Geoscience Reference
In-Depth Information
■
EXAMPLE 15.6
Problem:
Determine the velocity pressure (VP), when
TP
= -0.35 in. wg,
SP
= -0.5 in. wg
Solution:
TP
=
SP
+
VP
VP
=
TP
-
SP
= -0.35 - (-0.50) = 0.15 in. wg
15.6.5.2 Velocity Pressure
The velocity pressure is related to the velocity (
V
) of air in the duct. The relationship is given by
VP
d
V
= 4005
where
VP
= Velocity pressure (in. wg, measured with a Pitot tube).
d
= Density correction factor.
■
EXAMPLE 15.7
Problem:
The velocity pressure of an airstream in a lab fume hood duct is 0.33 in. wg What is the
velocity? (Density correction factor
d
is 1.)
Solution:
V
= 4005 × (
VP
/
d
)
1/2
= 4005 × (0.33/1)
1/2
= 2300 fpm
15.6.5.3 Static Pressure
Static pressure is the potential energy of the ventilation system. It is converted to kinetic energy
(VP) and other (less useful) forms of energy (heat, vibration, and noise). These are the losses of the
system.
Volume flow rate can be described by
Q
=
V
×
A
where
Q
= Volume flow rate (cfm).
A
= Cross-sectional area of duct (ft
2
).
V
= Velocity (fpm).
■
EXAMPLE 15.8
Problem:
The cross-sectional area of a duct is 0.7854 ft
2
. The velocity of air flowing in the duct is
2250 fpm. What is flow rate
Q
?
Solution:
Q
=
V
×
A
= 2250 fpm × 0.7854 ft
2
= 1770 scfm
■
EXAMPLE 15.9
Problem:
The static pressure is measured in a 10-in. square duct at -1.15 in. wg. The average total
pressure is -0.85 in. wg. Find the velocity and volume flow rate of the air flowing in the duct at STP
and
d
= 1.
Search WWH ::
Custom Search