1 / 80

PM3125: Lectures 4 to 6

PM3125: Lectures 4 to 6. Content of Lectures 1 to 6: Heat transfer: Source of heat Heat transfer Steam and electricity as heating media Determination of requirement of amount of steam/electrical energy Steam pressure Mathematical problems on heat transfer. Heat Transfer.

hansel
Download Presentation

PM3125: Lectures 4 to 6

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. PM3125: Lectures 4 to 6 • Content of Lectures 1 to 6: • Heat transfer: • Source of heat • Heat transfer • Steam and electricity as heating media • Determination of requirement of amount of steam/electrical energy • Steam pressure • Mathematical problems on heat transfer

  2. Heat Transfer is the means by which energy moves from a hotter object to a colder object

  3. Mechanisms of Heat Transfer Conduction is the flow of heat by direct contact between a warmer and a cooler body. Convection is the flow of heat carried by moving gas or liquid. (warm air rises, gives up heat, cools, then falls) Radiation is the flow of heat without need of an intervening medium. (by infrared radiation, or light)

  4. Mechanisms of Heat Transfer Latent heat Conduction Convection Radiation

  5. Conduction HOT (lots of vibration) COLD (not much vibration) Heat travels along the rod

  6. Conduction Conduction is the process whereby heat is transferred directly through a material, any bulk motion of the material playing no role in the transfer. Those materials that conduct heat well are called thermal conductors, while those that conduct heat poorly are known as thermal insulators. Most metals are excellent thermal conductors, while wood, glass, and most plastics are common thermal insulators. The free electrons in metals are responsible for the excellent thermal conductivity of metals.

  7. Conduction: Fourier’s Law ( ) ΔT k A t Q = L Cross-sectional area A L Q = heat transferred k = thermal conductivity A = cross sectional area DT = temperature difference between two ends L = length t = duration of heat transfer What is the unit of k?

  8. Thermal Conductivities

  9. Conduction through Single Wall . . . Q Q Q ( ) ΔT k A t Q = L Use Fourier’s Law: T1 k A (T1 – T2) T2 T1 = x Δx Δx

  10. Conduction through Single Wall . . . Q Q Q k A (T1 – T2) T1 = Δx T1 – T2 = Δx/(kA) T2 T1 x Δx Thermal resistance (in k/W) (opposing heat flow) 10

  11. Conduction through Composite Wall . . . Q Q Q B C A T1 T2 T3 T4 kA kB kC x ΔxA ΔxB ΔxC T1 – T2 T3 – T4 T2 – T3 = = = (Δx/kA)A (Δx/kA)C (Δx/kA)B 11

  12. Conduction through Composite Wall T1 – T2 T3 – T4 T2 – T3 = = = (Δx/kA)A (Δx/kA)C (Δx/kA)B [ ] + (Δx/kA)B + (Δx/kA)B + (Δx/kA)C + (Δx/kA)C (Δx/kA)A (Δx/kA)A = T1 – T2 + T2 – T3 + T3 – T4 . . . Q Q Q T1 – T4 = 12

  13. . Q Example 1 An industrial furnace wall is constructed of 21 cm thick fireclay brick having k = 1.04 W/m.K. This is covered on the outer surface with 3 cm layer of insulating material having k = 0.07 W/m.K. The innermost surface is at 1000oC and the outermost surface is at 40oC. Calculate the steady state heat transfer per area. Solution: We start with the equation Tin – Tout = + (Δx/kA)insulation (Δx/kA)fireclay

  14. . . Q Q Example 1 continued (1000– 40) A = + (0.03/0.07) (0.21/1.04) = 1522.6 W/m2 A

  15. . . Q Q Example 2 We want to reduce the heat loss in Example 1 to 960 W/m2. What should be the insulation thickness? Solution: We start with the equation Tin – Tout = + (Δx/kA)insulation (Δx/kA)fireclay (1000– 40) W/m2 = 960 = A + (Δx)insulation/0.07) (0.21/1.04) (Δx)insulation cm = 5.6

  16. Conduction through hollow-cylinder . Q ro Ti ri To L Ti – To = [ln(ro/ri)] / 2πkL

  17. . Q Conduction through the composite wall in a hollow-cylinder r3 r2 To Material A Ti r1 Material B Ti – To = + [ln(r3/r2)] / 2πkBL [ln(r2/r1)] / 2πkAL

  18. . Q Example 3 A thick walled tube of stainless steel ( k = 19 W/m.K) with 2-cm inner diameter and 4-cm outer diameter is covered with a 3-cm layer of asbestos insulation (k = 0.2 W/m.K). If the inside-wall temperature of the pipe is maintained at 600oC and the outside of the insulation at 100oC, calculate the heat loss per meter of length. Solution: We start with the equation Ti – To = + [ln(r3/r2)] / 2πkBL [ln(r2/r1)] / 2πkAL

  19. . . Q Q Example 3 continued 2 π L (600 – 100) = + [ln(5/2)] / 0.2 [ln(2/1)] / 19 = 680 W/m L

  20. Mechanisms of Heat Transfer Conduction is the flow of heat by direct contact between a warmer and a cooler body. Convection is the flow of heat carried by moving gas or liquid. (warm air rises, gives up heat, cools, then falls) Radiation is the flow of heat without need of an intervening medium. (by infrared radiation, or light) 

  21. Convection Convection is the process in which heat is carried from place to place by the bulk movement of a fluid (gas or liquid). Convection currents are set up when a pan of water is heated.

  22. Convection It explains why breezes come from the ocean in the day and from the land at night

  23. Convection: Newton’s Law of Cooling = h A (Tsurface – Tfluid) conv. . Q Flowing fluid at Tfluid Heated surface at Tsurface Area exposed Heat transfer coefficient (in W/m2.K)

  24. Convection: Newton’s Law of Cooling conv. . Q Flowing fluid at Tfluid Heated surface at Tsurface Tsurface – Tfluid = 1/(hA) Convective heat resistance (in k/W)

  25. . Q = h A (Tsurface – Tfluid) conv. . Q conv. A Example 4 The convection heat transfer coefficient between a surface at 50oC and ambient air at 30oC is 20 W/m2.K. Calculate the heat flux leaving the surface by convection. Solution: Use Newton’s Law of cooling : Flowing fluid at Tfluid = 30oC = (20 W/m2.K) x A x (50-30)oC Heated surface at Tsurface = 50oC Heat flux leaving the surface: = 20 x 20 = 400 W/m2 h = 20 W/m2.K

  26. . Q = h A (Tsurface – Tfluid) conv. Example 5 Air at 300°C flows over a flat plate of dimensions 0.50 m by 0.25 m. If the convection heat transfer coefficient is 250 W/m2.K, determine the heat transfer rate from the air to one side of the plate when the plate is maintained at 40°C. Solution: Use Newton’s Law of cooling : Flowing fluid at Tfluid = 300oC Heated surface at Tsurface = 40oC = 250 W/m2.K x 0.125 m2 x (40 - 300)oC = - 8125 W/m2 h = 250 W/m2.K A = 0.50x0.25 m2 Heat is transferred from the air to the plate.

  27. Forced Convection In forced convection, a fluid is forced by external forces such as fans. In forced convection over external surface: Tfluid = the free stream temperature (T∞), or a temperature far removed from the surface In forced convection through a tube or channel: Tfluid = the bulk temperature

  28. Free Convection In free convection, a fluid is circulated due to buoyancy effects, in which less dense fluid near the heated surface rises and thereby setting up convection. In free (or partially forced) convection over external surface: Tfluid = (Tsurface + Tfree stream) / 2 In free or forced convection through a tube or channel: Tfluid = (Tinlet + Toutlet) / 2

  29. Change of Phase Convection Change-of-phase convection is observed with boiling or condensation . It is a very complicated mechanism and therefore will not be covered in this course.

  30. Overall Heat Transfer through a Plane Wall . . . Q Q Q T1 – T2 T2– TB TA – T1 = = = 1/(hBA) 1/(hAA) Δx/(kA) Fluid A at TA > T1 T1 T2 Fluid B at TB < T2 x Δx

  31. Overall Heat Transfer through a Plane Wall . . Q Q T1 – T2 T2– TB TA – T1 = = = 1/(hBA) 1/(hAA) Δx/(kA) . TA – TB Q = 1/(hAA) + Δx/(kA) + 1/(hBA) (TA – TB) = U A where U is the overall heat transfer coefficient given by 1/U = 1/hA + Δx/k + 1/hB

  32. Overall heat transfer through hollow-cylinder ro Ti ri To . Q Fluid A is inside the pipe Fluid B is outside the pipe TA > TB L (TA – TB) = U A where 1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + 1/(hBAo)

  33. . Q Example 6 Steam at 120oC flows in an insulated pipe. The pipe is mild steel (k = 45 W/m K) and has an inside radius of 5 cm and an outside radius of 5.5 cm. The pipe is covered with a 2.5 cm layer of 85% magnesia (k = 0.07 W/m K). The inside heat transfer coefficient (hi) is 85 W/m2 K, and the outside coefficient (ho) is 12.5 W/m2 K. Determine the heat transfer rate from the steam per m of pipe length, if the surrounding air is at 35oC. Solution: Start with (120 – 35) (TA – TB) = U A = U A What is UA?

  34. Example 6 continued 1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + … + 1/(hBAo) + ln(5.5/5) / 2π(45)L 1/UA = 1/(85Ain) + 1/(12.5Aout) + ln(8/5.5) / 2π(0.07)L Ain = 2π(0.05)L and Aout = 2π(0.08)L 1/UA = (0.235 + 0.0021 +5.35 + 1) / 2πL

  35. . . Q Q Example 6 continued UA = 2πL / (0.235 + 0.0021 +5.35 + 1) (120 – 35) = U A air steel =2πL (120 – 35) / (0.235 + 0.0021 +5.35 + 1) insulation steam = 81 L / L = 81 W/m

  36. Mechanisms of Heat Transfer Conduction is the flow of heat by direct contact between a warmer and a cooler body. Convection is the flow of heat carried by moving gas or liquid. (warm air rises, gives up heat, cools, then falls) Radiation is the flow of heat without need of an intervening medium. (by infrared radiation, or light)  

  37. Radiation Radiation is the process in which energy is transferred by means of electromagnetic waves of wavelength band between 0.1 and 100 micrometers solely as a result of the temperature of a surface. Heat transfer by radiation can take place through vacuum. This is because electromagnetic waves can propagate through empty space.

  38. Q = εσ A T4 t The Stefan–Boltzmann Law of Radiation ε = emissivity, which takes a value between 0 (for an ideal reflector) and 1 (for a black body). σ = 5.668 x 10-8 W/m2.K4 is the Stefan-Boltzmann constant A = surface area of the radiator T = temperature of the radiator in Kelvin.

  39. Why is the mother shielding her cub? Ratio of the surface area of a cub to its volume is much larger than for its mother.

  40. What is the Sun’s surface temperature? The sun provides about 1000 W/m2 at the Earth's surface. Assume the Sun's emissivity ε = 1Distance from Sun to Earth =  R = 1.5 x 1011 m Radius of the Sun = r = 6.9 x 108 m    

  41. Q = εσ A T4 t What is the Sun’s surface temperature? (4 π 6.92 x 1016 m2) = 5.98 x 1018 m2 (4 π 1.52 x 1022 m2)(1000 W/m2) = 2.83 x 1026 W 2.83 x 1026 W T4 = (1) (5.67 x 10-8 W/m2.K4) (5.98 x 1018 m2) ε σ T = 5375 K

  42. Q = εσ A (T4 - To4 ) t If object at temperature T is surrounded by an environment at temperature T0, the net radioactive heat flow is: Temperature of the radiating surface Temperature of the environment

  43. Q = εσ A T4 t Example 7 What is the rate at which radiation is emitted by a surfaceof area 0.5 m2, emissivity 0.8, and temperature 150°C? Solution: [(273+150) K]4 0.5 m2 0.8 5.67 x 10-8 W/m2.K4 Q (0.8) (5.67 x 10-8 W/m2.K4) (0.5 m2) (423 K)4 = t = 726 W

  44. Q = εσ A (T4 - To4 ) t Example 8 If the surface of Example 7 is placed in a large, evacuated chamber whose walls are maintained at 25°C, what is the net rate at which radiation is exchanged between the surface and the chamber walls? Solution: [(273+25) K]4 [(273+150) K]4 Q (0.8) x (5.67 x 10-8 W/m2.K4) x (0.5 m2) x [(423 K)4 -(298 K)4 ] = t = 547 W

  45. Example 8 continued Note that 547 W of heat loss from the surface occurs at the instant the surface is placed in the chamber. That is, when the surface is at 150oC and the chamber wall is at 25oC. With increasing time, the surface would cool due to the heat loss. Therefore its temperature, as well as the heat loss, would decrease with increasing time. Steady-state conditions would eventually be achieved when the temperature of the surface reached that of the surroundings.

  46. Q = εσ A (T4 - To4 ) t Example 9 Under steady state operation, a 50 W incandescent light bulb has a surface temperature of 135°C when the room air is at a temperature of 25°C. If the bulb may be approximated as a 60 mm diameter sphere with a diffuse, gray surface of emissivity 0.8, what is the radiant heat transfer from the bulb surface to its surroundings? Solution: [(273+25) K]4 [(273+135) K]4 Q (0.8) x (5.67 x 10-8 J/s.m2.K4) x [π x (0.06) m2] x [(408 K)4 -(298 K)4 ] = t = 10.2 W (about 20% of the power is dissipated by radiation)

  47. Mathematical Problems on Heat Exchanger Q = mc cc (Tc,out – Tc,in) = mh ch (Th,in – Th,out) Tc,in Th,out Th,in Tc,out . . .

  48. Mathematical Problems on Heat Exchanger Th,in Th,out Tc,out Tc,in Tc,in Parallel-flow heat exchanger Th,out Th,in Tc,out high heat transfer low heat transfer

  49. Mathematical Problems on Heat Exchanger Th,in Th,out Tc,out Tc,in Parallel-flow heat exchanger ΔTa ΔTb a b . Q = U A ΔT ΔTa - ΔTb is the log mean temperature difference (LMTD) where ΔT = ln(ΔTa / ΔTb)

  50. Mathematical Problems on Heat Exchanger Q = mc cc (Tc,out – Tc,in) = mh ch (Th,in – Th,out) Tc,out Counter-flow heat exchanger Th,out Th,in Tc,in . . .

More Related