1 / 76

Topic 7 Heat Transfer

Topic 7 Heat Transfer. FOOD1360/1577 Principles of Food Engineering Robert Driscoll. Heat Transfer. S&H Chapter 4 Notes on electric circuits Sect 3.2. Movement of heat energy from one point to another by virtue of temperature gradient (difference). Heat flux. hot. cold.

jake
Download Presentation

Topic 7 Heat Transfer

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Topic 7Heat Transfer FOOD1360/1577 Principles of Food Engineering Robert Driscoll FST 2010

  2. Heat Transfer S&H Chapter 4 Notes on electric circuits Sect 3.2 • Movement of heat energy from one point to another by virtue of temperature gradient (difference) Heat flux hot cold FST 2010

  3. Use in unit operations: • Used in most unit operations in food industry • Why do we heat? • Cooking • Sterilising • Pasteurising • Blanching • Browning • Why do we cool? • Preservation • Texture (icecream!) • Stop cooking quickly (see S&H Sect 4.1: Heat Exchangers). FST 2010

  4. Modes of heat transfer • Conduction • Convection • Radiation These notes follow S&H Sect 4.3. So look at his examples! FST 2010

  5. 1. Conduction • Heat transfer between neighbouring molecules Esp. for solids, but also for fluids!·vibration of molecules, or·drift of electrons (in metals)·no net movement of molecules FST 2010

  6. Fourier’s equation • Rate of flow = driving force / resistance • The negative indicates heat flows from hot to cold. • Logical: rate of heat flow depends on • Temperature difference • Material properties • Distance to flow • Cross-sectional area for flow. FST 2010

  7. Terms used • dQ/dt = rate of heat flow (J/s or Watts) • dx = thickness (m). Use Dx for finite differences. • A = area normal to the flow direction (m2) • k = thermal conductivity of the material (in J/oC.s.m or W/m.K) • dT = temperature difference (oC) (use DT = T1-T2 for finite difference). • For steady state heat flow through a material, FST 2010

  8. Example: slab A T1 k T2 x FST 2010

  9. Worked Example • Use Fourier’s equation to findDT=T1-T2 (see pic). • Solution: • Why is Dx negative? A=0.1m2 T1 T2 k=0.2 x=25 mm FST 2010

  10. Thermal Conductivity, k • thermal conductivity: a property of a material • good conductors of heat: metals • bad conductors of heat: insulators • What are typical values? FST 2010

  11. Some examples Metals: 50-400 J/oC.s.m Glass: 0.51 J/oC.s.m Alloys: 10-120 J/oC.s.m Air: 0.024 J/oC.s.m Water: 0.7 J/oC.s.m Ice: 2.3 J/oC.s.m Insulations: 0.026-0.052 J/oC.s.m (cork, foamed plastics, expanded rubber etc.) • Air is a good insulator! FST 2010

  12. Thermal conductivity of Foods • Close to k for water for higher moisture foods • For fruits and vegs with W >60%: k = 0.148 + 0.00493 W W = water content in % • Meats with W=60-80%: k = 0.08 + 0.0052 W • For solid foods, low W, see Toledo, Singh and Heldman, Fellows etc. FST 2010

  13. Example • What is the thermal conductivity of a block of meat with a moisture content of 75%? • How would this information be useful? FST 2010

  14. Notes: • Heat flux = heat transfer rate per unit area = dQ/Adt (in J/s.m2) • Use dx for differential, x for difference. • Power is measured in Watt (W) • Energy is measured in Joule (J) • Power is energy per second (W = J/s) FST 2010

  15. Example • A cork slab 10 cm thick has one face at -12oC and the other at 21oC. If the thermal conductivity of cork is 0.042 W/mK, estimate the heat transfer per square meter. Solution: ‘Per square meter’ means find heat flux. So which gives 13.9 W/m2. FST 2010

  16. Similarity to electrical circuits • From physics - electrical circuits: • heat flow corresponds to electric current • x/kA corresponds to resistance • T corresponds to voltage (driving force) • This is a useful analogy. FST 2010

  17. Electrical Circuit • For a simple circuit, V A R I FST 2010

  18. dT R=dx/kA Apply to heat So we have defined a heat resistance, R FST 2010

  19. dT k1 k2 T1 T2 dx1 dx2 Flow through layers • If heat flows through a series of layers of material, we call it series heat transfer. REAL ANALOGY FST 2010

  20. With the electric circuit, total resistance is R1 + R2. • So total heat resistance is: • Setting RT=1/UA is convenient: • U is the overall heat transfer coefficient. Defines heat flowresistance. Defines heat flowin terms of U FST 2010

  21. 2. Convection • Heat transfer by the net movement of moleculesin a fluid is called a convection current: Fluid Flow Cold Body Hot Body FST 2010

  22. Hot Body Ts Fluid Flow T Just for one surface • Governing equation: FST 2010

  23. Two Types of Convection • Natural convection (slow heat transfer): • induced by temperature gradient • current driven by density difference • Forced convection (rapid heat transfer): • movement of molecules by external energy • e.g. a stirrer, fan, agitator, pump etc. FST 2010

  24. Which is faster? • In liquids, convection is much faster than conduction! Soup heats faster than dog food. ConvectionMixing HeatDiffuses FST 2010

  25. Newton’s law of cooling • where • h = convective heat transfer coefficient • A = area of heat transfer • T= temperature difference. • h is also called the surface heat transfer coefficient (J/oC.s.m2 or W/m2K) FST 2010

  26. Convective resistance • Since dQ/dt=hAT, then • Again comparing with our electrical circuit analogy: • we can identify a convective resistance as R=1/hA. FST 2010

  27. Examples of htc • ‘htc’ = heat transfer coefficient • Values determined experimentally • Air: Free convection 5-25 J/oC.s.m2 Forced convection 10-200 • Water: Free convection 20-100 Forced convection 50-10,000 Boiling water 3,000 –100,000 Condensing water vapour 5000 – 100,000 FST 2010

  28. Example • Estimate the rate of heat loss from a hot plate with a surface area of 1m2. The plate surface temperature is 120oC and the ambient temperature is 20oC. Convective heat transfer is estimated to be 10J/oC.s.m2 FST 2010

  29. 3. Radiation • heat transfer by electromagnetic waves • similar to transfer of light • Important in baking ovens • requires no physical medium for propagation • can occur in vacuum (unlike conduction and convection) FST 2010

  30. Radiation equation q = s e A T4 • where • s = Stefan-Boltzmann constant • = 5.669x10-8 J/m2.s.K4 • e = emissivity, relative to a black body • T = temperature of emitting body in K FST 2010

  31. Example calculation • Calculate the rate of heat energy emitted by 10 m2 of a polished stainless steel tube (emissivity=0.15). The temperature of the tube is 100oC. FST 2010

  32. R1 R2 R3 Combined modes • For resistances in series, add resistances. h2,A h1,A k,x,A T1 T2 FST 2010

  33. Finding U: • Adding the resistances: • Solving: • where U is the overall htc for the system. Note: UA is the reciprocal of RT FST 2010

  34. 100oC base 3 mm 250oC Worked Problem • Water boiling in a Cu saucepan (ID=200 mm). • For copper, k=400 W/m.K • Convective resistance (base to water) = 0.0105 K/W • Find rate of heat transfer. FST 2010

  35. Solution • Two resistances: • conduction through copper saucepan base • convection from base to water. • So find each resistance, then add: • So heat flow is: FST 2010 Later problem: other types

  36. Heat Flow Thru Pipes Exterior, T2 Length, L Product, T1 r2 r1 FST 2010

  37. End View • Apply Fourier’s equation to heat flows through a thin annulus, dr. dr T2 r2 r1 T1 FST 2010

  38. Derivation • Since FST 2010

  39. Solving: FST 2010

  40. Cylinder thermal resistance • Comparing with the electrical circuit analogy again, we can identify Rcyl: • This resistance acts like the slab and convective resistances seen before. FST 2010

  41. 150oC Example: Two Conductive Layers • Calculate the rate of heat loss through a fibreglass insulated steam pipe. Ignore convection. fibreglass T=35oC ID = 2.1 cm MD = 2.67 cm OD = 3.335 cm kfibreglass= 0.0351 W/m.K kpipe= 45 W/m.K iron pipe FST 2010

  42. Solving for heat flow Two layers, two resistances to add. Rpipe= ln(1.335/1.05)/(2 45) (per unit length) = 0.00085 Rinsul= ln(1.667/1.335)/(2 0.0351) = 1.007 Thus rate of heat flow = (150-35)/(1.008) = 114 W/m Total resistance FST 2010

  43. Example: Conduction / Convection • Data: h1=6.5 W/m2K h2=4.5 W/m2K T1=135oC T=20oC k1=202 W/mK ri=4.88 cm ro=5.08 cm h2, T k h1, T1 ri ro FST 2010

  44. Solution • Inner convective resistance: • Outer convective resistance: • Pipe wall conductive resistance: FST 2010

  45. So total resistance = 0.502+0.696+0=1.20 • So heat loss is: • Note that heat losses are written per unit length. Why? FST 2010

  46. Insulation • Insulation has low thermal conductivity • Often used for lagging – covering pipes and vessels to reduce heat flow • Important over • long lengths or • large surface areas. FST 2010

  47. Choice of insulation • Depends on: • Cost of materials vs. rate of heat loss • Conductivity of the insulation (may vary with age) • The finish (coating) of the insulation • Critical thickness! FST 2010

  48. Critical Thickness • Adding insulation may increase heat losses! • Why? – increase in surface area. • where r0 is outer radius – main variable. FST 2010

  49. Minimum RT • Differentiate wrt ro: • Solving, RT has a minimum at outer radius rc=k/h FST 2010

  50. Plotting RT K/W FST 2010

More Related