Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
Wa 18 Air Preheater Wa 17 t go 叶 + emin toc tgo emax t gc Twa t17 t18 Wa 18 Air Preheater Wo 100 Assuming the size of tube - 1 oc Two Design of Air Preheater t17 AP = 5.55 x 10-8 G² page 332 AP = 2.54 to 5.08 cm H₂O page 447 Use AP = 5 cm H₂O AP = 1.9685 in H₂O = 5955.54694 kg/m² - hr 1747.768014 NE (D) (H) T (0.0422 ) ( 3.75) NE = 3515.525 Say N₁ = 3516 tubes theoretical number of tubes Note: Assume the actual number of tubes so that there will be uniform spacing G = 5.55 x 10- @max G= 1.654 kg/m²-s to Over-all Coefficient of Heat Transfer, U U = A + BG No 59 For Smooth Tube Economizers No = 60 Size A = 1.95 page 330 Do Outside diameter 0.0422 B 0.00045 Di Inside diameter 0.0351 m Thickness 0.00355 m U = 1.95 + 0.00455955.55) Logarithmic Mean Temperature Difference U = 4.6300 tgc = 423.47 °C Omax = tab ta Qair preheater 3611 kJ/s from 3rd Checking E in Tabulation Ntactual tgD = 416.43 °C t17= 40°C = 416.43 - 32 384.43 °C Qair preheater 3106992.321 kcal/hr Q U ( A ) ( LMTD) Nta (Ntb) = 60 (59) Ntactual = 3540 Length Nta = tubes actual 60 tubes +18= 32 °C toc 423.47 - 40 A = 3106992.321 4.6300 (383.9509668) 1747.768014 m² = Area of Heating Surface L = 383.47 °C = Width Ntb= 59 Nea (Do) + (Nta + 1 ) S 60 (0.0422) + ( 60 + 1) (0.025) tubes LMTD @max In 8min max 384.43- 383.47 LMTD = In 384.43 383.47 LMTD =383.9509668 °C Assume Number of Tubes A = (D₂) (H) (N₂) H = 3.75 m S 0.025 m W = W W 4.057 m Ntb (Do) + (Nth + 1) S 59 (0.0422) + ( 59+ 1 ) (0.025) 3.9898 m Note: value from square root of N₁

Wa 18 Air Preheater Wa 17 t go 叶 + emin toc tgo emax t gc Twa t17 t18 Wa 18 Air Preheater Wo 100 Assuming the size of tube - 1 oc Two Design of Air Preheater t17 AP = 5.55 x 10-8 G² page 332 AP = 2.54 to 5.08 cm H₂O page 447 Use AP = 5 cm H₂O AP = 1.9685 in H₂O = 5955.54694 kg/m² - hr 1747.768014 NE (D) (H) T (0.0422 ) ( 3.75) NE = 3515.525 Say N₁ = 3516 tubes theoretical number of tubes Note: Assume the actual number of tubes so that there will be uniform spacing G = 5.55 x 10- @max G= 1.654 kg/m²-s to Over-all Coefficient of Heat Transfer, U U = A + BG No 59 For Smooth Tube Economizers No = 60 Size A = 1.95 page 330 Do Outside diameter 0.0422 B 0.00045 Di Inside diameter 0.0351 m Thickness 0.00355 m U = 1.95 + 0.00455955.55) Logarithmic Mean Temperature Difference U = 4.6300 tgc = 423.47 °C Omax = tab ta Qair preheater 3611 kJ/s from 3rd Checking E in Tabulation Ntactual tgD = 416.43 °C t17= 40°C = 416.43 - 32 384.43 °C Qair preheater 3106992.321 kcal/hr Q U ( A ) ( LMTD) Nta (Ntb) = 60 (59) Ntactual = 3540 Length Nta = tubes actual 60 tubes +18= 32 °C toc 423.47 - 40 A = 3106992.321 4.6300 (383.9509668) 1747.768014 m² = Area of Heating Surface L = 383.47 °C = Width Ntb= 59 Nea (Do) + (Nta + 1 ) S 60 (0.0422) + ( 60 + 1) (0.025) tubes LMTD @max In 8min max 384.43- 383.47 LMTD = In 384.43 383.47 LMTD =383.9509668 °C Assume Number of Tubes A = (D₂) (H) (N₂) H = 3.75 m S 0.025 m W = W W 4.057 m Ntb (Do) + (Nth + 1) S 59 (0.0422) + ( 59+ 1 ) (0.025) 3.9898 m Note: value from square root of N₁

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter6: Forced Convection Over Exterior Surfaces
Section: Chapter Questions
Problem 6.52P
See similar textbooks
icon
Related questions
Question

what is an air preheater, what are formulas, and their importance, define the diagram, and give me a script on how to explain the design of an air preheater, and how did values end up in that number. based on standards 

Wa 18 Air Preheater Wa 17 t go 叶 + emin toc tgo emax t gc Twa t17 t18
Transcribed Image Text:Wa 18 Air Preheater Wa 17 t go 叶 + emin toc tgo emax t gc Twa t17 t18
Wa 18 Air Preheater Wo 100 Assuming the size of tube - 1 oc Two Design of Air Preheater t17 AP = 5.55 x 10-8 G² page 332 AP = 2.54 to 5.08 cm H₂O page 447 Use AP = 5 cm H₂O AP = 1.9685 in H₂O = 5955.54694 kg/m² - hr 1747.768014 NE (D) (H) T (0.0422 ) ( 3.75) NE = 3515.525 Say N₁ = 3516 tubes theoretical number of tubes Note: Assume the actual number of tubes so that there will be uniform spacing G = 5.55 x 10- @max G= 1.654 kg/m²-s to Over-all Coefficient of Heat Transfer, U U = A + BG No 59 For Smooth Tube Economizers No = 60 Size A = 1.95 page 330 Do Outside diameter 0.0422 B 0.00045 Di Inside diameter 0.0351 m Thickness 0.00355 m U = 1.95 + 0.00455955.55) Logarithmic Mean Temperature Difference U = 4.6300 tgc = 423.47 °C Omax = tab ta Qair preheater 3611 kJ/s from 3rd Checking E in Tabulation Ntactual tgD = 416.43 °C t17= 40°C = 416.43 - 32 384.43 °C Qair preheater 3106992.321 kcal/hr Q U ( A ) ( LMTD) Nta (Ntb) = 60 (59) Ntactual = 3540 Length Nta = tubes actual 60 tubes +18= 32 °C toc 423.47 - 40 A = 3106992.321 4.6300 (383.9509668) 1747.768014 m² = Area of Heating Surface L = 383.47 °C = Width Ntb= 59 Nea (Do) + (Nta + 1 ) S 60 (0.0422) + ( 60 + 1) (0.025) tubes LMTD @max In 8min max 384.43- 383.47 LMTD = In 384.43 383.47 LMTD =383.9509668 °C Assume Number of Tubes A = (D₂) (H) (N₂) H = 3.75 m S 0.025 m W = W W 4.057 m Ntb (Do) + (Nth + 1) S 59 (0.0422) + ( 59+ 1 ) (0.025) 3.9898 m Note: value from square root of N₁
Transcribed Image Text:Wa 18 Air Preheater Wo 100 Assuming the size of tube - 1 oc Two Design of Air Preheater t17 AP = 5.55 x 10-8 G² page 332 AP = 2.54 to 5.08 cm H₂O page 447 Use AP = 5 cm H₂O AP = 1.9685 in H₂O = 5955.54694 kg/m² - hr 1747.768014 NE (D) (H) T (0.0422 ) ( 3.75) NE = 3515.525 Say N₁ = 3516 tubes theoretical number of tubes Note: Assume the actual number of tubes so that there will be uniform spacing G = 5.55 x 10- @max G= 1.654 kg/m²-s to Over-all Coefficient of Heat Transfer, U U = A + BG No 59 For Smooth Tube Economizers No = 60 Size A = 1.95 page 330 Do Outside diameter 0.0422 B 0.00045 Di Inside diameter 0.0351 m Thickness 0.00355 m U = 1.95 + 0.00455955.55) Logarithmic Mean Temperature Difference U = 4.6300 tgc = 423.47 °C Omax = tab ta Qair preheater 3611 kJ/s from 3rd Checking E in Tabulation Ntactual tgD = 416.43 °C t17= 40°C = 416.43 - 32 384.43 °C Qair preheater 3106992.321 kcal/hr Q U ( A ) ( LMTD) Nta (Ntb) = 60 (59) Ntactual = 3540 Length Nta = tubes actual 60 tubes +18= 32 °C toc 423.47 - 40 A = 3106992.321 4.6300 (383.9509668) 1747.768014 m² = Area of Heating Surface L = 383.47 °C = Width Ntb= 59 Nea (Do) + (Nta + 1 ) S 60 (0.0422) + ( 60 + 1) (0.025) tubes LMTD @max In 8min max 384.43- 383.47 LMTD = In 384.43 383.47 LMTD =383.9509668 °C Assume Number of Tubes A = (D₂) (H) (N₂) H = 3.75 m S 0.025 m W = W W 4.057 m Ntb (Do) + (Nth + 1) S 59 (0.0422) + ( 59+ 1 ) (0.025) 3.9898 m Note: value from square root of N₁
Expert Solution
steps

Step by step

Solved in 2 steps with 8 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Principles of Heat Transfer (Activate Learning wi…
Principles of Heat Transfer (Activate Learning wi…
Mechanical Engineering
ISBN:
9781305387102
Author:
Kreith, Frank; Manglik, Raj M.
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS