cem 350 hvac distribution systems sizing 10 2016
TRANSCRIPT
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HVAC Distribution & Sizing
CEM 350 Facility Systems DesignHVAC Distribution Systems
Diffuser Selection and LayoutDuctwork Sizing
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HVAC Distribution Systems
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Positive Pressure (supply)
Negative Pressure (return or exhaust)
Distribution System Distribution System Plans SymbolsPlans Symbols
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Arrow indicates air flow direction
Distribution System Distribution System Plans SymbolsPlans Symbols
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Distribution System Distribution System Plans SymbolsPlans Symbols Flow patterns
1-way
4-way
2-way
3-way
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Thermostat
Smoke/FireDamper
Distribution System Distribution System Plans SymbolsPlans Symbols
T
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Double Line
Single Line
Distribution System Distribution System Plans SymbolsPlans Symbols
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Plan
Section
Distribution System Distribution System Plans SymbolsPlans Symbols
z z
z z12 x 16
16 x 12
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Distribution Distribution
System System PlansPlansDouble Line
Single Line
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Distribution Distribution
System System PlansPlans
Double LineSingle Line
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CeilingCeilingPlenum Plenum
PlansPlansShows duct path from distribution network to supply diffuser or return register
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Diffuser Selection and Layout
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Diffuser SelectionDiffuser SelectionDiffuser Selection Criteria Air flow Throw Noise Criteria (NC) Level Appearance
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Diffuser SelectionDiffuser SelectionAir FlowThrowNC Level
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Throw:
Distance of air movement
Avoid Gaps and overlap Obstructions/deflectors
Diffuser SelectionDiffuser Selection
Velocity (fpm)150 100 50
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Diffuser LayoutDiffuser Layout1. Use Room Sensible Load (no latent, no ventilation) to determine air flow
Qs=1.08 x CFM x ΔT
where T=|Tsa-Tra|
thus
CFM= Qs (1.08 x ΔT)
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Diffuser LayoutDiffuser Layout2. Define Supply Air temperatures
Heating:Tsa range is 90-110ºFTra=68ºF
Cooling: Tsa range is 45-55ºFTra=78ºF
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Diffuser LayoutDiffuser Layout3. Define ΔT
Heating: ΔT=|110-68|=42ºF
Cooling: ΔT=|55-78|=23ºF
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Diffuser LayoutDiffuser Layout4. Determine Air Flow (CFM)
CFMhtg= Qs (1.08 x ΔThtg)
CFMclg= Qs (1.08 x ΔTclg)
Larger result determines air flow
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Diffuser LayoutDiffuser Layout5. Revise discharge air temperature to match required air flow
CFMpeak= Qs (1.08 x |Tsa-Tra|)
solve for Tsa
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Diffuser LayoutDiffuser Layout6. Select diffuser layout
Regular patternUniform coverage
Avoid “short circuiting” with exhaust/return registers
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Diffuser Layout ExampleDiffuser Layout ExampleOffice space with overhead heating and cooling supply
NC level 35
8’
16’
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Diffuser Layout ExampleDiffuser Layout ExampleHeating Qs= 11,800 Btuh @ 68ºFCooling Qs=8,600 Btuh @ 78ºF
CFMhtg= Qs (1.08 x ΔT) =11,800/(1.08 x 42)=260 CFM
CFMclg= Qs (1.08 x ΔT) =8,600/(1.08 x 23)=346 CFM
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Diffuser Layout ExampleDiffuser Layout ExampleRevise Heating Tsa
CFMpeak= Qs (1.08 x ΔT)
=346=11,800/(1.08 x |Tsa-68|)
Tsa=99.6ºF
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Diffuser Layout ExampleDiffuser Layout ExampleDefine Pattern
346 Cfm
Round up to 0 or 5 cfm
1@350 cfm2@175=350 cfm3@115=345 cfm4@90=360 cfm
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Define Pattern
346 Cfm
Round up to 0 or 5 cfm
1@350 cfm 2-way2@175=350 cfm 4-way3@115=345 cfm 3-way4@90=360 cfm 2-way
Diffuser Layout ExampleDiffuser Layout Example
8’
16’
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Diffuser SelectionDiffuser SelectionNC 35 Air FlowThrow
Select8” Rd4-way
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Define Pattern
346 Cfm
2@175=350 cfm 4-way
Diffuser Layout ExampleDiffuser Layout Example
8’
16’
4’4’
4’
4’
4’
4’
4’
4’
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Return Register SelectionReturn Register SelectionSelection Criteria Air flow Noise Criteria (NC) Level Appearance
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Return Register SelectionReturn Register SelectionAir FlowNC Level
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Avoid Short circuiting with supply diffusers Locating in visually obtrusive location
Return Register LayoutReturn Register Layout
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Define Pattern
Supply=350 cfm
Return1@350=350 cfm
Return Register LayoutReturn Register Layout
8’
16’
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Return Register SelectionReturn Register SelectionAir Flow 350 cfmNC Level 35
Select10” x 8”350 cfmNC 27db
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Define Pattern
Supply=350 cfm
Return1@350=350 cfm10” x 8”NC 27db
Return Register LayoutReturn Register Layout
8’
16’
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Ductwork Sizing
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Ductwork SizingDuctwork SizingVolume (Q) is a function of cross sectional area (A) and velocity (V)
Q=AV
however, momentum, friction and turbulence must also be accounted for in the sizing method
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MomentumMomentumAs air leaves fan, centrifugal motion creates momentum
FAN
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FrictionFrictionAs air moves along a duct, friction slows the velocity at the edges
FAN
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TurbulenceTurbulenceAs ducts change direction or cross-sectional dimensions, turbulence is created
FAN
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Static PressureStatic PressureForce required to overcome friction and loss of momentum due to turbulence
As air encounters friction or turbulence, static pressure is reduced
Fans add static pressure
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Pressure MeasurementPressure MeasurementStatic pressure is measured in inches of force against a water column
Inches-water gauge
Positive pressure pushes airNegative pressure draws air
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Pressure MeasurementPressure MeasurementStraight ducts have a pressure loss of
“w.g./100’
based on diameter and velocity
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Friction Friction Loss ChartLoss Chart
What is the pressure loss/100 ft in a 12” diameter duct delivering 1000 cfm of air? Velocity?
0.2”/100 FT
1325 fpm
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Equivalent LengthEquivalent LengthDescribes the amount of static pressure lost in a fitting that would be comparable to a length of straight duct
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Ductwork ComparisonDuctwork ComparisonRound ductwork is the most efficient but requires greater depth
Rectangular ductwork is the least efficient but can be reduced in depth to accommodate smaller clearances
Avoid aspect ratios greater than 5:1
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Equal Friction MethodEqual Friction MethodPresumes that friction in ductwork can be balanced to allow uniform friction loss through all branches
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Equal Friction MethodEqual Friction Method1. Find effective length (EL) of longest run2. Establish allowed static pressure loss/100’ΔP=100(SP)/EL3. Size ducts4. Repeat for each branch
Note: velocity must be higher in each upstream section
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Equal Friction Method Equal Friction Method ExampleExample
Size ductwork serving office diffusers from earlier example
Elbow equivalent length: 10’Straight fitting equiv. length: 5’AHU connection: 50’
AHU
4’ 4’
6’ 6’
30’ 12’ 8’
6’
175 cfm (typ.)
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Equal Friction Method Equal Friction Method ExampleExample
Supply Diffuser pressure loss: 0.038”Return Register pressure loss: 0.159”
Fan: 0.535”w.g. (75% for supply)
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Equal Friction MethodEqual Friction Method1. Find effective length of longest run
Identify longest runLabel duct sections
AHU
4’ 4’
6’ 6’
30’ 12’ 8’
6’
175 cfm (typ.)
1
234
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WorksheetWorksheetDuct Actual Equiv Effective Air ΔP Duct AirSection Length Length Length Vol. ”/100’ Diam Velocity
1 6’ 10’ 16’ 1752 8’ 5’ 13’ 1753 12’ 5’ 17’ 3504 30’ 50’ 80’ 700
56’ 70’ 126’
AHU
4’ 4’
6’ 6’
30’ 12’ 8’
6’
175 cfm (typ.)
1
234
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Equal Friction MethodEqual Friction Method2. Establish allowed static pressure loss/100’
Fan SP: 0.533”-Supply Diff: 0.038”-Return Reg: 0.159”Available: 0.336”x 0.75= 0.252”
ΔP/100’=100(SP)/EL= 100(.252)/126= 0.2”/100’
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WorksheetWorksheetDuct Actual Equiv Effective Air ΔP Duct AirSection Length Length Length Vol. ”/100’ Diam Velocity
1 6’ 10’ 16’ 175 0.22 8’ 5’ 13’ 175 0.23 12’ 5’ 17’ 350 0.24 30’ 50’ 80’ 700 0.2
56’ 70’ 126’
AHU
4’ 4’
6’ 6’
30’ 12’ 8’
6’
175 cfm (typ.)
1
234
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Equal Friction MethodEqual Friction Method3. Size ducts
RR-6
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Equal Friction MethodEqual Friction Method3. Size ducts1 175cfm 7” diam @ 620 fpm2 175cfm7” diam @ 620 fpm3 350 cfm 9” diam @ 800 fpm4 700 cfm 12” diam @ 900 fpm
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WorksheetWorksheetDuct Actual Equiv Effective Air ΔP Duct AirSection Length Length Length Vol. ”/100’ Diam Velocity
1 6’ 10’ 16’ 175 0.2 7” 620 fpm2 8’ 5’ 13’ 175 0.2 7” 620 fpm3 12’ 5’ 17’ 350 0.2 9” 800 fpm4 30’ 50’ 80’ 700 0.2 12” 900 fpm
56’ 70’ 126’
AHU
4’ 4’
6’ 6’
30’ 12’ 8’
6’
175 cfm (typ.)
1
234
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