogous to dT R is analogous to 𝑹𝒕𝒉
- 78. Resistances in seriesand parallel 𝑹 = 𝑹𝒄𝒐𝒏𝒗𝟏 + 𝑹𝟏 + 𝑹𝟐 + 𝑹𝑪𝒐𝒏𝒄𝟐 𝑹 = 𝟏 𝒉𝟏𝑨 + 𝑳𝟏 𝒌𝟏𝑨 + 𝑳𝟐 𝒌𝟐𝑨 + 𝟏 𝒉𝟐𝑨 𝑹 = 𝟏 𝑨 𝟏 𝒉𝟏 + 𝑳𝟏 𝒌𝟏 + 𝑳𝟐 𝒌𝟐 + 𝟏 𝒉𝟐 𝟏 𝑹𝒆𝒒. = 𝟏 𝑹𝟏 + 𝟏 𝑹𝟐 𝟏 𝑹𝒆𝒒. = 𝟏 𝑳𝟏 𝒌𝟏𝑨𝟏 + 𝟏 𝑳𝟐 𝒌𝟐𝑨𝟐 𝑹𝒆𝒒. = 𝟏 𝟏 𝑳𝟏 𝒌𝟏𝑨𝟏 + 𝟏 𝑳𝟐 𝒌𝟐𝑨𝟐
- 79. Resistances in combination 𝑹= 𝑹𝒆𝒒. + 𝑹𝟑 + 𝑹𝒄𝒐𝒏𝒗 𝑹𝒆𝒒. = 𝟏 𝟏 𝑳𝟏 𝒌𝟏𝑨𝟏 + 𝟏 𝑳𝟐 𝒌𝟐𝑨𝟐 𝟏 𝑹𝒆𝒒. = 𝟏 𝑳𝟏 𝒌𝟏𝑨𝟏 + 𝟏 𝑳𝟐 𝒌𝟐𝑨𝟐 𝑹 = 𝟏 𝟏 𝑳𝟏 𝒌𝟏𝑨𝟏 + 𝟏 𝑳𝟐 𝒌𝟐𝑨𝟐 + 𝑳𝟑 𝒌𝟑𝑨𝟑 + 𝟏 𝒉𝑨𝟑
- 80. Heat flow throughthe composite wall/slab
- 81. Heat flow throughthe composite wall/slab 𝑸 = 𝒉𝟏𝑨 𝑻∞𝟏 − 𝑻𝟏 = 𝒌𝟏𝑨 𝑻𝟏 − 𝑻𝟐 𝑳𝟏 = 𝒌𝟐𝑨 𝑻𝟐 − 𝑻𝟑 𝑳𝟐 = 𝒉𝟐𝑨 𝑻𝟑 − 𝑻∞𝟐 In steady state conduction heat transfer rate is constant through out this composite slab so we can write 𝑸 = 𝑻∞𝟏 − 𝑻𝟏 𝟏 𝒉𝟏𝑨 = 𝑻𝟏 − 𝑻𝟐 𝑳𝟏 𝒌𝟏𝑨 = 𝑻𝟐 − 𝑻𝟑 𝑳𝟐 𝒌𝟐𝑨 = 𝑻𝟑 − 𝑻∞𝟐 𝟏 𝒉𝟐𝑨 𝑻∞𝟏 − 𝑻𝟏 = 𝑸 ∗ 𝟏 𝒉𝟏𝑨------------------(1) 𝑻𝟏 − 𝑻𝟐 = 𝑸 ∗ 𝑳𝟏 𝒌𝟏𝑨-------------------(2) 𝑻𝟐 − 𝑻𝟑 = 𝑸 ∗ 𝑳𝟐 𝒌𝟐𝑨-------------------(3) 𝑻𝟑 − 𝑻∞𝟐 = 𝑸 ∗ 𝟏 𝒉𝟐𝑨------------------(4)
- 82. 𝑻∞𝟏 − 𝑻∞𝟐= 𝑸 𝟏 𝒉𝟏𝑨 + 𝑳𝟏 𝒌𝟏𝑨 + 𝑳𝟐 𝒌𝟐𝑨 + 𝟏 𝒉𝟐𝑨 Adding 1, 2, 3 & 4 we get 𝑸 = 𝑻∞𝟏 − 𝑻∞𝟐 𝟏 𝒉𝟏𝑨 + 𝑳𝟏 𝒌𝟏𝑨 + 𝑳𝟐 𝒌𝟐𝑨 + 𝟏 𝒉𝟐𝑨 𝑸 = 𝑻∞𝟏 − 𝑻∞𝟐 𝟏 𝒉𝟏𝑨 + 𝑳𝒏 𝒌𝒏𝑨 𝒏 𝟏 + 𝟏 𝒉𝟐𝑨 𝑸 = 𝑻∞𝟏 − 𝑻∞𝟐 𝟏 𝑨 𝟏 𝒉𝟏 + 𝑳𝒏 𝒌𝒏 𝒏 𝟏 + 𝟏 𝒉𝟐 𝑸 = 𝑻𝟏 − 𝑻𝒏+𝟏 𝟏 𝑨 𝑳𝒏 𝒌𝒏 𝒏 𝟏 𝑸 = 𝑨 𝑻∞𝟏 − 𝑻∞𝟐 𝟏 𝒉𝟏 + 𝑳𝒏 𝒌𝒏 𝒏 𝟏 + 𝟏 𝒉𝟐
- 83. 𝑸 = 𝑻∞𝟏 −𝑻∞𝟐 𝟏 𝒉𝟏𝑨 + 𝑳𝟏 𝒌𝟏𝑨 + 𝑳𝟐 𝒌𝟐𝑨 + 𝟏 𝒉𝟐𝑨 Overall Heat Transfer Coefficient 𝑸 = 𝑨 𝑻∞𝟏 − 𝑻∞𝟐 𝟏 𝒉𝟏 + 𝑳𝟏 𝒌𝟏 + 𝑳𝟐 𝒌𝟐 + 𝟏 𝒉𝟐 When there is fluid to fluid heat transfer it usual practice to adopt overall heat transfer coefficient If U is the overall heat transfer coefficient then 𝑸 = 𝑼. 𝑨. 𝑻∞𝟏 − 𝑻∞𝟐 𝑸 = 𝑼. 𝑨. 𝑻∞𝟏 − 𝑻∞𝟐 = 𝑨 𝑻∞𝟏 − 𝑻∞𝟐 𝟏 𝒉𝟏 + 𝑳𝟏 𝒌𝟏 + 𝑳𝟐 𝒌𝟐 + 𝟏 𝒉𝟐
- 84. 𝑼 = 𝟏 𝟏 𝒉𝟏 + 𝑳𝟏 𝒌𝟏 + 𝑳𝟐 𝒌𝟐 + 𝟏 𝒉𝟐 𝑼. 𝑨.𝑻∞𝟏 − 𝑻∞𝟐 = 𝑻∞𝟏 − 𝑻∞𝟐 𝟏 𝒉𝟏𝑨 + 𝑳𝟏 𝒌𝟏𝑨 + 𝑳𝟐 𝒌𝟐𝑨 + 𝟏 𝒉𝟐𝑨 𝑼𝑨 = 𝟏 𝑹 0r 𝑼 = 𝟏 𝑨 𝑹 For unit area overall heat transfer coefficient is reciprocal of total thermal resistance
- 85. Thermal Contact Resistance Idealor perfect contact Actual or imperfect contact 𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = 𝑄𝐶𝑜𝑛𝑡𝑎𝑐𝑡 + 𝑄𝑔𝑎𝑝 Depends upon Surface roughness Material properties Temperature Pressure In analogous to Newton’s law of cooling It can be also expressed as 𝑄 = ℎ𝑐. 𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
- 86. Thermal Contact Resistance 𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒= ℎ𝑐. 𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 ℎ𝑐 = 𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 Where ℎ𝑐 is thermal contact conductance 𝑊 𝑚2℃ Thermal contact resistance is the inverse of thermal contact conductance 𝑅𝑐 = 1 ℎ𝑐 = 𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 Unit of thermal contact resistance is 𝑚2℃ 𝑊 Thermal contact resistance can be minimized by By applying thermal grease like silicon oil Removing air by better conducting gas like Helium or Hydrogen Inserting soft metallic foil like Tin, Silver or copper Increasing the interface pressure
- 87. One Dimensional SteadyState Heat Conduction Through Cylinder Consider a hollow cylinder of homogeneous material through which heat is flowing in r direction Let L – length of Cylinder k – Thermal Conductivity 𝑟1and 𝑟2 be the inner and outer radius of cyl. respectively. T1 & T2 are the temperatures at inner and outer surface of cylinder respectively The general heat conduction equation in Cylindrical /Polar coordinate system is given by 𝝏𝟐𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 + 𝟏 𝒓𝟐 𝝏𝟐𝑻 𝝏𝝓𝟐 + 𝝏𝟐𝑻 𝝏𝒛𝟐 + 𝒒𝒈 ′′′ 𝒌 = 𝟏 𝜶 . 𝝏𝑻 𝝏𝝉 = 0 𝝏𝟐𝑻 𝝏𝒓𝟐 + 𝟏 𝒓 𝝏𝑻 𝝏𝒓 = 𝟎 or 𝒅𝟐𝑻 𝒅𝒓𝟐 + 𝟏 𝒓 𝒅𝑻 𝒅𝒓 = 𝟎
- 88. 𝒅𝟐𝑻 𝒅𝒓𝟐 + 𝟏 𝒓 𝒅𝑻 𝒅𝒓 = 𝟎 ⟹ 𝟏 𝒓 . 𝒅 𝒅𝒓 𝒓. 𝒅𝑻 𝒅𝒓 =0 𝒅 𝒅𝒓 𝒓. 𝒅𝑻 𝒅𝒓 = 0 On integrating both side we get 𝒓. 𝒅𝑻 𝒅𝒓 = C ⟹ 𝒅𝑻 𝒅𝒓 = C 𝒓 On integrating both side we get 𝑻 = 𝑪. 𝒍𝒏 𝒓 + 𝑪𝟏 −−−− −(𝒊) C & 𝐶𝟐 are the arbitrary constant and can be calculated by applying boundary condition At 𝒓 = 𝒓𝟏 ; 𝑻 = 𝑻𝟏 𝒓 = 𝒓𝟐 ; 𝑻 = 𝑻𝟐 𝑻𝟏 = 𝑪. 𝒍𝒏 𝒓𝟏 + 𝑪𝟏 𝑻𝟐 = 𝑪. 𝒍𝒏 𝒓𝟐 + 𝑪𝟏 - 𝑻𝟏 − 𝑻𝟐 = 𝑪 𝒍𝒏 𝒓𝟏 − 𝒍𝒏 𝒓𝟐
- 89. 𝑪 = 𝑻𝟏 −𝑻𝟐 𝒍𝒏 𝒓𝟏 𝒓𝟐 = − 𝑻𝟏 − 𝑻𝟐 𝒍𝒏 𝒓𝟐 𝒓𝟏 Again we can write 𝑻𝟏 = − 𝑻𝟏 − 𝑻𝟐 𝒍𝒏 𝒓𝟐 𝒓𝟏 . 𝒍𝒏 𝒓𝟏 + 𝑪𝟏 𝑪𝟏 = 𝑻𝟏 + 𝑻𝟏 − 𝑻𝟐 𝒍𝒏 𝒓𝟐 𝒓𝟏 . 𝒍𝒏 𝒓𝟏 By putting value of 𝑪 and 𝑪𝟏 in equation no. (i) 𝑻 = 𝑻𝟏 + 𝑻𝟏 − 𝑻𝟐 𝒍𝒏 𝒓𝟐 𝒓𝟏 . 𝒍𝒏 𝒓𝟏 − 𝑻𝟏 − 𝑻𝟐 𝒍𝒏 𝒓𝟐 𝒓𝟏 . 𝒍𝒏 𝒓 𝑻 − 𝑻𝟏 × 𝒍𝒏 𝒓𝟐 𝒓𝟏 = 𝑻𝟏 − 𝑻𝟐 . 𝒍𝒏 𝒓𝟏 − 𝑻𝟏 − 𝑻𝟐 . 𝒍𝒏 𝒓 = 𝑻𝟐 − 𝑻𝟏 . 𝒍𝒏 𝒓 − 𝑻𝟐 − 𝑻𝟏 . 𝒍𝒏 𝒓𝟏 = 𝑻𝟐 − 𝑻𝟏 𝒍𝒏 𝒓 − 𝒍𝒏 𝒓𝟏 𝑻 − 𝑻𝟏 𝑻𝟐 − 𝑻𝟏 = 𝒍𝒏 𝒓 𝒓𝟏 𝒍𝒏 𝒓𝟐 𝒓𝟏
- 90. For Heat Flowthrough Slab/Wall We have from Fourier’s law of heat conduction 𝑸 = −𝒌𝑨 𝒅𝑻 𝒅𝒓 𝒅𝑻 𝒅𝒓 = 𝒅 𝒅𝒓 𝑻𝟏 + 𝑻𝟏 − 𝑻𝟐 𝒍𝒏 𝒓𝟐 𝒓𝟏 . 𝒍𝒏 𝒓𝟏 − 𝑻𝟏 − 𝑻𝟐 𝒍𝒏 𝒓𝟐 𝒓𝟏 . 𝒍𝒏 𝒓 𝒅𝑻 𝒅𝒓 = − 𝑻𝟏 − 𝑻𝟐 𝒓. 𝒍𝒏 𝒓𝟐 𝒓𝟏 ∴ 𝑸 = −𝒌𝑨 − 𝑻𝟏 − 𝑻𝟐 𝒓. 𝒍𝒏 𝒓𝟐 𝒓𝟏 Or 𝑸 = 𝒌. 𝟐𝝅𝒓𝑳 𝑻𝟏 − 𝑻𝟐 𝒓. 𝒍𝒏 𝒓𝟐 𝒓𝟏 𝑸 = 𝑻𝟏 − 𝑻𝟐 𝒍𝒏 𝒓𝟐 𝒓𝟏 𝟐𝝅𝒌𝑳
- 91. Alternative Method We havefrom Fourier’s law of heat conduction 𝑄 = −𝑘𝐴 𝑑𝑇 𝑑𝑟 𝑄 = −𝑘. 2𝜋𝑟𝐿. 𝑑𝑇 𝑑𝑟 𝑄 𝑟 . 𝑑𝑟 = −𝑘. 2𝜋𝐿. 𝑑𝑇 Integrating on both side 𝑄 1 𝑟 𝑑𝑟 𝑟2 𝑟1 = −𝑘. 2𝜋𝐿 𝑑𝑇 𝑇2 𝑇1 𝑄 1 𝑟 𝑑𝑟 𝑟2 𝑟1 = 𝑘. 2𝜋𝐿 𝑑𝑇 𝑇1 𝑇2 𝑄. 𝑙𝑛 𝑟 𝑟1 𝑟2 = 2𝜋𝑘𝐿 𝑇 𝑇2 𝑇1 𝑄. 𝑙𝑛 𝑟2 𝑟1 = 2𝜋𝑘𝐿 𝑇1 − 𝑇2 𝑄 = 2𝜋𝑘𝐿 𝑇1 − 𝑇2 𝑙𝑛 𝑟2 𝑟1 𝑄 = 𝑇1 − 𝑇2 𝑙𝑛 𝑟2 𝑟1 2𝜋𝑘𝐿
- 92. Heat flow throughthe composite cylinder
- 93. EXTENDED SURFACES -FINS Mechanical Attachment to the base surface of a system for enhancing the heat transfer rate by increasing the effective area of the body is called as extended surfaces of fins Combine Conduction through the fin and Convection to/from the fin
- 94. Heat rate fromthe base surface is given by 𝑄𝐶𝑜𝑛𝑣. = ℎ 𝐴𝑠 𝑇𝑠 − 𝑇∞
- 95. Applications of ExtendedSurfaces or Fins Heat Exchangers used in Process Industries Economiser used to heat the feed water in steam power plant Radiators of automobiles Air Cooled Engine Cylinder and Head Cooling coils and condenser coils in refrigerator and air conditioners Small Capacity Compressor Electric motor bodies Transformer and electronic equipment's Handle of ladle used to pour the molten metal
- 96. Types of ExtendedSurfaces or Fins Rectangular Fins IC Engines and Electronic Component Trapezoidal Fins Rocket Engines and Fishes
- 97. Splines Electrical Generator Shaft AnnularFins Fin Tube heat exchanger
- 98. Longitudinal Fins Electrical motorbodies Pin Fins Electrical motor bodies