Post on 06-Apr-2018
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DUCT DESIGN
Overview.
Duct Design Objective
To design a duct system which will fit the available space which is economical in
both first cost and operating cost, and which conducts air at satisfactory noise
level, and at proper temperature to satisfy the job requirements.
Duct design follows after the room load and air quantities have been determined.
Table-2.1.2, shows the summary of calculated values. As calculations procedure
is typical, ZONE 1 , duct design and calculation is only presented on this
report.
Steps in Duct Design
1. Locate the supply air outlets, and then select the size and type required for
proper air distribution.
2. Determine how many zones of temperature control will be required for
both perimeter and interior zones.
3. Sketch the approximate path of the supply and return ducts on the plan,
taking into account obstruction, clearances, architectural limitations, and
all possible foreseeable contingencies.
4. Size the main and all branch ducts by the following common adapted
methodology by HVAC industry.
5. Last calculate Air Pressure losses and select the fan.
Points to avoid in Duct Sizing
1. Duct should be sized properly to ensure that the air system can be
balanced to deliver the required air volume to each space
2. An oversize duct system will be difficult to balance and will drive up the
installed cost of the system.
3. Undersize duct system will create higher air pressure drops, high noise
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levels and will not deliver required airflow quantities.
A. ADMINISTRATION BUILDING
The Machine Room ( AHUs ) are located in the third floor or covered roof
deck. The building is composed of 3 stories ( Ground, First and Second
floors ) and designed to 8 zones.
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TABLE 2.1.2
CALCULATED ROOM AIR FLOW RATES OF ZONE -1
SN ROOM NO. /FLOOR
RoomDimension
L X W
( Meter )
CeilingHeight
( Meter )
Air
Flow Rates
( CFM )
1 Ground Floor 1.1 Room - 01 3.5 x 3.5 3.0 5501.2 Room - 02 8.0 x 3.5 3.0 5801.3 Room - 03 8.0 x 7.0 3.0 10001.4 Room - 04 8.0 x 3.5 3.0 5801.5 Room - 05 3.5 x 3.5 3.0 5501.6 Corridor-1 3.0 x 3.0 3.0 2251.7 Corridor-2 3.0 x 3.0 3.0 225
Sub-total Air flow 3,7102 First Floor
2.1 Room - 06 3.5 x 3.5 3.0 5502.2 Room - 07 3.5 x 3.5 3.0 4502.3 Room 08 8.0 x 3.5 3.0 5802.4 Room 09 8.0 x 3.5 3.0 5802.5 Room 10 3.5 x 3.5 3.0 4502.6 Room 11 3.5 x 3.5 3.0 5502.7 Corridor-1 3.0 x 3.0 3.0 2252.8 Corridor-2 3.0 x 3.0 3.0 225
Sub-total Air Flow 3,6103 Second Floor
3.1 Room - 12 3.5 X 7.0 3.0 11403.2 Room 13 8.0 X 7.0 3.0 17403.3 Room 14 4.0 X 3.5 3.0 6003.4 Room - 15 3.5 X 3.5 3.0 6203.5 Corridor-1 3.0 X 3.0 3.0 3403.6 Corridor-2 3.0 X 3.0 3.0 340
Sub-total Air Flow 4,780TOTAL AIR FLOW RATES 12,100
STEP 1
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Locate the supply air outlets, and then select the size and type required for
proper air distribution.
This task is a collaboration between the designing Architect which is in-charge for
the type, also for the location of air outlets or terminals ( aesthetic point of
view ) . While the HVAC engineer is to make sure that the selected type met the
required factors as follows;
1. Outlet airflow
2. Discharge velocity and throw
3. Air distribution pattern
4. Pressure losses
5. Sound level
Its more appropriate to refer to Manufacturers of air terminals data catalog in the
selection process.
Terminology
Grille : A louvered or perforated covering for an air passage
opening which can be located on a wall, ceiling or floor
Air Diffuser : A circular , square, or rectangular air distribution outlet,
generally located in the ceiling and comprised of deflecting
members discharging supply air in various directions and
planes, and arrange to promote mixing primary of primary
air with secondary room air.
Register : A grille equipped with an integral damper.
Damper : A device used to vary the volume of air passing through
an air outlet, air inlet or duct.
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Face Area : The total plane area of the portion of a grille bounded by a
line tangent to the outer edges of the openings through
which air can pass.
Effective Area : The net area of an outlet or inlet device through which air
can pass, equal to the free area times the coefficient of
discharge.
Face Velocity : The velocity obtained by dividing the air quantity by the
component face area.
Terminal Velocity : The maximum air stream velocity at the end of the
throw.
Throw : The horizontal or vertical axial distance an air stream
travels after leaving an air outlet before the maximum
stream velocity is reduced to specified terminal level.
Drop : The vertical distance that the lower edge of a horizontally
projected airstream drops between the outlet and the end
of its throw.
Noise Criterion Curves ( NC Curves ) : Curves that define the limits which
octave band spectrum of a noise source must not exceed
if a certain level of occupant acceptance is to be
achieved.
SMACNA : Sheet Metal and Air Conditioning Contractor National
Association Inc.
Attachments 01, shows the location of air outlets, sizes and type for Ground
Floor . Selection basis are shown below;
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For Ground Floor Room -02
Length of Room : 8.0 meter ( 26.25 feet )
Outlet Capacity Desired : 580 CFM
Noise Level Requirement : Not to exceed 45 NC
( Refer to Attachment 02 ) for the given value of general offices room from
SMACNA manual )
Solution in Selecting the proper register and performance data;
1. The distance from the outlet to the opposite wall is 26.25 feet . Assume the
throw is being selected to equal of the distance or approximately equal
to ; ( Refer to Attachment 03 )
Throw = 26.25 feet x 0.75 = 19.68 feet say 19 feet
2. Now using the performance data chart ( Refer to Attachment 04 )
locate the 600 CFM ( go to the higher value as there is no 580 CFM ) on
the vertical column marked CFM. Next moving horizontally using THROW
of 19 feet , a 28 x 8 outlet with a 45 0 is selected. The selection will give
a throw of 19 feet and a Total Pressure of 0.025 In. w.g. , while NC is
given not to exceed 20 NC.
3. Solving for the Static Pressure ( this datum is needed on the Total Static
Pressure calculation ) ;
Core Area = ( Nominal Length ) ( Nominal Width )
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---------------------------------------------------------------
144 in2 / ft 2
= ( 28 0.25 ) ( 8 0.25 ) / 144
= 1.49 ft2
Core Velocity = CFM / Core Area
= 580 CFM / 1.49 ft2
= 389 FPM
From Table 1 in Attachment 05 , Velocity pressure @ 389 FPM is
0.010 In. w.g. ( take the value of 400 FPM )
Static Pressure = Total Pressure Velocity Pressure
= 0.025 0.010
= 0.015 In.w.g.
4. Selected Supply Register is 28 x 8, double deflected
Air Flow Rate = 580 CFM
Static Pressure Drop = 0.015 In. w.g.
Throw = 19 ft.
NC = rated at 20 or less than 45
Drop = Adjustable due to the selection of a double
deflection register
STEP 2
Determine how many zones of temperature control will be required for both
perimeter and interior zones.
Its economical ( less route of ducts ) to separate the zone / supply of perimeter
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areas and the interior. Like ZONE-1 supplies the perimeter at NE area from
ground , first and second floor , while another ZONE supplies the opposite
perimeter side. Other Zones for the interior . In total, Main Building has 8 ZONES
or 8 AHUs.
STEP 3
Sketch the approximate path of the supply and return ducts on the plan,
taking into account obstruction, clearances, architectural limitations, and
all possible foreseeable contingencies.
Attachment 6, 7 and 8 , are the final sketch of Zone -1 ( AHU-1 ) , supply and
return ducts lay-out.
STEP 4
Size the main and all branch ducts by the following common adapted
methodology by HVAC industry.
The selection of methodology for the duct sizing is of utmost important as the
first and operational costs are dependent on it.
The method adapted in the duct sizing is the Modified Equal Friction Method.
A method which combined the advantages of the equal friction and static regain
methods. Static regain ( or loss ) due to velocity changes, has been added to the
equal friction design procedure by using fitting pressure losses calculated with
new loss coefficient in Chapter 14 SMACNA.
Duct Sizing :
Take friction loss at 0.08 in. w.g. per 100 feet of duct run ( ASHRAE &
SMACNA recommended value ) .
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Example :
Consider the Ground Floor ducting at Room -01 , ( refer to Attachment -06)
The Air Quantity is computed as 550 CFM, what will be the proper supply duct
size?
Refer toAttachment -07 , A ductculator or ductsizer gives a 14 x 8 duct size
for a 550 CFM at 0.08 in. w.g. friction loss .
Sizing the supply main duct of Ground Floor ( Attachment -06), add all the
supply CFM which is 3,710 CFM at 0.08 in.w.g. gives a 30 x 16 duct size.
( Attachment -08 )
Attachment 09, 10 and 11 , shows the duct lay-out and sizes at Ground, First
and AHU-01 at Roof Deck Mechanical Room using the sizing method shown
above.
STEP 5
Calculate Air Pressure losses and select the fan
TABLE 2.1.3
AHU -1 DUCT SIZING WORK SHEET
A B C D E F G H I J
I
T
E
Duct
Run
S
E
C
T
ITEM FLOW FRICTION
LOSS
PER
V Vp Loss
Coef
Rect.
Size
Loss
Per
Item
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M
N
O.
I
O
N(CFM)
100 FEET
w.g.(fpm) w.g.
C (In.)
w.g.
1 Plenum B A1 - Damper 12,100 - 800 - - - 0.252 - do - A2 - Filter 12,100 - 400 - - - 0.30
3 -do- A3 - C. Coil 12,100 - 500 - - - 0.404 Run B- K C - Transition 12,100 - 1,600 0.16 0.05 60 x 40
30 x 40
0.008
5 -do- D 900 Elbow 12,100 0.08 1,600 0.16 0.16 30 x 40 0.025
6 -do- D-E 8.20 Duct 12,100 0.08 1,600 - - 30 x 40 0.0065
7 -do- E 450 Elbow 12,100 0.08 1,600 0.16 0.16 30 x 40 0.025
8 -do- E-F 9.8 Duct 12,100 0.08 1,600 - - 30 x 40 0.0079 -do- F 900 Elbow 12,100 0.08 1,600 0.16 0.16 30 x 40 0.025
10 -do- F-G 3.0 Duct 12,100 0.08 1,600 - - 30 x 40 0.00211 -do- H 900 Wye 12,100
/7,320 0.08 1,600 0.16 0.40 30 x40 0.064
12 -do- H-I 12 Duct-riser 7,320 0.08 1,318 - - 20 x 40 0.010
13 -do- I 900 Wye 7,320 /
3,710
0.08 1,318 0.108 0.032 20 x 40
24 x 20 0.003
14 -do- I J 12 Duct-riser 3,710 0.08 1,113 - - 24 x 20 0.010
15 -do- K 900 Elbow 3,710 0.08 1,113 0.858 0.17 24 x 20 0.14
16 -do- L - Transition 3,710 0.08 556 0.019 0.86 24 x 20
40 x 24
0.008
17 -do- M - Dovetail 1,855 0.08 954 0.056 0.30 20 x 14 0.01718 -do- M1 - Volume
Damper
0.08 954 0,056 2 20 x 14 0.112
19 -do- P Elbow 550 0.08 707 0.031 0.15 14 x 8 0.004
20
-do- Q Air
Terminal550 0.08 353 - - 28 x 8 0.015
Sub- Total PRESSURE LOSS ( From Point A to Q ) 1.445Allowance for off sets fittings in field actual execution 0.455Building Pressure allowance range 0.05 to 0.1 in.w.g. 0.10
Total Static Pressure in.w.g. ( air side ) 2.0
5.1 Supply Fan Plenum
The static pressure losses of mixed air dampers ( A1 ), Filters ( A2 ) and
Cooling Coil ( A3 ) either be taken from SMACNA Tables or from
manufacturers data sheets.
5.1.1Mixed Air Damper Section ( A1 )
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For mixed air damper ( A1 ) from SMACNA , HVAC DUCT DESIGN
Manual , 1990 edition, page 9.4 (Attachment -12) is given at 0.25 in.w.g.
Enter this figure in column J.
Velocity of 800 fpm is taken from Attachment 17, recommended
velocity
5.1.2 Filter Section ( A2 )
A high velocity cleanable viscous impingement , flat panel type filters is
given at 0.30 in.w.g. design static pressure loss (Attachment 13 )
5.1.3 Cooling Coil Section ( A3 )
A 4 row wet- cooling coil gives a 0.40 in.w.g. static pressure loss (
Attachment -14 ).
Velocity of 500 fpm median value of 400-650 is taken fromAttachment
17
5.2 The run duct that has the remotest and most demanding air flow rates
will be the reference of the total static pressure losses.
From point A1 in the AHU to the Ground Floor , section K ( air terminal
outlet ) , refer toAttachment 6 & 11.
5.2.1 For Item 4, Transition Fitting
Lets consider here a 0.08 in. w.g. friction loss per 100 feet.
Solving for the transition fitting pressure loss
TP = C x Vp ( Equation -1 )
Where :
TP : Total Pressure, in.w.g.
C : Coefficient of fitting loss
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Vp : Velocity Pressure, in.w.g.
A1 = 60 x 40 = 2400 sq.in.
A = 30 x 40 = 1200 sq.in.
A1/A = 2400 /1200 = 2
Consider angle of 15 400 , Table 14-12, Attachment -19 , Figure A,
gives C value as 0.05.
Vp = ( 1600 / 4005 )2
= 0.16 in.w.g.
Pressure loss of transition = C x Vp
= 0.05 x 0.16
= 0.008 in.w.g.
5.2.2 For Item 5, Elbow, 900
A 12,100 CFM at 0.08 in.w.g. friction loss ductculator will give a 1,600
fpm velocity and a 30 x 40 duct rectangular size. ( within the
recommended velocity range given in Attachment - 17, having a
maximum of 2000 fpm for general offices application.
InAttachment 16, Figure F, having a R/W ratio = 2, ( 80/ 40 ) and
H/W = 0.75 ( 30 x 40 ), gives loss coefficient of 0.16 . While Velocity
pressure ( Vp) for 1600 fpm is 0.16 in.w.g. taken from Table 14-6 ,
Attachment 18. Or can be solved by ;
Velocity = 4005 ( Vp )0.5 or
Vp = ( Velocity / 4005 )2
= ( 1600 / 4005 )2
= 0.16 in.w.g.
Pressure Loss of Elbow = C x Vp
= 0.16 x 0.16
= 0.025 in.w.g.
5.2.3 For Item 6, Duct section D-E
The static pressure loss for duct is;
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SP = 8.20 ft. x 0.08 / 100 feet = 0.0065 in.w.g
5.2.4 For Item 7, Elbow, 450
As this is no 450 elbow given in available tables of coefficient , consider
the value of a 90 0 elbow pressure loss at 0.025 in.w.g.
5.2.5 For item 8 , duct section E-F
The static pressure loss for duct is;
SP = 9.80 ft. x 0.08 / 100 ft. = 0.007 in.w.g.
5.2.6 For item 9, Elbow 900
Pressure Loss = 0.025 in.w.g.
5.2.7 For Item 10, Duct section F-G
SP = 3.0 ft. x 0.08 /100 ft. = 0.0024 in.w.g.
5.2.8
For Item 11, a 900
Wye , ( in riser not shown in drawing )
Attachment 15, Figure W, solving for ;
Qb/Qc = 4.780 cfm ( air supply to whole 2nd floor )/ 12,100 ( main )
= 0.40
Ab/Ac = 30 x 18 / 40 x 30 = 540 in2 / 1200 in2
= 0.45
Plot these value in the Table 14-14 , gives C in main = 0.40
Velocity at Ac = 1600 fpm ( 40 x 30 )
Vp = 0.16 in.w.g.
Pressure Loss = C x Vp
= 0.40 x 0.16
= 0.064 in.w.g.
5.2.9 For item 12, duct ( not shown riser )
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SP = 12 x 0.08 / 100 = 0.010 in.w.g.
5.2.10 For item 13, 900 Wye ( in riser not shown in drawing )
Qb/Qc = 3,710/7,320 = 0.50
Ab/Ac = 24 x 20 / 30 x 40
= 0.40
As theres no value 0.40 Ab/Ac ( from 0.25 then 0.50 ) , by interpolation;
0.25 / 0.40 = 0.02 / x
X = 0.40 ( 0.02 ) / 0.25 = 0.032
Vp = ( 1318 / 4005 )2
= 0.108 in.w.g.
Pressure Loss = C x Vp
= 0.032 x 0.108
= 0.003 in.w.g.
5.2.11 For Item 14 , Duct riser ( drawing not shown )
SP = 12 x 0.08 / 100 = 0.009 in.w.g.
5.2.12 For Item 15, 900 Elbow
H/W = 20 / 24 = 0.83
R/W = 48 / 24 = 0.50
From Attachment 16 , Figure F , as theres no direct value for 0.83 H/W
, interpolate ;
.75/0.16 = 0.83 / x
X = 0.16 x 0.83 / 0.75
C = 0.17
Vp = ( 3,710 / 4005 )2
= 0.858 in.w.g.
SP = C x Vp
= 0.17 x 0.858
= 0.14 in.w.g.
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5.2.13 For item 16, transition fitting, as the required velocity for the branches
is 800 to 1000 fpm, we will reduce the end riser velocity of 1,310 fpm by making a
converging fitting, 24 x 20 & 30 x 16 outlet area. Thus ;
As / A = 30 x 16 / 24 x 20 = 1
From Attachment 20, Figure D, say at 300 , theres no value for
As/A = 1 , hence , assume C = 0.43 ( half value of 0.86 )
V = 3,710 / 40 x 24 / 144
= 556 fpm ( now its reduce from 1,310 to 556 fpm )
Vp = ( 556 /4005 )2
= 0.019 in.w.g.
TP = C x Vp
= 0.43 x 0.019
= 0.008 in.w.g.
5.2.14 For Item 17 , Dovetail Fitting ,Attachment 15, Figure X,
A1b / Ac = 20 x 14 / 30 x 16
= 0.58
C = 0.30
Velocity = Qs / A = 1,855 cfm / 20 x 14/144
= 954 fpm
Vp = ( 954/4005 )2
= 0.056 in.w.g.
TP = C x Vp
= 0.30 x 0.056 = 0.017 in.w.g.
5.2.15 For Item 18 , Volume Damper , Attachment 21, Figure E, parallel
blades ;
L/R = NW/ 2 ( H + W )
Where :
N : number of blades ( say 4 pc )
W : duct dimension parallel to blade axis
L : sum of damper blades length
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R : perimeter of duct
H : duct dimension perpendicular to blade axis
= 3 ( 20 ) / 2 ( 14+ 20 )
= 0.88 say 1
C = 2 ( say a 200 deflection of blade )
Velocity = 1855 cfm / 20 x 14 /144
= 954 fpm
Vp = 0.056
TP = C x Vp
= 2 x 0.056
= 0.112 in.w.g.
For item 19 , elbow, same method presented above, while item 20 air terminal
pressure drop calculation is presented earlier .
ATTACHMENTS
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Attachment - 01
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Attachment - 02
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Attachment 03
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Attachment 04
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Attachment 05
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Attachment - 06
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Attachment - 07
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Attachment - 08
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Attachment - 09
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Attachment - 10
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Attachment - 11
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Attachment - 12
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Attachment - 13
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Attachment - 14
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Attachment 15
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Attachment - 16
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Attachment 17
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Attachment 18
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Attachment 19
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Attachment 20
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Attachment 21
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