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d = 10[ - 5 1,663785 × 10−3 × Q1,85 × Lt ΔΡ ΧΡ Where: d = internal diameter (mm) Q = flow rate of compressed air inside the pipe (m³/h) AP = Permissible Pressure Loss (kgf/cm²) P = System pressure (kgf/cm²) Lt = sum of the linear lengths of the pipe (L1) with the equivalents of the accidents (L2): Lt = L1 + L2

d = 10[ - 5 1,663785 × 10−3 × Q1,85 × Lt ΔΡ ΧΡ Where: d = internal diameter (mm) Q = flow rate of compressed air inside the pipe (m³/h) AP = Permissible Pressure Loss (kgf/cm²) P = System pressure (kgf/cm²) Lt = sum of the linear lengths of the pipe (L1) with the equivalents of the accidents (L2): Lt = L1 + L2

Elements Of Electromagnetics
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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Q568:

Develop a methodology for a workshop project report, aiming to calculate the diameter of the compressed air pipes in the greenhouse areas (area where only spray guns are used), polishing area (area where only polishing guns are used) and the length of the repair area piping (area where the full flow rate is applied)

The formula used is in the attached image:

d = 10[ - 5 1,663785 × 10−3 × Q1,85 × Lt ΔΡ ΧΡ Where: d = internal diameter (mm) Q = flow rate of compressed air inside the pipe (m³/h) AP = Permissible Pressure Loss (kgf/cm²) P = System pressure (kgf/cm²) Lt = sum of the linear lengths of the pipe (L1) with the equivalents of the accidents (L2): Lt = L1 + L2
Transcribed Image Text:d = 10[ - 5 1,663785 × 10−3 × Q1,85 × Lt ΔΡ ΧΡ Where: d = internal diameter (mm) Q = flow rate of compressed air inside the pipe (m³/h) AP = Permissible Pressure Loss (kgf/cm²) P = System pressure (kgf/cm²) Lt = sum of the linear lengths of the pipe (L1) with the equivalents of the accidents (L2): Lt = L1 + L2
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