MET330 202001E Applied Fluid Mechanics - Week 3 Review AssignmentP1. Questions, Concepts, and Definitions Minor Losses Question 1. Which of the following statements is true about the...

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Answered 1 days AfterFeb 07, 2023

Answer To: MET330 202001E Applied Fluid Mechanics - Week 3 Review AssignmentP1. Questions, Concepts, and...

Dr Shweta answered on Feb 09 2023
48 Votes
MCQ answers:
    1
    A
    2
    E
    3
    A
    4
    A
    5
    C
    6
    E
    7
    D
    8
    E
    9
    E
    10
    E
    11
    E
    12
    A
    13
    A
    14
    E
    15
    D
    16
    E
    17
    B
    18
    D
    19
    E
    20
    E
Review Assignment solutions:
Ans 1 Given: outer Diameter and thickness of smaller pipe = Do1 = 50 mm and t1 = 2.4mm
outer Diameter and thickness of large pipe = Do2 = 90 mm and t2 = 2.8mm, V1 = 3m/s
Calculations:
Firstly, we calculate inner diameter of smaller and larger pipe
Di1 = Do1 -2t1
= 50-2*2.4 = 45.2mm
Di2 = Do2 -2t2
=90-2*2.8 = 84.4 mm
Now using equation of continuity
A1V1= A2V2
π/4Di12 * V1 = π/4Di22 * V2
45.22 *3 =84.42*V2
V2 = 0.86m/s
Head loss due to sudden enlargement is
(Hl)se = (v1-v2)2/2g
= (3-0.86)2/2*9.8 = 0.2334m
Ans 2 Given: d1 = 1 in =0.0254m
D2 = 3.5 in = 0.0889 m
Calculations:
Q= A1V1
3 *10-3 = π/4D12 * V1
= V1 = 5.92m/s
Q= A2V2
3 *10-3 = π/4D22 * V2
= V2 = 0.4833m/s
Head loss due to sudden enlargement is
(Hl)se = (v1-v2)2/2g
(Hl)se = (5.92-0.4833)2/2*9.8
= 1.506 m
Ans 3. Given: Smaller pipe Dod = 25mm, t = 2mm
larger pipe Dod = 80mm, t = 2.8m, v1 = 3m/s, Ɵ= 20.
Firstly, we calculate inner diameter of smaller pipe
Di1 = Dod -2t
= 25-2*2 = 21mm = 0.021m
Then calculate inner diameter of larger pipe
Di2 = Dod -2t
= 80-2*2.8 = 74.4 mm = 0.0744m
Now, D2/D1 = 74.4/21 = 3.543
Head loss due to gradual...
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