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Piping and Ductwork Systems.
Piping systems are used to distribute liquids. Ductwork systems are used to distribute gases. The
mechanical system used to move the liquid is a pump and the mechanical system used to move the
gas is a fan. Pumps and fans are driven by electric motors.
Fluids
The term “Fluid” consists of gases and liquids. Fluids have a definite mass and volume at a given
temperature and pressure. They have no consistent shape when it is not confined a container. They
cannot sustain shear (lateral) stress under equilibrium conditions. They cannot be torn, fractured or
broken into smaller pieces.
Gases are fluids that do not have a definite volume. A gas has no shape and it assumes the volume
of the container that it is confined in. Gases can be compressed. They are affected by temperature
and pressure. The gas volume in an enclosed container is the container volume. Two containers of
different volumes can contain the same mass of gas.
Liquids are fluids that have a definite volume of their own that is independent of the shape and
volume of the container. When a liquid is placed in a container, it assumes the shape of the
container but the volume and mass remain the same under constant temperature and pressure.
Liquids can be considered non-compressible. The volume will not change significantly under
pressure. The volume of the liquid can change appreciably at different temperatures.
Pressure (P):
Pressure is the force per unit area. The pressure exerted by a force or weight of 100 lbs on 10 ft2 is
(100 lbs / 10 ft2) 10 pound per square foot (1 psf). The same weight resting on 40 ft2 is (100 lbs /
40 ft2) is 2.5 pounds per square foot (2.5 psf) or 2.5 lbs on 144 square inches = 0.01736 pounds
per square inch (psi).
S o lid
1 0 0 lb s w e ig h t
or
L iq u id in U rn
1 0 0 lb s w eig h t
A re a = 1 0 ft2
P ressu re = 1 0 0 /1 0 = 1 0 lb s/ft2
Instructor: Varkie C. Thomas, Ph.D., P.E.
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A rea = 4 0 ft2
P re ssu re = 1 0 0 /4 0 = 2 .5 lb s/ft2
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Density (d) is the mass per unit volume. Example: Density of water is approximately 62.4 pounds
per cubic foot (I-P) and 1000 kilograms per cubic meter (S-I) at normal pressure and temperature.
Density of Water = 62.4 lbs / ft3 = 1000 kgs / m3
Cube
Weight of Water
L = Length = 1 foot or 1 meter
1 ft3 = 62.4 lbs
W = Width = 1 foot or 1 meter
1 m3 = 1000 kgs
H = Height = 1 foot or 1 meter
A = Area of each of 6 surfaces =
1 square foot or 1 square meter
V - Volume of Cube = 1 cubic foot (ft3) , cubic meter (m3)
W
H
L
Pressure Units as Height of Column of Water
Figure ? shows that a measuring scale for pressure can be the height of a column of liquid. For
example the density of water is 62.4 lbs/ft3. The pressure at the bottom of a tank with any surface
areas and containing water to a height of 1 foot is 62.4 lbs/ft2 and at a height of 4 feet it is 4 times
the pressure that of 1 foot. The first tank in the figure contains 500 lbs of water but the pressure at
the bottom of the tank of height 2 feet is 2 x 62.4 = 128.4 lbs/ft2. The second tank contains 250 lbs
of water but the pressure at the bottom of the tank of height 4 feet is 4 x 62.4 = 249.6 lbs/ft2.
Pressure can therefore be expressed as the height of a column of water. In the case of liquids
(water) and pumping systems it is measured in feet of water. However, the pressure of steam is
measured in psi. In the case of gases (air) and fan systems it is meaured in inches of water.
Density of W ater
o
at 60 F ( lbs/ft3 ) =
at 17 o C ( kgs/m3 ) =
Press
H2O
H2O
H2O
Prss
Prss
In. Hg.
Col
Col
Col
lbs/in2
kg/cm2
(ft)
0.09
1.20
0.26
0.44
1'
62.4
1000
(m)
(psi)
Kpa
0.1
0.03
0.04
0.00
3.60
0.3
0.09
0.13
0.01
6.00
0.5
0.15
0.22
0.02
0.62
8.40
0.7
0.21
0.30
0.02
0.88
12.0
1.0
0.30
0.43
0.03
8.82
120
10
3.05
4.33
0.30
29.9
407
33.9
10.3
14.7
1.03
100
30.5
43.3
3.04
150
45.7
65.0
4.57
200
61.0
86.6
6.09
350
107
152
10.7
600
183
260
18.3
(in)
1'
2'
2'
4'
2'
Atmos.
Area (A) = 2' x 2' = 4 ft2
Volume (V) = 2' x 2' x 2' = 8 ft3
W eight (W ) = 62.4 x 8 = 499.2 lbs
Pressure (P) = W /A =
499.2/4 = 124,8 lbs/ft2
2' H2O = 124.8 lbs/ft2
1' H2O = 124.8 x 1/2 = 62.4
Instructor: Varkie C. Thomas, Ph.D., P.E.
Area (A) = 1' x 1' = 1 ft2
Volume (V) = 1' x 1' x 4' = 4 ft3
W eight (W ) = 62.4 x 4 = 250 lbs
Pressure (P) = W /A =
249.6/1 = 249.6 lbs/ft2
4' H2O = 249.6 lbs/ft2
1' H2O = 249.6 x 1/4 = 62.4
Skidmore, Owings & Merrill LLP
Press.
1 ft H2O to lbs/in2
1 ft H2O to kgs/cm2
1 ft H2O to inch Hg.
1 foot to meters
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Atmospheric Air Pressure
The term "air" when referred to in comfort air-conditioning consists of water vapor and dry air
(DA) which consists of the rest of gases in the mixture of gases. Steam (water, H2O) is not a
mixture of different substances but a stand-alone chemical compound.
The surface of the earth is covered with air. Like all object masses, the mass of air is attracted
towards the planet earth by gravity. Traces of air can exist up to 200 miles ( 1 million feet) above
the earth. However, 90 percent of the mass of atmospheric air is within 10 miles (53,000 feet,
16,100 meters) of the earth's surface.
This mass of air exerts a pressure on the surface of the earth and this is called atmospheric air
pressure. On average this pressure at sea level is approximately 14.7 pounds per square inch or psi
( 34 feet (10.3 meters) of water, 30 inches (750 millimeters ) of mercury, 1 kilogram per square
centimeter. Pressure is also measured in atmospheres where 1 atmosphere = 14.7 psi.
The highest point of the earth above sea level is Mount Everest in the Himalayan mountain range
separating India and Tibet in China. The height of land measured above sea level is called
elevation above sea level or altitude and the altitude of Mt. Everest is 29,029 feet (8,948 meters).
The highest capital city of a country is La Paz in Bolivia, South America which is at an altitude of
12,000 feet. Passenger aircraft fly at altitudes of 20,000 to 70,000 feet and this height is still well
within the earth's gravitational pull.
The density (lbs/ft3) of air therefore decreases with the elevation above sea level of the location.
Since we breathe in the same volume of air at any elevation, the mass amount (lbs) of air, and
consequently the amount of oxygen, that we breathe in decreases with increasing elevations. The
decrease in density also affects the operation of several types of equipment that use or handle air.
These equipment are therefore rated for zero elevation and they have to be reconfigured or
reselected for higher elevations.
The pressure of liquids is measured in feet of water and the pressure of gases is measured in inches
of water. Steam is also a gas, but the pressure of steam is measured in pounds per square inch (psi).
The pressure of atmospheric air is measured in inches of mercury (in.Hg.). The choice of
measuring fluid has to do with convenience and practicality. For instance all situations (air, water,
gases, liquids) could be based on one set of units such as psi. This would produce very large or
very small values for the different situations.
The density of air also varies with the temperature as shown in the table below. Outdoor air at -20
o
F is much denser or heavier than the air at 70 oF indoor temperature. Outdoor air in winter is
therefore at a higher pressure than the air indoors. This results in the infiltration of outdoor air
through the building envelope (cracks around the windows and doors and through porous walls)
into the indoor space.
Instructor: Varkie C. Thomas, Ph.D., P.E.
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Altitude
feet
30,000
12,000
10,000
8,000
6,000
5,000
4,000
2,000
1,000
0
Pressure (weight) due
to column of air
Altitude
above
Sea Level
Press Press
psi in.Hg
4.0
8.1
9.0 18.3
10.1 20.6
10.9 22.2
11.8 24.0
12.2 24.8
12.7 25.9
13.7 27.9
14.2 28.9
14.7 29.9
Illinois Institute of Technology, Chicago
Density
lbs/ft3
0.0205
0.0461
0.0517
0.0558
0.0604
0.0625
0.0651
0.0702
0.0205
0.0753
Altitude
meters
Pressure
Kpa
9,144
3,658
3,048
2,438
1,829
1,524
1,219
610
305
0
27.58
62.05
69.64
75.15
81.36
84.12
87.56
94.46
97.91
101.35
Typical
Location
Mt. Everest
LaPaz, SA
Lima, Peru
Boulder, CO
Denver, CO
Miami, FL
Figure T em p
O
D e n s ity
T em p
D e n s ity
T em p
D e n s ity
T em p
D e n s ity
F
lb s /ft3
O
F
lb s /ft3
O
F
lb s /ft3
O
F
lb s /ft3
-6 0
0 .0 9 9
20
0 .0 8 3
70
0 .0 7 5
130
0 .0 6 7
-2 0
0 .0 9 0
32
0 .0 8 1
90
0 .0 7 2
150
0 .0 6 5
0
0 .0 8 6
50
0 .0 7 8
110
0 .0 7 0
212
0 .0 5 9
Table Gauge Pressure
Every surface or object on the earth has the weight of the atmosphere resting on it. The absolute
weight or pressure exerted by a substance is therefore the weight of the substance resting on the
weighing scale plus the weight of air covering the weighing scale. The sum of these two pressures
is called absolute pressure. The pressure exerted by the atmosphere can be ignored for everyday
situations and we can measure pressures and weights above atmospheric pressure.
The pressure that excludes atmospheric pressure is called gauge pressure. The absolute pressure
cannot be ignored in scientific calculations since the atmospheric pressure varies with altitude. The
absolute units for pressure must be used in dealing with equations such as the Gas Laws.
Absolute Pressure (psia) = Gauge Pressure (psig) + 14.7 psi (atmospheric pressure)
Gauge Pressure (psig) = Absolute Pressure (psia) - 14.7 psi (atmospheric pressure)
It is impossible to make a pressure measurement on the earth's surface unless it is made relative to
atmospheric pressure. Pressure gauges, piezometers and all pressure measuring devices indicate
gauge pressure, that is pressure above 14.7 psia. Zero psia is supposed to be a perfect vacuum. This
cannot be achieved in practice.
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Pressure of Air =
14.7 lbs/in2
Atmospheric Pressure
50 lb
Weight
50 lbs
Apples
100 in2
Weight W = 10 lbs
The atmospheric pressure of 14.7 psi cancels out
on both sides of the weighing scale
Area A = 5 in2
Gauge Pressure (PSIG) on surface A =
W / A = 10 / 5 = 2 lbs / in2
Absolute Pressure PSIA
on LHS Scale =
Absolute Pressure PSIA
on RHS Scale =
Absolute Pressure (PSIA) =
Gauge Pressure + Air Pressure =
2 + 14.7 = 16.7 lbs/in2
14.7 psi + 50 lbs / 100 in2 =
14.7 + 0.5 = 15.2 psia
14.7 psi + 50 lbs / 100 in2 =
14.7 + 0.5 = 15.2 psia
Measuring Pressure
O pen to
atm osphere
T he pressure exerted from
any point in a liquid is the
sam e in all directions
vacuum
pressure =
0 psia
A tm ospheric
P ressure =
14.7 psi
29.92 in. H g.
33.94 ft.H 2O
T op of tube is
open to atm osphere
E qual
P ressure
Line
H eight =
29.92 in. H g.
33.94 ft.H 2O
T op of tube closed and
there is no air above liquid
The instrument used to measure atmospheric pressure is called a barometer. Mercury is used to
measure this pressure since an instrument using a column of water that is 34 feet high is not
practical. The density of mercury is 849.4 lbs/ft3 compared to 62.4 lbs/ft3 of water. The principle
of measuring atmospheric pressure is shown in Figure - ?. A tube that is closed at one end is first
inverted and filled with the liquid being used to measure pressure.
The open end is next held closed and turned over and then immersed in a tank containing the same
liquid. The end is then opened. The liquid in the tube will rise until it reaches a height that has the
same equivalent pressure of the atmosphere at the level of the tank. The space above the level in
the tube is a vacuum at 0 pressure. As with measuring temperature, the mercury barometer is not
suitable for measuring very high and low pressures and other types of instruments are used.
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Duct and Pipe Sizing.
Ducts are made typically of thin lightweight galvanized steel or aluminum (10 to 22 gauge) and
they are used to distribute air and gases. Duct shape can be round, rectangular or oval. Pipes are
made of are made of steel, copper, cast iron and other heavier materials and they are used to
distribute water and liquids. Pipes are always round.
Gases and liquids are both fluids and they differ in their properties of density, kinematic viscosity
and specific heat. The pipe sizing principles, theory and equations for both ducts (gases) and pipes
(liquids) are the same. Ducts are first sized as round and then the equivalent rectangular and oval
sizes are determined to produce the same pressure drop as the round duct.
The following section shows the pipe/duct sizing and heat gain/loss equations. Pipe and duct
design computer programs use these equations. However charts (graphs) generated by the
equations are used for sizing pipes and ducts manually. In the case of rectangular ducts, tables are
provided for converting round ducts to equivalent rectangular ducts.
Pipe (and Round Duct) Sizing
The general principles of pipe sizing are described in the ASHRAE Handbook: 1985 Fundamentals,
Chapter 34, p. 34.1. The Darcy-Weisbach and Colebrook-White equations are used to calculate the
pressure drop in a pipe section due to fluid friction. The Darcy-Weisbach equation is:
⎡L⎤ ⎡ 2 ⎤
Δ h= f ⎢ ⎥ ⎢V ⎥
⎣D⎦ ⎣2 g ⎦
where
f
D
L
V
g
h =
=
=
=
=
=
head loss due to friction (ft)
friction factor, dimensionless
inside diameter of pipe (ft)
length of pipe section (ft)
average velocity (ft/sec)
acceleration of gravity (ft/sec2)
The friction factor f is a function of the pipe roughness
parameter, the Reynolds number.
R e=
, inside diameter D and a dimensionless
Dvρ
μ
ρ = fluid density at given temperature (lb/cu ft)
μ = dynamic viscosity of fluid (lb/ft sec)
Laminar flow exists where Re < 2100. For this condition, the friction factor f is obtained from:
where
f=
64
Re
Where Re > 2100, the flow is assumed to be turbulent. The Moody diagram that relates the friction
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factor f with Reynolds number and the relative roughness /D is shown in ASHRAE Handbook: 1985
Fundamentals, p. 2.10, fig. 13. The Colebrook-White equation for turbulent flow, shown in Equation
20, is used for the friction factor f.
⎡
9.3
= 1.14 + 2 log 10 (D / ε ) - 2 log 10 ⎢1 +
f
⎢⎣ R e ( ε / D )
ε = absolute roughness of inside pipe wall (ft)
1
where
⎤
⎥
f ⎥⎦
For fully rough flow, the value of Reynolds number is high and the last term in Equation 20 can be
neglected. Equation 21 can be used in its place.
⎡D⎤
= 1.14 + 2 log10 ⎢ ⎥
⎣ε ⎦
f
1
Equation 20 is used to calculate the friction factor f for turbulent flow. The Newton-Raphson iterative
method is used to solve for f since f appears on both sides of the equation. The initial value of f for this
iteration is obtained from Equation 21. As Reynolds number increases, the values from Equation 20
approach those that would be obtained by applying Equation 21 directly for fully rough flow.
Pipe sizing and the size of each pipe section depend on your criteria. The criteria can be based on the
limits for pressure loss per 100 ft, maximum velocity or maximum flow. The sizing iteration consists of
comparing the pressure drop/100 ft, velocity or flow against the limits you specify. This is done for
each standard pipe size, beginning with the smallest size and continuing until a size is found that meets
the criteria. When the maximum pipe size limit is reached, you must use your engineering judgment to
decide whether to:
• maintain the sizing criteria and increase the pipe size above the maximum limit or
• maintain the pipe size limit and calculate the new criteria for this size.
Liquid
Temperature oF
Properties
-30
WATER
Density (lb/cu ft)
Kinematic viscosity (sq ft/sec)
Specific heat (Btu/lb oF)
GLYCOL
Density (lb/cu ft)
Kinematic viscosity (sq ft/sec)
Specific heat (Btu/lb oF)
AIR
Density (lb/cu ft)
Kinematic viscosity (sq ft/sec)
Specific heat (Btu/lb oF)
67.98
595.0
0.70
0
67.55
190.0
0.73
30
60
100
150
212
62.4220
.0
1.0
62.37
12.17
1.0
62.00
7.39
1.0
61.20
4.76
1.0
59.81
3.2
1.005
67.11
85.4
0.76
66.55
48.6
0.78
65.74
22.6
0.81
64.68
12.5
0.85
63.12
6.4
0.88
0.075
?
0.24
Fig.: Properties of Liquids (Water, Glycol and Brine)
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Nominal Diameter
Roughness Factor
Minim
Maxim
Closed
Open
Steel: Schedule 40
.250"
24"
.00015
S80
Steel: Schedule 80
.250"
24"
ST
Steel: Standard Weight
.250"
XS
Steel: Extra Strength
Cl 125
Pipe
Material
Description
Density
Conduct
-ivity
S40
.0018
489.02
2.5
.00015
.0018
489.02
2.5
36"
.00015
.0018
489.02
2.5
.250"
36"
.00015
.0018
489.02
2.5
Cast Iron: 125 psi
3"
48"
.00085
.0018
483.84
0.767
Cl 175
Cast Iron: 175 psi
3"
48"
.00085
.0018
483.84
0.767
Cl 250
Cast Iron: 250 psi
6"
36"
.00085
.0018
483.84
0.767
CK
Copper: Type K
.250"
12"
.000005
.000005
558.14
16.33
CL
Copper: Type L
.250"
12"
.000005
.000005
558.14
16.33
CM
Copper: Type M
.375"
12"
.000005
.000005
558.14
16.33
PVC
Plastic: PVC
.500"
12"
.000005
.000005
94.7
0.1
CPVC
Plastic: CPVC
.500"
6"
.000005
.000005
105.7
0.079
Fig: Properties of Pipe Materials used in Buildings
Thermal Analysis of Pipes
The heat gain/loss and temperature calculation options apply to liquids and steam only.
You can choose between two options for determining the fluid temperature in each pipe section. In the
first option, you can assume an average supply and return fluid temperature for all supply and return
sections. This data is used to calculate the fluid properties. An example of the use of average
temperatures is 200 oF supply and 160 oF return for a hot water heating system.
In the case of uninsulated pipes and high temperature steam and hot water, the supply temperature at
each terminal must be calculated. This is done by calculating the entering and leaving temperature of
each supply section, beginning with the initial temperature of the first section. The first section must be
identified. In the case of liquids, the first section is the section downstream of the pump station.
The entering temperature of any supply section is the leaving temperature of the upstream section. You
can reset the leaving section temperature for sections that have primary equipment.
The following equations are used to calculate liquid and steam heat gains and losses:
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T avg - T amb
Qs=
R s log e
K1
where
College of Architecture
Qs
Tavg
Tamb
Ri
Ro
Rs
K1
K2
1/h =
Ro
Rs
R s log e
Ri +
Ro + 1 / h
K2
= rate of heat transfer per square foot of outer surface (Btu/hr sq ft)
= average temperature of section ( F)
= temperature of ambient air ( F)
= inside radius of pipe, (in.)
= outside radius of pipe, (in.)
= outside radius of insulation (in.) - Ro + insulation thickness
= thermal conductivity of pipe (Btu in./hr sq ft F)
= thermal conductivity of insulation (Btu in./hr sq ft F)
outside surface resistance (hr sq ft F/Btu in. = 0.6)
Equation 12 is based on heat flow Equations 11 and 12 in ASHRAE Handbook: 1981 Fundamentals, p.
23.8. The average temperature of the section is the mean value of the temperatures entering and leaving
the section. Since the leaving temperature is unknown, the average temperature is calculated
iteratively.
Q s = q s • As
where
Qs = total rate of heat transfer from pipe section (Btu/hr)
As = outside surface area of pipe (sq ft)
The temperature of the liquid flowing through the pipe section is obtained from Equation 14. The
procedure for determining steam temperature changes is described in Steam Piping.
dT s =
where
Q s (Btu / hr)
⎡ cu ft ⎤
⎡ lb ⎤
⎡ gal. ⎤
⎡ min .⎤
⎡ btu ⎤
x 60 ⎢
x Df ⎢
xCp ⎢
• 0.13368 ⎢
Fs ⎢
⎥
⎥
⎥
⎥
⎣ min .⎦
⎣ hr ⎦
⎣ lb ° F ⎥⎦
⎣ gal. ⎦
⎣ cu ft ⎦
dTs =
Fs
Df
Cp
change in liquid temperature in section ( F)
= flow through section (GPM)
= density of liquid (lb/cu ft)
= specific heat of liquid (Btu/lb F)
Tl=Te - d Ts
where
Tl =
Te =
temperature of fluid leaving section ( F)
temperature of fluid entering section ( F)
T l +T e
2
The average temperature Tavg in Equation 12 depends on the leaving section temperature in Equation
16. The procedure consists of initializing the leaving temperature to the entering section temperature
T avg =
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and then iterating through Equations 12 through 16 until a steady state value of Tavg occurs.
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Frictional Losses for Noncircular Ducts
All friction loss calculations are based on the equivalent hydraulic diameter. With equal length of
round and rectangular ducts, constant flow in each duct, and equal resistance to flow in both the
round and rectangular ducts, the equivalent round of a rectangular duct is calculated by:
De = 1.30
(a b )0.625
(a + b )0.250
where
De = circular equivalent of a rectangular duct for equal length, fluid resistance and air flow
(in.)
a = length of one side of duct (in.)
b = length of adjacent side of duct (in.)
The mean velocity in a rectangular and oval duct will be less than its circular equivalent.
For oval ducts, the corresponding equations are:
1.55 A0.625
De =
0.25
p
A=
π b2
+ b (a - b)
4
P = π b + 2 (a - b)
where
p
a
b
= perimeter of oval duct (in.)
= length of major axis (in.)
= length of minor axis (in.)
For both rectangular and oval ducts, the length of the sides is initially determined by the
target aspect ratio. If the resulting dimensions fall outside the minimum and maximum
allowable limits you have set, the dimensions are recalculated without using the target
aspect ratio.
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Dynamic Losses
Dynamic losses are caused by restrictions and changes in direction to the flow through a
piece of equipment (volume damper, heating coil, etc.) and duct fittings. HVAC Systems
Duct Design, SMACNA, 1985 lists the fittings available for round and rectangular ducts.
Since little dynamic loss data for oval fittings are available, the data for rectangular
fittings are used as an approximation.
Fittings
A duct fitting can occur anywhere along the length of a duct section. The program does
not limit the number of fitting types or multiples thereof per duct section. If a fitting type
is not available in the tables, its dynamic loss has to be entered as a special loss.
All the necessary engineering performance information for fittings is provided in the
Ducts Program. The engineering design effort is to locate the appropriate fitting type in
the duct network system. The duct fitting type and shape type should be compatible.
Fittings are classified as junctions, transitions, and elbows.
Junctions
Junctions are fittings which split the air stream into two or more
branches. Converging junctions join two or more air streams into one
and are basically used in a return/extract duct system. Fittings called
take-offs, tees, and wyes are in this category. Loss coefficients for
junctions are functions of the duct dimensions, air velocities and
airflow rates.
Transitions Transitions are fittings which change the duct size or shape without
changing airflow direction or airflow rate. Transitions can be converging or diverging. Loss coefficients for transitions are functions of
upstream and downstream duct velocities, angle of transition,
transition length, and Reynolds number, Re.
Elbows
Elbows are fittings which change the direction of the air stream
without changing the air quantity or velocity. The loss coefficients of
elbows are functions of the elbow radius, duct dimensions, angle of
turn, and Reynolds number, Re.
By definition, a new duct section occurs when there is a change in air quantity, velocity,
shape, duct material or duct insulation. Every duct section, therefore, begins with a
junction or transition type fitting. These fittings are commonly referred to as take-off
fittings. There is always one, and only one, take-off fitting per duct section.
Instructor: Varkie C. Thomas, Ph.D., P.E.
(Spring-2003
Skidmore, Owings & Merrill LLP
ARCH-551 (Fall-2002) ARCH 552
S1-12
Energy Efficient Building Design
Chicago
College of Architecture
Illinois Institute of Technology,
Fitting Losses
Methods of computing the energy losses from the various fitting types are based on
information found in ASHRAE Handbook: 1981 Fundamentals p. 33.28 through 33.50
The fluid resistance coefficient represents the ratio of the total pressure loss to the
dynamic pressure at the referenced cross-section O:
Co =
Δ Pt
⎡V ⎤
2
ρ ⎢
⎥
⎣ c f ⎦o
=
Δ Pt
Pv,o
where
cf
= conversion factor (1097)
ΔPt = total losses of fitting in terms of total pressure (in. of water)
= overall fluid resistance coefficient referenced to section O, dimensionless
Co
V
= average velocity to which coefficient Co is referenced (ft/min)
Pv,o = velocity pressure (in. of water)
ρ
= fluid density (lbm/cu ft)
For entries, exists, elbows and transitions, the fitting total pressure loss at section is
calculated by:
Δ Pt = C o P v,o
where the subscript o is the cross section at which the velocity pressure is referenced.
For converging and diverging flow junctions, the total pressure loss through the main
section is calculated as:
Δ Pt = C c,s P v,c
For total pressure losses through the branch section
Δ Pt = C c,b P v,c
Instructor: Varkie C. Thomas, Ph.D., P.E.
(Spring-2003
Skidmore, Owings & Merrill LLP
ARCH-551 (Fall-2002) ARCH 552
S1-13
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
where
Cc,s = main local coefficient, dimensionless
Cc,b = branch local coefficient, dimensionless
Pv,c = velocity pressure at the common section, c
A tee nomenclature is shown in Fig. 1-7 for converging and diverging flow junction where,
Δ Pt (s to c) = C c,s Pv,c | Δ Pt (c to s) = C c,s Pv,c
Δ pt (b to c) = C c,b Pv,c | Δ Pt (c to b) = C c,b Pv,c
(reprodu
ced with
permissi
on from ASHRAE Handbook: 1981 Fundamentals,
Fig. 6, p. 33.8)
Duct Material
Uncoated Carbon Steel, Clean
Aluminum
Galvanized Steel, Hot Dipped
Stainless Steel
Fibrous Glass Duct, Rigid
Flexible Duct, Metallic
Fibrous Glass Duct Liner
Roughness factor ft
0.00015
0.0002
0.0005
0.0003
0.0003
0.007
0.015
Fig: Duct Material Absolute Roughness
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
Courses ARCH-551 and ARCH 552
S1-14
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
Sizing Methods
The sections of supply duct systems can be sized using one of the following methods:
equal friction
static regain
total pressure
velocity reduction
constant velocity
(Ducts & Pipes)
(Ductwork only)
(Based on judgment, Ducts & Pipes)
Equal friction and constant velocity methods are used for manual duct sizing. Static regain and
total pressure methods require the use of computer programs. Velocity reduction is a “rule of
thumb” method. Constant velocity is used for short runs of ductwork such as flexible ducts.
Flexible ducts are always considered round in shape.
Equal Friction Sizing Method
This is the most commonly used method for duct and pipe sizing. It can be done manually using
charts generated for a particular gas or liquid fluid (using the properties of the fluid) and a
particular duct or pipe material (using the properties of the pipe material).
In the equal friction method, the system is sized for a constant pressure loss per unit length of
duct. The equal friction method can be used for the design of supply and extract duct systems.
The equal friction sizing method works iteratively between the minimum and maximum velocity
limits to determine a duct size that results in the specified pressure loss per unit length.
Static Regain Sizing Method
For this method, a section of the duct system is sized so that the increase in static pressure due to
velocity reduction from its upstream section, offsets the friction loss in the section.
The advantage of this method is that all sections have approximately the same entering static
pressure, thereby simplifying outlet selection. One disadvantage might be seen in networks with a
large pressure drop in a section near the fan outlet. The velocity could be reduced to the minimum
within a few sections in such a way that all the ductwork downstream would be sized using
minimum velocity. Another disadvantage could stem from specifying a very low minimum
velocity. Ducts would then tend to be very large at the end of long branch runs. The sizing
method does not account for the total mechanical energy supplied to the air by the fan.
Total Pressure Sizing Method
The total pressure sizing method is a variation of the static regain method. The total pressure of
any point in the ductwork represents the actual energy of the moving air at that point. The
advantage of this method is that it accounts for all mechanical energy losses in a system. The
system design does not have to be dependent on an assumed velocity at the fan outlet.
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
Courses ARCH-551 and ARCH 552
S1-15
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
Piping Systems and Networks
Piping systems can be open or closed. An open system is affected by atmospheric pressure. The flow
in open condensate return and plumbing drainage systems is gravitational. The flow in an open
cooling tower water system is forced. The pipe sections of an open network system are either all
supply or all return. A closed network system includes both supply and return sections.
Open Network Systems
Pumps and static heads are used to force circulation in Open Systems except as noted
• Steam supply
• Open steam condensate return (gravitational)
• Closed condensate return
• Cooling tower water (partially gravitational)
• Fuel oil supply
• Fuel oil return
• Gasoline supply
• Fuel gas supply
• Domestic cold water supply
• Domestic hot water supply
• Storm sewer return (gravitational)
• Sanitary sewer return (gravitational)
• Sanitary vents (gravitational)
Closed Network Systems
Closed systems apply mainly to liquids. Examples of closed network systems include
• Chilled water
• HVAC hot water
• High temperature hot water
• Glycols, Brines
Supply-Return Systems
The flow arrangements in closed piping networks can be:
• Two-pipes, direct-return
• Two-pipes, reverse-return
• Primary-secondary
Primary-Secondary Systems
Network arrangements can be combinations of direct return and reverse return loops.
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
Courses ARCH-551 and ARCH 552
S1-16
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
CLOSED PIPING SYSTEM
Pipe-A
Height of
Water in
Pipe-A =
H1
OPEN PIPING SYSTEM
The pressure
due to the
column of
water in vertical
Pipe-A
cancels out
the pressure
due to the
column of
water in vertical
Pipe-B
Pressure due to
the water in these
2 pipes cancel out
The pump must
lift the water thru
this height (feet)
STATIC LIFT
Pipe-B
Height of
Water in
Pipe-B =
H2
Pressure due to
the water in these
2 pipes cancel out
Closed Piping System Pump Total Pressure consists of
(1) frictional losses (resistance) due to pipe inside surface
(2) dynamic losses (resistance) through fittings & valves
(3) equipment losses (resistance) through coils, cillers, etc.
Air Out
Open Piping System Pump Total Pressure consists of
(1) frictional losses (resistance) due to pipe inside surface
(2) dynamic losses (resistance) through fittings & valves
(3) equipment losses (resistance) through coils, chillers, etc.
(4) STATIC LIFT
Fan
COOLING TOWER
STATIC
LIFT
Air In
Air In
ROOF
Basin
95oF
Chiller Condenser
85oF
Pump
CWS
CWR
Condenser Water (CW) System (Open)
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
Evaporative Cooling
Courses ARCH-551 and ARCH 552
S1-17
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
Pipe and Duct Sizing using Charts
The flow (water or air) quantities in the pipe and duct sections are established by the room heating
and cooling loads. For example:
Room Sensible Heat Gain (SHG) in summer = 216,000 btu hr
Room Temp (Tr) = 75oF, Ducted Supply Air Temp (Ts) = 55oF
Supply Air CFM = SHG / [ 1.08 * ( Tr – Ts ) ] = 216,000 / [ 1.08 * ( 75 – 55 ) ] = 10,000.
Room Sensible Heat Loss (SHL) in winter = 500,000 btu hr. Room Temp (Tr) = 75oF,
Radiator: Entering Water Temp (EWT) = 200oF. Leaving Water Temp (LWT) = 180oF.
Supply Water GPM = SHL / [ 500 * ( EWT – LWT ) ] = 500,000 / [ 500 * ( 200 – 180 ) ] = 50.
GPM
Example: Below is the format of a duct (round) or pipe sizing chart. In this particular case it is a
pipe sizing chart. The flow through the pipe is 250 GPM. What is the pipe size in inches and
flow velocity in feet per second (FPS) if the pressure drop (ft) per 100 ft of pipe must not exceed
4.0. From the chart the pipe size is 4 inches and the velocity is 7 FPS.
7 ft/sec Vel
4"
250
250 gpm
Flow Velocity
4"
4" Pipe Diam
4' PD / 100' of Pipe
Pipe Size
PD' / 100'
Pressure Drop (ft) in Pipe per 100 feet of Pipe
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
4
PD' / 100'
Courses ARCH-551 and ARCH 552
S1-18
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
Pipe Sizing Criteri Schedule 40 Steel
Design: 3'/100' PD , 10 fps max vel
Nominl
Pipe
Size
Outside
Wall
Inside
Diameter Thickness Diameter
S-40 Steel
High: 5'/100' PD , 12 fps max vel
Design
Maxim: 7'/100' PD , 15 fps max vel
High
Maxim
P.D. per
Velocity
Flow
P.D. per
Velocity
Flow
P.D. per
Velocity
Flow
100 ft
(ft/sec)
(gpm)
100 ft
(ft/sec)
(gpm)
100 ft
(ft/sec)
(gpm)
(in)
(in)
(in)
0.38
0.675
0.091
0.493
3.0
0.9
0.5
5.0
1.7
1
7.0
2.5
1.5
0.50
0.840
0.109
0.622
3.0
1.6
1.5
5.0
2.1
2
7.0
2.6
2.5
0.75
1.050
0.113
0.824
3.0
2.1
3.5
5.0
2.7
4.5
7.0
3.3
5.5
1.00
1.315
0.133
1.049
3.0
2.4
6.5
5.0
3.2
8.5
7.0
3.7
10
1.25
1.660
0.140
1.380
3.0
2.6
12
5.0
3.7
17
7.0
4.5
21
1.50
1.900
0.145
1.610
3.0
3.2
20
5.0
4.3
27
7.0
5.1
32
2.00
2.375
0.154
2.067
3.0
3.8
40
5.0
4.8
50
7.0
5.7
60
2.50
2.875
0.203
2.469
3.0
4.3
65
5.0
5.7
85
7.0
6.5
97
3.00
3.500
0.216
3.068
3.0
4.8
110
5.0
6.3
145
7.0
7.6
175
3.50
4.000
0.226
3.548
3.0
5.3
160
5.0
7.0
200
7.0
8.5
250
4.00
4.500
0.237
4.026
3.0
5.8
230
5.0
7.6
300
7.0
8.8
350
5.00
5.563
0.258
5.047
3.0
6.4
400
5.0
8.3
520
7.0
10.3
640
6.00
6.625
0.280
6.065
3.0
7.7
690
5.0
10.0
900
7.0
12.2
1,100
8.00
8.625
0.322
7.891
3.0
9.0
1,400
5.0
12.0
1,900
7.0
14.1
2,200
10.00
10.75
0.365
10.02
2.7
10.0
2,500
3.8
12.0
3,000
5.8
15.0
3,700
12.00
12.75
0.406
11.94
2.1
10.0
3,500
3.0
12.0
4,200
4.6
15.0
5,200
14.00
14.00
0.437
13.13
1.9
10.0
4,200
2.7
12.0
5,100
4.1
15.0
6,300
16.00
16.00
0.500
15.00
1.7
10.0
5,500
2.3
12.0
6,600
3.6
15.0
8,300
18.00
18.00
0.562
16.88
1.5
10.0
7,000
2.0
12.0
8,400
3.0
15.0
10500
20.00
20.00
0.593
18.81
1.3
10.0
8,900
1.8
12.0
10400
2.6
15.0
13000
22.00
22.00
1.250
20.75
1.1
10.0
10500
1.6
12.0
12600
2.4
15.0
15700
24.00
24.00
1.360
22.64
1.0
10.0
12500
1.4
12.0
15000
2.2
15.0
18700
26.00
26.00
0.750
25.25
0.9
10.0
15500
1.3
12.0
18600
2.1
15.0
23300
28.00
28.00
0.750
27.25
0.8
10.0
18100
1.2
12.0
21700
2.0
15.0
27100
30.00
30.00
0.750
29.25
0.7
10.0
20800
1.1
12.0
25000
1.9
15.0
31300
32.00
32.00
0.750
31.25
0.6
10.0
23800
1.0
12.0
28500
1.8
15.0
35700
34.00
34.00
0.750
33.25
0.5
10.0
26900
0.9
12.0
32300
1.7
15.0
40400
36.00
36.00
0.750
35.25
0.4
10.0
30300
0.8
12.0
36300
1.6
15.0
45400
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
Courses ARCH-551 and ARCH 552
S1-19
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
Water Distribution Piping Systems Analysis
The following is a summary procedure for designing a piping distribution system.
Supply and return water quantities for each room have to be calculated first based on heating and
cooling loads. GPM = SHL (or SHG) / [ 500 * ( EWT –LWT ) ]
Establish the design criteria limits for designing the water distribution system. This includes:
(1) Pressure drop per length of pipe, (2) Maximum velocity limit.
Locate the terminal devices (radiators, fan-coil-units, etc.) in the rooms and assign the calculated
room water flow quantities to them.
Route piping from the pumps to the terminal devices.
Determine the water flow quantities in each pipe section.
Size the piping based on the design criteria (low, medium or high pressure/velocity systems).
PIPE SIZING
Direct Return
Room Sensible Heat Loss (SHL) = 400,000 btu/hr. EWT = 200oF, LWT = 180oF.
10 gpm
40 gpm
HWS
10 gpm
1
10 gpm
2
30 gpm
40 gpm
HWR
10 gpm
3
20 gpm
30 gpm
Sizing Criteria: 4' PD per 100' pipe length
GPM = 400,000 / [ 500* ( 200 - 180 ) ] = 40
4 Radiators at 10 GPM each
GPM
10
20
30
40
4
10 gpm
20 gpm
10 gpm
SIZE
1.25
1.50
2.00
2.00
Vel FPS
3.00
3.15
2.87
3.82
Pipe Sizing Examples
Pipe Sizing: Closed System, Schedule 40 Steel
Fill in the Blanks
ANSWERS
FLOW
GPM
10
30
100
500
1,000
3,000
DESIGN CRITERIA
Nominal
ACTUAL FLOW
PD' per
Max Vel
Pipe Size
PD' per
Max Vel
100'
ft/sec
inches
100'
ft/sec
3
5
1.5
2
5
5
5
20,000
2,000
3
6
6
6
FLOW
GPM
10
30
60
100
500
1,000
3,000
8
10
24
6
0.75
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
20,000
2,000
3.3
DESIGN CRITERIA
Nominal
ACTUAL FLOW
PD' per
Max Vel
Pipe Size
PD' per
Max Vel
100'
ft/sec
inches
100'
ft/sec
3
5
5
6
6
1.25
1.5
2
3
5
6
8
10
12
24
6
0.75
1.7
6
7
2.5
4.5
1.7
1.6
4
1.6
2.5
2.3
3
2.2
4.7
6
4.5
8.2*
5.8
6.7
12.5*
8.5
16
23
2
5
5
3
8
10
COMMENTS
Velocity too high
Try Next Size
Velocity too high
Try Next Size
Courses ARCH-551 and ARCH 552
S1-20
Energy Efficient Building Design
Heating capacity (H) of all four radiator terminals = 150,000 btu/hr.
Leaving Water Temp (LWT) = 170oF
Entering Water Temp (EWT) = 200oF
Required flow through each terminal = H / (500 * ( EWT - LWT ) = 10 gpm
CLOSED PIPING SYSTEM
Direct Return
HWS
College of Architecture, Illinois Institute of Technology, Chicago
10 gpm
10 gpm
10 gpm
10 gpm
1
2
3
4
40
30
20
40
20
10
HWR
The piping circuit loop to terminal 1 is smaller than the circuit loop to terminal 4. More water will try
to flow through terminal 1. Balancing valves have to be installed in the branches to terminals 1, 2 and 3
so that their circuit pressure drops are equal to that of terminal 4.
Pump
Reverse Return
Pipe Flow (gpm)
HWS
10 gpm
10 gpm
10 gpm
10 gpm
1
2
3
4
40
30
20
10
10
20
30
40
The piping circuit loops to all four terminals are the same. The system is self-balanced
Pump
HWR
Reverse Return
Pipe Sizes
HWS
10
30
40
10 gpm
10 gpm
10 gpm
1
2
3
4
1.25"
1.25"
1.25"
1.25"
2"
2"
1.5"
1.25"
10 gpm
1.25"
1.5"
2"
2"
Pump
HWR
2"
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
Courses ARCH-551 and ARCH 552
S1-21
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
Air Distribution Ductwork Systems Analysis
The following is a summary procedure for designing an air distribution system.
Supply or extract (return and exhaust) air quantities for each room have to be calculated first
based on heating and cooling loads and indoor air quality standards. CFM = SHG / [ 1.08 * ( Tr –
Ts ) ]
• Establish the design criteria limits for designing the air distribution system. This includes:
(1) Sizing method and associated velocity limits, (2) Ductwork dimensional criteria, (3)
Static and total pressure limits to be used in selecting fans and sizing ductwork.
• Locate the terminal devices (diffusers, registers and grilles) in the rooms and assign the
calculated room air quantities to them.
• Route ductwork from the fans to the terminal devices.
• Determine the air quantities in each duct section.
• Size the ducts based on the design criteria (low, medium or high pressure/velocity systems).
1600 cfm High Velocity Round
DUCT SIZING
Terminal Box
1600 cfm Low Velocity Rectang
200 cfm
400 cfm
400 cfm
800 cfm
Room SHG = 34,560 btu/hr, Tr = 75oF, Ts = 55oF
CFM = 34,560 / [ 1.08 * (75 - 55) ] = 1,600
No. of Diffusers = 8
CFM / Diffuser = 200
CFM
Systm Size RND Vel FPM Size RCT
9" x 6"
527
8
200
Low
12" x 6"
833
9
400
Low
22" x 8"
688
14
800
Low
18
833
30" x 10"
1,600
Low
12
1874
1,600
High
• Calculate the pressure drop in each duct section. This consists of frictional losses in the
ductwork and dynamic losses in the fittings (bends, splitters, dampers, etc.). A duct
section has a constant air quantity, constant velocity (size does not change) and constant
shape (round, rectangular or oval). A new section is created when one of these factors
change
1
2
3
4
Case-1
The ductwork airflow
circuits to all four
diffusers are exactly
the same
`
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
1
2
3
4
Case-2
The ductwork airflow
circuits to diffusers
1 and 2 are the same
and longer than the
circuits to diffusers
3 and 4
Courses ARCH-551 and ARCH 552
S1-22
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
Duct Sizing Examples
Duct Sizing : Galvanized Steel
DESIGN CRITERIA
CFM
Round
EQUIVALENT RECTANGULAR DUCT SIZING
ACTUAL FLOW
PD" per
Max Vel
Duct Size
PD" per
Max Vel
Nearest
Height
Width
Asp Ratio
Velocity
100'
ft/min
Inches
100'
ft/min
Round"
H"
W"
W/H
Rect Duct
0.1
0.1
0.5
0.5
0.1
1,000
1,000
2,000
3,600
2,500
500
2,000
5,000
20,000
50,000
60,000
25,000
200
8
12
21.9
3
1.5
30
24
60
2
7
Note: Rectangular Duct Velocity (fpm) = (CFM x 144)/(W x H)
Answers
DESIGN CRITERIA
CFM
Fill in the Blanks
Round
EQUIVALENT RECTANGULAR DUCT SIZING
ACTUAL FLOW
COMMENTS
PD" per
Max Vel
Duct Size
PD" per
Max Vel
Nearest
Height
Width
Asp Ratio
Velocity
(Basis of
100'
ft/min
Inches
100'
ft/min
Round"
H"
W"
W/H
Rect Duct
Sizing)
0.1
0.1
0.5
0.5
0.1
1,000
1,000
2,000
3,600
2,500
12
20
22
22
60
0.06
0.06
0.20
0.45
0.10
0.65
0.35
1.50
650
925
1,900
3,600
2,500
5,400
3,500
780
12.2
20.2
21.9
32.0
59.6
45.7
36.6
7.0
8
12
12
24
42
30
24
4
16
30
36
36
72
60
48
11
2
2.5
3
1.5
1.72
2
2
2.75
562
800
1,667
3,333
2,381
4,800
3,125
655
500
2,000
5,000
20,000
50,000
60,000
25,000
200
7
PD
Vel
Vel
Vel
PD & Vel
AHU : Air Handling Unit
10
Relief Air
Mixed
Air
Recirculation Ai
OA : Outdoor Air
1
2
F
3
PHC
4
CC
F
CHW
HW
PHC
CC
HC
Hum
Steam
HW, Steam
or Electric
CHW
HW, Steam
or Electric
Air System
5
HC
Supply
Air Fan
6
7
Filters
Chilled Water
Hot Water
PreHeat Coil
Cooling Coil
Heating Coil
Humidifier
7 SA : Supply Air
Hum
9
8
RA : Return Air
Return
Air Fan
8
RA : Return Air
Infiltration
(Winter)
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
Room / Space
11
Exhaust
Air Fan
11
EA : Exhaust Air
TA : Transfer Air
(from adjacent rooms)
12
Courses ARCH-551 and ARCH 552
S1-23
Energy Efficient Building Design
College of Architecture, Illinois Institute of Technology, Chicago
Toilets, Lobby, Elevators, Closets
CORE
Stairwells, Mechanical Shafts
High Velocity Ducts
Low Velocity Ducts
Terminal Box
Air-Light Diffuser
Comparison of Duct and Terminal VAV Box Layout Options
Duct Design Criteria :
Low Velocity (Downstream of Terminal Box) :
High Velocity (Upstream of Terminal Box) :
0.10
0.15
PD"/100' =
PD"/100' =
Max Vel (fpm) =
Max Vel (fpm) =
Duct
Number
1500
3000
Option-1 : Tree Layout to Single Terminal Box (TB)
10'
160'
2
1
3
4
5
6
7
Terminal Box
8000 cfm max
4000 cfm min
Shortest Run
Longest Run
8
100'
AHU
16 Diffusers:
500
250
Max cfm =
Min cfm =
PD' =
PD' =
0.3
0.1
Option-2 : Loop Layout to Multiple Terminal Boxes (TB)
Shortest Run
160'
8000 cfm max
4000 cfm min
2
3
4
5
6
150'
Longest Run
1
AHU
Terminal Box
1000 cfm max
500 cfm min
9
8
7
1010
11
Instructor: Varkie C. Thomas, Ph.D., P.E., CEM
Courses ARCH-551 and ARCH 552
S1-24