Conveying Air

1. Computing pressure drops :ducts and fittings

2. Designing a duct system

3. Choose a fan

4. Designing the distribution of air in a conditioned space

Fan motor – large consumer of energy

Duct system – occupy considerable space in building

Because there are many decisions to design fan-and-duct system So it’s designed to achieve a workable system only,should optimize of life time energy cost, duct-system cost,

and building-space cost of fan and ducts.

Topic6-2 Fan and Duct Systems

การสง่ลมเย็น• ปรมิาณลมเย็น

• วิธีการจดัสง่

• ลกัษณะการกระจายของลมเย็นภายในหอ้งปรบัอากาศ

→ ภาระความเย็น

→ ออกแบบการวางระบบท่อ

ลกัษณะการใชง้าน

การออกแบบดา้นสถาปัตยกรรม การตกแตง่ภายใน

ขอ้ก าหนด (ความสบาย, กฎหมาย)

ลกัษณะการใชง้าน ลกัษณะภมูิศาสตร์

การออกแบบภายใน, ภายนอก

→ เลือกระบบปรบัอากาศ

Bernoulli’s equation: steady, irrotational, inviscid and incompressible flow

Flow of air through ducts

3

p1 + (V12/2) + gz1 = p2 + (V2

2/2) + gz2 = pT = Total pressure

All fluids have viscosity: pressure loss in overcoming friction, head loss hL

Modified B-Eq; p1 + (V12/2) + gz1 = p2 + (V2

2/2) + gz2 + ghL

To overcome the fluid friction and give sufficient air distribution, a fan is located rise as Fan Total Pressure (FTP, Pa) with air flow rate Qair (m3/s)

required power input to the fan; Pfan = QairFTP/fan

FTP = (p2 - p1) + (V22 - V1

2)/2 + g(z2 - z1) + ghL

If the static pressures p1= p2 = patm , uniform cross section (v1=v2), then FTP is equal to pressure loss due to friction: FTP = ghL = p

1→ →2

Pressure drop; straight circular duct

Pressure drop in straight ducts

2

2V

D

Lfp =

p = pressure drop, Paf = friction factor, dimensionless f(Re,/D)L = length, mD = inside diameter (ID) of duct, m = roughness of inside surface of tube , mV = velocity of fluid, m/s = density of fluid, kg/m3

Re = Reynolds number, VD/

= viscosity, Pas

2

39122141

−

+−+=

fD

Df

)/Re(

.loglog.

Colebrook’s formula (implicit)- turbulent flow

Modified formula (explicit)

2

90

745

73250

−

+=

.Re

.

.log.

Df

Moody’s chart

Example6-1 Compute the pressure drop in 15 m of straight circular sheet metal duct 300 mm in diameter when the flow rate of 20 C air is 0.5 m3/s.

1.Data: straight circular duct sheet metal0.3 m = D15 m= L0.5 m3/s = flow rate, QAir 20 CCompute pressure drop p = ?

2.Assumption- Fully developed flow- turbulent flow

4.Methods & Equationsp = f (L/D)(V2/2)f = f(Re,/D) Moody chart/formulaRe = VD/

V = flow rate/( D2/4)

5.Properties of air @20C = 1.2041 kg/m3, = 18.178E-6 Pa.s/D of sheet metal = 0.00015/0.3 = 0.0005

6.CalculationV = 0.5/(*0.32/4) = 7.07 m/s,

Re = VD/ = 140500 > 2300f = 0.25[log{/3.7D+5.74/Re0.9}]-2= 0.0196 p = f (L/D)(V2/2) = 29.5 Pa

7. Analysis & check -Re > 2300 turbulent flow, p 29.3 Pa is possible for 15 m length of 0.3m-duct

Q =0.5 m3/sD =0.3 m, L = 15 m

Moody’s chart or f = 0.25[log{/3.7D+5.74/Re0.9}]-2

6

Standard chart for estimating Pressure drop in straight, circular, sheet-metal ducts, 20 C air, absolute roughness 0.00015 m.

Graph p/L & Q @D, V

Known (Q, D) -- find p, V

Q = 0.5 m3/sD = 0.3 m

p/L = 1.97 Pa/m, V = 7.05 m/sL = 15 m -- p = 29.4 Pa

Q = 0.5 m3/s

Rectangular ducts are widely used in air-conditioning practice

Pressure drop in rectangular ducts

2

2V

D

Lfp

eq

=Equivalent diameter, Deq = 4A/ perimeter

Circular: Deq = 4( D2/4)/( D) = D

Rectangular: Deq = 4(ab)/2(a+b)graph p/L by [Deq ,V]Q = VA = V(ab) Qgraph for round duct

By pround duct = prect. duct = f(L/D)V2/2using f = CRe-0.2 = C(VD/)-0.2

Rect: Deq = 4(ab)/2(a+b), Vrect = Q/A = Q/(ab)Find Deq,f for Round: Vround = 4Q/(D2

eq,f)Deq,f = 1.3(ab)0.625/(a+b)0.25

graph p/L by [Deq,f ,Q]V = Q/A = Q/(ab) Vgraph for round duct

Example6-2 An air flow rate of 1.5 m3/s passes through a rectangular duct 0.3 by 0.5 m. Calculate the pressure drop in 40 m of straight duct using (a) Deq and (b) Deq,f .

1.Data: Rectangular duct 1.5 m3/s = Q0.3 by 0.5 m = ab40 m = LCompute p (a) Deq ,(b) Deq,f

2.Assumption-sheet metal-Air 20 C- turbulent flow

4.Methods & Equationsgraph p/L by (a) [Deq ,V] , (b) [Deq,f ,Q]Deq = 4(ab)/2(a+b) [V = Q/(ab)]Deq,f = 1.3(ab)0.625/(a+b)0.25

5.Properties – graph for circular ducts

6.Calculation (a) Deq = 0.375 m, Vactual = 1.5/(0.3*0.5) = 10 m/s[Deq ,V] → p = 3.0 Pa/mL = 40m -- p = 120 Pa

7. Analysis & check -p/L 3 Pa/m is possible for 0.3 by 0.5 m rectangular duct.

(b) Deq,f = 0.42 m, Q = 1.5 m3/s[Deq,f ,Q] → p = 3.0 Pa/mL = 40m -- p = 120 Pa

Q =1.5 m3/sab =0.30.5 m2, L = 40 m

Pressure drop in straight, circular, sheet-metal ducts, 20 C air, absolute roughness 0.00015 m.

Graph p/L & Q @D, V

graph p/L by [Deq ,V]Deq = 4(ab)/2(a+b) (a) Deq = 0.375 m, Vactual = 1.5/(0.3*0.5) = 10 m/s[Deq ,V] → p = 3.0 Pa/mQ = V(ab) Qgraph for round duct

graph p/L by [Deq,f ,Q]Deq,f = 1.3(ab)0.625/(a+b)0.25

(b) Deq,f = 0.42 m, Q = 1.5 m3/s[Deq,f ,Q] → p = 3.0 Pa/mV = Q/(ab) Vgraph for round duct

Q = 1.5 m3/s

Fitting: change in area and direction

Example : enlargements, contractions, elbows, branches, dampers, filters, register

Pressure drop in fitting 3-12 m – up to 20 m of straight duct

Type and quality of construction → influence on p

Air-pressure drop must be known for properly design

p in fitting → drag along surface (less, short length)→ momentum exchange (mostly) → sudden expansion

Pressure drop in fittings

11

Pressure loss for an incompressible fluid= Geometry of duct or fitting + Product of V2/2 group

2

2V

D

Lfp

Geometry

=

The V2/2 term

Air flows frictionless:

22

2

22

2

11 VpVp+=+

Bernoulli equation

Steady flow:2211 VAVA =

Geometry

−

=− 1

2

2

2

1

2

121

A

AVpp

p1 – p2 = pressure conversion, not pressure loss

12

Sudden enlargement

Friction loss, Revised Bernoulli equation:

losspVpVp

++=+22

2

22

2

11

Momentum equation:

( ) ( ) 1112222221 AVVAVVApAp −=−

Difference in force = rate of change of momentum; dF = d(mV)

Control volume

loss

2

2

1

2

121 1

2p

A

AVpp +

−

=−

geometry

loss

2

2

1

2

1 12

−=

A

AVp

( ) ( )

−=−

2

12

1

2

221A

AVVpp

2

2

12

1

2

2

=

A

AVV

( )

−

=−

2

1

2

2

12

121A

A

A

AVpp

Borda-Carnot equationAgree sufficiently well with experimentUsed for duct-system design 13

( )

−

=+

−

2

1

2

2

12

1loss

2

2

1

2

1 12 A

A

A

AVp

A

AV

Sudden enlargement

Example 6-3 Air at standard atmosphere pressure and a temperature of 20 C flowing with a velocity of 12 m/s enters a sudden enlargementwhere the duct area doubles. What is the increase in static pressure of the air as it passes through the enlargement?

1.Data: Air @1 atm, 20 C12 m/s = V1

Duct area doubles = A2/A1 = 2Compute p2 - p1 = ?

2.Assumption : fully developed turbulent flow

4.Methods & Equations- Sudden enlargement- Borda-Carnot equation – ploss = f(, V1, A1/A2) - Revised Bernoulli equation

losspA

AVpp +

−

=− 1

2

2

2

1

2

121

5.Properties -- Air @1 atm, 20 C = 1.204 kg/m3

6.Calculation

ploss = 0.5(12)2(1.204)(1-0.5)2 = 21.7 Pap2 - p1 = 0.5(12)2(1.204)(22-1) - ploss

= 65 – 21.7 = 43.3 Pa

7. Analysis & checkPressure rising by ploss

14

2

2

1

2

1loss 1

2

−=

A

AVp

Duct size is abruptly reduced in direction of flow

Sudden contraction

Flow pattern: vena contracta at 1’

Concept in predicting pressure loss:- No loss from 1 to 1’- Treat flow from 1’ to 2 as a sudden enlargement

No loss

Sudden enlargement

This logic is quite valid:- Accelerating flow (converging) is efficient - Decelerating flow is difficult without losses

2

2

1

2

1 12

−=

A

AVp ''

loss

contraction coefficient Cc

'

'

1

2

2

1

V

V

A

ACc ==

22

2loss 1

1

2

−=

cC

Vp

15

contraction coefficient Cc = f(A2/A1) determined experimentally by Weisbash (1855)

Sudden contraction

ncontractiosudden factorgeometry

22

2loss 1

1

2

−=

cC

Vp

Sudden contractionMax geometry factor (GF)→ 1/3

tenlargemen suddenfactor geometry

loss

2

2

1

2

1 12

−=

A

AVp

Sudden enlargementMax geometry factor (GF)→ 1

Sudden change: (ploss)enlargement > (ploss)contraction for Area ratio>0.6

16

Most common elbow used in duct system – 90 turns,(circular or rectangular in cross section)

Turns

Weisbash : ploss -- contracted region in plane 1’ to plane 2

No loss ploss

Reynold nuber (Re = VDh/) influence in ploss but not a dominant factor.

17

Turns

Madison and Parker (1836) – ploss in rectangular elbows

Flat 90 elbow (large W/H) suffers less pressure drop than deep 90 elbow (small W/H)

W/H = 4

Reduce ploss – subdivision of elbow into multiple elbow of large W/H by installing turning vanes

18

Pressure loss in elbows of circular cross section

Turns

19

A main duct supplies air to several branch ducts – supply air (SA)

Branch takeoffs

upstream u → downstream d →

22

1402

−=

u

dd

V

VVp ).(loss

ploss in straight is small compared to other loss in system, neglected in low V

ploss in straight (pu - pd)

ploss in branch takeoffgraph [ploss / (Vb

2/2)]& Vd/Vu

20

Example 6-4 A 60, 30- by 30-cm branch takeoff leaves a 30- by 50-cm trunk duct. The size of the downstream section is also 30 by 50 cm. The upstream flow is 1.5 m3/s, and the branch flow rate is 0.5 m3/s. The upstream pressure is 500 Pa and the air temperature is 15 C. (a) What is the pressure following the straight-through section, and (b) what is the pressure in the branch line?.

1.Data:60, 3030-cm branch takeoff 3050-cm trunk duct

1.5 m3/s = Qu

0.5 m3/s = Qb

500 Pa = pu

Air @15 CCompute (a) pd ,(b) pb

2.Assumption – high V flow

4.Methods & EquationsRevised Bernoulli equation: (a) pd = pu + (Vu

2/2) - (Vd2/2) – ploss,u-d

ploss,u-d = (Vd2/2)(0.4)(1- Vd/Vu)2

Qd = Qu - Qb

Au = 0.3 0.5 = 0.15 m2

Vu = Qu/Au, Vd = Qd/Ad

(b) pb = pu + (Vu2/2) - (Vb

2/2) – ploss,u-b

ploss,u-b = (Vb2/2)*graph value

branch takeoff [60, Vb/Vu]→[ploss /(Vb2/2)]

Ab = 0.3 0.3 = 0.09 m2

Vb = Qb/Ab

5.Properties - Air @15 C, = 1.225 kg/m3

21

22

5.CalculationsQd = Qu - Qb = 1.5 – 0.5 = 1.0 m3/sAu = 0.3 0.5 = 0.15 m2

Vu = Qu/Au = 1.5/0.15 = 10 m/sVd = Qd/Ad = 1.0/0.15 = 6.67m/sRevised Bernoulli equation: ploss,u-d = (Vd

2/2)(0.4)(1- Vd/Vu)2

ploss,u-d = 27.25(0.4)(1-0.667)2 =1.21 Pa(a) pd = pu + (Vu

2/2) - (Vd2/2) – ploss,u-d

pd = 500+61.25-21.25-1.21 = 533 Pa

3050-cm Qu = 1.5 m3/spu = 500 Pa

(a) pd = ?

3030-cm Qb = 0.5 m3/s (b) pb = ?

ploss,u-b = (Vb2/2)*graph value

Ab = 0.3 0.3 = 0.09 m2

Vb = Qb/Ab = 0.5/0.09 = 5.56 m/sbranch [60,Vb/Vu=0.556]→[ploss /(Vb

2/2)=2.5]ploss,u-b = 2.5(Vb

2/2) = 47.3 Pa(b) pb = pu + (Vu

2/2) - (Vb2/2) – ploss,u-b

pb = 500+61.25-18.93-47.3= 495 Pa 6. Analysis & check - ploss in straight < ploss in branch pu - pd = -33 Pa, pu - pb = +5 Pa

= 533 Pa

= 495 Pa

ploss,u-d = 1.21 Pa

ploss,u-b = 47.3 Pa

- bring air from various branch to main duct – return air (RA)

Branch entries

23

Momentum equation:

ddubbdudd AppAVAVAV )(cos −=−− 222

If = 90 , the momentum eq. +Revised Bernoulli eq.

22

12

−=

d

ud

V

VVp

loss

−

= 151

2

22

b

dd

A

AVp .loss

ploss in straight (pu - pd) ploss from branch(pb - pd) for Ad/Ab > 4

24

Data on pressure loss in elbows and branch fittings are available in term of equivalent length, Leq.

2

2V

D

Lfp

eqloss =

25

ท่อลม (Ductwork)วสัด:ุ สงักะสแีผน่เรยีบ (Galvanized-iron sheets) ขึน้รปูตามขนาดทอ่ลม ใชร้อยต่อเทคนิคพเิศษ เชน่ ตะเขบ็ (Seams), สอดใส ่(Slip), เกีย่วทบั (Locks)หรอืแผน่อลมูนิมัขึน้รปูตามขนาดทอ่ลมทอ่ลมส าเรจ็รปูสรา้งจาก plastic fiberglass fiberglass-metal เป็นฉนวนในตวั น ้าหนกัเบา เกบ็เสยีงได ้ราคาแพง แต่คา่ตดิตัง้ถูกกวา่ระบบทอ่แบบแรก

ทอ่ลม จะถกูประกอบกบั Fittings ขนาดมาตรฐาน หรอืขึน้รปูตามทีก่ าหนดระบบทอ่ลมจะถกูตดิตัง้ตามมาตรฐานทอ่ลม มกีารหุม้ฉนวน หรอืใชว้สัดุทีเ่ป็นฉนวนในตวั

26

Ductwork of heating Circular, rectangular ducts and fitting, oval ducts are used to meet space limitations

27

Duct system = straight duct + elbows + branch outlets + branch inlets + dampers + terminal (registers and diffusers)

Ductwork of cooling

Requirements of a duct system:

Design of duct systems

28

(1) Convey specified airflow rates to prescribed locations

(2) Economical in first cost + fan cost + building space cost

(3) Not transmit or generate noise

Simple methods – find duct sizes to match reasonably space & velocities

(1) Velocity method

(3) Equal-friction method

(2) Static-regain method

1. Select -- velocities in mains and branches

Velocity method

29

2. Calculate -- pressure drop in all runs

3. Select -- fan for highest pressure drop (pdrop)

4. Damper installation → open highest pdrop line

→ other lines-- throttled dampers to Vdesign

Damper – throttling devices for air, consist of pivot metal plates

30

Velocity method: multi-brance duct system

→ Air-flow rate in duct section A to I: can be computed

1. Select -- velocities in mains and branches

2. Calculate -- pressure drop in all runs

pdrop in straight ducts, elbows, branch takeoffs

Air-flow rate at outlets 1 to 5 : known from load calculationQsupply = qSensible /1.23(ts – ti)

3. Select -- fan for highest pdrop 92 Pa (A-C-G-H)

pdrop in other component, i.e. coils, filters

4. Damper installation → open damper A-C-G-H line

→ other lines– partially closed dampers

→ To provide highest pdrop 92 Pa at Vdesign

Improved design:- enlarge duct A-C-G-H lineto reduce pdrop

- other duct – also reduced

Q1 Q2 Q3

Q4

Q5

Often used equal-friction method:

Equal-friction method

31

1. Decide - pdrop

2. Compute – Leq in all runs (Lstraight duct + Leq, fittings)

3. Compute pressure gradient (dp/dx) -- pdrop / Leq,longest

4. Select duct size for Leq,longest run (duct line) → by dp/dx & flow rate

5. Select duct size for other sections → select size for available pdrop

-- with velocity appropriate for noise restrictions

Leq,elbow 3 – 10 m, Leq,branch takeoffs up to 20 m

If A-C-G-H -- Leq,longest

4.Select duct size of section A

5.Select duct size of section Bby chosen to dissipate available pdrop

* Equal-friction method results in better design: reduced duct size and cost

1.Data: 2-branch duct system2.6 m3/s = QB

1.0 m3/s = QC-D

4 m = Leq, branch, A-C

2 m = Leq, elbow, C-D

4 Pa/m = selected (p/Leq)A-B

Compute DC-D= ?

2.Assume- ploss, branch, A-B = 0 32

Duct A

branch

elbow

Duct B

Duct DDuct C

4.Methods - known p/L & Q → find D by iteration or Graph

Without dampering: pC-D = pA-B Equal-friction method

(p/Leq)C-D = (p/Leq)A-B *Leq,A-B/Leq,C-D

Leq,C-D = Leq, branch, A-C +LC+ Leq, elbow, C-D + LD

Iteration: Assume Di, compute Di+1 = fV2/(2p/L)f = f(Re,/D),Re = VD/,V = Q/( D2/4)Recalculate until Di+1 convert to Di

P6-8 A two-branch duct system of circular duct is shown. The fittings have the following equivalent length of straight duct: upstream to branch, 4 m; elbows, 2 m. There is negligible pressure loss in the straight-through section of the branch. The designer selects 4 Pa/m as the pressure gradient in the 12- and 15-m straight sections. What diameter should be selected in the branch section to use the available pressure without dampering?

33

P6-8 2-branch duct system [AB, ACD], QB , QC-D , Leq, branch, A-C, Leq, elbow, C-D, (p/Leq)A-B , ploss, branch, A-B → DC-D= ?

6.Calculations; find D by Graph pA-B = (p/Leq)A-B *Leq,A-B = (4 Pa/m)*(12+15) = 108Leq,C-D = Leq, branch, A-C +LC+ Leq, elbow, C-D + LD =4+5+2+7 = 16 m(p/Leq)C-D = (p/Leq)A-B*Leq,A-B/Leq,C-D = 108/16 = 6 Pa/m

(1)Graph [p/L,Q] = [6 Pa/m,1 m3/s] → D = 0.31 m roughly!

Duct A

branch

elbow

Duct B

Duct DDuct C

(2) known p/L & Q → find D by iteration: properties air 15C, = 1.225 kg/m3, = 18.178 E-6 Pa.s, of sheet metal = 0.00015 m,

Known Q = 1 m3/s, p/L = 6 Pa/m, assume D1 = 0.3 mV1 = 4Q/(D2) = 14.15, Re1 = VD/ = 281000f = F(Re,/D) or f = 0.25[log{/3.7D+5.74/Re0.9}]-2= 0.01984compute D2 = fV2/(2p/L) = 0.3698 mRecalculate until Di+1 convert to Di → not converge!

0.2

0.3

0.4

0.5

0 10 20 30

34

P6-8 2-branch duct system [AB, ACD], QB , QC-D , Leq, branch, A-C, Leq, elbow, C-D, (p/Leq)A-B , ploss, branch, A-B → DC-D= ?

6.Calculations known p/L & Q(1)Graph [p/L,Q] = [6 Pa/m,1 m3/s] → D = 0.31 m roughly!(2)Iteration: Assume Di →Vi = 4Q/(D2

i) → Rei = ViDi/→fi = 0.25[log{/3.7Di+5.74/Rei

0.9}]-2 → Di+1 = fiVi2/(2p/L)

Recalculate until Di+1 convert to Di → not converge!0.2

0.3

0.4

0.5

0 10 20 30

(3)Numerical method, Nonlinear equation solver: f(D) = 0: f(D) = D -fV2/(2p/L) V= 4Q/(D2) , Re = VD/ = 4Q/(D) f = 0.25[log{/(3.7D)+5.74/Re0.9}]-2 = 0.25[log{/(3.7D)+5.74(D/(4Q))0.9}]-2

f(D) = D -{0.25[log{/(3.7D)+5.74(D/(4Q)) 0.9}]-2}(16Q2/(2D4)/(2p/L)f(D) = D -2Q2/{(D42p/L)/[log{/(3.7D)+5.74(D/(4Q))0.9}]2}f(D) = D –C1/{D4[log{/(3.7D)+5.74{C2D}0.9}]2}, C1 = 2Q2/(2p/L), C2 = /(4Q)

If rectangular duct is used with b = 0.2 m, a =?

35

Classification of duct systems

1. Low pressure system: residential buildingsVelocity 10 m/s ( 2000 fpm) Static pressure SP 50 mm.wg ( 2 in.wg) {1 m/s = 196.85 fpm} {wg = water, gage} {1 in.wg = 249.1 Pa}

2. Medium pressure system: commercial buildingsVelocity 10 m/s ( 2000 fpm)Static pressure SP 150 mm.wg ( 3 in.wg)

3. High pressure system: industrial buildingsVelocity 10 m/s ( 2000 fpm)Static pressure SP 150 - 250 mm.wg (3 - 5 in.wg)

Choice of velocity selection: economics, space limitation, noise

36

Recommended air velocities: applications and noise criteria

velocity → pdrop → fan power & noise

velocity → duct size → first cost & space

Residences: 3 – 5 m/s(600 – 1000 fpm)

Theatres: 4 – 6.5 m/s(800 – 1300 fpm)

Restaurants: 7.5 – 10 m/s(1500 – 2000 fpm)

Ex. Public building with no extensive acoustic treatment:Main Vsupply 1500 fpmMain Vreturn 1300 fpmBranch Vsupply 1200 fpmBranch Vreturn 1000 fpm

General rules for duct design

37

1. Air conveyed as directly as possible to save space, power and material

2. Sudden changes avoided. if not possible, turning vanes used to reduce pressure loss

3. Diverging sections should be gradual. Angle of divergence ≤ 20o

4. Aspect ratio as close to 1.0 as possible. Normally, not exceed 4

5. Air velocities within permissible limits to reduce noise and vibration

6. Duct material as smooth as possible to reduce frictional losses

38

อุปกรณ์หลกัของระบบสง่ลมเยน็

1. พดัลมหรอืชุดสง่ลมเยน็ (Fan or Air Handling Unit)

2. ทอ่ลมที ่(Duct work) และ สว่นบรรจบ (Fittings) เชน่ Elbow, Tee

3. หวัจา่ย (Terminal Distribution) เชน่ หน้ากาก หวัจา่ยเพดาน

4. อุปกรณ์เสรมิ (accessories) เชน่ ลิน้ปรบัลม (Dampers), แผน่เปลีย่นทศิทางลม (Turning vanes), ชอ่งจดัลมลม (Equalizing grid), ขอ้กนักระเทอืน (Flexible connectors), จุดตรวจสอบ (Test points)

39

การเลอืกขนาดทอ่ลม

Aspect ratio (AR) ต ่าเทา่ทีช่อ่งทางเดนิทอ่ลม จะท าได้สว่นมาก ทอ่ลมมคีวามสงูในฝ้าเพดาน 200, 250, 300 mm (8”, 10”, 12”)

AR สงู มผีล - static pressure สงู – พดัลมตวัใหญ่- ใชว้สัดุหนา น ้าหนกัมาก - ชว่งต่อสัน้ลง ใชเ้หลก็ฉากเสรมิใหญ่ขึน้ จุดแขวนทอ่มาก- Heat loss มาก ใชฉ้นวนหนา

สว่นมาก AR 2:1 to 4:1

Design of duct systems

40

Simple methods – find duct sizes to match reasonably space & velocities

(1) Velocity method

(3) Equal-friction method

(2) Static-regain method

Select -- velocity in mains and decrease velocities in branches

ท่อขนาดเลก็ ประหยดั แตป่รบัความเรว็ลมไดย้าก

Increase static head in branches – High velocity system > 2000 fpm

ท่อขนาดใหญ่ ปรบัความเรว็ลมไดส้ะดวก แตร่าคาสงู

Use equal friction per length in all branches

ท่อขนาดกลาง ปรบัความเรว็ลมได ้ราคาพอดีๆนิยมใชม้ากท่ีสดุ

Owning cost of a duct system : duct and installation, insulation,

Optimization of duct systems

41

sound attenuation, energy to drive fan and space requirements

Minimize owning cost

Optimization study may be difficult

For small duct systems – engineering cost > owning cost

For large duct systems – optimization is ok with value of building

Simple of optimization procedure: a duct system = Lstraight duct + fan

Select Dduct to minimize initial cost + operating cost

Total cost = C = initial cost + lifetime operating cost

Estimators: initial cost = mass of metal in duct system + installation cost (6 times)

initial cost = C1*D*L = (thickness)(D)(L)(metal)(installed cost/kg)

Operating cost = C2*p*Q*H = energy cost/hour * operating hour

Optimization of duct systems

42

2162 42

22

)/( D

Q

D

Lf

V

D

Lfp

==

Assume: f, = constant

Operating cost = C3 *L*H*Q3/D5

C = C1 D*L + C3 L*H*Q3/D5

C eq. – differentiated and equated to zero

61

1

3

35/

=

C

HQCDopt

Assume: fan cost and motor cost = constant

Analysis of assumption1: initial cost = C1 *D*L -- not valid for small duct sizes

Because for small duct sizes -- D → larger fan and motor → initial cost

Simple duct system = Lstraight duct + fanAnalysis of assumption2:

Actual duct system – elbows, branches, other fittings – may affect greatly

Several different duct design -- investigated in order to select an optimum.

System balancing

43

Load calculation → Thermal distribution system

Refrigeration systems → Compressor, condenser, evaporator, valve, piping systems, controls, etc.

Air-conditioning (A/C) systems → Fan and duct systems

Installation → Measurement of actual airflow rates- be taken→ Adjust damper correspond to the designed flow rate

To reduce cost for balancing an air-handling system in a large building: → Duct system designed nearly balanced when installed

Designers is at the mercy of installers

However, quality of construction →affect pressure drop, i.e. fitting

Investment in engineering time in the design of duct system → result in a better-operating system

Centrifugal fan and their characteristics

Air enters fan axially → turns and moves radially into blades

→ enters scroll→ outlet

Fan inlet – single or double (both sides)

4-types Blades:

Air-foil or backward-curved blade fan – used in high volume/high pressure

forward-curved blade fan – used in low pressure A/C system

1.Radial – low volume & high pressure2.Forward-curved – proper for constant flow, silent3.Backward-curved – easy to control volume and p4.Airfoil – high volume & high pressure

For some Sky-A/C : Indoor unit1. Backward curved blade fans - Cassette type 2. Forward curved blade fans - Duct & Ceiling type

Fan in A/C

Radial or straight blade– low volume & high pressure

forward-curved– proper for constant flow, silent

backward-curved – easy to control volume and pressure

Outdoor unit : Propeller fan– low speed & high volume & low static pressure

Large A/C systemAir duct system

Indoor unit:- Duct type - Ceiling type

Indoor unit:- Cassette type

46

forward-curved-blade centrifugal fan

Dip in static pressure at low flow rate eddies flow in blade channel

Power = pressure rise + kinetic energy

Pideal = Q*(p2 – p1)+ w*V2/2

Efficiency, = Pideal / Pactual

47

Example 6-5 Compute the efficiency of the fan whose characteristics are shown in Fig.6-15 when it operate at 20 r/s and deliver 1.5 m3/s.

1.Data: Fan - air20 r/s – rotative speed (n)1.5 m3/s = QFig.6-15 Characteristic fan curve: D 270 mm, outlet 0.517 by 0.289 mCompute fan efficiency,

4.Methods & Equations

= Pideal / Pactual

Pactual → graph -- known Q, n

Pideal = Q*(p2 – p1)+ w*V2/2

Characteristic fan curve: P & prise = f(Q,n)

1.5 m3/s, 20 r/s → (p2 – p1) = 500 Pa→ Pactual = 1.2 kW

w = *Q

V = Q/A, Aoutlet = 0.517 * 0.289 = 0.149 m2

5. Properties: air = 1.2 kg/m3

6. Calculations: w*V2/2 = 1.8*10.12/2 = 91 WQ*(p2 – p1) = 1.5*500 = 750 PaPideal = 750 + 91 = 841 W = 100%*841/1200 = 70%

w = 1.2*1.5 = 1.8 kg/s

V = 1.5/0.149 = 10.1 m/s

48

Propeller-type & forward-curved-blade centrifugal fan

In comparison with propeller fans, centrifugal fans (blowers) require less horse power at higher SP and have a much less steep SP curve, low noise

Performance of a centrifugal fan, forward-curved-blade type

Performance of a propeller-type fan

propeller fans, large volume at low static pressure (SP) – high noise

air-cooled condensers: require large air volume at small SP of finned-tupe condenser

Large Q at low SP

Less hp at high SP

80 cfm, 22%SP

80 cfm, 47%SP

Fan curve

: group of relationships that predict fan performanceFan laws

49

of changing -- condition of air, operating speed, size

Law of constant system (duct and fittings – no changed)

= varies as = is proportional to

Fan:

Duct and fittings:

Power:

P = Q(SP)+ QV2/2P Q3 or V3

Q V

SP V2/2SP Q2, P Q3

50

Balance-point diagram for Fan-and-duct system curve

SPbp

SPe = 0.75*SPbp

HPe = 1.5*HPbp

If SP1 = 0.4 in.wg at 4000 cfm, Fan law 1: SP Q2 → (SP2/SP1) = (Q2/Q1)2

SP2 = SP1 (Q2/Q1)2 → SP2 = 0.4(6000/4000) 2 = 0.9 in.wg → plot duct system curve

Intersection of duct-system curve and fan-curve is balance point bp: SPbp, HPbp

HPbp

If SPe were in error 25%: SPe/SPbp= 0.75

→ new bp at 7600 cfm = 1.6*Qbp

and HPe = 1.24*HPbp

Such balance diagrams are instructivesince they readily show the effects of changing the operation system or effect of error in estimating the resistance of the duct system

Not obey Fan law 1: Variation in rotative speed

51

Balance-point at two different speeds

The fan-duct-system balance points change as the fan speed changes.

Fan law 1: Variation in rotative speed SP Q2 , HP Q3

balance point A: SPA = 69%, QA = 65%, HPA = 55%

(SPA/SPB) = (QA/QB)2

→ (69/55) (65/57)2 1.3

balance point B: SPB = 55%, QB = 57%, HPB = 36%

(HPA/HPB) = (QA/QB)3

→ (55/36) (65/57)3 1.6

Obey Fan law 1: Variation in rotativespeed

The balance points at two speeds show the results of Fan law 1 on the operating of the complete system.

52

Modulation of volume flow rate of VAV air systems

1. Damper modulation: air damper to vary the opening of the air flow resistance.

Fan curve

System curve

3. Inlet cone modulation varies the

peripheral area of the fan impeller and therefore its performance curve.

2. Inlet vanes modulation varies the opening/angle of inlet vanes at the fan inlet

and then gives different fan performance curves.

4. Blade pitch modulation varies the blade

angle of the axial fan and its performance curve.

5. Fan speed modulation using adjustable

frequency AC drives varies the fan speed by supplying a variable-frequency and variable-voltage power source. Pulse width modulation (PWM) is universally applicable.

*lowest cost *easy control *wastes energy

*low cost *more energy efficient than damper, but not AC drives/inlet cone

*not expensive *backward curved centrifugal fans

* energy efficient * vane and tubular axial fans

* Most energy efficient for large centrifugal fans

53

Example 6-6 The motor driving a fan is rate at 15 A and is currently drawing 11 A while providing a rotative speed of the fan of 15 r/s. The airflow rate delivered by the fan is to be increased as much as possible. What is the permissible rotative speed of the fan while staying within the rating of the motor, and what percentage increase airflow rate is possible?

1.Data: Fan - air11 A = I1 →15 r/s – rotative speed (1)If 15 A = Irating → rating =? And %Qincrease =?

4.Methods & EquationsVariation in rotative speed, constant → Law 1 (Q , SP 2, P 3)

P = VI → P I P 3 → I 3 → (Irating / I1) = (rating/1)3

Q → (Qrating /Q1) = rating/1)

%Qincrease =(rating -1)*100/1

→ rating = 1(Irating / I1)1/36. Calculations: rating = 15(15 / 11)1/3 = 16.6 r/s%Qincrease =(16.6-15)*100/15 = 10.6%

Handling air properly within conditioned space (room):

Air distribution in rooms

54

1.Flow rate with (treturn- tsupply) of air – compensated for heat gain in the room

2. Vsupply air should not higher than 0.25 m/s in the room

3. Some air motion – to break up temperature difference in the room

Select/design-- location of supply-air diffuser and return-air grille for 3 objectives

Understanding -- Prediction of velocity and temperature distributions in the room:

1. Behavior of a free-stream jet

2. Velocity distribution at air inlet, i.e. supply-air diffuser and return-air grille

3. Buoyancy

4. Deflection

Return-air grille controls velocity effects →Low velocity – prevent excessive air noise

Warmer air stream – rise, cooler air stream – drop

Avoid – discharge of cool air to drop and strike at occupants

stream of cool air strike solid surface – deflects as light would

Avoid – stream of cool air deflects onto occupants before diffused

Continuity & momentum equations

→Velocity in circular jet:

2225571

417

)/(.

.

xrx

Auu oo

+=

Circular and plan jet

55

1. Centerline velocity uo decay as x increases -- uo x

2. Jet spreads as it moves from outlet-- r x

3. Air is entrained as jet moves from opening

4. Jet of large diameter (Ao) sustains its velocity better than a small diameter jet

56

Example 6-7 An air jet issues from a 100-mm-diameter opening with a velocity of 2.1 m/s. What is the centerline velocity 1 and 2m from the opening?

1.Data: circular air jet100 mm = Do

2.1 m/s = uo

Compute u at r = 0; x = 1, 2m

4.Methods & Equationscenterline velocity in circular jet:

x

Du

xrx

Auu oooo

cl)2/(41.7

)/(5.571

41.7222

=

+=

Velocity in plane jet from a long, narrow slot:

−=

x

y

x

buu o 6771

402 2 .tanh.

b = width of opening, my = normal distance from centerplane to point where u is being computed, m

centerline velocity decrease more rapidly in circular jet than in plane jet

plane jet entrains less air than circular jet → not decelerated as rapidly

6. Calculations: x = 1 m; ucl = (7.41*2.1**0.1/2)/1 = 1.38 m/sx = 1 m; ucl = (7.41*2.1**0.1/2)/2 = 0.69 m/s

Actual installation –long narrow slots

Diffusers and induction

57

Large circular opening – rarely used long distance of circular jet into room

diffuser – provide air pattern → velocity decays before air reach occupied region

– expand influence of supply outlet over a large region

– breaks up temperature gradients

Entrainment and its undesirable results

Pocket of low-p

circular diffuser

Cool supply air dumping down immediately Entrainment of

upper surface

58

การกระจายลมในหอ้งปรบัอากาศหวัจา่ยลมเพดาน (Ceiling Diffusers)หน้ากากจา่ยลมฝาผนงั (Grilles or Registers)• เพือ่สรา้งเน้ือทีใ่ชง้าน (Occupied zone):

→ 6 ft (1.8 m) สงูจากพืน้ และ ½ ft (15 cm) หา่งจากฝาผนงั

• เพือ่ใหเ้กดิสภาพอากาศ:* อุณหภูมสิม ่าเสมอกนัในเนื้อทีใ่ชง้าน

* มกีระแสลม (ไมใ่หม้จีุดอบั)

* สง่กระแสลมไปยงัต าแหน่งทีม่ีภาระความรอ้นใหส้มดุลกนั* รกัษาระดบัเสยีงไมใ่หเ้กนิพกิดั

Temp. profile, T < 6C at 6”-6 ft high

Air stream: 30-50 fpm at 6”-6 fthigh and 2 ft from wallAir volume: > 8 ACH

Load area

Noise control: NC25-NC45

15 cfm/m2 (7 (L/s)/m2) at 9 ft high room

59

ลกัษณะการกระจายลมภายในหอ้ง1. การคลุกเคลา้ผสมกนั (Diffusion)

→ Air duct vface = 400-800 FPM → vcomfort = 25-50 FPM (0.15–0.25 m/s)2. Diffusion Temperature

3. Induction, Entrainment or Aspiration

4. ความเรว็ลมทีห่วัจา่ย Outlet velocity of Face velocity

5. รศัมพีน่ลมเยน็ Throw or Radius of Diffusion or Spread

6. เกณฑข์องระดบัเสยีง Noise Criteria, NC

FPM = ft/min = 196.85*(m/s)

Room DB – Supply DB = Diffusion Temperature 20-30 F, 11-16 C Room DB 78 F, 25.6 C; Supply DB 50-60 F, 10-15.6 C

→ 25-50 dB

The Bernoulli effect -- low pressure created by high-velocity discharge

Distance from diffuser, grille or register to a point of terminal velocity80 FPM

N25-Studios N30-Residences N35-School,hotel N40-Office N45-Cafetarias

→ vface or vk = 400-800 FPM → ไมเ่รว็เกนิไป จนเสยีงดงั

60

จ าแนกวธิจีา่ยลมเยน็เขา้สูห่อ้งปรบัอากาศ1. จา่ยทีเ่พดาน (Ceiling Distribution)

2. จา่ยทีฝ่าผนงั (Wall Distribution)

3. จา่ยตามขอบพืน้หอ้ง (Perimeter Distribution)

→ Ceiling diffusers or linear diffusers→ ลมเยน็มคีวามหนาแน่นมากกวา่อากาศในหอ้ง กระแสลมจะคลุกเคลา้ไดง้า่ยกวา่

→ Horizontal air stream – to “throw” the air toward a terminal point → ตดิตัง้ทีฝ่าผนงัทีม่ภีาระความรอ้นสงูหรอือากาศรอ้นเรว็กวา่จดุอื่น (หน้าต่างกระจก) → เพือ่เหน่ียวน าใหอ้ากาศรอ้นถกูดดูเขา้คอยลเ์ยน็เรว็กวา่

→ เหมาะกบัโครงสรา้งทีม่ภีาระมากทีข่อบหอ้งเนื่องจาก exposed walls, large glass area, severe outside condition

→ เหมาะกบัการสง่ลมรอ้น ทีม่กัจะมอีากาศเยน็ทีข่อบหอ้ง→ ใหค้วามสบายกวา่แต่จะมี heat loss มากกวา่แบบ ceiling diffusers

61

ลกัษณะการกระจายลมภายในหอ้ง

Ceiling diffuser Side wall grille

Perimeter diffuser Linear diffuser

Induction

Aspiration

Throw

vface = 400-800 FPM → vcomfort = 25-50 FPM)

Selection of Ceiling Diffusers:- แบบกลม Circular- แบบจตุัรสั Square- แบบครึง่ซกี Half-round- แบบผนืผา้ Directional- Perforated plate- LSD Linear Slot Diffuser

62

การเลอืกขนาดของหวัจา่ยเพดาน

From Catalog:- ขนาดเสน้ผา่ศนูยก์ลาง Size (in)- พืน้ทีห่วัจา่ย Neck area (ft2)- ความเรว็ลม Velocity (fpm)- ปรมิาณลมทีผ่า่น (CFM)- ความดนัตกครอ่ม

Static Pressure, SP (in. wg)- รศัมพีน่ลมได ้RAD (ft)- ระดบัเสยีง NC (dB)

63

ตวัอยา่งการเลอืกขนาดของหวัจา่ยเพดานEx 7.1: 30*50 ft*ft,3000 CFM,H=10ftหา Ceiling diffuserSol: 1.แบง่พืน้ทีห่อ้งไมเ่กนิ 1.5:150:30 = 1.7:1 ไมเ่หมาะสมเลอืก (50/2):30 = 25:30 < 1:1.5 o.k.2.เลอืกหวัจา่ยจาก catalogData: 2 zones of 25*30ขอ้ก าหนด: Private office NC 35 dBขอ้ก าหนด: พืน้ที ่25*30, 1500 CFMได้ D 18”, 1.767 ft2, 1590 CFMSP 0.05 in wg, RAD 13-26 ft, NC32 30 ft

50 ft

64

ขอ้ก าหนด: Private office NC 35 dBขอ้ก าหนด: พืน้ที ่25*30, 1500 CFMได้ D 18”, 1.767 ft2, 1590 CFMSP 0.05 in wg, RAD 13-26 ft, NC32

65

การเลอืกขนาดของหน้ากากลม

ปัจจยัในการพจิารณาเลอืกหน้ากากลม:- Throw T (ft)- ความเรว็ลมทีห่น้ากาก Vk (fpm)- กระแสลม VTerminal (fpm)- Total Pressure PT (in. wg)- ขนาดของหน้ากาก W H (in in)- Deflection 0 20 40 55

กระแสลมเยน็ออกในแนวนอนเขา้ไปคลุกเคลา้กบัลมในหอ้ง

66

ตวัอยา่งการเลอืกขนาดของหน้ากากลมEx 7.2: 20*15 ft*ft,300 CFM,H=11ftหา Grille or RegisterSol: 1.ตดิดา้น 20 ft, Throw = 15 ft2.เลอืกหวัจา่ยจาก catalogเลอืก Deflection 40,300CFM,T15ftขอ้ก าหนด: ได้ 340 CFM, 900 FPMPT 0.05 in wg, T15ftWxH = 24*4,20*5,16*6,12*8,10*10** เลอืก Deflection 40 ดจูากภาพการกระจายในหอ้งขนาด 20ft*15ft ซึง่ควรพน่หา่งผนงั ½ ft สงู 6 ft ตาม occupied zone

15 ft

20 ft

67

เลอืก Deflection 40,300CFM,T15ftขอ้ก าหนด: ได้ 340 CFM, 900 FPMPT 0.05 in wg, T15ftWxH = 24*4,20*5,16*6,12*8,10*10** เลอืก Deflection 40 ดจูากภาพการกระจายในหอ้งขนาด 20ft*15ft ซึง่ควรพน่หา่งผนงั ½ ft สงู 6 ft ตาม occupied zone

68

ปัจจยัส าคญัของการส่งลมเยน็ของการปรบัอากาศ1. ส่งลมเยน็ เกดิความรอ้นรัว่เขา้ต ่า ลมเยน็รัว่ออกต ่า ระดบัเสยีงต ่า ราคาก่อสรา้งและคา่พลงังานพดัลมต ่า

2. ปริมาณลมท่ีจ่ายพอเพียง เพือ่รกัษากระแสลมและอุณหภมูิ

3. จ่ายลมเยน็ให้สมดลุกบับริเวณท่ีภาระความร้อนแตกต่างกนั

4. ปรบัปริมาณลมท่ีท่อแยก เพือ่จดัแบง่ลมเยน็ใหส้มดุลกบัภาระความรอ้นทีอ่าจผนัแปรได้

5. เตรียมจดุวดั อณุหภมิู ความดนั หรือความเรว็ ตามความจ าเป็น เพือ่ตรวจวดัแกไ้ข

69

การออกแบบ return-air ducts (RA)1. Negative pressure: ความดนัต ่ากวา่หอ้งปรบัอากาศ

2. สร้างให้สัน้ท่ีสดุ อปุกรณ์น้อยท่ีสดุ เพือ่ลดเสยีงลมดดูกลบั

3. return velocity < supply velocity

4. Ventilation ระบายอากาศทิง้ และมกีารน าอากาศภายนอกมาผสม

5. Plenum chamber การสรา้งทอ่ลมผสมก่อนเขา้กลอ่งผสมลม

6. Return pressure Loss 75-80% of Supply duct หรอื PLReturn < 0.25*PLSupply

7. Return grill มีขนาดใหญ่กว่า Supply diffuser

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การก าหนดต าแหน่งและการเลือกชนิดของ Diffuser

ถูกตอ้งตามหลกัวศิวกรรม สวยงามตามหลกัสถาปัตยกรรม เหมาะสมกบัการใชง้าน และ การบ ารงุรกัษา โดย1. เป็นระเบยีบ ระยะหา่งสม ่าเสมอ เป็น module แต่ละชว่งเสา2. พืน้ทีเ่ดยีวกนั ควรม ีFace size เท่าๆกนั ปรบัปรมิาณลมดว้ย Neck size3. ไมจ่า่ยลมเขา้หากระจก หน้าต่าง รอบๆอาคาร (condensation)4. ไมจ่า่ยลมเขา้ผนงัทบึ เพราะคราบเขมา่ ระยะควรมากกวา่ Throw ที ่150 fpm5. ถา้ตอ้งจดัวางหวัจา่ยใกลห้น้าต่าง กระจก ผนงัทบึ ใชห้วัจา่ย 1-3 ทาง เป่าออก 6. ระยะหา่ง 2 หวัจา่ย ไมค่วรน้อยกวา่ 2 เท่าของ Throw, หา้มลมกระจาย overlap7. Perimeter zone ทีร่บั radiation ควรจา่ยลมมากกวา่บรเิวณอื่น8. จดั Return air ใกล ้Perimeter zone เพือ่น าลมรอ้นเขา้คอลย์เยน็ทนัที9. หน้ากากลมกลบั (return air grille) ควรอยูท่ีล่มน่ิงหรอืใกลแ้หลง่ความรอ้น10. พจิารณาเสยีง อุณหภมูขิองอากาศ ความเรว็ลมปลายทีบ่รเิวณผูใ้ชง้าน