I need handwritten of this just do it handwritten u just have to copy paste on white page Equating eq, (1) and (2)%3D+2v², = v², + pV3ghi.NOW , find avg. velocity at point (3)-(3)V3 =A3v,%3DPUT, 5/12ft for DV2 == 7. 33V3ft/sFROM Data book, obtaion value of loss coefficients Kµfor 90°value for bend pipe, (kL) = 0. 3value of sharp edge, (kL) = 1now, find net loss coefficientEKL = 3 × 0. 3 + 1Head loss, hį = (f + EKL)put, f = 0.021 , L = 15ft ,D = 5/12 , EKĻ = 1.9hL = |0.0+ 1.9| 28hL = 2.655PUT , hL = 2. 655– . V3 = 7.33Vaftlsin eq. (3)(7.33V,)+ V3 x g × 2. 655-= 96. 382V3(v.)²V3 =V22x95.382put V2 = 1. 2ftls ,V2 == 8. 80ft/sV3 = (1.2 8.802x95.382= 0. 784ft³lsThus , 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 ofthe 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 airp = pressurep = density of airv = velocity of fluida = kinetic energy correction factorW fan = fan powerWturbine = turbine powerhL = Head lossg = Acceleration due to gravityNote – pressure at both ends is atmospheric, velocity of a fluid atpoint (1) is zero and loss of head are negligible.take, a = 1W fan =put pvz for mW fan = pv;(1)now find avg. velocityV2 = *V2whereV2 = Avg. velocityV2 = volumetric flow rateA2 = Cross sectional areaD = Dia. of pipenow put , 1. 2ft²/s for v , (5in = 5/12ft) for D= 8. 80f t/s1.2(5/12)2V2 =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 + mghtmw fan = m + mgh,put pv3 for mw fan = pv.pis글 +prs ghu........ (2)

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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

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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