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# Thermodynamics Chapter 3- Heat Transfer

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### Thermodynamics Chapter 3- Heat Transfer

1. 1. Heat Transfer Heat Transfer Applied Thermodynamics & Heat Engines S.Y. B. Tech. ME0223 SEM - IV Production EngineeringME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
2. 2. Heat Transfer Outline • One – Dimensional Steady State Heat Transfer by conduction through plane wall, Radial Heat Transfer by Conduction through hollow Cylinder / Sphere. Conduction through Composite Plane and Cylindrical Wall. • Heat flow by Convection. Free and Forced Convection. Nusselt, Reynolds and Prandtl Numbers. Heat Transfer between two fluids separated by Composite Plane and Cylindrical wall. Overall Heat Transfer Coefficient. • Heat Exchangers, types of Heat Exchangers, Log Mean Temperature Difference. • Radiation Heat Transfer, Absorptivity, Reflectivity and Transmissivity, Monochromatic Emissive Power, Wein’s Law, Stefan-Boltzman’s Law and Kirchoff’s Law.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
3. 3. Heat Transfer Heat Transfer HEAT TRANSFER is a science that seeks to predict the energy transfer that may take place between material bodies, as a result of temperature difference. Heat Transfer RATE is the desired objective of an analysis that points out the difference between Heat Transfer and Thermodynamics. Thermodynamics is dealt with equilibrium, and does not predict how fast the change will take place. Heat Transfer supplements the First and Second Laws of Thermodynamics, with additional rules to analyse the Energy Transfer RATES.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
4. 4. Heat Transfer Conduction Heat Transfer Fourier Law : Heat Transfer (HT) Rate per unit cross – sectional (c/s) area is proportional to the Temperature Gradient. q T A x T q k A x Q = HT Rate, ∂T/∂T = Temperature Gradient in the direction of Heat Flow. k = Constant of Proportionality, known as THERMAL CONDUCTIVITY, (W/mºC) NOTE : Negative sign is to indicate that Heat flows from High – Temperature to Low – Temperature region, i.e. to satisfy Second Law of Thermodynamics.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
5. 5. Heat Transfer Conduction Heat Transfer Heat Conduction through Plane Wall : Generalised Case : 1. Temperature changes with time. 2. Internal Heat Sources. Energy Balance gives; qgen = qi A dx Energy conducted in left face + Heat generated within element A = Change in Internal Energy + Energy conducted out right face. qx qx+dx T Energy in left face = q kA x x Energy generated within element = qi A dx T x dx Change in Internal Energy = cA dx Energy out right face = T T T qx dx kA A k k dx x x dx x x xME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
6. 6. Heat Transfer Conduction Heat Transfer Heat Conduction through Plane Wall : Combining the terms; T T T T kA qi A dx cA dx A k k dx x x x x T T qgen = qi A dx k qi c x x A This is One – Dimensional Heat Conduction Equation qx qx+dx T T T T kx ky kz qi c x x y y z z For Constant Thermal Conductivity, kx = ky = kz = k 2 2 2 x dx T T T qi 1 T x2 y2 z2 k Where, α = ( k / ρc ) is called Thermal Diffusivity. (m2/sec) (↑) α ; (↑) the heat will diffuse through the material.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
7. 7. Heat Transfer Conduction Heat Transfer 2 2 2 T 1 T 1 T T qi 1 THeat Conduction through Cylinder : r2 r r r2 2 z2 k 1 2 1 T 1 2 T qi 1 THeat Conduction through Sphere : rT sin r r2 r 2 sin r 2 sin 2 2 k Special Cases : 1. Steady State One – Dimensional (No Heat Generation) : d 2T 0 dx2 2. Steady State One – Dimensional, Cylindrical co-ordinates (No Heat Generation) : d 2T 1 dT 0 dr 2 r dr 3. Steady State One – Dimensional, with Heat Generation : d 2T qi 0 dx2 kME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
8. 8. Heat Transfer Thermal Conductivity GAS : Kinetic Energy of the molecules of gas is transmitted from High – Temperature region to that of Low – Temperature through continuous random motion, colliding with one another and exchanging Energy as well as momentum. LIQUIDS : Kinetic Energy is transmitted from High – Temperature region to that of Low – Temperature by the same mechanism. BUT the situation is more complex; as the molecules are closely spaced and molecular force fields exert strong influence on the Energy exchange. SOLIDS : (a) Free Electrons : Good Conductors have large number of free electrons, which transfer electric charge as well as Thermal Energy. Hence, are known as electron gas. EXCEPTION : Diamond ! (b) Lattice Vibrations : Vibrational Energy in lattice structure of the material. NOTE : Mode (a) is predominant than Mode (b).ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
9. 9. Heat Transfer Thermal Conductivity SOLIDS :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
10. 10. Heat Transfer Multilayer InsulationAlternate Layers of Metal and Non-Metal Metal having higher Reflectivity.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
11. 11. Heat Transfer Thermal Conductivity LIQUIDS :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
12. 12. Heat Transfer Thermal Conductivity GASES :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
13. 13. Heat Transfer Thermal Conductivity Comparison :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
14. 14. Heat Transfer Conduction through Plane Wall Fourier’s Law, Generalised Form T q k A x qgen = qi A dx For Const. k; Integration yields ; A qx qx+dx kA q (T2 T1 ) x For k with some linear relationship, like k = k0(1+βT); x dx kA 2 2 q (T2 T1 ) (T2 T1 ) x 2ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
15. 15. Heat Transfer Conduction through Composite Wall Since Heat Flow through all sections must be SAME ; T2 T1 T3 T2 T4 T3 q kA A kB A kC A xA xB xC Thus, solving the equations would result in, T1 T4 q q q xA xB xC kA A kB A kC A A B C ELECTRICAL ANALOGY : 1. HT Rate = Heat Flow 2. k, thickness of material & area = Thermal Resistance 1 2 3 4 3. ΔT = Thermal Potential Difference. q Therm alPotentialDifference RA RB RC HeatFlow Therm al sis tance ReT1 T4 xA xB xC Toverall T2 T3 q kA A kB A kC A RthME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
16. 16. Heat Transfer Conduction through Composite Wall B F q C E A G D 1 2 3 4 5 RB RF q RA RC RE T1 T2 RD T3 T4 RG T5ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
17. 17. Heat Transfer Example 1An exterior wall of a house is approximated by a 4-in layer of common brick (k=0.7W/m.ºC) followed by a 1.5-in layer of Gypsum plaster (k=0.48 W/m.ºC). What thickness ofloosely packed Rockwool insulation (k=0.065 W/m.ºC) should be added to reduce the Heatloss through the wall by 80 % ? TOverall Heat Loss is given by; q RthBecause the Heat loss with the Rockwool q with insulation Rth without insulationinsulation will be only 20 %, of that before 0.2 q withoutinsulation Rth with insulationinsulation, x 4 0.0254For brick and Plaster, for unit area; Rb 0.145 m 2 . C / W k 0.7 x 1.5 0.0254 Rp 0.079 m 2 . C / W k 0.48 2So that the Thermal Resistance without insulation is; R 0.145 0.079 0.224 m . C / WME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
18. 18. Heat Transfer Example 1…contd 0.224 Now; R with Insulation 1.122 m 2 . C / W 0.2 This is the SUM of the previous value and the Resistance for the Rockwool. 1.122 0.224 Rrw x x Rrw 0.898 k 0.065 xrw 0.0584m 2.30 in …ANSME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
19. 19. Heat Transfer Conduction through Radial Systems Cylinder with; 1. Inside Radius, ri. 2. Outside Radius, ro. 3. Length, L 4. Temperature Gradient, Ti-To q 5. L >> r; → Heat Flow in Radial direction only. dr ro ri r Area for Heat Flow; Ar = 2πrL T T Fourier’s Law will be, q k Ar 2 krL r r q Boundary Conditions : RA T = Ti at r = ri Ti To T = To at r = ro ln ro / ri 2 k L Ti To Rth q 2 kL Solution to the Equation is; ln ro / riME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
20. 20. Heat Transfer Conduction through Radial Systems q Thermal Resistance is, T4 ln ro / ri Rth T3 2 kL R1 T2 For Composite Cylinder; T1 R2 2 L T1 T4 A q R3 ln r2 / r1 / k A ln r3 / r2 / k B ln r4 / r3 / k B B C R4 For Spheres; q RA RB RC 4 k Ti ToT1 T4 q 1 1 ln r2 / r1 T2 ln r3 / r2 T3 ln r4 / r3 ri ro 2 kAL 2 kB L 2 kC LME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
21. 21. Heat Transfer Example 2 A thick-walled tube of stainless steel (18% Cr, 8% Ni, k=19 W/m.ºC) with 2 cm inner diameter (ID) and 4 cm outer diameter (OD) is covered with a 3 cm layer of asbestos insulation (k=0.2 W/m.ºC). If the inside wall temperature of the pipe is maintained at 600 ºC, calculate the heat loss per meter of length and the tube-insulation interface temperature. Heat flow is given by; q q 2 (T1 T3 ) 2 (600 100) T3=100 ºC 680W / m L ln r2 / r1 / kS ln r3 / r2 / k A ln 2 / 1 / 19 ln 5 / 2 / 0.2 This Heat Flow is used to calculate the tube-insulation R1 T2 interface temperature as; R2 q (T2 T3 ) T1=600 ºC 680W / m L ln r3 / r2 / 2 k A R3 T2 = 595.8 ºC…ANSStainless Steel Asbestos S. Y. B. Tech. Prod Engg.
22. 22. Heat Transfer Critical Thickness of Insulation Consider a layer of Insulation around a circular pipe. Inner Temperature of Insulator, fixed at Ti Outer surface exposed to convective environment, T∞ h, T∞ From Thermal Network; R1 2 L Ti T q ln ro / ri 1 Ti R2 k ro h Expression to determine the outer radius of Insulation, ro for maximum HT; 1 1 2 L Ti T 2 dq kro kro q 0 2 dro ln ro / ri 1Ti T∞ k ro h which gives; ln ro / ri 1 k r0 2 kL 2 ro Lh hME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
23. 23. Heat Transfer Example 3 Calculate the critical radius of asbestos (k=0.17 W/m.ºC) surrounding a pipe and exposed to room air at 20 ºC with h=3 W/m2. ºC . Calculate the heat loss from a 200 ºC, 5 cm diameter pipe when covered with the critical radius of insulation and without insulation. k 0.17 r0 0.0567 m 5.67 cm h 3.0 Inside radius of the insulation is 5.0/2 = 2.5 cm. Heat Transfer is calculated as; q 2 (200 20) 105.7 W / m L ln(5.67 / 2.5) 1 0.17 (0.0567)(3.0) Without insulation, the convection from the outer surface of the pipe is; q h(2 r )(Ti T0 ) (3.0)( 2 )( 0.025 )( 200 20) 84.8W / m LME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
24. 24. Heat Transfer Example 3…contd Thus, the addition of (5.67-2.5) = 3.17 cm of insulation actually increases the Heat Transfer by @ 25 %. Alternatively, if fiberglass (k=0.04 W/m.ºC) is employed as the insulation material, it would give; k 0.04 ro 0.0133 m 1.33 cm h 3.0 Now, the value of the Critical Radius is less than the outside radius of the pipe (2.5 cm). So, addition of any fiberglass insulation would cause a decrease in the Heat Transfer.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
25. 25. Heat Transfer Thermal Contact Resistance Two solid bars in contact. q q A B Sides insulated to assure that Heat flows in Axial direction only. ΔxA ΔxB Thermal Conductivity may be different. But Heat Flux through the materials T under Steady – State MUST be same. T1 T2A Actual Temperature profile approx. as T2B shown. T3 The Temperature Drop at Plane 2, the Contact Plane is said to be due to 1 2 3 x Thermal Contact Resistance.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
26. 26. Heat Transfer Thermal Contact Resistance T Energy Balance gives; T1 T1 T2 A T2 A T2 B T2 B T3 T2A q kA A kB A xA 1 / hC A xB T2B T1 T3 q T3 xA 1 xB kA A hC A kB A Surface Roughness is exaggerated. 1 2 3 x No Real Surface is perfectly smooth. HT at joints can be contributed to : 1. Solid – Solid conduction at spots of contact. A 2. Conduction through entrapped gases through the void spaces. Second factor is the major Resistance to Heat B Flow, as the conductivity of a gas is much lower than that of a solid.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
27. 27. Heat Transfer Thermal Contact Resistance Designating; 1. Contact Area, Ac A 2. Void Area, Av 3. Thickness of Void Space, Lg 4. Thermal Conductivity of the Fluid in Void space, kf T2 A T2 B T2 A T2 B T2 A T2 B q k f Av B Lg / 2k A AC Lg / 2k B AC Lg 1 / hC A Total C/s. Area of the bars is A. Solving for hc; T2A 1 AC 2k A k B Av hC kf Lg Lg A k A kB A T2B Most usually, AIR is the fluid in void spaces. Hence, kf is very small compared to kA and kB.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
28. 28. Heat Transfer Example 4 Two 3.0 cm diameter 304 stainless steel bars, 10 cm long have ground surface and are exposed to air with a surface roughness of about 1 μm. If the surfaces are pressed together with a pressure of 50 atm and the two-bar combination is exposed to an overall temperature difference of 100 ºC, calculate the axial Heat Flow and Temperature Drop across the contact surface. Take 1/hc=5.28 X 10-4 m2.ºC/W. The overall Heat Flow is subject to three resistances, 1. One Conducting Resistance for each bar 2. Contact Resistance. x (0.1)(4) For the bars; Rth 8.679 C / W kA (16.3) (3 X 100 2 ) 2 1 (5.28X 10 4 )(4) Contact Resistance; RC 0.747 C / W hC A (3 X 10 2 ) 2 Total Thermal Resistance; Rth (2)(3.679) 0.747 8.105 C / WME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
29. 29. Heat Transfer Example 4…contd T 100 Overall Heat Flow is; q 12.338 W …ANS Rth 8.105 Temperature Drop across the contact is found by taking the ratio of the Contact Resistance to the Total Resistance. RC (0.747) TC T (100) 6.0544 C …ANS Rth 12.338 i.e. 6 % of the total resistance.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
30. 30. Heat Transfer Radiation Heat Transfer Physical Interpretation : Thermal Radiation is the electromagnetic radiation as result of its temperature. There are many types of electromagnetic radiations, Thermal Radiation is one of them. Regardless of the type, electromagnetic radiation is propagated at the speed of light, 3 X 108 m/sec. This speed is the product of wavelength and frequency of the radiation. c = Speed of light c=λυ λ = Wavelength υ = Frequency NOTE : Unit of λ may be cm, A˚ or μm.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
31. 31. Heat Transfer Radiation Heat Transfer Thermal Radiation log λ, m 1 μm 1 A˚3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 X-RaysRadioWaves Infrared Ultraviolet γ-Rays VisibleThermal Radiation → 0.1 – 100 μm.Visible Light Portion → 0.35 – 0.75 μm. S. Y. B. Tech. Prod Engg.
32. 32. Heat Transfer Radiation Heat Transfer Physical Interpretation : Propagation of Thermal radiation takes place in the form of discrete quanta. Each quantum having energy; E = hυ h is Planck’s Constant, given by; h = 6.625 X 10-34 J.sec A very basic physical picture of the Radiation propagation → Considering each quantum as a particle having Energy, Mass and momentum. E = mc2 = hυ m = hυ c2 & Momentum = c (hυ) = hυ c2 cME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
33. 33. Heat Transfer Stefan – Boltzmann Law Applying the principles of Quantum-Statistical Thermodynamics; Energy Density per unit volume per unit wavelength; 8 hc 5 u e hc / kT 1 Where, k is Boltzmann’s Constant; 1.38066 X 10-23 J/molecule.K When Energy Density is integrated over all wavelengths; Total Energy emitted is proportional to the fourth power of absolute temperature. Eb = ζT4 Equation is known as Stefan – Boltzmann Law. Eb = Energy radiated per unit time per unit area by the ideal radiator, W/m2 ζ = Stefan – Boltzmann Constant; 5.667 X 10-8 W/m2.K4ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
34. 34. Heat Transfer Stefan – Boltzmann Law Subscript b in this equation → Radiation from a Blackbody. Materials which obey this Law appear black to the eye, as they do not reflect any radiation. Thus, a Blackbody is a body which absorbs all the radiations incident upon it. Eb is known as the Emissive Power of the Blackbody. i.e. Energy radiated per unit time per unit area. Eb = ζT4 qemitted = ζ.A.T4 NOTE : “Blackness” of a surface to Thermal Radiation can be quite deceiving. e.g. Lamp-black…..and Ice….!!!ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
35. 35. Heat Transfer Radiation Properties Radiant Energy incident on a surface; (a) Part is Reflected, (b) Part is Absorbed, (c) Part is Transmitted. Incident Radiation Reflection Reflected Energy Reflectivity = ρ = Incident Energy Absorbed Energy Absorption Absorptivity = α = Incident Energy Transmitted Energy Transmissivity = η = Transmission Incident Energy ρ+α+η=1 NOTE : Most solids do not transmit Thermal Radiations. → ρ+α=1ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
36. 36. Heat Transfer Radiation Properties Types of Reflection : 1. Angle of Incidence = Angle of Reflection → Specular Reflection 2. Incident beam distributed uniformly in all directions → Diffuse Reflection Source Source θ1 θ2 Specular Reflection Diffuse Reflection Offers the Mirror Image to the ObserverME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
37. 37. Heat Transfer Radiation Properties No REAL surface cab be perfectly Specular or Diffuse. Ordinary mirror → Specular for visible region; but may not be in complete spectrum of Thermal Radiation. Rough surface exhibits the Diffused behaviour. Polished surface exhibits the Specular behaviour.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
38. 38. Heat Transfer Kirchhoff’s Law Assume perfect Black enclosure. Radiant flux incident is qi W/m2. A Sample body placed inside Enclosure, in Black Enclosure Thermal Equilibrium with it. EA qiAα Sample Energy Absorbed by the Sample = Energy emitted. i.e. E A = qi A α …(I) Replacing the Sample Body with a Blackbody; Eb A = qi A (1) …(II) Dividing (I) by (II); E / Eb = αME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
39. 39. Heat Transfer Kirchhoff’s Law Ratio of Emissive Power of a body Black Enclosure to that of a Perfectly Black body, at the same temperature is known as EA qiAα Sample Emissivity, ε of the body. ε = E / Eb Thus; ε=α This is known as Kirchhoff’s Identity.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
40. 40. Heat Transfer Gray Body Gray Body can be defined as the body whose Monochromatic Emissivity, ελ is independent of wavelength. Monochromatic Emissivity is defined as the ratio of Monochromatic Emissive Power of the body to that of a Blackbody at the SAME wavelength and temperature. ελ = Eλ / Ebλ Total Emissivity of the body is related to the Monochromatic Emissivity as; E Eb d And Eb Eb d T4 0 0 Eb d Thus, E 0 EB T4ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
41. 41. Heat Transfer Gray Body When the GRAY BODY condition is applied, ελ = ε A functional relation for Ebλ was derived by Planck by introducing the concept of QUANTUM in Electromagnetic Theory. The derivation is by Statistical Thermodynamics. u c By this theory, Ebλ is related to the Energy Density by; Eb 4 5 C1 Eb Where, λ = Wavelength, μm e C2 / T 1 T = Temperature, K C1 = 3.743 X 108 ,W.μm/m2 C2 = 1.4387 X 104, μm.KME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
42. 42. Heat Transfer Wien’s Displacement Law A plot of Ebλ as a function of T and λ. Peak of the curve is shifted to SHORTER Wavelengths at HIGHER Temperatures. Points of the curve are related by; λmax T = 2897.6 μm.K This is known as Wien’s Displacement Law.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
43. 43. Heat Transfer Wien’s Displacement LawShift in the maximum point of the Radiation Curve helps to explain the change in colourof a body as it is heated.Band of wavelengths visible to the eye lies between 0.3 – 0.7 μm.Very small portion of the radiant energy spectrum at low temperatures is visible to eye.As the body is heated, maximum intensity shifts to shorter wavelengths.Accordingly, first visible sight of increase in temperature of a body is DARK REDcolour.Further heating yields BRIGHT RED colour.Then BRIGHT YELLOW.And, finally WHITE…!!Material also looks brighter at higher temperatures, as large portion of the totalradiation falls within the visible range.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
44. 44. Heat Transfer Radiation Heat Transfer General interest in amount of Energy radiated from a Blackbody in a certain specified Wavelength range. Eb d Eb 0 The fraction of Total Energy Radiated between 0 to λ is given by; 0 Eb 0 Eb d Eb C1 0 Diving both sides by T5; 5 T5 T e C2 / T 1 This ratio is standardized in Graph as well as Tabular forms; with parameters as; 1. λT 2. Ebλ / T5 3. Eb 0-λT / ζT4ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
45. 45. Heat Transfer Radiation Heat Transfer If the Radiant Energy between Wavelengths λ1 and λ2 are desired; Eb 0 Eb 0 Eb Eb 0 2 1 1 2 Eb 0 Eb 0 NOTE : Eb 0-∞ is the Total Radiation emitted over all Wavelengths = ζT4ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
46. 46. Heat Transfer Radiation Heat Transfer EXAMPLE : Solar Radiation has a spectrum, approx. to that of a Blackbody at 5800 K. Ordinary window glass transmits Radiation to about 2.5 μm. From the Table for Radiation Function, λT = (2.5)(5800) = 14,500 μm.K The fraction of then Solar Spectrum is about 0.97. Thus, the regular glass transmits most of the Radiation incident upon it. On the other hand, the room Radiation at 300 K has λT = (2.5)(300) = 750 μm.K Radiation Fraction corresponding to this value is 0.001 per cent. Thus, the ordinary glass essentially TRANSPARENT to Visible light, is almost OPAQUE for Thermal Radiation at room temperature.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
47. 47. Heat Transfer Heat Exchangers Overall Heat Transfer Coefficient : Heat Transfer is expressed as; TA kA q h1 A (TA T1 ) (T1 T2 ) h2 A (T2 TB ) h2 x Applying the Thermal Resistance; Fluid B T1 (TA TB ) q q T2 (1 / h1 A) ( x / kA) (1 / h2 A) Overall Heat Transfer by combined Fluid A Conduction + Convection is expressed in h1 TB terms of Overall Heat Transfer Coefficient , U, defined as; q RA RB RC q = U A ΔToverallTA TB 1 x 1 1 T1 T2 U h1 A kA h2 A (1 / h1 ) ( x / k ) (1 / h2 )ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
48. 48. Heat Transfer Heat Exchangers Overall Heat Transfer Coefficient : Hollow Cylinder exposed to Fluid B in Convective environment on its inner and outer surfaces.Fluid A in Area for Convection is NOT same for both fluids. → ID and thickness of the inner tube. q Overall Heat Transfer would be; RA RB RC TA TB (TA TB ) 1 1 q Ti ln (ro / ri ) To 1 ln (r0 / ri ) 1 hi Ai 2 kL ho Ao hi Ai 2 kL ho AoME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
49. 49. Heat Transfer Heat Exchangers Overall Heat Transfer Coefficient : Overall Heat Transfer Coefficient can be based on either INNER side or OUTER area of the tube. 1 Based in INNER Area; Ui 1 Ai ln (r0 / ri ) Ai 1 hi 2 kL Ao ho 1 Based in OUTER Area; Uo Ao 1 Ao ln (r0 / ri ) 1 Ai hi 2 kL ho 1 In general, for either Plane Wall or Cylinder, UA RthME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
50. 50. Heat Transfer Example 5 Water flows at 50 °C inside a 2.5 cm inside diameter tube such that hi = 3500 W/m2.°C. The tube has a wall thickness of 0.8 mm with thermal conductivity of 16 W/m.°C. The outside of the tube loses heat by free convection with ho = 7.6 W/m2.°C.Calculate the overall heat transfer coefficient and heat air at 20 °C. 3 Resistances in series. L = 1.0 mtr, di = 0.025 mtr and do = 0.025 + (2)(0.008) mtr = 0.0266 mtr. 1 1 Ri 0.00364 C / W hi Ai (3500 (0.025)(1.0) ) ln (d o / di ) ln (0.0266/ 0.025) Rt 0.00062 C / W 2 kL 2 (16)(1.0) 1 1 Ro 1.575 C / W ho Ao (7.6) (0.0266 1.0) )(ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
51. 51. Heat Transfer Example 5….contd This clearly states that the controlling resistance for the Overall Heat Transfer Coefficient is Outside Convection Resistance. Hence, the Overall Heat Transfer Coefficient is based on Outside Tube Area. Toverall q U 0 Ao T Rth 1 1 U0 A0 Rth (0.0266 1.0) 0.00364 0.00062 1.575 )( 7.577 W / m 2 . C ….ANS Heat Transfer is obtained by; q Uo Ao T (7.577) (0.0266 1.0)(50 20) 19 W ….ANS )(ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
52. 52. Heat Transfer Heat Exchangers Fouling Factor : After a period, the performance of the Heat Exchanger gets degraded as; 1. HT surface may become coated with various deposits. 2. HT surface may get corroded due to interaction between fluid and material. This coating offers additional Resistance to the Heat Flow. Performance Degradation effect is presented by introducing Fouling Factor or Fouling Resistance, Rf. 1 1 Fouling Factor, Rf is defined as; Rf U dirty U cleanME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
53. 53. Heat Transfer Types of Heat Exchangers Shell-And-Tube Heat Exchanger :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
54. 54. Heat Transfer Types of Heat Exchangers Shell-And-Tube Heat Exchanger :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
55. 55. Heat Transfer Types of Heat Exchangers Shell-And-Tube Heat Exchanger :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
56. 56. Heat Transfer Types of Heat Exchangers Shell-And-Tube Heat Exchanger :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
57. 57. Heat Transfer Types of Heat Exchangers Miniature / Compact Heat Exchanger :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
58. 58. Heat Transfer Types of Heat Exchangers Cross-Flow Heat Exchanger :ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
59. 59. Heat Transfer Types of Heat ExchangersCross-Flow Heat Exchanger : S. Y. B. Tech. Prod Engg.
60. 60. Heat Transfer Log Mean Temperature Difference TEMPERATURE PROFILES :T T Th1 Hot Fluid Th1 Hot Fluid dq dq Th Tc1 Th Th2 Th2 Tc Tc Tc2 dA dA Cold Fluid Cold Fluid Tc2 Tc1 A A 1 2 1 2 Parallel Flow Counter FlowME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
61. 61. Heat Transfer Log Mean Temperature Difference q = U A ΔTmT dq = U dA (Th-Tc) U = Overall Heat Transfer Coefficient. Hot Fluid A = Surface Area for Heat Transfer Th1 dq consistent with definition of U. Th ΔTm = Suitable Mean Temperature Th2 Difference across Heat Exchanger. Tc Tc2 As can be seen, the Temperature dA Difference between the Hot and Cold Tc1 Cold Fluid fluids vary between Inlet and Outlet. A Average Heat Transfer Area for the 1 2 above equation is required.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
62. 62. Heat Transfer Log Mean Temperature Difference Heat transferred through elemental area dA;T dq = U dA (Th-Tc) dq mh Ch dTh mc Cc dTc Th1 Hot Fluid dq U dA(Th Tc ) dq Th dq dq Th2 dTh And; dTc m h Ch m c Cc Tc Tc2 dA 1 1 Cold Fluid dTh dTc d (Th Tc ) dq Tc1 m h Ch m c Cc Solving for dq; A 1 2 d (Th Tc ) 1 1 U dA Th Tc m h Ch m c CcME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
63. 63. Heat Transfer Log Mean Temperature Difference Equation can be integrated between conditions 1 and 2 to yield;T dq = U dA (Th-Tc) (Th 2 Tc 2 ) 1 1 ln UA Th1 Hot Fluid (Th1 Tc1 ) m h Ch m c Cc dq Th Again; Th2 q q m h Ch & mc Cc (Th1 Th 2 ) (Tc 2 Tc1 ) Tc Tc2 This substitution gives; dA Tc1 Cold Fluid (Th 2 Tc 2 ) (Th1 Tc1 ) q UA ln (Th 2 Tc 2 ) /(Th1 Tc1 ) A 1 2 (Th 2 Tc 2 ) (Th1 Tc1 ) OR; Tm ln (Th 2 Tc 2 ) /(Th1 Tc1 )ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
64. 64. Heat Transfer Log Mean Temperature Difference This Temperature Difference, ΔTm , is known as Log Mean Temperature Difference. Tone end of HE Tother end of HE LMTD natural log Ratio of both Ts Main Assumption : 1. Specific Heats (Cc and Ch) of fluids do not vary with Temperatures. 2. Convective HT Coefficients (h) are constant throughout the Heat Exchanger. Serious concerns for validity due to : 1. Entrance Effects. 2. Fluid Viscosity. 3. Change in Th. Conductivity.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
65. 65. Heat Transfer Log Mean Temperature DifferenceME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
66. 66. Heat Transfer Log Mean Temperature DifferenceME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
67. 67. Heat Transfer Log Mean Temperature DifferenceME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
68. 68. Heat Transfer Log Mean Temperature DifferenceME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
69. 69. Heat Transfer Example 6 Water at the rate of 68 kg/min is heated from 35 to 75 °C by an oil having specific heat of 1.9 kJ/kg.°C. The fluids are used in a counterflow double-pipe heat exchanger, and the oil enters in the exchanger at 110 °C and leaves at 75 °C. The overall heat transfer coefficient is 320 W/m2.°C. Calculate the heat exchanger area. Total Heat Transfer is calculated by Energy absorbed by water; q m w Cw Tw (68)(4.18)(75 35) 11.37 MJ / min 189.5 kW Since all fluid temperatures are known, LMTD T can be calculated. 110 °C Hot Fluid Th (Th 2 Tc 2 ) (Th1 Tc1 ) dq Tm ln (Th 2 Tc 2 ) /(Th1 Tc1 ) 75 °C 75 °C (75 35) (110 75) Tc 37.44 C ln (75 35) /(110 75) dA 35 °C And; q U A Tm Yields; q 189.5 X 103 Cold Fluid A 15.82 m2….ANS A U Tm (320)(37.44) 1 2ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
70. 70. Heat Transfer Example 7 In stead of double-pipe heat exchanger, of Example 6, it is desired to use a shell-and- tube heat exchanger with water making one shell pass and oil making two tube passes. Calculate the heat exchanger area assuming other conditions same. T1 = 35 °C T2 = 75 °C t1 = 110 °C t2 = 75 °C t2 t1 75 110 P 0.467 T1 t1 35 110 T1 T2 35 75 R 1.143 t2 t1 75 110 Correction Factor from Chart = 0.8 q U A F Tm yields; q 189.5 X 103 A 19.53 m2 ….ANS U F Tm (320)(0.8)(37.44)ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
71. 71. Heat Transfer Effectiveness-NTU Method LMTD approach is suitable when both the inlet and outlet temperatures are known, or can be easily computed. However, when the temperatures are to be evaluated by an iterative method, analysis becomes quite complicated as it involves the Logarithmic function. In this case, the method of analysis is based on the Effectiveness of the Heat Exchanger in transferring the given amount of Heat. Effectiveness of the Heat Exchanger is defined as; Actual Heat Transfer Effectiveness Maxim umPossible Heat Transfer Actual Heat Transfer is calculated by; 1. Energy lost by HOT fluid. OR 2. Energy gained by COLD fluid.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
72. 72. Heat Transfer Effectiveness-NTU Method For Parallel-Flow Heat Exchanger; q mh Cph (Th1 Th 2 ) mc Cpc (Tc 2 Tc1 ) For Counter-Flow Heat Exchanger; q mh Cph (Th1 Th 2 ) mc Cpc (Tc1 Tc 2 ) Maximum possible Heat Transfer Maximum possible Temperature Difference Difference in INLET Temperatures of Hot and Cold fluids. Maximum Temperature Difference Minimum (m C ) value. Thus; Maximum Heat Transfer is given by; qmax (m Cp) min (Thinlet Tcinlet ) The ( m Cp )fluid may be Hot or Cold, depending on their respective mass flow rates and Specific Heats.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
73. 73. Heat Transfer Effectiveness-NTU Method mh Cph (Th1 Th 2 ) Th1 Th 2 h mh Cph (Th1 Tc1 ) Th1 Tc1 For Parallel-Flow Heat Exchanger; mc Cpc (Tc 2 Tc1 ) Tc 2 Tc1 c mc Cpc (Th1 Tc1 ) Th1 Tc1 mh Cph (Th1 Th 2 ) Th1 Th 2 h mh Cph (Th1 Tc 2 ) Th1 Tc 2 For Counter-Flow Heat Exchanger; mc Cpc (Tc1 Tc 2 ) Tc1 Tc 2 c mc Cpc (Th1 Tc 2 ) Th1 Tc 2ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
74. 74. Heat Transfer Effectiveness-NTU Method Effectiveness, ε, can be derived in a different way; For Parallel-Flow Heat Exchanger; ln (Th 2 Tc 2 ) UA 1 1 (Th1 Tc1 ) m h Cph m c CpcT dq = U dA (Th-Tc) UA m c Cpc Th1 Hot Fluid 1 dq m c Cpc m h Cph Th Th2 OR (Th 2 Tc 2 ) UA m c Cpc exp 1 Tc2 (Th1 Tc1 ) m c Cpc m h Cph Tc dA Cold Fluid If Cold fluid is min ( m Cp ) fluid; Tc1 Tc 2 Tc1 A c Th1 Tc1 1 2ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
75. 75. Heat Transfer Effectiveness-NTU Method We know; dq m h Cph dTh mc Cpc dTc mc Cpc mh Cph (Th1 Th 2 ) mc Cpc (Tc 2 Tc1 ) Th 2 Th1 (Tc1 Tc 2 ) mh Cph This yields; (Th 2 Tc 2 ) Th1 (mc Cpc / m h Cph )(Tc1 Tc 2 ) Tc 2 (Th1 Tc1 ) (Th1 Tc1 ) (Th1 Tc1 ) (m c Cpc / m h Cph )(Tc1 Tc 2 ) (Tc1 Tc 2 ) m c Cpc 1 1 c (Th1 Tc1 ) m h Cph 1 exp UA / m c Cpc 1 (m c Cpc / m h Cph ) c 1 (m c Cpc / m h Cph )ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
76. 76. Heat Transfer Effectiveness-NTU Method It can be shown that the SAME expression results if Hot fluid is min ( m Cp ) fluid; EXCEPT that (mc Cpc )and (mh Cp h ) are interchanged. 1 exp UA / Cmin 1 Cmin / Cmax In a General Form; parallel 1 Cmin / Cmax where, C = (m C ) ; defined as CAPACITY RATE. Similar analysis for Counter-Flow Heat Exchanger yields; 1 exp UA / Cmin 1 Cmin / Cmax counter 1 (Cmin / Cmax ) exp UA / Cmin 1 Cmin / Cmax The group of terms, (UA/Cmin ) is known as Number of Transfer Units (NTU). This is so, since it is the indication of the size of the Heat Exchanger.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
77. 77. Heat Transfer Effectiveness-NTU Method Heat Exchanger Effectiveness Relations : N = NTU = UA/Cmin C = Cmin/Cmax Flow Geometry Relation Double Pipe : 1 exp[ N (1 C )] Parallel Flow 1 C 1 exp[ N (1 C )] Counter Flow 1 C exp[ N (1 C )] N Counter Flow, C = 1 N 1 Cross Flow : exp( NCn) 1 0.22 Both Fluids Mixed 1 exp where n N Cn 1 1 C 1 Both Fluids Unmixed 1 exp( N ) 1 exp( NC ) N Cmax mixed, Cmin Unmixed (1/ C){ exp[ C(1 e N )]} 1 Cmax Unmixed, Cmin Mixed 1 exp{ (1 / C )[1 exp( NC )]} Shell-and-Tube : 1 2 1/ 2 1 exp[ N (1 C 2 )1/ 2 ] 1 Shell-pass; 2/4/6 Tube-pass 2 1 C (1 C ) X 1 exp[ N (1 C 2 )1/ 2 ] All Exchangers, C = 0 : 1 e NME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
78. 78. Heat Transfer Effectiveness-NTU Method Heat Exchanger NTU Relations : N = NTU = UA/Cmin C = Cmin/Cmax ε = Effectiveness Flow Geometry Relation Double Pipe : ln[1 (1 C ) ] Parallel Flow N 1 C Counter Flow 1 1 N ln C 1 C 1 Counter Flow, C = 1 N 1 Cross Flow : 1 Cmax mixed, Cmin Unmixed N ln 1 ln (1 C ) C 1 Cmax Unmixed, Cmin Mixed N ln [1 C ln (1 )] C Shell-and-Tube : 2 1/ 2 (2 / ) 1 C (1 C 2 )1/ 2 1 Shell-pass; 2/4/6 Tube-pass N (1 C ) X ln (2 / ) 1 C (1 C 2 )1/ 2 All Exchangers, C = 0 : N ln (1 )ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
79. 79. Heat Transfer Example 8 A cross-flow heat exchanger is used to heat an oil in the tubes (C=1.9 kJ/kg.ºC) from 15 ºC to 85 ºC. Blowing across the outside of the tubes is steam which enters at 130 ºC and leaves at 110 ºC with a mass flow rate of 5.2 kg/sec. The overall heat transfer coefficient is 275 W/m2.ºC and C for steam is 1.86 kJ/kg.ºC. Calculate the surface area of the heat exchanger. Total Heat Transfer is calculated from Energy Balance of Steam; q m s Cs Ts (5.2)(1.86)(130 110) 193 kW T 130 °C Hot Fluid Th ∆Tm is calculated by treating as a dq Counter-Flow Heat Transfer; 85 °C 110 °C Tc (130 85) (110 15) Tm 66.9 C (130 85) dA 15 °C ln (110 15) Cold Fluid A 1ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
80. 80. Heat Transfer Example 8….contd t1 and t2 represent unmixed fluid (i.e. Oil) and T1 and T2 represent the mixed fluid (i.e. Steam). Hence; T1 = 130 ºC; T2 = 110 ºC; t1 = 15 ºC; t2 = 85 ºC t2 t1 85 15 T1 T2 130 110P 0.609 And R 0.286 T1 t1 130 15 t2 t1 85 15From LMTD Correction Chart; F = 0..97Heat Transfer Area is; q 193 X 103 A U F Tm (275)(0.97)(66.9) 10.82 m 2 ….ANS S. Y. B. Tech. Prod Engg.
81. 81. Heat Transfer Example 9 Calculate the heat exchanger performance in Example 8; if the oil flow rate is reduced to half while the steam flow rate is kept constant. Assume U remains same as 275 W/m2.ºC. Calculating the Oil flow rate; 193X 103 q mo Co To mo 1.45 kg / sec (1.9)(85 15) New Flow rate is half of this value. i.e. 0.725 kg/sec. We assume the Inlet Temperatures remain same as 130 ºC for Steam and 15 ºC for Oil. Hence, q mo Co (Te,o 15) ms Cs (130 Te,s ) But, both the Exit Temperatures Te,o and Te,s are unknown. The values of R and P can not be calculated without these temperatures. Hence, ∆Tm can not be calculated. ITERATIVE procedure MUST be used to solve this example. However, this example can be solved with Effectiveness-NTU Approach.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
82. 82. Heat Transfer Example 10 Solve Example 9 by Effectiveness-NTU Method. For Steam; Cs m s Cps (5.2)(1.86) 9.67 kW / C For Oil; Co mo Cpo (0.725)(1.9) 1.38 kW / C Thus, the fluid having minimum ( m Cp ) is Oil. Cmin / Cmax 1.38/ 9.67 0.143 NTU U A / Cmin (275)(10.82) / 1380 2.156It is observed that unmixed fluid (i.e. Oil) has Cmin and mixed fluid (i.e. Steam) has Cmax.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
83. 83. Heat Transfer Example 10….contdHence; from the Table; we get; (1 / C ){1 exp[ C (1 e N )]} 2.156 (1 / 0 / 143) 1 exp[ (0.143)(1 e )] 0.831Using the Effectiveness, we can calculate the Temperature Difference for Oil as; To ( Tmax ) (0.831 130 15) 95.5 C )( Thus, the Heat Transfer is; q mo Cpo To (1.38)(95.5) 132 kWThus,Reduction in Oil flow rate by 50 % results in reduction in Heat Transfer by 32 % only. S. Y. B. Tech. Prod Engg.
84. 84. Heat Transfer Example 11 Hot oil at 100 ºC is used to heat air in a shell-and-tube heat exchanger the oil makes 6 tube passes and the air makes one shell-pass. 2 kg/sec of air are to be heated from 20 ºC to 80 ºC. The specific heat of the oil is 2100 kJ/kg.ºC, and its flow rate is 3.0 kg/sec. Calculate the area required for the heat exchanger for U = 200 W/m2.ºC. Energy Balance is; q mo Cpo T0 ma Cpa Ta (3.0)(2100 100 Te,o ) (2.0)(1009 80 20) )( )( Te,0 80.27 C We have; Cmax Ch mo Cpo (3.0)(2100 6300 W / C ) Cmin Cc m a Cpa (2.0)(1009) 2018 W / C And; Cmain 2018 C 0.3203 Cmax 6300ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
85. 85. Heat Transfer Example 11….contd Tc (80 20) Effectiveness is; 0.75 Tmax (100 20) From the NTU Table; 2 1/ 2 (2 / ) 1 C (1 C 2 )1/ 2 N (1 C ) X ln (2 / ) 1 C (1 C 2 )1/ 2 2 1/ 2 (2 / 0.75) 1 0.3203 (1 0.32032 )1/ 2 (1 0.3203 ) X ln (2 / 0.75) 1 0.3203 (1 0.32032 )1/ 2 1.99 UA NTU Thus; Cmin Cmin (2018 ) 2 A NTU (2.1949) 22.146 m….ANS U (200)ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
86. 86. Heat Transfer Convection Heat Transfer T∞ Consider a heated plate shown in Fig. u∞ Temperature of the plate is Tw and that of surrounding is T∞ u q Velocity profile is as shown in Fig. Tw Velocity reduces to Zero at the plate surface as a result of Viscous Action.Since no velocity at the plate surface, Heat is transferred by Conduction only.Then , WHY Convection ?ANS : Temperature Gradient depends on the rate at which fluid carries away the Heat.Overall Effect of Convection is given by Newton’s Law of Cooling. q = h A ΔT = h A (Tw - T∞)h is known as the CONVECTIVE HEAT TRANSFER COEFFICIENT. (W/m2.K) S. Y. B. Tech. Prod Engg.
87. 87. Heat Transfer Convection Heat Transfer T∞ Convective Heat Transfer has dependence u∞ on Viscosity as well as Thermal properties of u the fluid. q Tw 1) Thermal Conductivity, k 2) Specific Heat, Cp 3) Density, ρHeated plate exposed to room air; without any external source of motion of fluid, themovement of air will be due to the Density Gradient.This is called Natural or Free Convection.Heated plate exposed to air blown by a fan; i.e. with an external source of motion of fluid.This is called Forced Convection. S. Y. B. Tech. Prod Engg.
88. 88. Heat Transfer Convection Energy Balance on Flow Channel The same analogy can be used for evaluating the Heat Loss / Gain resulting from a fluid flowingTe Ti inside a channel or tube, as shown in Fig. Heated wall at temperature Tw loses heat to the m cooler fluid through the channel (i.e. pipe). Temperature rise from inlet (Ti) to exit (Te). q q m Cp (Te Ti ) h A (Tw,avg T fluid,avg ) Te, Ti and Tfluid are known as Bulk or Energy Average Temperatures. S. Y. B. Tech. Prod Engg.
89. 89. Heat Transfer Convection Boundary Condition We know that; qconv h A (Tw T ) With Electrical Analogy, as in case of Conduction; (Tw T ) qconv 1 hA The term 1 is known as Convective Resistance; hAME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
90. 90. Heat Transfer Conduction – Convection SystemHeat conducted through a body, frequently needs to be removed by Convection process.e.g. Furnace walls, Motorcycle Engine, etc.Finned Tube arrangement is the most common for such Heat Exchange applications. Consider a One – Dimensional fin. dqconv =h P dx (T-T∞) t Surrounding fluid at T∞. Base of fin at T0. A qx Qx+x Energy Balance of element of fin with thickness dx ; dx Energy in left face = Energy out right face + L Energy lost by ConvectionBase S. Y. B. Tech. Prod Engg.
91. 91. Heat Transfer Conduction – Convection System Convection Heat Transfer; qconv h A (Tw T ) Where, Area of fin is surface area for Convection. Let the C/s. area be A and perimeter be P. dT Energy in left face; qx kA dx dT dT d 2T Energy out right face; qx dx kA kA 2 dx dx x dx dx dx Energy lost by Convection; qconv h P dx (T T ) NOTE : Differential area of the fin is the product of Perimeter and the differential length dx.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
92. 92. Heat Transfer Conduction – Convection System d 2T hPCombining the terms; we get; T T 0 dx 2 kA d2 hPLet, θ = (T - T∞) 0 dx 2 kALet m2 = hP/kAThus, the general solution of the equation becomes; mx C1 e C2 emxOne boundary condition is; θ = θ0 = (T - T∞) at x = 0. S. Y. B. Tech. Prod Engg.
93. 93. Heat Transfer Conduction – Convection System Other boundary conditions are; CASE 1 : Fin is very long. Temperature at the fin end is that of surrounding. CASE 2 : Fin has finite length. Temperature loss due to Convection. CASE 3 : Fin end is insulated. dT/dx = 0 at x = L. For CASE 1, Boundary Conditions are : θ = θ0 = (T - T∞) at x = 0. θ=0 at x = ∞. T T mx And, the solution becomes; e 0 T0 TME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
94. 94. Heat Transfer Conduction – Convection System For CASE 3, Boundary Conditions are : θ = θ0 = (T - T∞) at x = 0. dθ/dx = 0 at x = L. This yields; 0 C1 C2 mL 0 m ( C1 e C2 e mL ) e mx emx cosh[m ( L x)] Solving for C1 and C2, we get; 0 1 e 2mL 1 e2mL cosh (mL) where, the hyperbolic functions are defined as; ex e x ex e x ex e x sinh x cosh x tanh x 2 2 ex e xME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
95. 95. Heat Transfer Conduction – Convection System Solution for CASE 2 is; T T cosh[m ( L x)] (h / mk )sinh[m ( L x)] T0 T cosh (mL) (h / mk )sinh (mL) All the Heat loss by the fin MUST be conducted to the base of fin at x = 0. dT Thus, the Heat loss is; qx dx kA dx x 0 Alternate method of integrating Convection Heat Loss; L L q h P (T T ) dx h P dx 0 0ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
96. 96. Heat Transfer Conduction – Convection System Application of Conduction equation is easier than that for Convection. m(0) For CASE 1 : q k A[ m 0e ] h Pk A 0 1 1 q kA 0m 1 e 2 mL 1 e 2 mL For CASE 3 : hPk A 0 tanh (mL) sinh (mL) (h / mk ) cosh (mL) For CASE 2 : q hPk A 0 cosh (mL) (h / mk )sinh (mL)ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
97. 97. Heat Transfer Viscous Flow A) Flow over a Flat Plate : Laminar Transition Turbulent y du dy u∞ x u u∞ Laminar Sublayer u At the leading edge of the plate, a region develops, where the influence of Viscous Forces is felt. These viscous forces are described in terms of Shear Stress, η, between the fluid layers.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
98. 98. Heat Transfer Viscous Flow Shear Stress is proportional to normal velocity gradient. du dyThe constant of proportionality, μ, is known as dynamic viscosity. (N-sec/m2)Region of flow, developed from the leading edge, in which the effects of Viscosity areobserved, is known as Boundary Layer.The point for end of Boundary Layer is chosen as the y co-ordinate where the velocitybecomes 99 % of the free – stream value.Initial development of Boundary Layer is Laminar.After some critical distance from leading edge, small disturbances in flow get amplified.This transition is continued till the flow becomes Turbulent. S. Y. B. Tech. Prod Engg.
99. 99. Heat Transfer Viscous Flow Laminar Transition Turbulent Turbulent Transition Laminar Development of Flow RegimesME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
100. 100. Heat Transfer Viscous FlowME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
101. 101. Heat Transfer Viscous FlowME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
102. 102. Heat Transfer Viscous FlowME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
103. 103. Heat Transfer Viscous Flow Transition from Laminar to Turbulent takes place when; u x u x 5 X 105 where, u∞ = Free – Stream Velocity (m/sec) x = Distance from leading edge (m) ν = μ / ρ = Kinematic Viscosity (m2/sec) This particular group of terms is known as Reynold’s Number; and denoted by (Re). It is a dimensionless quantity. u x u x ReME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
104. 104. Heat Transfer Viscous Flow Reynolds Number (Re) = Ratio of Momentum Forces ( α ρu∞2 ) to Shear Stress ( α μu∞ / x ) . Range for Reynlold’s No. (Re) transition from Laminar to Turbulent lies between 2 X 105 to 106; depending on; 1. Surface Roughness. 2. Turbulence Level. NOTE : Generally, Transition ends at twice the Re where it starts.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
105. 105. Heat Transfer Viscous Flow Laminar Transition Turbulent y du dy u∞ x u u∞ Laminar Sublayer u Laminar profile is approximately Parabolic. Turbulent profile has a initial part, close to plate, is very nearly Linear. This is due to the Laminar Sublayer that adheres to the surface. Portion outside this Sublayer is relatively Flat.ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.
106. 106. Heat Transfer Viscous Flow Physical mechanism of Viscosity Momentum Transfer Laminar flow Molecules move from one lamina to another, carrying Momentum α Velocity Net Momentum Transfer from High Velocity region to Low Velocity Region. Force in direction of flow, i.e. Viscous shear Stress, η Rate of Momentum Transfer α Rate of movement of molecules α T Turbulent flow has no distinct fluid layers. Macroscopic chunks of fluid, transporting Energy and Momentum, in stead of microscopic molecular motion. Larger Viscous shear Stress, ηME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.