1. Page  1 Study of Heat and Work transfer (quantitatively) Thermodynamics & Heat Transfer Thermodynamics Heat Transfer Study of “How heat flows” every activity involves heat transfer

2. Page  2 Driving Potential & Resistance Teaching – Learning Process in a Class Room Teacher’s Interest & Knowledge Students’ Interest & Capability Class Room Ambience Knowledge Transfer DRIVING POTENTIAL Difference between Teacher’s interest + knowledge & students’ interest + capability However, higher potential difference leads to increase in entropy RESISTANCE FACTORS 1.Time of lecture 2.Nature of subject 3.Ambience (conduciveness)

3. Page  3 Conduction Solids > Lattice vibrations Fluids > Molecular collisions The transfer of energy in a solid or fluid via molecular contact without bulk motion MODE TT T   T 0 x PHYSICAL PHENOMENON MATHEMATICAL EQUATION

4. Page  4 Conduction (contd.) Fourier Law of Heat Conduction The heat flux, q is directly proportional to temperature gradient The proportionality constant, k, is defined as the thermal conductivity, a thermo physical property.

5. Page  5 Conduction (contd.) Thermal Conductivity, k Silver = 410 Wm -1 K -1 k/k silver Silver1 Gold0.7 Copper0.93 Aluminum0.86 Brass (70% Cu:30% Ni)0.33 Platinum, Lead0.25 Mild steel (0.1% Cu), Cast iron0.12 Bismuth0.07 Mercury0.04 METALS k/k silver Air0.19 Water0.0014 Granite, Sandstone0.011 Average rock0.012 Limestone0.007 Ice0.015 Glass (crown)0.0058 Concrete (1:2:4)0.0042 Brick0.0038 Snow (fresh or average)0.005 Soil (sandy, dry)0.002 Soil (8% moist)0.0033 Wood0.0045 NON-METALS

6. Page  6 Convection Convection occurs in liquids and gases. Energy is carried with fluid motion when convection occurs. PHYSICAL PHENOMENON MATHEMATICAL EQUATION

7. Page  7 Convection (contd.) The quantity h is called the convective heat transfer coefficient (W/m2-K). It is dependent on the type of fluid flowing past the wall and the velocity distribution. Thus, h is not a thermo physical property. Newton’s Law of Cooling Convection Processh(W/m 2 -K) Free convection Gases2–25 Liquids 50–1000 Forced convection Gases25–250 Liquids 50–20,000 Convection phase change2,500–200,000

8. Page  8 Convection (contd.)  Single phase fluids (gases and liquids) –Forced convection –Free convection, or natural convection –Mixed convection (forced plus free)  Convection with phase change –Boiling –Condensation Convective Processes

9. Page  9 Radiation Energy transfer in the form of electromagnetic waves PHYSICAL PHENOMENON MATHEMATICAL EQUATION

10. Page  10 Radiation (contd.) Stefan-Boltzman Law The emissive power of a black body over all wave lengths is proportional to fourth power of temperature

11. Page  11 One Dimensional Heat Conduction Net rate of heat gain by conduction Rate of energy generation Rate of increase of internal energy = + [Aq] x – [Aq] x+ Δx A Δx g + = Heat flow in [Aq] x Heat flow out [Aq] x+Δx g k ΔxΔx

12. Page  12 One Dimensional Heat Conduction (contd.) As Δx  0, the first term on the LHS, by definition, becomes the derivative of [Aq] with respect to x Heat flow in [Aq] x Heat flow out [Aq] x+Δx g k ΔxΔx

13. Page  13 One Dimensional Heat Conduction (contd.) Rectangular Coordinates Cylindrical Coordinates Spherical Coordinates A Compact Equation n = 0 n = 1 n = 2

14. Page  14 Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind)

15. Page  15 Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) 0 L x T1T1 T2T2 T (x,t) | x=0 = T (0,t) = T 1 T (x,t) | x=L = T (L,t) = T 2

16. Page  16 Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) 0 L x Heat Supply Conduction flux Heat Supply Conduction flux W/m 2 Plate

17. Page  17 Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Hollow Cylinder or hollow sphere b r Heat Supply W/m 2 a Conduction flux Heat Supply W/m 2

18. Page  18 Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Plate Conduction Convection T1, h1T1, h1 Fluid Flow Conduction T2, h2T2, h2 Fluid Flow Convection heat flux from the fluid at T 1 to the surface at x = 0 Conduction heat flux from the surface at x= 0 into the plate

19. Page  19 Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Plate Conduction Convection T1, h1T1, h1 Fluid Flow Conduction T2, h2T2, h2 Fluid Flow Convection heat flux from the fluid at T 2 to the surface at x = L Conduction heat flux from the surface at x = L into the plate

20. Page  20 Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Hollow Cylinder or hollow sphere b r Heat Supply a Fluid Flow T1, h1T1, h1 T2, h2T2, h2 Convection heat flux from the fluid at T 1 to the surface at r = a Conduction heat flux from the surface at r= a into the plate

21. Page  21 Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Hollow Cylinder or hollow sphere b r Heat Supply a Fluid Flow T1, h1T1, h1 T2, h2T2, h2 Convection heat flux from the fluid at T 2 to the surface at r = b Conduction heat flux from the surface at r= b into the plate

22. Page  22 Steady State One Dimensional Heat Conduction Rectangular Coordinates T = T 1 0 L T = T 2 x Governing Equation

23. Page  23 Steady State One Dimensional Heat Conduction Cylindrical Coordinates (Solid Cylinder) T = T 1 0 r T = T 2 b Governing Equation Solving,

24. Page  24 Steady State One Dimensional Heat Conduction Cylindrical Coordinates (Solid Cylinder) Solved Example T = T 1 0 r T = T 2 b Solution T(0) = 350 °C q(r) = 10 6 W/m 2 For r=1cm g 0 = 2 x 10 8 W/m 3 k = 20 W/(m.°C) T2 = 100 °C What will be the 1.Centre temperature T(0) 2.Heat flux at the boundary surface (r=1cm) Equations to use (derive)

25. Page  25 Steady State One Dimensional Heat Conduction Cylindrical Coordinates (Hollow Cylinder) Determination of Temperature Distribution Solving, a b r 0 T1T1 T2T2 k Mathematical formulation of this problem is in a < r < b

26. Page  26 Steady State One Dimensional Heat Conduction Cylindrical Coordinates (Hollow Cylinder) Expression for radial heat flow Q over a length H Since, a b r 0 T1T1 T2T2 k The heat flow is determined from, Rearranging, where,

27. Page  27 Steady State One Dimensional Heat Conduction Cylindrical Coordinates (Hollow Cylinder) Expression for thermal resistance for length H here, A 0 = 2πaH =area of inner surface of cylinder A 1 = 2πbH =area of outer surface of cylinder A m = logarithmic mean area t = b – a = thickness of cylinder a b r 0 T1T1 T2T2 k Above equation can be rearranged as, where,

28. Page  28 Steady State One Dimensional Heat Conduction Spherical Coordinates (Hollow Sphere) Expression for temperature distribution in a < r < b The mathematical formulation is given by, where, r a 0 b

29. Page  29 Steady State One Dimensional Heat Conduction Spherical Coordinates (Hollow Sphere) Expression for heat flow rate Q and thermal resistance R Heat flow rate is determined using the equation, r a 0 b using, from last slide where,

30. Page  30 Composite Medium Example (Furnace Wall) FURNACE REFRACTORY LINING 2REFRACTORY LINING 1 FURNACE WALL BRICK WALL Ambient

31. Page  31 Composite Medium Example (Condenser Water Tube) Condensing Medium (Steam) Cooling Water Scale Tube Wall

32. Page  32 Composite Medium Composite Slab (resistance in series) L1L1 L2L2 L3L3 TaTa T0T0 T1T1 T2T2 T3T3 TbTb TaTa T0T0 T1T1 T2T2 T3T3 TbTb RaRa R1R1 R2R2 R3R3 RbRb Q Q Ta, haTa, ha Fluid Flow Tb, hbTb, hb Q Q

33. Page  33 Composite Medium Composite Slab (resistance in parallel) A E B C D Insulated T1T1 T2T2 RARA RDRD RcRc RBRB RERE T1T1 T2T2

34. Page  34 Composite Medium Composite Cylinder H haha k1k1 k2k2 k3k3 hbhb TaTa T0T0 T1T1 T2T2 T3T3 TbTb RaRa R1R1 R2R2 R3R3 RbRb Q Q

35. Page  35 Composite Medium Composite Spheres TaTa T0T0 T1T1 T2T2 T3T3 TbTb RaRa R1R1 R2R2 R3R3 RbRb Q Q hbhb haha k1k1 k2k2 k3k3

36. Page  36 Composite Medium Critical Thickness of Insulation T1T1 r1r1 roro H Convection into an ambient at T ∞,h 0 Insulation Radius, r rcrc Heat Loss, q

37. Page  37 Composite Medium Critical Thickness of Insulation T1T1 r1r1 roro H Convection into an ambient at T ∞,h 0 The rate of heat loss Q from the tube is given by For CylinderFor Sphere

38. Page  38 Composite Medium Solved Example (Composite Cylinder) Calculate, 1.Heat loss from tube for length H=10m 2.Temperature drops resulting in thermal resistances 5 cm 7.6 cm K = 15 W/(m °C) Insulation t =2 cm K=0.2 W(m.°C) T a =330°C h a =400 W/(m 2.°C) Ambient air T b =30°C h b = 60 W/(m 2.°C) Determination of heat loss

39. Page  39 Composite Medium Solved Example (Composite Cylinder) Calculate, 1.Heat loss from tube for length H=10m 2.Temperature drops resulting in thermal resistances 5 cm 7.6 cm K = 15 W/(m °C) Insulation t =2 cm K=0.2 W(m.°C) T a =330°C h a =400 W/(m 2.°C) Ambient air T b =30°C h b = 60 W/(m 2.°C) Determination of heat loss

40. Page  40 Composite Medium Solved Example (Composite Cylinder) Calculate, 1.Heat loss from tube for length H=10m 2.Temperature drops resulting in thermal resistances 5 cm 7.6 cm K = 15 W/(m °C) Insulation t =2 cm K=0.2 W(m.°C) T a =330°C h a =400 W/(m 2.°C) Ambient air T b =30°C h b = 60 W/(m 2.°C) Determination of temperature drops

41. Page  41 Composite Medium Solved Example (Composite Wall) 2 x 4 wood studs have actual dimensions of 4.13 x 9.21 cm with k = 0.1 W/m.°C Calculate, 1.Overall heat transfer coefficient 2.R value of the wall Thermal resistance model Note: ‘k’ is expressed in W/m °C T air inside R sheath outside R sheath inside R insul R sheath outside R sheath inside R stud R convection inside R convection outside R brick T air outside Two parallel heat flow paths are possible 1.Through the studs 2.Through the insulation COMMON BRICK, k =0.69 INSULATION, k=0.04 GYPSUM SHEATH k=0.96 k=0.48 2x 4 STUDS 40.6 cm 8cm 1.9cm Outside Air Convection h=15 W/m 2 C Inside Air Convection h=7.5 W/m 2 C

42. Page  42 Composite Medium Solved Example (Composite Wall) Calculate, 1.Overall heat transfer coefficient 2.R value of the wall Heat flow through the studs Note: ‘k’ is expressed in W/m °C Area = 0.0413m 2 /unit depth Heat flow occurs through 6 thermal resistances 1.Convection Resistance outside of brick 2.Conduction resistance in brick 3.Conduction resistance through outer sheet 4.Conduction resistance through wood stud 5.Conduction resistance through inner sheet 6.Convection resistance on inside Recall, COMMON BRICK, k =0.69 INSULATION, k=0.04 GYPSUM SHEATH k=0.96 k=0.48 2x 4 STUDS 40.6 cm 8cm 1.9cm Outside Air Convection h=15 W/m 2 C Inside Air Convection h=7.5 W/m 2 C

43. Page  43 Composite Medium Solved Example (Composite Wall) Calculate, 1.Overall heat transfer coefficient 2.R value of the wall Heat flow through the insulation Note: ‘k’ is expressed in W/m °C The five of the materials are same, but the resistances involve different area terms, i.e., 40.6 - 4.13 cm instead of 4.13 cm. Thus the total resistance of the insulation section is given below COMMON BRICK, k =0.69 INSULATION, k=0.04 GYPSUM SHEATH k=0.96 k=0.48 2x 4 STUDS 40.6 cm 8cm 1.9cm Outside Air Convection h=15 W/m 2 C Inside Air Convection h=7.5 W/m 2 C

44. Page  44 Composite Medium Solved Example (Composite Wall) Calculate, 1.Overall heat transfer coefficient 2.R value of the wall 1. Overall heat transfer coefficient Note: ‘k’ is expressed in W/m °C Overall resistance is obtained by combining the parallel resistances as calculated earlier. Overall heat transfer coefficient is found by, (here, A = 0.406m 2 ) COMMON BRICK, k =0.69 INSULATION, k=0.04 GYPSUM SHEATH k=0.96 k=0.48 2x 4 STUDS 40.6 cm 8cm 1.9cm Outside Air Convection h=15 W/m 2 C Inside Air Convection h=7.5 W/m 2 C

45. Page  45 Composite Medium Solved Example (Composite Wall) Calculate, 1.Overall heat transfer coefficient 2.R value of the wall 2. R Value of the wall Note: ‘k’ is expressed in W/m °C The resistance of the wall is calculated using the overall heat transfer coefficient, as given below: COMMON BRICK, k =0.69 INSULATION, k=0.04 GYPSUM SHEATH k=0.96 k=0.48 2x 4 STUDS 40.6 cm 8cm 1.9cm Outside Air Convection h=15 W/m 2 C Inside Air Convection h=7.5 W/m 2 C

46. Page  46 Composite Medium Solved Example (Critical Thickness of Insulation) Calculate, the critical thickness of rubber and the maximum heat transfer rate per metre length of conductor. The temperature of rubber is not to exceed 65 °C (due to heat generated within). Critical thickness r = 5mm Rubber k = 0.155 W/mK Ambient at 30°C, 8.5 W/m 2 K Maximum heat transfer rate

47. Page  47 Heat Source Systems Plane wall with heat generation Expression for mid plane temperature is given by, The temperature distribution can also be written in alternative form as: q = heat generated per unit volume TwTw TwTw x L L x=0

48. Page  48 Conduction-Convection Systems Fins / Extended Surfaces Necessity for fins Biot Number = LONGITUDINAL RECTANGULAR FIN Internal Conductive resistance Surface Convective resistance FIN TYPES RADIAL FIN

49. Page  49 Conduction-Convection Systems Governing Equation (Rectangular Fin) Net Heat Conducted – Heat Convected = 0 qxqx q x+dx dx L Z x base A t where, &

50. Page  50 Conduction-Convection Systems Boundary Conditions LONG FIN SHORT FIN (end insulated) SHORT FIN ( end not insulated)

51. Page  51 Conduction-Convection Systems Types of Fin Boundaries Type of FIN boundary Temperature DistributionHeat transferred by fin Q Long Fin (T L = T ∞ ) e -mx (Tb-T ∞ )(hPkA) 0.5 Short Fin (end insulated) (hPkA) 0.5 (T b -T ∞ ) tanh (mL) * Short Fin (end not insulated) Specified End Temperature At x=L; T=T L * For higher values of mL (i.e., m=4), tanh mL = 0.999 ≈ 1. Thus Q short fin  Q long fin for higher values of mL

52. Page  52 Conduction-Convection Systems Performance Parameters Fin Efficiency In practical applications, a finned heat transfer surface is composed of the finned surfaces and the unfinned portion. In such cases total heat transfer is used. Q total = Q fin + Q unfinned = η a f h θ 0 + ( a – a f ) h θ 0 Where, a = total heat transfer area (i.e., fin surface + unfinned surface) a f = heat transfer area of fins only Q total = [ηβ+(1-β)] a h θ 0 ≡ η‘ a h θ 0 Where, η‘ = βη +1 – β = area – weighted fin efficiency β = a f / a

53. Page  53 Conduction-Convection Systems Performance Parameters Fin Efficiency Fin Efficiency, η L (2h/kt) 0.5 Fin efficiency curves are available for fins of various configuration (eg. Axial, circular disk fins of various length, thickness etc) Each curve is specific for specific fin configuration

54. Page  54 Conduction-Convection Systems Performance Parameters Fin Effectiveness Although the addition of fins on a surface increases surface area, it also increases thermal resistance over the portion of the surface where fins are attached. Therefore there may be situations in which the addition of fins does not improve heat transfer. Pk / (Ah) > 1 (to justify usage of fins)

55. Page  55 Conduction-Convection Systems Solved Example A steel rod is exposed to ambient air. If one end of the rod is maintained at a temperature of 120 °C, calculate the heat loss from the rod Diameter = 2cm Length = 25 cm k = 50 W / m. °C T amb = 20°C h = 64 W / m 2. °C T base =120°C The condition for other end of the rod is not specified explicitly. By considering L/D ratio, it appears that a long fin assumption is applicable. Using the simplest analysis to solve, computing mL: Therefore, expression for Q long fin can be used.

56. Page  56 Conduction-Convection Systems Solved Example (Fin Efficiency) CIRCULAR DISK FIN Circular disk fins of constant thickness are attached on a 2.5 cm OD tube with a spacing of 100 fins per 1m length of tube. Fin Properties: Aluminium k = 160 W / m.°C, t = 1mm L = 1 cm Tube wall temperature = 170 °C; Ambient temperature = 30 °C Heat transfer coeff. of ambient, h = 200 W/m 2. °C. Calculate, 1.Fin Efficiency and area weighted fin efficiency 2.Heat lost to the ambient air per 1m length of tube 3.Heat loss with that if there were no fins on tube t L Fin Efficiency, η L (2h/kt) 0.5 r o /r i Fin Efficiency Fin efficiency is determined using the graph shown aside. The following parameters are calculated, firstly:

57. Page  57 Conduction-Convection Systems Solved Example (Fin Efficiency) CIRCULAR DISK FIN Calculate, 1.Fin Efficiency and area weighted fin efficiency 2.Heat lost to the ambient air per 1m length of tube 3.Heat loss with that if there were no fins on tube t L Area Weighted Fin Efficiency Ratio of heat transfer area for fin to the total heat transfer area, β Fin Surface per cm of tube length = 2π(r 0 2 -r i 2 ) = 2π[2.25 2 -1.25 2 ] = 21.99 cm 2 Total heat transfer surface per cm of tube length = 2π (r 0 2 -r i 2 ) + 2πr i (1 – t) = 2π[2.25 2 -1.25 2 ] + 2π(1.25)(1 – 0.1) = 29.06 cm 2 β = a f / a = 21.99 / 29.06 = 0.757 Area Weighted Fin Efficiency, η’ = βη +1 – β = 0.757(0.9) + 0.243 = 0.924 Tube OD = 2.5 cm 100 fins per 1m tube length k fin = 160 W/m°C t = 1mm; L = 1cm T tube = 170°C; T amb = 30°C h amb = 200 W/m 2. °C

58. Page  58 Conduction-Convection Systems Solved Example (Fin Efficiency) CIRCULAR DISK FIN Calculate, 1.Fin Efficiency and area weighted fin efficiency 2.Heat lost to the ambient air per 1m length of tube 3.Heat loss with that if there were no fins on tube t L Heat lost to ambient per 1m length of tube Total heat transfer surface a per 1m of tube length a = 29.06 x 100 cm 2 = 0.29 m 2 Q = η’ahθ 0 = 0.924 x 0.29 x 200 (170 – 30) = 7503 W Tube OD = 2.5 cm 100 fins per 1m tube length k fin = 160 W/m°C t = 1mm; L = 1cm T tube = 170°C; T amb = 30°C h amb = 200 W/m 2. °C Heat lost per 1m length of tube with no fins Q no fin = 2πr i hθ 0 = 2π x 0.0125 x 200 x (170 – 30) = 2199 W Clearly, the addition of fins increases the heat dissipation by a factor of about 3.4

59. Page  59 Transient Conduction If the surface temperature of a solid body is suddenly altered, the temperature within the body begins to change over time. Variation of temperature both with position and time makes determination of temperature distribution under transient condition more complicated. In some situations, variation of temperature with position is negligible under transient state, hence the temperature is considered to vary only with time. The analysis under the above assumption is called lumped system analysis. Biot Number, Bi = (hx) / k Lumped System Analysis is applicable only when Bi < 0.1

60. Page  60 Systems with Negligible Internal Resistance The convective heat loss from the body (shown aside) has its magnitude equal to decrease in internal energy of solid. On Integration, Solving and rearranging, Lumped Heat Analysis Volume V Area A Q T∞T∞ T T=T 0 at t=0 1/hA C th =ρcV S T0T0 T∞T∞

61. Page  61 Systems with Negligible Internal Resistance It is a non-dimensional parameter used to test the validity of the lumped heat capacity approach. The characteristic length (L c ) for some common shapes is given below: Plane Wall (thickness 2L)Long cylinder (radius R) Sphere (radius R)Cube (side L) The lumped heat capacity approach for simple shapes such as plates, cylinders, spheres and cubes can be used if Bi < 0.1 Biot Number

62. Page  62 Systems with Negligible Internal Resistance For a rapid response of temperature measuring device, the index, (hAt/ρcV) should be large to make the exponential term reach zero faster. This can be achieved by decreasing wire diameter, density and specific heat or by increasing value of ‘h’. The quantity (ρcV/hA) has the units of time and is called ‘time constant’ of system. Hence at time t=t * (one time constant), At the end of time period t * the temperature difference between the body and ambient would be 0.368 of the initial temperature difference. In other words, the temperature difference would be reduced by 63.2 percent. This reduction in 63.2 percent of initial temperature difference is called ‘sensitivity’ Lower the value of time constant, better the response of instrument. Response time of a Temperature measuring Instrument

63. Page  63 Systems with Negligible Surface Resistance When convective heat transfer coefficient at the surface is assumed to be infinite, the surface temperature remains constant at all the time (t>0) and its value is equal to that of ambient temperature. The systems exhibiting above said conditions are considered to have ‘negligible surface resistance’ An important application of this process is in heat treatment of metals by quenching, viz., the dropping of a metallic sphere initially at 300 °C into a 20 °C oil bath. Mathematical formulation of this case is : T s = T ∞ (t>0) TsTs T 0 (x) for t = 0 x L Large Flat Plate with Negligible Surface Resistance Boundary Conditions

64. Page  64 Heat flow in an Infinitely Thick Plate A semi-infinite body is one in which at any instant of time there is always a point where the effect of heating / cooling at one of its boundaries is not felt at all. At this point the temperature remains unchanged. Mathematical formulation is : with initial and boundary conditions, Semi-Infinite Plate Semi-infinite body QoQo TsTs x ToTo at t=0

65. Page  65 Systems with Finite Surface and Internal Resistance Mathematical formulation : Infinitely Large Flat Plate of Finite Thickness (2L) T∞T∞ h x=0 h x=L x=-L at t=0 x -x T∞T∞

66. Page  66 Chart Solutions of Transient Heat Conduction Problems Time History Mid Plane Heisler Charts (by Heisler, 1947) Infinite Plate 1 0.9 0.8 0.6 0.4 0.2 0.1 0.5 T(x,t) - T ∞ T i - T ∞ Fourier number, ατ/L 2 hL/k

67. Page  67 Chart Solutions of Transient Heat Conduction Problems Time History Any Position, x Heisler Charts (by Heisler, 1947) Infinite Plate Biot Number, hL/k x/L 1 0.9 0.8 0.6 0.4 0.2 0.1100 T(x,t) - T ∞ T i - T ∞ 0 1

68. Page  68 Chart Solutions of Transient Heat Conduction Problems Heat Flow Heisler Charts (by Heisler, 1947) Infinite Plate Q/Q o hL/k 10 1 0.5 0.1 0.05 0.01 0.001 50 2040

69. Page  69 Lumped System Analysis Solved Example Determination of Time required to cool Aluminium Ball ρ = 2700 kg/m 3 c = 900 J/kg K k = 205 W/mK 5.5 kg T fluid = 15°C h = 58 W / m 2. °C Time required to cool the aluminium ball to 95°C ? T initial =290°C Volume Radius Characteristic Length

70. Page  70 Lumped System Analysis Solved Example The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1mm diameter sphere (shown aside) Determine how long it will take for the thermocouple to read 99% of initial temperature difference Gas T ∞ h=210 W/m 2 °C How long will it take for the thermocouple to read 99 % of Initial Temperature difference ? L c = V/A s = (1/6)D = (1/6)x0.001= 1.67x10 -4 m Bi = hL/k= (210x 1.67x10 -4 ) x 35= 0.001 < 0.1 Therefore, lumped system analysis is applicable. In order to read 99% of initial temperature difference T i – T ∞ between the junction and the gas, we must have Temperature Measurement by Thermocouples Junction (Sphere) D= 1mm ρ = 8500 kg/m 3 k = 35 W/mK c = 320 J/kg K Thermocouple Wire

71. Page  71 Lumped System Analysis Solved Example The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1mm diameter sphere (shown aside) Determine how long it will take for the thermocouple to read 99% of initial temperature difference Gas T ∞ h=210 W/m 2 °C How long will it take for the thermocouple to read 99 % of Initial Temperature difference ? t = 10s Temperature Measurement by Thermocouples Junction (Sphere) D= 1mm ρ = 8500 kg/m 3 k = 35 W/mK c = 320 J/kg K Thermocouple Wire Time

72. Page  72 Transient Conduction in Semi-infinite Solids Solved Example A water pipe is to be buried in soil at sufficient depth from the surface to prevent freezing in winter. What minimum depth is required to prevent the freezing of pipe when soil is at uniform temperature of T i = 10 °C, the surface is subjected to a uniform temperature of T 0 = -15 °C continuously for 50 days. Also the pipe surface temperature should not fall below 0 °C

73. Page  73 Transient Conduction in Semi-infinite Solids Solved Example What burial depth is needed to prevent freezing of the pipe ? SOIL Water Pipe (to be buried) ? T surface = -15 °C Condition : T pipe wall should not fall below 0 °C T soil = 10 °C

74. Page  74 T(x,t) - T surface T initial - T surface Temperature Distribution in Semi-infinite Solid

75. Page  75 Determination of Burial depth

76. Page  76 Determination of Burial depth The pipe should be buried at least to a depth of 1.12 m to prevent freezing.

77. Page  77 Application of Heisler Charts Aluminium Slab Thickness=10cm α = 8.4x10 -5 m 2 /s ρ = 2700 kg/m 3 c = 900 J/kg K k = 215 W/mK T fluid = 100°C h = 1200 W / m 2. °C Mid-plane Temperature and Surface Temperature after 1 min? T initial =500°C

78. Page  78 Determination of Mid plane Temperature 2L=10 cm ; L = 5 cm ; t = 1min = 60 s αt/L 2 = (8.4x10 -5 x 60) / 0.05 2 = 2.016 Bi = hL/k= (1200x0.05) x 215= 0.28 Using above two parameters in Heisler Chart,

79. Page  79 Determination of Surface Temperature For x/L = 1 and Bi = 0.28,

80. Page  80 Energy Loss h 2 αt/k 2 = (1200 2 x8.4x10 -5 x 60) / 215 2 = 0.157 Bi = hL/k= (1200x0.05) x 215= 0.28 Using above 2 parameters in Heisler Chart for Heat flow, Q/Q 0 = 0.32

81. Page  81 Heat removed per unit surface area