All installations and services must be performed by qualified service personnel.
70
Next, the branch of the system, which includes segments “D” and “E” can be
evaluated. To balance the system, the pressure drop for segments “D” and “E”
combined must be approximately equal to the pressure drops of segments “A”,
“B” and “C” combined. Thus,
.
CBADE
PPPPP
+∆
Using the previous pressure drop equation, the pressure drop through segment
“A” is 0.0321 in. W.G. Then, the pressure drop for the “D” and “E” branch is,
....0643.0
...0257.0...0644.0...0321.0
GWin
GWinGWinGWinPP
DE
=
++=∆+∆
The duct diameter required to pass 196 CFM @ 0.0643 in. W.G. pressure drop
is,
()
.,6.73.5
0643.0
196*17*00123174.0
2058.0
82.1
ininD
E
≅=
⎥
⎦
⎤
⎢
⎣
⎡
=
Including a 6 in., 90-degree elbow in segment “E”, the required diameter would
be,
()
.,7.14.7
0643.0
196*5.3217*00123174.0
2058.0
82.1
ininD
E
≅=
⎥
⎦
⎤
⎢
⎣
⎡
+
=
or approximately 1 inch larger than without the elbow.
The pressure drop through segment “D” should be calculated for a 5-ft. straight
section with a 7-in., 90-degree elbow. From our previous equation, the pressure
drop is about,
)
....0614.0
7
196*9.375*00123174.0
86.4
82.1
GWinP
D
=
+
=∆
The flowrate through section “D” is 392 CFM (196 CFM + 196 CFM). With the
required pressure drop, the duct diameter is,
()
..9.07.9
0614.0
392*9.375*00123174.0
2058.0
82.1
ininD
D
≅=
⎥
⎦
⎤
⎢
⎣
⎡
+
=
Recalculating the duct size considering the effective length of a 9 in., 90 degree
elbow,
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