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 Homework Statement:

A compressor delivers 300 s–1 of free air into a pipe at a pressure of
6 bar gauge. Using the pressure drop formula: Pressure drop = 800lQ^2/Rd^5.31. Calculate the minimum diameter of pipe if the pressure drop in a system is
to be limited to 0.3 bar when is delivered through a pipe of equivalent
length 160 m .
 Relevant Equations:
 pressure drop = 800lQ^2/Rd^5.31
Hello.
My attempt at the solution is as follows:
l = 160m
Q= 300 ls1
R= p2/p1= 6+1.01/1.01 = 6.94 (2dp)
d = Unknown
Pressure drop = 0.3 Bar
0.3 = 800*160*300^2/6.94*d^5.31
0.3 = 1.152x10^10/6.94*d^5.31
0.3 ( 6.94*d^5.31) = 1.152x10^10
6.94*d^5.31 = 1.152x10^10/0.3
d^5.31 = (1.152x10^10/0.3)/6.94
d= 5.31 root (1.152x10^10/0.3)/6.94
d= 394984.81 mm (I think mm is correct?)
d= 394.98 M
I'm struggling to see how this answer can be correct as an internal diameter of a pipe being so large seems ridiculous to me. Any help as to where I've gone wrong would be appreciated. Thank you!
My attempt at the solution is as follows:
l = 160m
Q= 300 ls1
R= p2/p1= 6+1.01/1.01 = 6.94 (2dp)
d = Unknown
Pressure drop = 0.3 Bar
0.3 = 800*160*300^2/6.94*d^5.31
0.3 = 1.152x10^10/6.94*d^5.31
0.3 ( 6.94*d^5.31) = 1.152x10^10
6.94*d^5.31 = 1.152x10^10/0.3
d^5.31 = (1.152x10^10/0.3)/6.94
d= 5.31 root (1.152x10^10/0.3)/6.94
d= 394984.81 mm (I think mm is correct?)
d= 394.98 M
I'm struggling to see how this answer can be correct as an internal diameter of a pipe being so large seems ridiculous to me. Any help as to where I've gone wrong would be appreciated. Thank you!