This project will incorporate the concepts that apply to the study of fluid mechanics. During this project, you will calculate flow and loss across the system at various points in the system....

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Answered 7 days AfterFeb 08, 2023

Answer To: This project will incorporate the concepts that apply to the study of fluid mechanics. During this...

Dr Shweta answered on Feb 10 2023
50 Votes
Ans 1: Given: pipe having heat exchanger and by-pass pipe are parallel therefore pressure drop across both pipes should be same.
Using GPM = 0.00006309 cubm/s and 1ft = 1/3.28 m
And formulas:
A = π/4D2 -----[1]
V = Q/A------[2]
Re = VD/v-----[3]
A = k/3.7D ------[4]
B = 5.74/Re0.9----[5]
f = 1.375/[ln(A+B)]2-----[6]
hf = f*(L +Le) *V2/2gd-----[7]
and hm = kV2/2g
here, Q = flow m3/s
k = inside roughness in m
Re = Reynold’s number
V = kinematic viscosity m2/s
F = friction factor
A = area
The calculations are performed as below:
Total losses in the pipe system = Sum of friction loss in straight pipes + Minor losses due to fittings and valves
[H = hf +hm] ----[9]
And [R = H/Q2] ----[10]
Here, R = overall resistance...
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