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CHAPTER 6

transition Reynolds number. is . =. D. tube diameter. =. u. mean velocity. n. =. kinematic viscosity. CHAPTER 6. HEAT TRANSFER IN CHANNEL FLOW. 6.1 Introduction. Important factors: . (1) Laminar vs. turbulent flow. where. (2) Entrance vs. fully developed region .

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CHAPTER 6

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  1. transition Reynolds number is = D tube diameter = u mean velocity n = kinematic viscosity CHAPTER 6 HEAT TRANSFER IN CHANNEL FLOW 6.1 Introduction • Important factors: (1) Laminar vs. turbulent flow where

  2. (2) Entrance vs. fully developed region Based on velocity and temperature distribution: two regions: (i) Entrance region (ii) Fully developed region (3) Surface boundary conditions Two common thermal boundary conditions: (i) Uniform surface temperature (ii) Uniform surface heat flux (4) Objective Depends on the thermal boundary condition: (i) Uniform surface temperature. Determine: axial variation of (1) Mean fluid temperature (2) Heat transfer coefficient (3) Surface heat flux

  3. (ii) Uniform surface flux. Determine axial variation of: (1) Mean fluid temperature (2) Heat transfer coefficient (3) Surface temperature 6.2 Hydrodynamic and Thermal Regions: General Features • Developing boundary velocity and thermal boundary layers • Two regions: • (1) Entrance region • (2) Fully developed region 6.2.1 Velocity Field

  4. x > L : fully developed flow h = ( v 0 ) Streamlines are parallel r = r u / x 0 For 2 - D, constant : • Hydrodynamic entrance region (2) Fully Developed Flow Region

  5. 6.2.2 Temperature Field • Thermal entrance region (2) Fully Developed Temperature Region

  6. (1) Scale analysis (2) Analytic or numerical methods 6.3 Hydrodynamic and Thermal Entrance Lengths 6.3.1 Scale Analysis

  7. u L u D L L = = = h h h Re Re L D v v D D h (6.2) - - 1/2 1/2 d ~ L Re Pr (4.24) t L - - 1/2 1/2 D ~ L Re Pr (a) t t (b) into (a)

  8. u L u D L L = = = t t t Re Re (b) L D t v v D D (6.3) L t ~ Pr (6.4) L h L = h C Re (6.5) h D D e (b) into (a) 6.3.2 Analytic/Numerical Solutions: Laminar Flow

  9. D = equivalent diameter e 4 A f = D e P A = flow area f P = perimeter C = coefficient Table 6.1 h 2 4 Compare with scaling: Rewrite (6.5)

  10. (b) L = t C PrRe (6.6) t D D e (6.3)

  11. (b) (c) (6.7a) (6.7b)

  12. ¢ ¢ ¢ ¢ = = q q A q P x (6.8) s s s s ¢ ¢ q P s = + T ( x ) T x (6.9) m mi c m p (1) Steady state (2) No energy generation (3) Negligible changes in kinetic and potential energy (4) No axial conduction or

  13. Newton’s law of cooling

  14. Example 6.2: Maximum Surface Temperature

  15. hD = = Nu 4 . 364 (A) D k (1) Observations

  16. (2) Problem Definition

  17. " p = - ( ) DL q mc T T (a) s p mo mi (4) Plan Execution (i) Assumptions • Steady state • Constant properties • Axisymmetric flow • Uniform surface heat flux • Negligible changes in kinetic and potential energy • Negligible axial conduction • Negligible dissipation (ii) Analysis Conservation of energy:

  18. - mc ( T T ) p mo mi L = (b) ¢ ¢ p D q s 2 = p r m ( / 4 ) D u (c) From (a) Conservation of mass: where

  19. é ù Px 1 ¢ ¢ = + + T ( x ) T q ê ú (6.10) s m s i c h ( x ) m ë û p p P D = (d) Surface temperature: Apply (6.10) Perimeter P: Maximum surface temperature: set x = L in (6.10)

  20. é ù L P 1 ¢ ¢ = + + T ( L ) T q ê ú (e) s m s i c h ( ) m L ë û p u D = Re (f) D n + T T mi mo T = (g) m 2 For water:

  21. L = h C Re (6.5) h D D e L = t C PrRe (6.6) t D D e Use (g)

  22. Compute L using (b). Use (c) to compute m Substitute into (b)

  23. hD = Nu = 4.364 (A) D k Equation (A) is applicable (iii) Computations. Apply (A) Use (e) (iv) Checking.Dimensional check:

  24. [ ] ¢ ¢ = - q h ( L ) T ( L ) T (i) s s mo (2) Compare value of h with Table 1.1 (3) Comments. • In laminar flow local h depends on local flow condition: entrance vs. fully developed (i) If flow is laminar or turbulent (ii) Entrance or fully developed

  25. Surface temperature: • Inlet temperature: (1) Mean temperature variation (2) Total heat 6.5 Channels with Uniform Surface Temperature • Section length: L Determine

  26. (3) Surface flux variation (a) Analysis Apply conservation of energy to element dx Assumptions (1) Steady state (2) No energy generation (3) Negligible changes in kinetic and potential energy (4) No axial conduction Newton's law: (b)

  27. (c) to x Integrate from x = 0 ) ( ( ) (6.11) Must determine h(x). Introduce (6.12) Combine (a) and (b) (6.12) into (6.11)

  28. (6.13) (6.14) Total heat: Apply conservation of energy:

  29. (6.15) • Uniform surface temperature, Surface flux: Apply Newton’s law: Properties: At mean of inlet and outlet temperatures Example 6.3: Required Tube Length • Air flows through tube • Mean velocity = 2 m/s, • D = 1.0 cm

  30. (A) Determine: tube length to raise temperature to • Laminar or turbulent flow? Check • Nusselt number for laminar fully developed flow (1) Observations • Uniform surface temperature • Uniform Nusselt number (h is constant) for fully • developed laminar flow • Length of tube is unknown

  31. Compute . Laminar or turbulent? (2) Problem Definition. Determine tube length needed to raise temperature to specified level (3) Solution Plan. • Use uniform surface temperature analysis (4) Plan Execution (i) Assumptions • Steady state • Fully developed flow • Constant properties

  32. (6.13) = specific heat, = average h, = flow rate, kg/s • Uniform surface temperature • Negligible changes in kinetic and potential energy • Negligible axial conduction • Negligible dissipation (ii) Analysis P = perimeter, m

  33. Tm(x) = mean temperature at x, = mean inlet temperature = 35 = surface temperature = 130 (a) Properties: at = outlet temperature = 105 (b) = x = distance from inlet, m Apply (a) at the outlet (x = L), solve for L

  34. P =  D (c) (d) = mean flow velocity = 2 m/s (e) = heat transfer coefficient, = thermal conductivity of air , D = inside tube diameter = 1 cm = 0.01 m  = density, kg/m3 For fully developed laminar flow

  35. (f) , for laminar fully developed (g) = = 1008.7 0.707 Compute: Reynolds number Use (b) Properties:

  36. , flow is laminar = 3.657 = 10.69 = 0.65 m Use (f) (iii) Computations P =  0.01(m) = 0.03142 m Substitute into (a)

  37. (i) for Set in (a) gives . (ii) for Set in (a) gives L =  Energy added at the surface = Energy gained by air (h) Energy added at surface Energy gained by air = (iv) Checking. Dimensional check Quantitative checks: (i) Approximate check: (i) (j) (j) and (k) into (i), solve for L (k)

  38. = 0.57 m (ii) Value of h is low compared with Table 1.1. Review solution. Deviation from Table 1.1 are expected (5) Comments. This problem is simplified by two conditions: fully developed and laminar flow

  39. 6.6 Determination of Heat Transfer Coefficient and Nusselt Number Two Methods: (1) Scale analysis (2) Analytic or numerical solutions 6.6.1 Scale Analysis Fourier’s law and Newton’s law (6.16)

  40. (a) (b) ~ (6.17) ~ Scales: (a) and (b) into (6.16) or

  41. (6.18) Entrance region: , ~D (fully developed) (6.19) Nusselt number: (6.17) into the above Special case: fully developed region (6.18) gives

  42. in the entrance region: For all Pr (4.24) ~ (c) Express in terms of (d) (6.20a) (4.24) into (6.18) (d) into (c)

  43. (6.20b) or

  44. (a) , , (6.21) , , 6.6.2 Basic Considerations for the Analytical Determination of Heat Flux, Heat Transfer Coefficient and Nusselt Number (1) Fourier’s law and Newton’s law Define dimensionless variables

  45. (6.22) (6.23) (6.24) Dimensionless mean temperature : (6.25) (6.21) into (a) Newton’s law Combine (6.22) and (6.23)

  46. (6.26) (6.27) Determine: and : Find . Apply energy equation Nusselt number: (6.24) into (6.26) (2) The Energy Equation Assumptions • Steady state

  47. (2.24) (6.28) • Laminar flow • Axisymmetric • Negligible gravity • Negligible dissipation • Negligible changes in kinetic and potential energy • Constant properties Replace z by x, express in dimensionless form

  48. (2.29) , Peclet number (6.30) (6.31) (3) Mean (Bulk) Temperature Need a reference local temperature. Use • Third term: radial conduction • Fourth term: axial conduction • Neglect axial conduction for: Simplify (6.28)

  49. (a) (b) (6.32a) (6.32b) where (b) into (a), assume constant properties In dimensionless form:

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