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CONVECTION presence of bulk fluid motion. FORCED CONNECT lon a - mechanism of heat transfer through a fluid in the SA - when the fluid is forced to flow by pressure differences, a fan, and so on. pump, SAMPLE PROBLEMS + GIVEN: REQ'D: Q Cover the surface) SOLUTION: Q = hA (Ts-T₂) hl k NuL = Properties of Air at IF (p, M₁K, Cp) Is + To 350 +276. ∙313 k IF 2 __P=1 atm = 0.101325 MPg. pat 3 10 K, 0.1 MPa: 1₁K 300 500 313K = 300 300 313. e at 313 K, 1 MPa: 7, K mol/dm ²³. fr 0.040103 0.024046 kg/m³ 1.1327 kg/m³ 6x29 foxy f+moll dm ³ 0.040205 0.03974 11.35.35 Kg/m ²³. FLAT PLATE (CRANK CASE) 0.6m x 0.2m x 0.1 m Ts = 350 K ATMOSPHERIC AIR = 276 K- 30 m/s Тоб P = 1 atm pat 013 K. 0.101325 MPa. P. MPa 0.1 0.101325 for Ср, и, 0.101325 -P, kg l m ³ 1.1327 11 at 313 K, 0.101325 MPa P, MPa Pars 0.1 K, at 0. 10135 MPa K = P, MPa 0-1 0.101325 11- 3535 1.1477 kg/m³ ik ( same process as ė 싸 19.09.29 X 10 -6 6 19.22496X10 19.0931 X10-pa-s K₁ W/m - R 27.2654 X10 27.5567 X 10 27.2658X10 Pa-s Spat 313 K, 0.101325 MPa Lp, Jl Kg. R P, MPa. 0.1 1006-6643 1 0.101325 Rey= • Lup u - Rel₁ = 1081 992.9717 5X105 Turbulent Nu 1020 2902 1006.6844J/kg. K Pr = CpW (1006-6844) (19.0931x 166) сри k 7.2658 X16 hL (0.6) (30) (1.1477) 19.0431 X10 Pr= 0.7049 HEAT FLOW OVER A FLAT PLATE, TURBULENT, FORCED CONVECTION. h(0₂6) 27.2658716-3 * 0.6 = Pr = 60 0.037 Ret 7 5X10² ≤ Ref = 10² as Pr 11/3 h = 100, 5597 W/m².K Q = hA (Is-T₂) = (100.5 = 97 W/m²³.K.) (0.6 X(0-2) (350-276²) Q = 892.9701 11) (spread over the * IF TURBULENT FLOW IS NOT. of 8 -...
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Alternative transcript:
(0.037) (1051992-971) | (0.7049) 110 SPREAD OVER THE SURFACE 113 Ny₁ = 0.664 Recr² Pr ³ + 0.0.34 Rex P₁043 [1-(Recr/ Rex) 0.8 ] equi. ૦.૫૩ Pr 113 = 0.664 (5X10³) " (0.7049) + (0-036) (1051992-971) 0+ (0-7049) 0-43 [- ઝાS/k।.૧૧.૧૩) * N₁ = = 1076-8941 0-8 5-67 b(0.6) 272658X103 +370-1944 h = 62.5702 W/m².K Q = (63² 5702) (0.6 X(0₁2) (350-276) Q=555-6234 W 2. GIVEN: AIR NUL AIR I₁ = 336. a R 1-0.305m REQ'D h assuming laminar flow. SOL'N FLAT PLATE, Laminar flows, external flow hl = P=101.0 kPa = to = 288.8 K V=3.0smk3 р, мра 0.1 ·0.1010 MPO * IF IF = 311 R 0.1013 Ts + To_ 338.2K + 288.8 K 2 • PROPERTIES OF AIR pat 311 K₁ 0.1013 MPa (NOTE: SAME PROCESS LIKE UD. for the determination OF CP, P, K, M) P+ kg/m³ 1-1374 11.4006 7.1522 kg/m³ Matoul K₁ 0.1013 MPa P, MPa 0.1 1 0.1010 0.1 1 0.1013 R. at 311 R₁ 0.1013 MPa. P, MPa 0.1013 _Lup Rexμ = LUP: Pr • LAMINAR Срм Cp at 311 K, a 1013 MPg P, MPa 0-1 = 4₁ Pa-s * Pr = 0.7051 19-0079 X 10 = -6 19.1399x10 19.0076X10"Pa-S (0.305) (3.05) (1.1522) 19:0076×106 Rex = 56389-7625 £ 5X 105 -6 K, W/m. K 27-1298X10² 27-4224 × 10 27.1302 x 103 Cp₁ Jl kg K 1006- 4295 1620.1553 1006. 4490J/kg-k (1006-4493) (19.0076 X 10 27-1002×10=3 • USING THE EQN FROM THE HANDBOOK 1 Nux = K h 1.128 P¹12 Rex ¹12 [17/0.0955/17) ²13744 = 19.2622 W/m².K. 1-128 (0.7051) ¹2 (56-389-7625 [1+ (0.0468) 0.7051)213] 114 USING CHURCHILL & ROSE h(0.305) 27.1302X10 4. GIVEN = CASTOR OIL To = 31 °C V= 0.06m/6 HEATED PLATE 1=6m is = 9310 (0.3387) (5613897625) ¹1/2 to TOSI)!! 113 [1 + (0.0468/0.7051) 43) SOU'N: Be= cara 161655 Kinematic viscosity 6x10-5m²/s u P 27 REQ'D. a. hydrodynamic boundary. Lvp α = 7.22Y 10² m²ls K=0-2131₁10-k layer thickness at the end of the plate, 8 b. h at the end of the plate c. Q/w - 11/2/ 6(0.06) 6x10-5 Re = 6000 L 5×105 :. Laminar lvat, the body b = 6.1306 W/m ² - K LOCAL HEAT TRANSFER COEFF. any point of Pr = 831 0249 Pr= & DATE: h = 2 (6.1306) = 12. 2612 Wilm²-K₁ h K Сор Cpu ()u * AVE. HEAT TRANSFER COEEF 8 - a.) Use Blasius Ean: 8 5.0 X over the entire bady √ Rek 5.0 √ 6000 8 = 0.3873 m 6X165 ר צבר.ב DATE: b. Egn. from the HB: _N₁ = n² = +128 Pr'¹/² Rex 1/2 Ilt 0.04683 Pr (5) (6) -0.213 Churchill h² Nu = h² I = $9.3842 W/m²³ - K 44-6921 W/m ² - K h (6) 0.213 = Q (1-125) (831-0249) ¹2 (6000)¹2 (tood)x 497 213-744 G. W 0.0965 (17 (831-0249) G W h = 8.7532 W/m²- K 5 & Rose 0.3387 Rey¹¹2 P₁113 1+ (0.0468 Pr C. Q₁ = hA(T₁-To) = 5 LW (Is-T₂) hL (Ts-T₂) +374 + (0.3387) (6000) ¹2 (831-0249)||3 0.0468 {1+ (4614-0249) ²2/37114 W Using h from the ean taken from the hand book (89.3842) (6) (93-38) 29, 496.786 W/m •Usingh from churchill & Rose espe 1 = (8:15:32)(2) (6) (93-38) 5773.112 W/m Q₂ W 5. GINEN ATMOSPHERIC AIR I∞o = 25°C V=smks VI FLAT PLATE. Is = 75°0 REQ'D': a S, B/A at the trailing edge b. Q SOLIN TE 25+ 73 = 50°C = 323.15 K = P = 0.101325 MPa (same process as no.1 for the A, CRE 44 k) pat 303.15 K 9 0.1013.25 MPa. P, MPa p₁ kg/m ³ 0.1 1.1091 11. 1406 11238 kg/m³ 0.10325 11 at 32.8.15k $ 0.101325. MPa. & MPa Pa-s 0.101325 0-1 1 0-101325. Cpat $23.15.R₁ 9 0.1013.25 MPa P; MPa Cp, J/kg. K 1607.8561 0-101325 6 19.5270X16 19.6567X16 19.5272X10 Pas Pr -6 k. at 3.29. 15 R. 9.0-1013.25 MPa. __P, MPa K₁ W/m-K- = 1020-9744 1007-8574J/kg. K •27:9536 X 10 -3. LVP Rex - (1)( 5)(1-1238) 19.572X10-16 M Rey = 287 752.4684 < 5×10.5. · LAMINAR Сри Pr=0.7040 28-2381 X 10 27.9540x-10-3W/m-K (1007.8574) (19.5272x164) 27.4540 X 10 La . ħi (1) સૅ . ૧3૫0×10 Q 8 Q A 요 8 = 9.3210 X10 M Using eqn from the HB Ny=hl 1.125 Pr 1/2 Rex!/2 к A Q h My A = = S b = 6₁8316 W/M ² -K =h(T₂-TA) = (6-8316) (75-25) 341.58 W/m ² Using Churchill & Rose 5.0 √ Rex th(1) 7•2s 40xp* DATE: 5.0 √287153 4684 [1+ (0.0410 0:0468-8213744 (1-12 8)(0.7040)¹2 (287 752.4684) ¹/2 [1 + (0.01681213] 149 0.7040 13: 6632 W/m²³- K hL 0.3387 Rex ¹/² Pr ¹3 [1+. (0.0963 013]1/4 -0. 338-31 (287 752.4684) + (0-1046) [1+ (104) 25/1/2 h = 4.3497 W/m² - K =(4-34 97) (75-25) 217.485 W/m ² C. • Wising h from HB! Q=(18-6632) (1) (7.5-25) Q₁ = 653-16 W •Using I from Churchill & case, Q = (4.3497) (2) (75-25) Q = 4.34.97 W 6- GIVEN H₂O v=onts_ Ti =80°C REQ'D: h. SOL'N 'Ib вык I = D=16mm Tg => 24° 0. 80136 Sat'd liquid Properties of $₂0 @ 3.3.1.15 K MW = 18.015 g/mol pat 331. 15 k... T, K 330 340 331. ISK -To - 36°C Sp at 0.31. 15 R T, K 8.30 340 331:15. 58°C = 331-15 K mol/dm³ P₁ 54:662) 54.371 984-133) Kg/m³ R0/mol-R ср 0.075373 0.075456 1.41.84.4321J/kg-K.. kat 331-15 K I.R. 330 340 331.15 Up at 331.15 R T, K Bab 340 381.15 R Ms at 24°C (297.15 K) T.K. 290 300 297.15 651-18X10 -3 L040-55%.10` 652-2576 XX-3 Pr сри K Pr = 3.6903 по Ub, Pats 489 49×10 421-97 X10 481-7252x10 Pats Dup Re=· M -481-7252 x 10- Re = 98 060 8.629 74000 ・・TURBULENT h(0.016) 1652.2576X1033 illis Pats 1084X106 853.84×10 919-4356 X10 Pas -6 (0.016)(3)(984-1831) (4184.4321)(481-7252x10²) -652-2576 X 103 segn 5-57. -0.027 Re 4's Prills (16) 0.14 Ms Nu accdg. to GEANKOPINS egn is valid : Re 76000 0.7 / Pr L 6000 760 = (0.027) (98060 8629) (0.093) 3 481-7252x06 0-14 919.43868106 ·h. = 1.4,1.417. 5668 W/m ² .K...
Physics Study Guide: Understanding Convection and Heat Transfer
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This study guide is an introductory resource on convection, designed specifically for our high school physics class. It aims to provide a clear and direct understanding of convection and its role in heat transfer, including how it differs from conduction and radiation. The guide includes: - An overview of the basic principles of convection, offering a foundational understanding for students. - A series of sample problems with computations focusing on forced convection, helping to apply theoretical concepts to practical situations. - Worksheets and exercises on conduction, convection, and radiation, enhancing the grasp of these three modes of heat transfer. - Accessible PDF worksheets with answers, allowing for self-assessment and thorough practice. - Detailed exploration of convection currents through tailored worksheet activities, aiding in visualizing and understanding this phenomenon. - A comprehensive worksheet dedicated to convection, reinforcing key concepts and terminologies. - An ‘Energy Worksheet’ section covering conduction, convection, and radiation, complete with an answer key for guided learning. - Specific exercises that differentiate between radiation, convection, and conduction, highlighting their unique characteristics and applications. This guide is structured to provide a thorough yet easy-to-understand approach to the study of convection and related heat transfer methods, making it a valuable tool for our current unit in physics.
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Physics Study Guide: Understanding Convection and Heat Transfer
Chemistry
Study note
Comments (1)
This study guide is an introductory resource on convection, designed specifically for our high school physics class. It aims to provide a clear and direct understanding of convection and its role in heat transfer, including how it differs from conduction and radiation. The guide includes: - An overview of the basic principles of convection, offering a foundational understanding for students. - A series of sample problems with computations focusing on forced convection, helping to apply theoretical concepts to practical situations. - Worksheets and exercises on conduction, convection, and radiation, enhancing the grasp of these three modes of heat transfer. - Accessible PDF worksheets with answers, allowing for self-assessment and thorough practice. - Detailed exploration of convection currents through tailored worksheet activities, aiding in visualizing and understanding this phenomenon. - A comprehensive worksheet dedicated to convection, reinforcing key concepts and terminologies. - An ‘Energy Worksheet’ section covering conduction, convection, and radiation, complete with an answer key for guided learning. - Specific exercises that differentiate between radiation, convection, and conduction, highlighting their unique characteristics and applications. This guide is structured to provide a thorough yet easy-to-understand approach to the study of convection and related heat transfer methods, making it a valuable tool for our current unit in physics.
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CONVECTION presence of bulk fluid motion. FORCED CONNECT lon a - mechanism of heat transfer through a fluid in the SA - when the fluid is forced to flow by pressure differences, a fan, and so on. pump, SAMPLE PROBLEMS + GIVEN: REQ'D: Q Cover the surface) SOLUTION: Q = hA (Ts-T₂) hl k NuL = Properties of Air at IF (p, M₁K, Cp) Is + To 350 +276. ∙313 k IF 2 __P=1 atm = 0.101325 MPg. pat 3 10 K, 0.1 MPa: 1₁K 300 500 313K = 300 300 313. e at 313 K, 1 MPa: 7, K mol/dm ²³. fr 0.040103 0.024046 kg/m³ 1.1327 kg/m³ 6x29 foxy f+moll dm ³ 0.040205 0.03974 11.35.35 Kg/m ²³. FLAT PLATE (CRANK CASE) 0.6m x 0.2m x 0.1 m Ts = 350 K ATMOSPHERIC AIR = 276 K- 30 m/s Тоб P = 1 atm pat 013 K. 0.101325 MPa. P. MPa 0.1 0.101325 for Ср, и, 0.101325 -P, kg l m ³ 1.1327 11 at 313 K, 0.101325 MPa P, MPa Pars 0.1 K, at 0. 10135 MPa K = P, MPa 0-1 0.101325 11- 3535 1.1477 kg/m³ ik ( same process as ė 싸 19.09.29 X 10 -6 6 19.22496X10 19.0931 X10-pa-s K₁ W/m - R 27.2654 X10 27.5567 X 10 27.2658X10 Pa-s Spat 313 K, 0.101325 MPa Lp, Jl Kg. R P, MPa. 0.1 1006-6643 1 0.101325 Rey= • Lup u - Rel₁ = 1081 992.9717 5X105 Turbulent Nu 1020 2902 1006.6844J/kg. K Pr = CpW (1006-6844) (19.0931x 166) сри k 7.2658 X16 hL (0.6) (30) (1.1477) 19.0431 X10 Pr= 0.7049 HEAT FLOW OVER A FLAT PLATE, TURBULENT, FORCED CONVECTION. h(0₂6) 27.2658716-3 * 0.6 = Pr = 60 0.037 Ret 7 5X10² ≤ Ref = 10² as Pr 11/3 h = 100, 5597 W/m².K Q = hA (Is-T₂) = (100.5 = 97 W/m²³.K.) (0.6 X(0-2) (350-276²) Q = 892.9701 11) (spread over the * IF TURBULENT FLOW IS NOT. of 8 -...
CONVECTION presence of bulk fluid motion. FORCED CONNECT lon a - mechanism of heat transfer through a fluid in the SA - when the fluid is forced to flow by pressure differences, a fan, and so on. pump, SAMPLE PROBLEMS + GIVEN: REQ'D: Q Cover the surface) SOLUTION: Q = hA (Ts-T₂) hl k NuL = Properties of Air at IF (p, M₁K, Cp) Is + To 350 +276. ∙313 k IF 2 __P=1 atm = 0.101325 MPg. pat 3 10 K, 0.1 MPa: 1₁K 300 500 313K = 300 300 313. e at 313 K, 1 MPa: 7, K mol/dm ²³. fr 0.040103 0.024046 kg/m³ 1.1327 kg/m³ 6x29 foxy f+moll dm ³ 0.040205 0.03974 11.35.35 Kg/m ²³. FLAT PLATE (CRANK CASE) 0.6m x 0.2m x 0.1 m Ts = 350 K ATMOSPHERIC AIR = 276 K- 30 m/s Тоб P = 1 atm pat 013 K. 0.101325 MPa. P. MPa 0.1 0.101325 for Ср, и, 0.101325 -P, kg l m ³ 1.1327 11 at 313 K, 0.101325 MPa P, MPa Pars 0.1 K, at 0. 10135 MPa K = P, MPa 0-1 0.101325 11- 3535 1.1477 kg/m³ ik ( same process as ė 싸 19.09.29 X 10 -6 6 19.22496X10 19.0931 X10-pa-s K₁ W/m - R 27.2654 X10 27.5567 X 10 27.2658X10 Pa-s Spat 313 K, 0.101325 MPa Lp, Jl Kg. R P, MPa. 0.1 1006-6643 1 0.101325 Rey= • Lup u - Rel₁ = 1081 992.9717 5X105 Turbulent Nu 1020 2902 1006.6844J/kg. K Pr = CpW (1006-6844) (19.0931x 166) сри k 7.2658 X16 hL (0.6) (30) (1.1477) 19.0431 X10 Pr= 0.7049 HEAT FLOW OVER A FLAT PLATE, TURBULENT, FORCED CONVECTION. h(0₂6) 27.2658716-3 * 0.6 = Pr = 60 0.037 Ret 7 5X10² ≤ Ref = 10² as Pr 11/3 h = 100, 5597 W/m².K Q = hA (Is-T₂) = (100.5 = 97 W/m²³.K.) (0.6 X(0-2) (350-276²) Q = 892.9701 11) (spread over the * IF TURBULENT FLOW IS NOT. of 8 -...
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Alternative transcript:
(0.037) (1051992-971) | (0.7049) 110 SPREAD OVER THE SURFACE 113 Ny₁ = 0.664 Recr² Pr ³ + 0.0.34 Rex P₁043 [1-(Recr/ Rex) 0.8 ] equi. ૦.૫૩ Pr 113 = 0.664 (5X10³) " (0.7049) + (0-036) (1051992-971) 0+ (0-7049) 0-43 [- ઝાS/k।.૧૧.૧૩) * N₁ = = 1076-8941 0-8 5-67 b(0.6) 272658X103 +370-1944 h = 62.5702 W/m².K Q = (63² 5702) (0.6 X(0₁2) (350-276) Q=555-6234 W 2. GIVEN: AIR NUL AIR I₁ = 336. a R 1-0.305m REQ'D h assuming laminar flow. SOL'N FLAT PLATE, Laminar flows, external flow hl = P=101.0 kPa = to = 288.8 K V=3.0smk3 р, мра 0.1 ·0.1010 MPO * IF IF = 311 R 0.1013 Ts + To_ 338.2K + 288.8 K 2 • PROPERTIES OF AIR pat 311 K₁ 0.1013 MPa (NOTE: SAME PROCESS LIKE UD. for the determination OF CP, P, K, M) P+ kg/m³ 1-1374 11.4006 7.1522 kg/m³ Matoul K₁ 0.1013 MPa P, MPa 0.1 1 0.1010 0.1 1 0.1013 R. at 311 R₁ 0.1013 MPa. P, MPa 0.1013 _Lup Rexμ = LUP: Pr • LAMINAR Срм Cp at 311 K, a 1013 MPg P, MPa 0-1 = 4₁ Pa-s * Pr = 0.7051 19-0079 X 10 = -6 19.1399x10 19.0076X10"Pa-S (0.305) (3.05) (1.1522) 19:0076×106 Rex = 56389-7625 £ 5X 105 -6 K, W/m. K 27-1298X10² 27-4224 × 10 27.1302 x 103 Cp₁ Jl kg K 1006- 4295 1620.1553 1006. 4490J/kg-k (1006-4493) (19.0076 X 10 27-1002×10=3 • USING THE EQN FROM THE HANDBOOK 1 Nux = K h 1.128 P¹12 Rex ¹12 [17/0.0955/17) ²13744 = 19.2622 W/m².K. 1-128 (0.7051) ¹2 (56-389-7625 [1+ (0.0468) 0.7051)213] 114 USING CHURCHILL & ROSE h(0.305) 27.1302X10 4. GIVEN = CASTOR OIL To = 31 °C V= 0.06m/6 HEATED PLATE 1=6m is = 9310 (0.3387) (5613897625) ¹1/2 to TOSI)!! 113 [1 + (0.0468/0.7051) 43) SOU'N: Be= cara 161655 Kinematic viscosity 6x10-5m²/s u P 27 REQ'D. a. hydrodynamic boundary. Lvp α = 7.22Y 10² m²ls K=0-2131₁10-k layer thickness at the end of the plate, 8 b. h at the end of the plate c. Q/w - 11/2/ 6(0.06) 6x10-5 Re = 6000 L 5×105 :. Laminar lvat, the body b = 6.1306 W/m ² - K LOCAL HEAT TRANSFER COEFF. any point of Pr = 831 0249 Pr= & DATE: h = 2 (6.1306) = 12. 2612 Wilm²-K₁ h K Сор Cpu ()u * AVE. HEAT TRANSFER COEEF 8 - a.) Use Blasius Ean: 8 5.0 X over the entire bady √ Rek 5.0 √ 6000 8 = 0.3873 m 6X165 ר צבר.ב DATE: b. Egn. from the HB: _N₁ = n² = +128 Pr'¹/² Rex 1/2 Ilt 0.04683 Pr (5) (6) -0.213 Churchill h² Nu = h² I = $9.3842 W/m²³ - K 44-6921 W/m ² - K h (6) 0.213 = Q (1-125) (831-0249) ¹2 (6000)¹2 (tood)x 497 213-744 G. W 0.0965 (17 (831-0249) G W h = 8.7532 W/m²- K 5 & Rose 0.3387 Rey¹¹2 P₁113 1+ (0.0468 Pr C. Q₁ = hA(T₁-To) = 5 LW (Is-T₂) hL (Ts-T₂) +374 + (0.3387) (6000) ¹2 (831-0249)||3 0.0468 {1+ (4614-0249) ²2/37114 W Using h from the ean taken from the hand book (89.3842) (6) (93-38) 29, 496.786 W/m •Usingh from churchill & Rose espe 1 = (8:15:32)(2) (6) (93-38) 5773.112 W/m Q₂ W 5. GINEN ATMOSPHERIC AIR I∞o = 25°C V=smks VI FLAT PLATE. Is = 75°0 REQ'D': a S, B/A at the trailing edge b. Q SOLIN TE 25+ 73 = 50°C = 323.15 K = P = 0.101325 MPa (same process as no.1 for the A, CRE 44 k) pat 303.15 K 9 0.1013.25 MPa. P, MPa p₁ kg/m ³ 0.1 1.1091 11. 1406 11238 kg/m³ 0.10325 11 at 32.8.15k $ 0.101325. MPa. & MPa Pa-s 0.101325 0-1 1 0-101325. Cpat $23.15.R₁ 9 0.1013.25 MPa P; MPa Cp, J/kg. K 1607.8561 0-101325 6 19.5270X16 19.6567X16 19.5272X10 Pas Pr -6 k. at 3.29. 15 R. 9.0-1013.25 MPa. __P, MPa K₁ W/m-K- = 1020-9744 1007-8574J/kg. K •27:9536 X 10 -3. LVP Rex - (1)( 5)(1-1238) 19.572X10-16 M Rey = 287 752.4684 < 5×10.5. · LAMINAR Сри Pr=0.7040 28-2381 X 10 27.9540x-10-3W/m-K (1007.8574) (19.5272x164) 27.4540 X 10 La . ħi (1) સૅ . ૧3૫0×10 Q 8 Q A 요 8 = 9.3210 X10 M Using eqn from the HB Ny=hl 1.125 Pr 1/2 Rex!/2 к A Q h My A = = S b = 6₁8316 W/M ² -K =h(T₂-TA) = (6-8316) (75-25) 341.58 W/m ² Using Churchill & Rose 5.0 √ Rex th(1) 7•2s 40xp* DATE: 5.0 √287153 4684 [1+ (0.0410 0:0468-8213744 (1-12 8)(0.7040)¹2 (287 752.4684) ¹/2 [1 + (0.01681213] 149 0.7040 13: 6632 W/m²³- K hL 0.3387 Rex ¹/² Pr ¹3 [1+. (0.0963 013]1/4 -0. 338-31 (287 752.4684) + (0-1046) [1+ (104) 25/1/2 h = 4.3497 W/m² - K =(4-34 97) (75-25) 217.485 W/m ² C. • Wising h from HB! Q=(18-6632) (1) (7.5-25) Q₁ = 653-16 W •Using I from Churchill & case, Q = (4.3497) (2) (75-25) Q = 4.34.97 W 6- GIVEN H₂O v=onts_ Ti =80°C REQ'D: h. SOL'N 'Ib вык I = D=16mm Tg => 24° 0. 80136 Sat'd liquid Properties of $₂0 @ 3.3.1.15 K MW = 18.015 g/mol pat 331. 15 k... T, K 330 340 331. ISK -To - 36°C Sp at 0.31. 15 R T, K 8.30 340 331:15. 58°C = 331-15 K mol/dm³ P₁ 54:662) 54.371 984-133) Kg/m³ R0/mol-R ср 0.075373 0.075456 1.41.84.4321J/kg-K.. kat 331-15 K I.R. 330 340 331.15 Up at 331.15 R T, K Bab 340 381.15 R Ms at 24°C (297.15 K) T.K. 290 300 297.15 651-18X10 -3 L040-55%.10` 652-2576 XX-3 Pr сри K Pr = 3.6903 по Ub, Pats 489 49×10 421-97 X10 481-7252x10 Pats Dup Re=· M -481-7252 x 10- Re = 98 060 8.629 74000 ・・TURBULENT h(0.016) 1652.2576X1033 illis Pats 1084X106 853.84×10 919-4356 X10 Pas -6 (0.016)(3)(984-1831) (4184.4321)(481-7252x10²) -652-2576 X 103 segn 5-57. -0.027 Re 4's Prills (16) 0.14 Ms Nu accdg. to GEANKOPINS egn is valid : Re 76000 0.7 / Pr L 6000 760 = (0.027) (98060 8629) (0.093) 3 481-7252x06 0-14 919.43868106 ·h. = 1.4,1.417. 5668 W/m ² .K...