Heat and Mass Transfer - HMT
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This document provides an overview of heat and mass transfer. It defines heat and mass transfer as dealing with the rate of transfer of thermal energy and discusses the three main modes of heat transfer: conduction, convection, and radiation. It also outlines some key topics that will be covered, including Fourier's law of conduction, Newton's law of cooling, Stefan-Boltzmann's law of radiation, thermal conductivity, heat transfer in gases, and heat exchangers. Equations for one-dimensional and radial conduction are presented.
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Heat and Mass Transfer - HMT
- 1. Workshop Technology OR Machine Shop Theory Heat And Mass Transfer β HMT Production planning and control β PPC Applied Mechanics Or Engineering Mechanics Engineering Materials or Material Science Engineering Thermodynamics Or Applied Thermodynamics IC Engines Engineering Mathematics - Numerical Analysis & more Strength of material OR Mechanics of solid @thelearninghub2019 @essentialamalgame @jaatishrao
- 2. HEAT AND MASS TRANSFER HEAT AND MASS TRANSFER
- 3. SCOPE OF THE COURSE αΆ π Q THERMODYNAMIC ANALYSIS HEAT TRANSFER ANALYSIS Deals with system in equilibrium. That is to bring a system from one equilibrium state to another, how much amount of heat is required. Evaluates at what rate the change of state occurs by calculating rate of heat transfer in Joule/Sec = Watt. HEAT AND MASS TRANSFER
- 4. SCOPE OF THE COURSE SN Section/Units Contents 1. Introduction Modes of heat transfer and Governing laws of heat transfer 2. Conduction Thermal conductivity, Heat conduction in gases, Intepretation Of Fourier's law, Electrical analogy of heat transfer, Critical radius of insulation, Heat generation in a slab and cylinder, Fins, Unsteady/Transient conduction 3. Convection Forced convection heat transfer, Reynoldβs Number, Prandtl Number, Nusselt Number, Incompressible flow over flat surface, HBL, TBL, Forced convection in flow through pipes and ducts, Free/Natural convection 4. Radiation Absorbtivity, Reflectivity, Transmitivity, Laws of thermal radiation, Shape factor, Radiation heat exchange 5. Heat Exchangers Types of heat exchangers, First law of thermodynamics, Classification of heat exchangers, LMTD for parallel and counter flow, NTU, Fouling factor. HEAT AND MASS TRANSFER
- 5. MODES OF HEAT TRANSFER MODES Conduction Convection Radiation HEAT AND MASS TRANSFER
- 6. MODES OF HEAT TRANSFER Conduction β The mode of heat transfer which generally (mark the word) occurs in a solid body due to temperature difference associated with molecular latticeβs vibrational energy as well by transfer of free electrons. This is the reason behind why all electrically good conductors are also in general a good conductor of heat as well. Viz the presence of abundant FREE electrons. EXCEPTION = Diamond HT LT HEAT AND MASS TRANSFER
- 7. MODES OF HEAT TRANSFER HEAT AND MASS TRANSFER
- 8. MODES OF HEAT TRANSFER Convection β The mode of heat transfer which generally occurs between solid surface and the surrounding fluid due to temperature difference associated with macroscopic bulk motion of the fluid transporting thermal energy. Natural Convection Forced Convection HEAT AND MASS TRANSFER
- 9. MODES OF HEAT TRANSFER Radiation β All matter with a temperature greater than absolute zero (viz 0 K)emits thermal radiation. The rate of emission occurs in form of electromagnetic waves which can propagate even through the vacuum ! Radiation is the mode of heat transfer which does not require any material medium and occurs at electromagnetic wave propagation travelling with speed of light. HEAT AND MASS TRANSFER
- 10. GOVERNING LAWS OF HEAT TRANSFER 1. Fourierβs law of conduction β The law states that the rate of heat transfer by conduction along a direction (given) is directly proportional to the temperature gradient along that direction and is also proportional to the area of heat transfer lying perpendicular to the direction of heat transfer. Qx ο΅ - ππ ππ₯ Qx ο΅ A Qx = K A ( ππ ππ₯ ) HEAT AND MASS TRANSFER K = Thermal conductivity
- 11. GOVERNING LAWS OF HEAT TRANSFER 2. Newtonβs law of cooling (For Convection) β The law states that the rate of heat transfer by convection between a solid body and surrounding fluid is directly proportional to the temperature difference between them and is also directly proportional to the area of contact or area of exposure between them. Qconv ο΅ (Tw - Tο₯) Qconv ο΅ A Qconv = h A (Tw - Tο₯ ) HEAT AND MASS TRANSFER h = Heat transfer coefficient of convection.
- 12. GOVERNING LAWS OF HEAT TRANSFER HEAT AND MASS TRANSFER K = Thermal conductivity = Property of the material h = Heat coefficient for convection οΉ Property of the material In forced convection h = f (v,D,ο²,ο,Cp,k) In free convection h = f (g,ο’,οο,L,ο,ο²,Cp,k)
- 13. GOVERNING LAWS OF HEAT TRANSFER HEAT AND MASS TRANSFER Ranges of βhβ SN Convection Value (Watt/m2K) 1 Free convection in gases. 3 - 25 2 Forced convection in gases. 25 β 400 3 Free convection in liquids. 250 β 600 4 Forced convection in liquids. 600 β 4000 5 Condensation heat transfer. 3000 β 25000 6 Boiling heat transfer. 5000 - 50000
- 14. GOVERNING LAWS OF HEAT TRANSFER 3.Stefan-Boltzmanβs law of Radiation β The law states that the radiation energy emitted from the surface of a black body (per unit time, per unit area) is directly proportional to the 4th power of the absolute temperature of the black body. Eb ο΅ T4 βTβ is in Kelvin Eb = ο³ T4 HEAT AND MASS TRANSFER ο³ = Stefan-Boltzman constant = 5.67 X 10-8 Watt/m2K
- 15. THERMAL CONDUCTIVITY (K) HEAT AND MASS TRANSFER It is a thermo-physical property of a material which tells about the ability of material to allow heat energy to get conducted through the material more rapidly/quickly. Insulators (a substance which does not readily allow the passage of heat) have very low thermal conductivity and hence prevents the conduction heat transfer rate. SN Material (Insulators) K (Watt/mK) 1. Asbestos 0.2 2. Refractory bricks 0.9 3. Glass wool 0.075 4. Polyurethane foam 0.02
- 16. HEAT CONDUCTION IN GASES HEAT AND MASS TRANSFER Heat conduction occurs in gases by molecular momentum transfer when high velocity and high temperature molecules collides with the low velocity low temperature molecules. However, in general gases are bad conductor of heat. Kair =0.026 w/mK
- 17. HEAT CONDUCTION IN GASES HEAT AND MASS TRANSFER As the temperature of gases increase, their thermal conductivity also increases. This is because at high temperature of gases, increased molecular activity may result in more no of collisions per unit time and hence more momentum transfer rate & hence more thermal heat conductivity. K T (Celsius)
- 18. HEAT AND MASS TRANSFER K = 8.43 w/mK HIGHEST A Thermometric Fluid. Low vapour Pressure HEAT CONDUCTION IN GASES
- 19. FOURIERβS LAW INTEPRETATION HEAT AND MASS TRANSFER ASSUMPTIONS :- β Steady state heat transfer conditions. T οΉ f (Time) β One dimensional heat conduction. T = f (x) β Uniform/Constant value of βKβ BOUNDARY CONDITIONS At x = 0, T = T1 At x = b, T = T2
- 20. FOURIERβS LAW INTEPRETATION HEAT AND MASS TRANSFER ΰΆ± π₯=0 π₯=π ππ₯ ππ₯ = ΰΆ± π1 π2 βπΎπ΄ ππ To satisfy steady state conditions, qx οΉ f (x) Viz qx = qx+dx qx X b = KA (T1-T2) qx = π²π¨ (π»π βπ»π ) π & qx π¨ = π² (π»π βπ»π ) π
- 21. ELECTRICAL ANALOGY OF HEAT TRANSFER HEAT AND MASS TRANSFER Electrical Thermal i (Ampere) q (Watt) οV (Volts) οT (Celsius/Kelvin) Relectrical RThermal i q Relectrical RThermal οV οT Relectrical = οV π Ohms RThermal= οπ π K/Watt
- 22. ELECTRICAL ANALOGY OF HEAT TRANSFER HEAT AND MASS TRANSFER (RTh)Conduction = π1 βπ2 π = π πΎπ΄
- 23. CONDUCTION HEAT TRANSFER THROUGH A COMPOSITE SLAB HEAT AND MASS TRANSFER T1 T2 T3 q RTh(1-2) = π1 πΎ1 π΄ RTh(2-3) = π2 πΎ2 π΄ RTh(1-3) = π1 πΎ1 π΄ + π2 πΎ2 π΄ RTh(1-3) = οπ π = π1 βπ3 π Rate of conduction heat Transfer βqβ is given by - q = οπ RTh(1β3) = π1 βπ3 π1 πΎ1 π΄ + π2 πΎ2 π΄ π π΄ = π»πππ‘ πππ’π₯ = π1 β π3 π1 πΎ1 + π2 πΎ2
- 24. NUMERICAL HEAT AND MASS TRANSFER Calculate the value of thermal conductivity of insulator in the composite slab as shown in the adjoining figure. 500 K 360 K q = 10,000 w/m2 20 1000π10ππ΄ 1 1000ππΎππ΄ 20 1000π10ππ΄
- 25. NUMERICAL HEAT AND MASS TRANSFER 500 K 360 K q/A = 10,000 w/m2 20 1000π10ππ΄ 1 1000ππΎππ΄ 20 1000π10ππ΄ Heat flux = q/A = οπ π πβ q/A = 500β360 20 10000 + 1 1000πΎ + 20 10000 10000 = 500β360 20 10000 + 1 1000πΎ + 20 10000 K = 0.1 w/mK
- 26. CONVECTION THERMAL RESISTANCE HEAT AND MASS TRANSFER qconv = h A (Tw - Tο₯) (RTh)Covection = Tw β Tο₯ π = 1 βπ΄
- 27. CONDUCTION - CONVECTION HEAT TRANSFER THROUGH A COMPOSITE SLAB HEAT AND MASS TRANSFER
- 28. CONDUCTION - CONVECTION HEAT TRANSFER THROUGH A COMPOSITE SLAB HEAT AND MASS TRANSFER TG Tο₯ T1 T2 T3 1 β1π΄ π1 πΎ1π΄ π2 πΎ2π΄ 1 β2π΄ q q = ππΊ βπβ 1 β1 π΄ + π1 πΎ1 π΄ + π2 πΎ2 π΄ + 1 β2 π΄ = Rate of heat transfer. q/A = ππΊ βπβ 1 β1 + π1 πΎ1 + π2 πΎ2 + 1 β2 = Heat flux (w/m2) q = UA (ππΊ β πβ) Remember Resistance = οπ» π π πΌ = π ππ + ππ π²π + ππ π²π + π ππ
- 29. RADIAL CONDUCTION THROUGH HOLLOW CYLINDER HEAT AND MASS TRANSFER
- 30. RADIAL CONDUCTION THROUGH HOLLOW CYLINDER HEAT AND MASS TRANSFER Area of conduction heat transfer = 2ο°rL q = - K A ( ππ ππ ) = - K 2ο°rL ( ππ ππ ) At r = r1, T = T1 At r = r2, T = T2 β«Χ¬β¬ π1 π2 π ππ π = - β«Χ¬β¬ π1 π2 2 πΎπΏ ππ q ln ( π2 π1 ) = 2ο°πΎπΏ (T1-T2) q = 2ο°π²π³ (T1βT2) ln (ππ ππ ) & RTh = οπ» π = ln (ππ ππ ) 2ο°πΎπΏ
- 31. RADIAL CONDUCTION THROUGH COMPOSITE CYLINDER HEAT AND MASS TRANSFER
- 32. RADIAL CONDUCTION THROUGH COMPOSITE CYLINDER HEAT AND MASS TRANSFER T1 T2 T3 q ππ π2 π1 2ππΎ1πΏ ππ π3 π2 2ππΎ2πΏ q = π1 βπ3 ππ π2 π1 2ππΎ1 πΏ + ππ π3 π2 2ππΎ2 πΏ
- 33. RADIAL CONDUCTION-CONVECTION HEAT TRANSFER THROUGH COMPOSITE CYLINDER HEAT AND MASS TRANSFER h1 = Heat transfer coefficient of convection (inside) h2 = Heat transfer coefficient of convection (outside) TG Tο₯ T1 T2 T3 q (Convection)inner Conduction (Convection)Outer Conduction 1 β1(2ππ1πΏ) ππ π2 π1 (2ππΎ1πΏ) ππ π3 π2 (2ππΎ2πΏ) 1 β2(2ππ3πΏ) q = 2ππ1πΏ (ππΊ βπβ ) 1 β1 (2ππ1 πΏ) + ππ π2 π1 (2ππΎ1 πΏ) + ππ π3 π2 (2ππΎ2 πΏ) + 1 β2 (2ππ3 πΏ) q = UiAiοT = U0A0οT = Ui (2ππ1πΏ)(ππΊ β πβ) = U0 (2ππ3πΏ)(ππΊ β πβ) A B Also,
- 34. RADIAL CONDUCTION-CONVECTION HEAT TRANSFER THROUGH COMPOSITE CYLINDER HEAT AND MASS TRANSFER 1 ππ = 1 β1 + π1 πΎ1 ln π2 π1 + π1 πΎ2 ln π3 π2 + π1 π3 1 β2 Comparing equation A and B 1 ππ = 1 β2 + π3 πΎ1 ln π2 π1 + π3 πΎ2 ln π3 π2 + π3 π1 1 β1 π π = π π‘π + π π‘π Thermal resistance of conduction gets neglected since ri = ro (approx)
- 35. CRITICAL RADIUS OF INSULATION HEAT AND MASS TRANSFER
- 36. HEAT AND MASS TRANSFER CRITICAL RADIUS OF INSULATION Conduction Convection ππ ππ ππ 2ππΎπΏ 1 β (2ππ0πΏ) q Ti Tο₯ q = ππ βπο₯ ππ ππ ππ 2ππΎπΏ + 1 β (2ππ0 πΏ) q = f(ro) & dq/dr = 0 (For maximum heat transfer) π πππ ππ βπο₯ ππ ππ ππ 2ππΎπΏ + 1 β (2ππ0 πΏ) = 0 GIVES ro = πΎπΌππ π’πππ‘πππ β Critical radius of insulation
- 37. CRITICAL RADIUS OF INSULATION HEAT AND MASS TRANSFER Electric power transmission cables Semiconductor devices Practical application of critical radius of insulation rcritical = K/h rcritical = 2K/h
- 38. RADIAL CONDUCTION THROUGH HOLLOW SPHERE HEAT AND MASS TRANSFER A = 4ο°2
- 39. RADIUS CONDUCTION THROUGH HOLLOW SPHERE HEAT AND MASS TRANSFER q = - K A ππ ππ (From Fourierβs law of conduction) = - K 4ο°r2 ( ππ ππ ) π β π π π£ππ§ ππ = ππ + ππ β«Χ¬β¬π1 π2 π ππ π2 = β«Χ¬β¬π1 π2 β 4ππΎππ q = 4ππΎ π1 βπ2 π1 π2 π2 βπ1 RTh q T1 T2 RTh = βπ π = π2 βπ1 4ππΎπ1 π2
- 40. DERIVATION OF GENERALIZED HEAT CONDUCTION EQUATION HEAT AND MASS TRANSFER qx = heat conducted into the element in X-direction = - K A ( ππ ππ₯ ) watt qx+dx = Heat conducted out of the element along X β direction. = qx + π ππ₯ ππ₯ ππ₯ qgenerated = αΆ π A dx watt
- 41. GENERALIZED HEAT CONDUCTION EQUATION DERIVATION HEAT AND MASS TRANSFER Writing energy balance equation of X-direction conduction qx + qgenerated = qx+dx + Rate of change of internal energy qx + αΆ π A dx = qx + π ππ₯ ππ₯ ππ₯ + π ππ‘ ( mcpT) t = sec, m = elementβs mass = ο² X (A dx) αΆ π A dx = π ππ₯ ππ₯ ππ₯ + π ππ‘ (ο² X (Adx) cp T) K π2 π ππ₯2 + αΆ π = ο² cp ππ ππ‘ Writing energy balance equation for all 3 dimensions. K π2 π ππ₯2 + K π2 π ππ¦2 + K π2 π ππ§2 + αΆ π = ο² cp ππ ππ‘
- 42. GENERALIZED HEAT CONDUCTION EQUATION DERIVATION HEAT AND MASS TRANSFER π2 π ππ₯2 + π2 π ππ¦2 + π2 π ππ§2 + αΆ π/K = ο² cp X 1 πΎ ππ ππ‘ Here πΎ ο² cp = ο‘ = Thermal diffusivity, A thermophysical property of material (The ratio between thermal conductivity of the material and its thermal capacity.) π2 π ππ₯2 + π2 π ππ¦2 + π2 π ππ§2 + αΆ π/K = 1 ο‘ ππ ππ‘ ππ ππ‘ = 0, T οΉ f(t) (Steady conditions) αΆ π = 0 (No heat generation) π2 π ππ₯2 + π2 π ππ¦2 + π2 π ππ§2 = 0 ο‘gases > ο‘liquid π»2 T = 0 Laplaceβs equation in T
- 43. HEAT GENERATION IN A SLAB HEAT AND MASS TRANSFER Assumptions (1) Steady state heat transfer conditions. T οΉ f(t) (2) One dimensional heat transfer T = f(x) (3) Uniform heat generation rate. (4) Constant value of βKβ π2 π ππ₯2 + π2 π ππ¦2 + π2 π ππ§2 + αΆ π/K = 1 ο‘ ππ ππ‘
- 44. HEAT GENERATION IN A SLAB HEAT AND MASS TRANSFER π2 π ππ₯2 + π2 π ππ¦2 + π2 π ππ§2 + αΆ π/K = 1 ο‘ ππ ππ‘ π2 π ππ₯2+ αΆ π/K = 0 π2 π ππ₯2 = - αΆ π/K ππ ππ₯ = - αΆ π/K x + C1 T = - αΆ π/K π₯2 2 + C1x + C2 Boundary conditions At x = +L and x = -L Gives T = Tw (Also renders C1 = 0)
- 45. HEAT GENERATION IN A SLAB HEAT AND MASS TRANSFER Temperature of the slab is max when ππ ππ₯ = 0 which gives x = 0 Viz max temperature of slab would be at centerline Now in case both sides of the slab are at different temperature then C1 would not be zero and hence max temp will not be at midplane (centerline) For such case, Let max temperature of slab as To At x = 0, T = T0 Viz C2 = 0 Which gives; T = - αΆ π/K π₯2 2 + T0 T β T0 = - αΆ π/K π₯2 2 T = - αΆ π/K π₯2 2 + C1x + C2 Parabolic temperature distribution A
- 46. HEAT GENERATION IN A SLAB HEAT AND MASS TRANSFER At x = +L Or x = -L ; T = Tw T0 β Tw = π πΏ2 2πΎ Dividing equation A & B we have; π΄ π΅ = T β T0 T0 β Tw = ( π₯ πΏ )2 B Non-dimensional temperature distribution Heat generated in slab = Heat out through convection αΆ π X (2L X A) = 2h A (Tw β Tο₯) Tw = αΆ ππΏ β + Tο₯
- 47. HEAT GENERATION IN A SLAB HEAT AND MASS TRANSFER Also from equation B we have T0 or Tmax = π πΏ2 2πΎ + αΆ ππΏ β + Tο₯
- 48. HEAT GENERATION IN A CYLINDER HEAT AND MASS TRANSFER r = 0 Tw Tw i R qconvection Tο₯ h L αΆ π = Heat generation rate = π2 π πππππ‘πππ ο°π 2 πΏ
- 49. HEAT GENERATION IN A CYLINDER HEAT AND MASS TRANSFER π2 π ππ₯2 + π2 π ππ¦2 + π2 π ππ§2 + αΆ π/K = 1 ο‘ ππ ππ‘ 1 π π ππ (π ππ ππ ) + αΆ π πΎ = 1 πΌ ππ ππ‘ 1 π2 π ππ (π2 ππ ππ ) + αΆ π πΎ = 1 πΌ ππ ππ‘ Generalized heat conduction coordinate systems. Rectilinear coordinate system Cylindrical coordinate system Spherical coordinate system
- 50. HEAT GENERATION IN A CYLINDER HEAT AND MASS TRANSFER 1 π π ππ (π ππ ππ ) + αΆ π πΎ = 1 πΌ ππ ππ‘ π2π ππ2 + 1 π ππ ππ = β αΆ π πΎ r π2 π ππ2 + ππ ππ = β αΆ ππ πΎ π ππ (π ππ ππ ) = β αΆ ππ πΎ Integrating w.r.t βrβ π ππ ππ = β αΆ ππ2 2πΎ + C1 ππ ππ = β αΆ ππ 2πΎ + πΆ1 π ππ ππ At r = R = β αΆ ππ 2πΎ + πΆ1 π Equation 1
- 51. HEAT GENERATION IN A CYLINDER HEAT AND MASS TRANSFER Again integrating T = β αΆ ππ2 4πΎ + C1 log r + C2 From boundary condition (2) Heat generated in the rod = Heat conducted radially at the surface = Heat convected from the surface to the fluid. αΆ π X ο°R2L = - K (2ο°RL) ππ ππ At r = R ππ ππ At r = R = β αΆ ππ 2πΎ From equation 1 and 2 C1 = 0 Boundary conditions (1) AT r = R, T = Tw (2) Steady state of rod Equation 2
- 52. HEAT GENERATION IN A CYLINDER HEAT AND MASS TRANSFER Temperature of the rod will be maximum at ππ ππ = 0 Gives r = 0 (viz at the axis of the rod/cylinder) Let max temp = T0 (At axis) i.e. At r = 0, T = T0 Gives T0 = C2 T = β αΆ ππ2 4πΎ + T0 Or T0 β T = αΆ ππ2 4πΎ AT r = R, T = Tw Or T0 β Tw = αΆ ππ 2 4πΎ Dividing equation 3 by 4 we have T0 β T T0 β Tw = π π 2 Equation 3 (Parabolic temp distribution) Equation 4 Temperature distribution in non-dimensional format
- 53. HEAT GENERATION IN A CYLINDER HEAT AND MASS TRANSFER The surface temperature Tw can also be obtained from energy balance equation for steady state condition of rod. αΆ π X ο°R2L = h 2ο°RL (Tw - Tο₯) Tw = αΆ ππ 2β + Tο₯ Similarly, Maximum temperature T0 can also be found out. T0 Or Tmax = αΆ ππ 2 4πΎ + αΆ ππ 2β + Tο₯ Surface temperature Maximum temperature
- 54. FINS HEAT AND MASS TRANSFER Fins are the projections protruding from a hot surface into ambient fluid. They are meant for increasing heat transfer rate by increasing surface area.
- 55. ANALYSIS OF RECTANGULAR FINS HEAT AND MASS TRANSFER A = Profile area = zt L = Length of fin z = Width of fin t = Thickness of fin
- 56. ANALYSIS OF RECTANGULAR FINS HEAT AND MASS TRANSFER Objectives (1) T = f(x) (2) qfin = ? qx = Heat conducted into the element = - KA ( ππ ππ₯ ) qx+dx = Heat conducted out of the element = qx + π ππ₯ ππ₯ ππ₯ Heat convected from the surface of fin = hP dx (T - Tο₯), P = Perimeter of fin = (2z + 2t) qx = qx+dx + qconvected (Energy balance equation)
- 57. ANALYSIS OF RECTANGULAR FINS HEAT AND MASS TRANSFER qx = qx + π ππ₯ ππ₯ ππ₯ + hP dx (T - Tο₯) 0 = π ππ₯ ππ₯ ππ₯ + hP dx (T - Tο₯) 0 = π ππ₯ β KA ( ππ ππ₯ ) ππ₯ + hP dx (T - Tο₯) π2 π ππ₯2 - βπ πΎπ΄ (T β Tο₯) = 0 Let (T - Tο₯) = ο± = f(x) viz. ππ ππ₯ = πο± ππ₯ π2 π ππ₯2 = πο± ππ₯ and put m2 = βπ πΎπ΄
- 58. ANALYSIS OF RECTANGULAR FINS HEAT AND MASS TRANSFER π2 ο± ππ₯2 - m2ο± = 0 T - Tο₯ = ο± = C1e-mx + C2emx Where, m = βπ πΎπ΄ C1 and C2 are constant of integration Boundary condition 1 At x = 0, T = T0 ο± = ο±0 = T0 - Tο₯ Boundary condition 2 (Has 3 unique cases) Standard format of 2nd order differential equation
- 59. ANALYSIS OF RECTANGULAR FINS HEAT AND MASS TRANSFER Case 1 β Fin is infinitely long (A very long fin) At x = ο₯, T = Tο₯ viz ο± = 00 i.e. π βπο₯ π0 βπο₯ = e-mx qfin = βππΎπ΄ (T0 - Tο₯) Case 2 β Fin is finite in length but tip is insulated (A adiabatic tip) qconvected from tip = h A (Tx=L - Tο₯) = Very small (negligible value) Heat conducted into the tip of fin = 0 β KA ( ππ ππ₯ )At x=L = 0 (viz. insulated tip) Temperature distribution Heat transfer
- 60. ANALYSIS OF RECTANGULAR FINS HEAT AND MASS TRANSFER ( ππ ππ₯ )At x=L = 0 πο± ππ₯ At x = L = 0 π βπο₯ π0 βπο₯ = cos β π (πΏβπ₯) cos β ππΏ qfin = βππΎπ΄ (T0 - Tο₯) Tanh mL Case 3 β Fin is finite and tip is not insulated (Uninsulated tip) π βπο₯ π0 βπο₯ = cos β π (πΏπβπ₯) cos β ππΏπ qfin = βππΎπ΄ (T0 - Tο₯) Tanh mLc m = ππ· π²π¨ Temperature distribution Heat transfer Temperature distribution Heat transfer Lc = Corrected length of fin = L + t/4 (Rectangular) = L + d/4 (Pin)
- 61. NUMERICAL HEAT AND MASS TRANSFER An uninsulated fin of 0.3 mm length is shown in the adjoining figure. Calculate the heat convected from the tip of the fin.
- 62. NUMERICAL HEAT AND MASS TRANSFER Lc = L + t/4 = 0.30 P = 2z + 2t = 2 (0.3 + 2/1000) = 0.604 A = z X t = 0.3 X 2 = 0.6 m2 m = ππ· π²π¨ = 8.603 m-1 π βπο₯ π0 βπο₯ = cos β π (πΏπβπ₯) cos β ππΏπ Putting x = 0.3 m T At x = 0.3 = 70.30C qfin = βππΎπ΄ (T0 - Tο₯) Tanh mLc = 281.1 Watt
- 63. FIN EFFICIENCY & FIN EFFECTIVENESS HEAT AND MASS TRANSFER Fin efficiency is defined as the ratio between actual heat transfer rate taking place through the fin and the maximum possible heat transfer that can occur through the fin. ο¨fin = ππππ‘π’ππ ππππ₯ Fin effectiveness is defined as the ratio between heat transfer with fins and heat transfer without fins. ο₯fin = ππ€ππ‘β ππππ ππ€ππ‘βππ’π‘ ππππ ο₯fin ο΅ 1/ β ο₯fin ο΅ π π΄ ο₯fin ο΅ πΎ
- 64. FIN EFFICIENCY & FIN EFFECTIVENESS HEAT AND MASS TRANSFER Moderately short Number of fins Closely spaced fins Thin High conductivity material As such Al Number of fins regulated = n = πππ‘ππ βπππ‘ π‘π ππ πππ π ππππ‘ππ π»πππ‘ πππ‘π π‘βπππ’πβ πππβ πππ
- 65. UNSTEADY STATE/TRANSIENT CONDUCTION HEAT TRANSFER HEAT AND MASS TRANSFER T = f(t) Ti = Initial temperature of the body at the moment at which t=0 seconds. Viz when the body is just exposed to fluid. T = Temperature of body at any time βt secβ later The rate of convection heat transfer between body and fluid = The rate of decrease of internal energy of body w.r.t time hA (T - Tο₯) = - m Cp ( ππ ππ‘ ) Joules/Sec = - ο² vCp ( ππ ππ‘ ) J/S Treating all the parameters including βhβ as constant and separating the variable T and t and then integrating, we have;
- 66. UNSTEADY STATE/TRANSIENT CONDUCTION HEAT TRANSFER HEAT AND MASS TRANSFER ΰΆ± π‘=0 π‘ ( βπ΄ ο² vCp ) ππ‘ = ΰΆ± ππ π β ππ π β πο₯ ( βπ΄ ο² vCp) t = ln π β πο₯ = ln ππ βπο₯ π βπο₯ e ( βπ΄ ο² vCp ) t = ππ βπο₯ π βπο₯ dT dt dT dt dT/dt t (Time) t (Time) T T -dT/dt = Rate of cooling = f(t) - Ve + Ve
- 67. CONVECTION HEAT TRANSFER HEAT AND MASS TRANSFER q = h A οT Convection Forced convection heat transfer Free convection heat transfer Velocity is evident Velocity is not evident r X R Y X u u u = f(r) u = f(x,y)
- 68. FORCED CONVECTION HEAT TRANSFER HEAT AND MASS TRANSFER h = f (v, D, ο², ο, Cp, K) v = velocity of fluid = [LT-1] D = characteristics dimensions of the body = [L] ο² = Density = [ML-3] ο = Dynamic viscosity (Pascal-Sec) = [ML-1T-1] Cp = Specific heat at constant pressure = [L2T-2ο±-1] K = Thermal conductivity = [MLT-3ο±-1] h = Convective heat transfer coefficient = [MT-3 ο±-1]
- 69. FORCED CONVECTION HEAT TRANSFER HEAT AND MASS TRANSFER Buckingham ο° theorem of dimensional analysis states that if there are βnβ number of variables in any functional relationship both dependent and independent. And if all the variables put together contains ββmβ number of fundamental dimensions, then the functional relationship among the variables can be expressed in term of βn-mβ number of dimensionless ο° terms. h = f (v, D, ο², ο, Cp, K) Here n = 7 m = 4 [M,L,T,ο±] According to Buckingham ο° theorem n β m = Number of dimensionless ο° terms = 7 β 4 = 3 Let ο°1, ο°2, ο°3 be these dimensionless terms.
- 70. FORCED CONVECTION HEAT TRANSFER HEAT AND MASS TRANSFER ο°1 = ( ha1 , vb1, Dc1 , ο²d1 ) ο ο°2 = ( ha2 , vb2, Dc2 , ο²d2 ) Cp ο°3 = ( ha2 , vb2, Dc2 , ο²d2 ) K Now ο°1 = [M0L0T0ο±0] = [MT-3ο±-1]a1 [LT-1]b1 [L]c1 [ML-3]d1 [ML-1T-1] For mass βMβ, 0 = a1 + d1 + 1 For length βLβ, 0 = b1 + c1 + - 3d1 β 1 For time βTβ, 0 = -3a1 β b1 β 1 For temperature βο±β, 0 = -a1 h = f (v, D, ο², ο, Cp, K) ο°1 = ο ππ«ο²
- 71. FORCED CONVECTION HEAT TRANSFER HEAT AND MASS TRANSFER h = f (v, D, ο², ο, Cp, K) Similarly ο°2 = ο²π£Cp β ο°3 = πΎ βπ· Now, 1 ο°1 = Reynoldβs Number = Re = ππ«ο² ο 1 ο°2 = Stantonβs Number = Sn = β ο²π£Cp 1 ο°3 = Nusselt Number = Nu = βπ· π² ο°1ο°2 ο°3 = Prandtlβs Number = Pr = οCp π²
- 72. PHYSICAL SIGNIFICANCE OF Re IN FORCED CONVECTION HEAT TRANSFER HEAT AND MASS TRANSFER 1.Reynoldβs Number Re β Defined as the ratio between inertia forces and viscous forces to which a flowing fluid is subjected to. Re = πΌππππ‘ππ πΉπππππ πππ πππ’π ππππππ = ππ«ο² ο = ππ« ο/ο² = ππ« ο΅ ο΅ = Kinematic viscosity Re is the criteria to tell whether the fluid flow is laminar or turbulent for incompresible flow through pipes as well ducts. Re < 2000 (Lower critical) β Flow is laminar Re > 4000 (Upper critical) β Flow is turbulent
- 73. HEAT AND MASS TRANSFER Axi-symmetric parabolic distribution of velocity. u = f(r) Logarithmic distribution of velocity. u = f(r) LAMINAR TURBULENT PHYSICAL SIGNIFICANCE OF Re IN FORCED CONVECTION HEAT TRANSFER
- 74. INCOMPRESSIBLE FLOW OVER FLAT SURFACE HEAT AND MASS TRANSFER
- 75. INCOMPRESSIBLE FLOW OVER FLAT SURFACE HEAT AND MASS TRANSFER Hydrodynamic boundary layer (HBL) is defined as a thin region formed over a flat surface (plate) inside which velocity gradient are seen in the normal direction to the plate. These velocity gradient are formed due to viscos nature or more precisely due to momentum diffusion through the fluid layer in the normal direction to the plate. Outside the HBL, free stream velocity exists in which no viscos influence can be seen At any given βxβ At y = 0, u = 0 At y = ο€, ππ ππ = 0 At y = 0, ππ π πππ = 0 At any given βxβ ππ ππ = f(y) so for ππ ππ At y = 0 ο€ = f(x)
- 76. INCOMPRESSIBLE FLOW OVER FLAT SURFACE HEAT AND MASS TRANSFER Rex < 5 X 105 (Flow over plate is laminar) Rex > 6.7-7 X 105 (Flow over plate is turbulent)
- 77. PHYSICAL SIGNIFICANCE OF Nu & PrIN FORCED CONVECTION HEAT TRANSFER HEAT AND MASS TRANSFER 2. Nusselt number Nu - Nusselt number is the ratio of convective to conductive heat transfer at a boundary in a fluid. Sometimes also known as dimensionless heat transfer coefficient. The Biot number is the ratio of the internal resistance of a body to heat conduction to its external resistance to heat convection. Therefore, a small Biot number represents small resistance to heat conduction, and thus small temperature gradients within the body. Nu = βπ· πΎπππ’ππ = π· πΎπ΄ 1 βπ΄ = πΆππππ’ππ‘πππ πππ ππ π‘ππππ πππππππ ππ¦ πππ’ππ (π π‘ππ‘ππππππ¦) ππ’πππππ ππππ£πππ‘ππ£π πππ ππ π‘ππππ But fluids are bad conductor of heat Numerator > Denominator Nu is always greater than 1 Bi = βπ· πΎπππππ
- 78. PHYSICAL SIGNIFICANCE OF Nu & PrIN FORCED CONVECTION HEAT TRANSFER HEAT AND MASS TRANSFER 3. Prandtl Number (Pr) β The only dimensionless number which is a property of the fluid. Defined as the ratio between kinematic viscosity and thermal diffusivity of fluid. Pr = πΎππππππ‘ππ π£ππ πππ ππ‘π¦ πβπππππ πππππ’π ππ£ππ‘π¦ = ο΅ ο‘ οπΆπ πΎ = ο ο² πΎ ο²πΆπ Pr for air = 0.65 to 0.73 Pr for water = 2 to 6 For engine oils = up to 100 ! Very low Pr
- 79. THERMAL BOUNDARY LAYER (TBL) HEAT AND MASS TRANSFER
- 80. THERMAL BOUNDARY LAYER (TBL) HEAT AND MASS TRANSFER Thermal Boundary Layer (TBL) is similar to HBL inside which velocity gradient are seen in the normal direction to the plate. Thermal boundary layer is also a thin region in which temperature gradients are present in the normal direction to the flat surface (Plate). These temperature gradient are formed due to heat transfer between plate and flowing fluid. At any given βxβ measured from leading edge. At y = 0, T = Tw At y = ο€t, T = Tο₯ , ππ» ππ = 0 At y = 0, ππ π» πππ = 0 At any given βxβ ππ» ππ = f(y) with maximum value of ππ» ππ At y = 0
- 81. ENERGY BALANCE FOR DIFFERENTIAL CONTROL VOLUME OF TBL HEAT AND MASS TRANSFER Heat conducted into the control volume = Heat convected from the hot plate at Tw to free stream fluid at Tο₯ - K dA ( ππ ππ¦ ) AT y = 0 = hx dA (Tw - Tο₯) hx = β πΎ(ππ ππ¦ ) AT y = 0 (Tw β Tο₯) Local convective heat transfer coefficient. hx x hx ο΅ π Fourierβs law Newtonβs law
- 82. MOMENTUM EQUATION OF HBL AND ENERGY EQUATION OF TBL HEAT AND MASS TRANSFER ο = f (x,y) & ο΅ = f (x,y), ο² = c
- 83. MOMENTUM EQUATION OF HBL AND ENERGY EQUATION OF TBL HEAT AND MASS TRANSFER CONTROL VOLUME Viscos force = ο [ ππ’ ππ¦ + π ππ¦ ( ππ’ ππ¦ ) ππ¦ ] (ππ₯ π 1) dx dy Pressure force = p X dy X 1 Pressure force = (p + ππ ππ₯ ππ₯) ππ¦ X 1 Viscos force Gravity force
- 84. MOMENTUM EQUATION OF HBL AND ENERGY EQUATION OF TBL HEAT AND MASS TRANSFER Applying Newtonβs II law of motion Ο πΉπ₯ = max (Conservation of linear momentum) Resulting momentum equation of HBL is ; u ππ’ ππ₯ + v ππ’ ππ¦ = ( ο ο² ) π2 π’ ππ¦2 - 1 ο² ( ππ ππ₯ ) u v V = u Τ¦ π + v Τ¦ π dp/dx = 0 for flow over plates & ( ο ο²) = ο΅ u ππ ππ + v ππ ππ = ο΅ ππ π πππ Momentum equation of HBL
- 85. MOMENTUM EQUATION OF HBL AND ENERGY EQUATION OF TBL HEAT AND MASS TRANSFER ο = f (x,y) & ο΅ = f (x,y), T = f(x,y), ο² = c
- 86. MOMENTUM EQUATION OF HBL AND ENERGY EQUATION OF TBL HEAT AND MASS TRANSFER Consider negligible heat conduction in x-direction. Thermal heat energy convected by fluid = mCpT J/Sec Viscos heat is neglected CONTROL VOLUME Heat convected = ο² (dx X 1) (v + ππ£ ππ¦ dy) Cp (T + ππ ππ¦ dy) Heat conducted = - K (dx X 1) ( ππ ππ¦ + π ππ¦ ππ ππ¦ dy) Heat conducted = - K (dx X 1) ππ ππ¦ Heat convected = ο² (dx X 1) v Cp T Heat convected = ο² (dy X 1) u Cp T Heat convected = ο² (dx X 1) (u + ππ’ ππ₯ du) Cp (T + ππ ππ₯ dx)
- 87. MOMENTUM EQUATION OF HBL AND ENERGY EQUATION OF TBL HEAT AND MASS TRANSFER Heat conducted through bottom face + Heat convected through left face + Heat convected through bottom face Heat conducted through top face + Heat convected through right face + Heat convected through top face 6 energies u ππ» ππ + v ππ» ππ = ο‘ ππ π» πππ Energy equation for TBL u ππ ππ + v ππ ππ = ο΅ ππ π πππ Momentum equation of HBL
- 88. FORCED CONVECTION IN FLOW THROUGH PIPES AND DUCTS HEAT AND MASS TRANSFER αΆ π = ο² X ο°R2 X Vmean Kg/s
- 89. FORCED CONVECTION IN FLOW THROUGH PIPES AND DUCTS HEAT AND MASS TRANSFER Tb (Bulk mean temperature) β Defined as {At a given cross section of pipe} the temperature which takes into account the variation of temperature of fluid layers with respect to βrβ at the cross section of the pipe and thus indicates the total thermal energy transported by the fluid through the cross section. Thermal energy/enthalpy transported by fluid through the cross section of pipe = αΆ π CpTb = ο² X ο°R2 X Vmean CpTb And the Bulk mean temperature is given by. Tb = 2 β«Χ¬β¬0 π π π’ π ππ R2 X Vmean
- 90. NUMERICAL HEAT AND MASS TRANSFER Find out the bulk mean temperature at exit. dq = qw X Area = qw ο°π· ππ₯ Differential heat rate through differential heat transfer area of length βdxβ = dq = αΆ π Cp dTb β«Χ¬β¬0 3 2500 π₯ π· ππ₯ = β«Χ¬β¬πππππ‘ ππ₯ππ‘ αΆ π Cp dTb (Tb)Exit = 620C
- 91. FREE/NATURAL CONVECTION HEAT AND MASS TRANSFER
- 92. FREE/NATURAL CONVECTION HEAT AND MASS TRANSFER h = f ( g, ο’, οο, L, ο, ο², Cp, K ) g = acceleration due to gravity ο’ = Isobaric volume expansion coefficient
- 93. RADIATION HEAT AND MASS TRANSFER Any body at a given temperature emit thermal radiation at all probable wavelength on electromagnetic spectrum and in all possible hemispherical direction. Total hemispherical emissive power (E) β Defined as radiation energy emitted from the surface of a body per unit time per unit area in all possible hemispherical directions integrated over all the wavelengths.
- 94. RADIATION HEAT AND MASS TRANSFER Total emissivity (ο₯) β Defined as the ratio between total hemispherical emissive power of a non-black body to that of a black body at same temperature. Non-black Black body E Eb ο₯ = π¬ π¬π ο₯ ο£ 1 ο₯b = 1
- 95. RADIATION HEAT AND MASS TRANSFER Black body is a body which absorbs all the thermal radiation incident or falling upon the body. Perfect absorber Ideal emitter Diffusive in nature
- 96. RADIATION HEAT AND MASS TRANSFER Monochromatic/Spectral hemispherical emissive power (Eο¬) - Eο¬ at a particular wavelength βο¬β is defined as the quantity which when multiplied by βdο¬β (Differentially small increment in wavelength) shall give the radiation energy emitted from the surface of a body per unit time per unit area in the wavelength range ο¬ - ο¬ + dο¬ dο¬ ο¬ ο¬ + dο¬ Eο¬ = f(ο¬)
- 97. RADIATION HEAT AND MASS TRANSFER Monochromatic/Spectral emissivity (ο₯ο¬) β Ratio of monochromatic hemispherical emissive power of a non-black body to that of a black body at same temperature & wavelength. ο₯ bο¬ ο¬ ο₯ ο¬ ο¬ For grey body ο₯ ο¬ = Constant
- 98. ABSORBTIVITY (ο‘), REFLECTIVITY (ο²), TRANSMITIVITY (ο) HEAT AND MASS TRANSFER ο‘ = Fraction of radiation energy incident upon a surface which is absorbed by it. ο² = Fraction of radiation energy incident upon a surface which is reflected by it. ο = Fraction of radiation energy incident upon a surface which is transmitted through it. For any surface, ο‘ + ο² + ο = 1 For opaque surface, ο‘ + ο² = 1 ( ο = 0 ) For black body, ο‘ = ο‘b = 1
- 99. ABSORBTIVITY (ο‘), REFLECTIVITY (ο²), TRANSMITIVITY (ο) HEAT AND MASS TRANSFER Sun Earth Short wavelength Long wavelength Co2
- 100. LAWS OF THERMAL RADIATION HEAT AND MASS TRANSFER 1.Kirchhoff's law of thermal radiation β States that whenever a body is in thermal equilibrium with its surroundings, its emissivity is equal to its absorptivity. ο₯ = ο‘ βA good absorber is always a good emitterβ For black body ο‘b = 1 ο₯b = 1
- 101. LAWS OF THERMAL RADIATION HEAT AND MASS TRANSFER 2.Plankβs law of thermal radiation β The law states that the monochromatic emissive power of a black body is dependent both upon the absolute temperature of the body and the wavelength of radiation energy emitted. Ebο¬ = f (ο¬,T)
- 102. LAWS OF THERMAL RADIATION HEAT AND MASS TRANSFER 3.Stefan-Boltzmanβs law β The law states that the total hemispherical emissive power of a black body is directly proportional to the fourth power of the absolute temperature of the black body. Eb ο΅ T4 Eb = ο³ T4 watt/m2 ο³ = Stefan-Boltzmanβs constant = 5.67 X 10-8 w/m2K4
- 103. LAWS OF THERMAL RADIATION HEAT AND MASS TRANSFER 4.Weins displacement law β For a non black body, whose emissivity is ο₯ , The total hemispherical emissive power of non-black body E = ο₯Eb E = ο₯ ο³ T4 If βAβ is the total surface area of non-black body, The total radiation energy emitted from entire non-black body = EA watt = ο₯ ο³ T4A
- 104. SHAPE FACTOR HEAT AND MASS TRANSFER 1 2 Radiation energy leaving 1 Radiation energy leaving 2 F12 = Fraction of radiation energy leaving surface 1 and that of reaching surface 2. F21 = Fraction of radiation energy leaving surface 2 and that of reaching surface 1. 0 ο£ Fmn ο£ 1
- 105. SHAPE FACTOR HEAT AND MASS TRANSFER Shapes with size difference Flat surface Concave surface Convex surface Reciprocity relationship A1 F12 = A2 F21 F21 > F12 F11 = 0 F11 = 0 F11 > 0
- 106. RADIATION HEAT EXCHANGE HEAT AND MASS TRANSFER CASE 1 Two infinitely large parallel plates Assumptions β 1. Steady state heat transfer conditions TοΉ f(t) 2. Surfaces are diffuse and grey. ο₯ο¬ = f (ο¬) Net radiation heat exchange between 1 and 2 per unit area is (q/A)1-2 = ο³ ( π1 4 βπ2 4 ) 1 ο₯1 + 1 ο₯2 β1 Watt/m2 = Radiation flux In case both the surfaces are black, ο₯1 = ο₯2 = 1 (q/A)1-2 = ο³ ( π1 4 β π2 4) w/m2
- 107. RADIATION HEAT EXCHANGE HEAT AND MASS TRANSFER CASE 2 Two infinitely long concentric cylindrical surfaces
- 108. RADIATION HEAT EXCHANGE HEAT AND MASS TRANSFER CASE 2 Two infinitely long concentric cylindrical surfaces Assumptions β 1. Steady state heat transfer conditions TοΉ f(t) 2. Surfaces are diffuse and grey. ο₯ο¬ = f (ο¬) 3. Let T1 > T2 Net radiation heat exchange between 1 and 2 is (q)1-2 = ο³ π1 4 βπ2 4 π΄1 1 ο₯1 + π΄1 π΄2 ( 1 ο₯2 β1) Watt F12 = 1 F21 < 1 π΄1 π΄2 = π 1 π 2
- 109. TYPES OF HEAT EXCHANGERS HEAT AND MASS TRANSFER Heat exchanger is a steady flow adiabatic open system device in which two flowing fluids exchange or transfer heat without loosing or gaining any heat from the ambient.
- 110. TYPES OF HEAT EXCHANGERS HEAT AND MASS TRANSFER Surface (steam) condenser Steam to cold water Economizer Hot flue gas to cold water Superheater Hot flue gas to dry saturated steam Air preheater Hot flue gas to combustion air Cooling tower Hot water to atmospheric air Oil cooler Hot oil to cool water/air
- 111. FIRST LAW OF THERMODYNAMICS HEAT AND MASS TRANSFER Q β W = οH + οKE + οPE Q = 0 (adiabatic),W = 0 , KE = 0, PE = 0 (no datum change) (οH)HE = 0 (οH)Hot fluid + (οH)Cold fluid = 0 - (οH)Hot fluid = +(οH)Cold fluid β Rate of enthalpy decrease of hot fluid = Rate of enthalpy increase of cold fluid β αΆ πh Cph (Thi β The) = αΆ πc Cpc (Tce β Tci)
- 112. CLASSIFICATION OF HEAT EXCHANGER HEAT AND MASS TRANSFER HE Direct transfer Parallel Counter Cross Direct contact Regenerative/ storage
- 113. CLASSIFICATION OF HEAT EXCHANGER HEAT AND MASS TRANSFER Direct transfer Parallel Counter Cross Parallel flow HE β An automobile radiator 1 π = 1 β1 + 1 β2 οT = Th β Tc = f (x) dq = U οT dA Tce > The
- 114. CLASSIFICATION OF HEAT EXCHANGER HEAT AND MASS TRANSFER (οS)Parallel flow HE > (οS)Counter flow HE Counter flow HE is thermodynamically more efficient than parallel flow HE.
- 115. Mean Temperature Difference MTD (οTm) HEAT AND MASS TRANSFER A parameter which takes into account the variation of temperature with respect to direction of flow of hot fluid by taking average all along the length of HE from inlet to exit. Q = U A οTm Q = Total heat transfer rate. U = Overall heat transfer coefficient. A = Total heat transfer area of HE. οTm = Mean Temperature Difference. Equation - 1
- 116. Mean Temperature Difference MTD (οTm) HEAT AND MASS TRANSFER dq = U οT dA β«Χ¬β¬πΌππππ‘ πΈπ₯ππ‘ ππ = β«Χ¬β¬πΌππππ‘ πΈπ₯ππ‘ ποT dA Q = U β«Χ¬β¬πΌππππ‘ πΈπ₯ππ‘ οT dA From equation 1 and 2, we have οTm = π π¨ β«Χ¬β¬πΌππππ‘ πΈπ₯ππ‘ οT dA Equation - 2
- 117. Logarithmic Mean Temperature Difference LMTD (οTm) HEAT AND MASS TRANSFER Parallel flow HE
- 118. Logarithmic Mean Temperature Difference LMTD (οTm) HEAT AND MASS TRANSFER Parallel flow HE dq = U οT dA dA = B dx οT = Th β Tc = f (x) At x = 0, οT = οTi = Thi β Tci At x = L, οT = οTe = The β Tce Also dq = - αΆ π h Cph dT = + αΆ π h Cpc dT Now οT = Th β Tc d(οT) = dTh β dTc d(οT) = - ππ αΆ πβ πΆπβ - ππ αΆ ππ πΆππ
- 119. Logarithmic Mean Temperature Difference LMTD (οTm) HEAT AND MASS TRANSFER Parallel flow HE d(οT) = - dq 1 αΆ πβ πΆπβ + 1 αΆ ππ πΆππ d(οT) = - U οT dA 1 αΆ πβ πΆπβ + 1 αΆ ππ πΆππ Separating the variables, we have ΰΆ± βππ βππ β d(οT) οT = ΰΆ± 0 πΏ π π΅ 1 αΆ πβ πΆπβ + 1 αΆ ππ πΆππ ln βππ βππ = U B L 1 αΆ πβ πΆπβ + 1 αΆ ππ πΆππ Q = Total heat rate of entire HE = αΆ πβ πΆπβ (Thi β The) = αΆ ππ πΆππ (Tce β Tci)
- 120. Logarithmic Mean Temperature Difference LMTD HEAT AND MASS TRANSFER Parallel flow HE ln βππ βππ = U A πβπ βπβπ π + πππ βπππ π = ππ΄ π ( οTi - οTe ) Q = U A οTi β οTe ln βππ βππ , Comparing this with Q = U A οTm , We have οTm = (οT)LMTD = οTi β οTe ln βπ»π βπ»π Logarithmic mean temperature difference for parallel flow HE
- 121. Logarithmic Mean Temperature Difference LMTD HEAT AND MASS TRANSFER Counter flow HE
- 122. Logarithmic Mean Temperature Difference LMTD HEAT AND MASS TRANSFER Counter flow HE (οT)LMTD = οTi β οTe ln βπ»π βπ»π
- 123. SPECIAL CASES HEAT AND MASS TRANSFER When one of the fluids in the HE is undergoing phase change like a steam condenser or evaporator or steam generator. Steam Condenser LMTD Parallel = LMTD Counter
- 124. SPECIAL CASES HEAT AND MASS TRANSFER Steam Generator LMTD Parallel = LMTD Counter
- 125. SPECIAL CASES HEAT AND MASS TRANSFER When both hot and cold fluids have equal capacity rates in a counter flow heat exchanger. αΆ πβ πΆπβ = αΆ ππ πΆππ Then from energy balance equation, αΆ πβ πΆπβ ( Thi βThe ) = αΆ ππ πΆππ ( Tce β Tci ) ( Thi βThe ) = ( Tce β Tci ) Thi β Tce = The β Tci οTi = οTe (οT)LMTD = οTi β οTe ln βπ»π βπ»π = π π = Undefined = οTi Or οTe αΆ π πΆπ = Heat capacity rate LβHospitalβs rule
- 126. Effectiveness of HE (ο₯) HEAT AND MASS TRANSFER It is defined as the ratio between actual heat transfer rate taking place between hot and cold fluids and the maximum possible heat transfer rate between them. ο₯HE = π ππππππ π πππ π ππππππ = Rate of enthalpy change of either fluid = αΆ πβ πΆπβ ( Thi βThe ) = αΆ ππ πΆππ ( Tce β Tci ) π πππ = αΆ π πΆπ ( Thi βTci ) αΆ π πΆπ = Heat capacity rate
- 127. NUMBER OF TRANSFER UNITS NTU HEAT AND MASS TRANSFER The number of transfer units (NTU = UA/(mcp)) itself is a combination of overall heat transfer coefficients, transfer area, fluid flow rate and heat capacity. It summarizes these dimensional parameters into one dimensionless parameter. The performance becomes a monotone function of this dimensionless parameter. It indicates overall size of HE. NTU = πΌπ¨ αΆ π πͺπ πΊππππ
- 128. FOULING FACTOR (F) HEAT AND MASS TRANSFER It takes into account the thermal resistance offered by any chemical scaling or deposits that are formed on the heat transfer surface on both hot side and cold side. Generally ranges from 0.0004 β 0.0006 π πΌ = π ππ + F1 + π ππ + F2
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