MET330 20xx0xE Applied Fluid MechanicsInstructor: Instructors nameWeek 2 Review Assignment - 2I pledge to support the Honor System of ECPI. I will refrain from any form of...

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Answered Same DayJan 30, 2023

Answer To: MET330 20xx0xE Applied Fluid MechanicsInstructor: Instructors nameWeek 2 Review...

Dr Shweta answered on Jan 31 2023
51 Votes
Ans 1.
Known values: Q = 0. 20 ft3/s, sg = 1.26 at 100 ͦ F
Governing equations: continuity equati
ons Q = A * V, ρglycerine = ρwater * S.G., Reynolds number = ρVD/µ
Calculations: ρglycerine = ρwater * S.G.
= 62.4 *1.26 = 78.624 lb./ft3
Area = π/4(4/12)2 = 0.0873ft3
Now., 0.20 = 0.0873 * V
So, V = 2.2909 ft/s
Reynolds number = ρVD/µ = 78.624 * 2.2909 *(4/12)/0.25125
= 239 as NRe is less than 2000 so flow is laminar
Ans 2.
Known values: 45C = T, DN 100 = pipe diameter = 0.1 m, 80 steel pipe, sg = 0.895, dynamic viscosity = 4.00 x 10-2 Pa∙s., laminar flow NRe =2000
Governing equations: ρfluid = ρwater * S.G., Q = A * V
Calculations: ρfluid = 0.895 *1000kg/m3
Velocity of oil in pipe is calculated as:
V = N * ȵ/ D * ρ
= 2000 * 4 * 10-2 /.01 * 895 = 0.894m/s
And Q = [3.142 * (0.1)2/4] *0.894 =...
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