EXAMPLE 8-1 Laminar Flow in Horizontal and Inclined Pipes Consider the fully developed flow of glycerin at 40°C through a 70-m-long, 4-cm-diameter, horizontal, circular pipe. If the flow velocity at the centerline is measured to be 6 m/s, determine the velocity profile and the pressure dif- ference across this 70-m-long section of the pipe, and the useful pumping power required to maintain this flow. For the same useful pumping power input, determine the percent increase of the flow rate if the pipe is inclined 15° downward and the percent decrease if it is inclined 15° upward. The pump is located outside this pipe section. D-2 em cavcerine Hello sir, I want the same şolution, but in a simplified way without the words of simplifying the solution. -15 ur) - ua Substituting, the velocity profile is determined to be mis(1 – - 6(1 - 2500r²) u(r) - (6 m/s)I- (0.02 m)) where u is in m/s and r is in m. The average velocity, the flow rate, and th Reynolds number are 6 m/s V - Vs - 3 m/s 2 V - VvA-V(mD'14) (3 m/s)[ 7(0.04 m) /4] 3.77 x 10- m% (1252 kg/mX3 m/s)(0.04 m) 0.3073 kg/m-s pVD Re - 488.9 which is less than 2300. Therefore, the flow is indeed laminar. Then th riction factor and the head loss become 64 64 - 0.1309 Re 488.9 LV? (70 m) (3 m/s)2 - 0.1309 D 2g - 105.1m (0.04 m) 2(9.81 m/s³) + ai 28 P8 + z, + hmp, . - P8 + a, 2g + 2z + hurtine, e + h For fully developed flow in a constant diameter pipe with bines, it reduces to pumps or tun AP - P, - P,- pg(z, - + h,) Then the pressure difference and the required useful pumping power for the horizontal case become AP - pg(z; - + h) I kPa 1000 kg/m-s - (1252 kg/mPX9.81 m/s³)(0 + 105.1 m) - 1291 kPa I kW kPa-m/s W - VAP - (3.77 x 10' m/s(1291 kPa) - 4.87 kW The elevation difference and the pressure difference for a pipe incline upwards 15° is Az - - z, - Lsin15° -- (70 m)sin15° - 18.1 m I kPa AP - (1252 kg/mX9.81 m/s-(18.1 m + 105.1 m 1000 kg/m-s - 1366 kPa

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Transcribed Image Text:- EXAMPLE 8-1 Laminar Flow in Horizontal and Inclined Pipes Consider the fully developed flow of glycerin at 40°C through a 70-m-long, 4-cm-diameter, horizontal, circular pipe. If the flow velocity at the centerline is measured to be 6 m/s, determine the velocity profile and the pressure dif- ference across this 70-m-long section of the pipe, and the useful pumping power required to maintain this flow. For the same useful pumping power input, determine the percent increase of the flow rate if the pipe is inclined 15° downward and the percent decrease if it is inclined 15° upward. The pump is located outside this pipe section. D-2 em 6 m/s Glycerine Hello sir, I want the same solution, but in a simplified way without the words of simplifying the solution. ur) - u I- Substituting, the velocity profile is determined to be u(r) - (6 m/s)1- - 6(1-2500r) (0.02 m) where u is in m/s and r is in m. The average velocity, the flow rate, and the Reynolds number are 6 m/s V - Vs - 3 m/s V - VvA- V#D³14) = (3 m/s)[7(0.04 m)/4] 3.77 x 10-m/s pVD (1252 kg/m)(3 m/s(0.04 m) Re - - 488.9 0.3073 kg/m-s which is less than 2300. Therefore, the flow is indeed laminar. Then the friction factor and the head loss become 64 f= 64 0.1309 Re 488.9 LV? h -f D 2g (70 m) (3 m/s)? - 0.1309 - 105.Im (0.04 m) 2(9.81 m/s²) P2 + z, + hump, . + az 28 Pg a 28 + zz + hurbine, e + hj Pg For fully developed flow in a constant diameter pipe with no pumps or tur- bines, it reduces to AP - P, - P, - pg(z, - 2 + h) Then the pressure difference and the required useful pumping power for the horizontal case become AP - pg(z2 - 4 + h) I kPa 1000 kg/m-s², = (1252 kg/mx9.81 m/s(0 + 105.1 m) - 1291 kPa W - VAP = (3.77 × 10' m/s)X1291 kPa)( kW kPa-m/s - 4.87 kW pump. The elevation difference and the pressure difference for a pipe inclined upwards 15° is Az - - 2 - Lsin15° - (70 m)sin15° - 18.1 m I kPa 1000 kg/m-s APd- (1252 kg/m(9.81 m/s?X18.1 m + 105.1 m) - 1366 kPa

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