Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
Equating eq, (1) and (2) pv², = pv?; + pV3ghu %3D (3) .......... NOW, find avg. velocity at point (3) V3 = %3D %3D PUT, 5/12ft for D V, V2 = = 7. 33V3ft/s FROM Data book, obtaion value of loss coefficients Kifor 90° value for bend pipe, (kµ) = 0. 3 value of sharp edge, (kL) = 1 now, find net loss coefficient EKL = 3 × 0. 3 + 1 Head loss, hL = (f + EKL) %3D put, f = 0.021 , L = 15ft , D = 5/12 , EK, = 1.9 hi = 1. hi = 2.655 PUT , h = 2. 655 . V3 = 7.33Vaftls in eq. (3) (7.33V,)* + V3 x g × 2.655 7.33V,)* %3D 2 = 96. 382V³, V3 = ( vz (v.)* 2x95.382 put V2 = 1.2ft ls ,V2 == 8. 80ft/s V3 = (1.28.80) 2x95.382 = 0. 784ft Is Thus, volumetric flow rate at point (3) will - 0. 785ft/s 1

Equating eq, (1) and (2) pv², = pv?; + pV3ghu %3D (3) .......... NOW, find avg. velocity at point (3) V3 = %3D %3D PUT, 5/12ft for D V, V2 = = 7. 33V3ft/s FROM Data book, obtaion value of loss coefficients Kifor 90° value for bend pipe, (kµ) = 0. 3 value of sharp edge, (kL) = 1 now, find net loss coefficient EKL = 3 × 0. 3 + 1 Head loss, hL = (f + EKL) %3D put, f = 0.021 , L = 15ft , D = 5/12 , EK, = 1.9 hi = 1. hi = 2.655 PUT , h = 2. 655 . V3 = 7.33Vaftls in eq. (3) (7.33V,)* + V3 x g × 2.655 7.33V,)* %3D 2 = 96. 382V³, V3 = ( vz (v.)* 2x95.382 put V2 = 1.2ft ls ,V2 == 8. 80ft/s V3 = (1.28.80) 2x95.382 = 0. 784ft Is Thus, volumetric flow rate at point (3) will - 0. 785ft/s 1

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
See similar textbooks
icon
Related questions
Question
I need handwritten of this just do it handwritten u just have to copy paste on white page
Equating eq, (1) and (2) %3D + 2 v², = v², + pV3ghi. NOW , find avg. velocity at point (3) -(3) V3 = A3 v, %3D PUT, 5/12ft for D V2 = = 7. 33V3ft/s FROM Data book, obtaion value of loss coefficients Kµfor 90° value for bend pipe, (kL) = 0. 3 value of sharp edge, (kL) = 1 now, find net loss coefficient EKL = 3 × 0. 3 + 1 Head loss, hį = (f + EKL) put, f = 0.021 , L = 15ft ,D = 5/12 , EKĻ = 1.9 hL = |0.0 + 1.9| 28 hL = 2.655 PUT , hL = 2. 655– . V3 = 7.33Vaftls in eq. (3) (7.33V,) + V3 x g × 2. 655- = 96. 382V3 (v.)² V3 = V2 2x95.382 put V2 = 1. 2ftls ,V2 == 8. 80ft/s V3 = (1.2 8.80 2x95.382 = 0. 784ft³ls Thus , volumetric flow rate at point (3) will – |0. 785ft³/s|
Transcribed Image Text:Equating eq, (1) and (2) %3D + 2 v², = v², + pV3ghi. NOW , find avg. velocity at point (3) -(3) V3 = A3 v, %3D PUT, 5/12ft for D V2 = = 7. 33V3ft/s FROM Data book, obtaion value of loss coefficients Kµfor 90° value for bend pipe, (kL) = 0. 3 value of sharp edge, (kL) = 1 now, find net loss coefficient EKL = 3 × 0. 3 + 1 Head loss, hį = (f + EKL) put, f = 0.021 , L = 15ft ,D = 5/12 , EKĻ = 1.9 hL = |0.0 + 1.9| 28 hL = 2.655 PUT , hL = 2. 655– . V3 = 7.33Vaftls in eq. (3) (7.33V,) + V3 x g × 2. 655- = 96. 382V3 (v.)² V3 = V2 2x95.382 put V2 = 1. 2ftls ,V2 == 8. 80ft/s V3 = (1.2 8.80 2x95.382 = 0. 784ft³ls Thus , volumetric flow rate at point (3) will – |0. 785ft³/s|
select point (1) toward be inside the drier and point ( 2) toward be the exit of the drier. consider the energy eg. b/w these two places. m (4 + aj + gzı) + wifan = m ¡( + az + gz2) + wrurbine + mght, Where, m = mass fow rate of air p = pressure p = density of air v = velocity of fluid a = kinetic energy correction factor W fan = fan power Wturbine = turbine power hL = Head loss g = Acceleration due to gravity Note – pressure at both ends is atmospheric, velocity of a fluid at point (1) is zero and loss of head are negligible. take, a = 1 W fan = put pvz for m W fan = pv; (1) now find avg. velocity V2 = * V2 where V2 = Avg. velocity V2 = volumetric flow rate A2 = Cross sectional area D = Dia. of pipe now put , 1. 2ft²/s for v , (5in = 5/12ft) for D = 8. 80f t/s 1.2 (5/12)2 V2 = now , consider a point (3) at exit of the duct , energy eq. b/w (1) and(3) i( + a1글 + 8zz) + wyan %3Di (E + az + gz3) + wurbine + mght m w fan = m + mgh, put pv3 for m w fan = pv. pis글 +prs ghu ........ (2)
Transcribed Image Text:select point (1) toward be inside the drier and point ( 2) toward be the exit of the drier. consider the energy eg. b/w these two places. m (4 + aj + gzı) + wifan = m ¡( + az + gz2) + wrurbine + mght, Where, m = mass fow rate of air p = pressure p = density of air v = velocity of fluid a = kinetic energy correction factor W fan = fan power Wturbine = turbine power hL = Head loss g = Acceleration due to gravity Note – pressure at both ends is atmospheric, velocity of a fluid at point (1) is zero and loss of head are negligible. take, a = 1 W fan = put pvz for m W fan = pv; (1) now find avg. velocity V2 = * V2 where V2 = Avg. velocity V2 = volumetric flow rate A2 = Cross sectional area D = Dia. of pipe now put , 1. 2ft²/s for v , (5in = 5/12ft) for D = 8. 80f t/s 1.2 (5/12)2 V2 = now , consider a point (3) at exit of the duct , energy eq. b/w (1) and(3) i( + a1글 + 8zz) + wyan %3Di (E + az + gz3) + wurbine + mght m w fan = m + mgh, put pv3 for m w fan = pv. pis글 +prs ghu ........ (2)
Expert Solution
steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Layout
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY
SEE MORE TEXTBOOKS