10.9-Nonround Conduits, fluid mech
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Nonround Conduits
10.9 onround Conduits
Previous sections have considered round pipes. This section extends this information by describing how to
account for conduits that are square, triangular, or any nonround shape. This information is important for
applications such as sizing ventilation ducts in buildings and for modeling flow in open channels.
When a conduit has a section area that is noncircular, then engineers modify the DarcyWeisbach equation, Eq.
(10.12), to use hydraulic diameter
D
h
in place of diameter.
(10.5)
Equation (10.5) is derived using the same approach as Eq. (10.12), and the hydraulic diameter that emerges from
this derivation is
(10.6)
where the “wetted perimeter” is that portion of the perimeter that is physically touching the fluid. The wetted
perimeter of a rectangular duct of dimension
L
×
w
is 2
L
+ 2
w
. Thus, the hydraulic diameter of this duct is:
Using Eq. (10.6), the hydraulic diameter of a round pipe is the pipe's diameter
D
. When Eq. (10.5) is used to
calculate head loss, the resistance coefficient
f
is found using a Reynolds number based on hydraulic diameter.
Use of hydraulic diameter is an approximation. According to White 21, this approximation introduces an
uncertainty of 40% for laminar flow and 15% for turbulent flow.
(10.7)
In addition to hydraulic diameter, engineers also use hydraulic radius, which is defined as
(10.8)
Notice that the ratio of
R
h
to
D
h
is 1/4 instead of 1/2. While this ratio is not logical, it is the convention used in
the literature and is useful to remember. Chapter 15, which focuses on openchannel flow, will present examples
of hydraulic radius.
To model flow in a nonround conduit, the approaches of the previous sections are followed with the only
difference being the use of hydraulic diameter in place of diameter. This is illustrated by Example 10.8.
EXAMPLE 10.8 PRESSURE DROP I A HVAC DUCT
Air (
T
= 20°C and
p
= 101 kPa absolute) flows at a rate of 2.5 m
3
/s in a horizontal, commercial steel,
HVAC duct. (Note that HVAC is an acronym for heating, ventilating, and air conditioning.) What is
the pressure drop in inches of water per 50 m of duct?
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Nonround Conduits
PROBLEM DEFINITION
Situation:
Air is flowing through a duct.
Find:
Pressure drop (inch H
2
O) in a length of 50 m.
Sketch:
Assumptions:
1. Fully developed flow, meaning that
V
1
=
V
2
and the velocity head terms in the energy equation
cancel out.
2. No sources of component head loss.
Properties:
1.
Air (20°C, 1 atm), Table A.2: ρ = 1.2 kg/m
3
, ν = 15.1 × 10
6
m
2
/s.
2. Steel pipe, Table 10.4:
k
s
= 0.046 mm.
PLAN
This is a case 1 problem because flow rate and duct dimensions are known. Thus, the solution is
straightforward.
1. Derive an equation for pressure drop by using the energy equation (7.29).
2. Calculate parameters needed to find head loss.
3. Calculate head loss by using the DarcyWeisbach equation (10.12).
4. Calculate pressure drop
p
by combining steps 1, 2, and 3.
SOLUTION
1. Energy equation (after termbyterm analysis)
2. Intermediate calculations
·
Flow rate equation
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Nonround Conduits
·
Hydraulic diameter
·
Reynolds number
Thus, flow is turbulent.
·
Relative roughness
·
Resistance coefficient (Moody diagram):
f
= 0.015
3. DarcyWeisbach equation
4. Pressure drop (from step 1)
Copyright ¨ 2009 John Wiley & Sons, Inc. All rights reserved.
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