Data: 1. pipe line diagram 2.Qmain & all Qbranch (m3/s or cfm)

Topic6-4 Duct Calculations by Chart

Step 1. Select air velocity: Vmain (m/s or fpm) [cost, noise, install, mainten]building; Vmain 4 – 12 m/s (800 – 2400 fpm), Vbrance 2.5 – 10 m/s (500 – 2000 fpm)

Industrial; Vmain 12–25 m/s (2400 – 5000 fpm), Vbrance 10–20 m/s (2000 – 4000 fpm)

Step 2. Main ductGraph pL(Q,V,D) → Known (Q,V) -- find pL,D

Step 3. Branch ductsGraph pL(Q,V,D) → Known (Q,pL) – find D,V

Step 4. Conversion

round - ab ductsGraph D(a,b) →Select b(D) – find a

Pressure drop; straight circular duct

Pressure drop in straight ducts

2

2V

D

Lfp =

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

Re = Reynolds number, VD/ = viscosity, Pas [ = / , m2/s]

1. Eck’s formula, 1973 f = [-2log{/(3.715D)+15/Re}]-2

Modified formula (explicit)

Colebrook’s formula, 1939 (implicit) Moody’s chart - turbulent flow f = [-2log{/(3.7D)+2.51/(Ref 0.5)}]-2

2. Smamee&Jain’s formula, 1976f = 0.25[log{/(3.7D)+5.74/Re0.9}]-2

3. Haaland’s formula, 1983f = [-1.8log{(/(3.7D))1.11+6.9/Re}]-2

Data: 1. pipe line diagram 2.Qmain & all Qbranch (m3/s)

Duct Calculations by equations

Step 1. Select air velocity: Vmain (m/s) [cost, noise, install, maintenance]building; Vmain 4 – 12 m/s, Vbrance 2.5 – 10 m/s

Industrial; Vmain 12–25 m/s, Vbrance 10–20 m/s

Step 2. Dmain (m) = (4Qmain/Vmain)0.5 → Dm(Q,V)

Step 3. fm = 0.25[log{/3.7D+5.74/Re0.9}]-2, Re = VD/

Air 20C; = 1.2041 kg/m3, = 18.178E-6 Pas, sheet metal = 15E-5 mfm = 0.25[log{1/(24667D)+1/(3802.7(VD)0.9)}]-2 → fm(V,D)

Step 4. pL (pa/m) = fmV2/(2D) = fmV2/(1.661D) → pL(f,V,D)

Step 5. Find fb for each branch Qb by solve implicit eq.fb = 0.25[log{(pL/fbQb

2)0.2/24547 +(fb/pLQb3)0.18/4747}]-2 → fb(f,Q,pL)

Step 6. Db = (fbQb2/(1.0246pL))

0.2 → Db(f,Q,pL)

Step 7. Vb = 4Qb/(Db2)→ Vb(D,Q) check for appropriate, noise etc.

Step 8. Deq,f = 1.265(ab)0.6/(a+b)0.2 → D(a,b)

SI unit:

D (m), V (m/s),

Q (m3/s), pL(Pa/m)

4

Duct A

branch

elbow

Duct B 3 m3/s

Duct C

Ex1. A two-branch duct system of rectangular air duct with the duct height not greater than 40 cm. 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 10 m/s as the velocity in the duct section A. in the 12- and 15-m straight sections. What diameter and ab of duct A, B and C should be selected to use the available pressure without dampering (equal-friction method)? Show calculations and fill in the table.

1 m3/s

8 m

Data: 1. PLD

2.Qmain & all Qbranch

Step 1. Vmain = 10 m/s (large building)

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

Graph p/L & Q @D, V

Step 2.Known (Q, D) -- find p, V

QA = 4 m3/sVA = 10 m/s

pL = 1.4-1.5 Pa/m

DA = 0.7-0.72 m

Step 3.QB = 3 m3/sDB = 0.63-0.64 mVB = 9.2-9.4 m/s

Step 3.QC = 1 m3/sDC = 0.41-0.42 mVC = 7.1-7.2 m/s

Step 4.DA = 0.7 mbA = 40 cmaA = 110 cm

0.7-0.72 1.4-1.5

1.4-1.5

1.4-1.5

0.63-0.649.2-9.4

0.41-0.427.1-7.2

110

90

36

6

Step 3. fm = 0.25[log{/3.7D+5.74/Re0.9}]-2, Re = VD/

Air 20C; = 1.2041 kg/m3, = 18.178E-6 Pas, sheet metal = 15E-5 mfm = 0.25[log{1/(24667D)+1/(3802.7(VD)0.9)}]-2 → fm(V,D)

Step 4. pL (pa/m) = fmV2/(2D) = fmV2/(1.661D) → pL(f,V,D)

Step 5. Find fb for each branch Qb by solve implicit eq.fb = 0.25[log{(pL/fbQb

2)0.2/24547+(fb/pL)0.18/(4747Qb

0.54)}]-2 → fb(f,Q,pL)

Step 6. Db = (fbQb2/(1.0246pL))

0.2 → Db(f,Q,pL)

Step 7. Vb = 4Qb/(Db2) → Vb(D,Q) check for appropriate, noise etc.

Step 2. Dmain (m) = (4Qmain/Vmain)0.5 → Dm(Q,V)

0.714

Step 8. Deq,f = 1.265(ab)0.6/(a+b)0.2 → D(a,b)

113 470700 0.0157 1.32 15.9

1.32

1.32

0.0161395500400.64 900.329.3

0.0182

19.8

17.21997000.42 40 370.147.1

35.7

41

Data: 1. pipe line diagram 2.Qmain & all Qbranch (cfm)

Duct Calculations by equations

Step 1. Select air velocity: Vmain (fpm) [cost, noise, install, maintenance]building; Vmain 800 – 2400 fpm, Vbrance 500 – 2000 fpm

Industrial; Vmain 2400 – 5000 fpm, Vbrance 2000 – 4000 fpm

Step 2. Dmain (in) = 12(4Qmain/Vmain)0.5 → Dm(Q,V)

Step 3. fm = 0.25[log{/3.7D+5.74/Re0.9}]-2, Re = VD/fm = 0.25[log{0.001596/D+0.86118/(VD)0.9}]-2 → fm(V,D)

Step 4. pL (in.wt/100 ft) =100fm(V/4009)2(12/D) =fm(V/401)212/D) →pL(f,V,D) Hw = (V/4009)2 = (V/60)2/(2gw/a) = V2 /(602*2*32.2/*(62.4/(12* 0.075))

Step 5. Find fb for each branch Qb by solve implicit eq.fb = 0.25[log{0.001328(pL/fbQb

2)0.2 +0.009334(fb/pLQb3)0.18}]-2 → fb(f,Q,pL)

Step 6. Db = 1.20255(fbQb2/pL)

0.2 → Db(f,Q,pL)

Step 7. Vb = 4Qb(12/Db)2/→ Vb(D,Q) check for appropriate, noise etc.

Step 8. Deq,f = 1.265(ab)0.6/(a+b)0.2 → D(a,b)

British unit:

D (in), V (fpm),

Q (cfm),

pL (in.wt/100 ft)

8

Sample 18-8 (N.C. Harris, Modern Air Conditioning Practice). Size the simple duct system shown as below if the duct is for general office space and the maximum duct depth possible is 16 in. Each diffuser has a friction loss (pdiff) of 0.04 in.wg. Find the total static-pressure loss for the system.

Data: 1. PLD 2.Qmain & all Qbranch

3. hloss 0.04 in.wg. Of each diffuser

Find duct sizes and PL (in.wg.)

Step 1. Select Vmain = 2000 fpm (office) Step 2. Dmain (in) = 12(4Qmain/Vmain)0.5 → Dm(Q,V)

Step 3. fm = 0.25[log{0.001596/D+0.86118/(VD)0.9}]-2 → fm(V,D)

Step 4. pL (in.wt/100 ft) = fm(V/401)212/D) → pL(f,V,D)

9

Sample 18-8 1. PLD 2.Qmain & all Qbranch 3. hloss 0.04 in.wg. Of each diffuser

Find duct sizes and PL (in.wg.)

Step 5. Find fb by solve implicit eq. F(fb) = 0F(fb) = fb - 0.25[log{0.001328(pL/fbQb

2)0.2 +0.009334(fb/pLQb3)0.18}]-2 = 0 → fb(f,Q,pL)

Step 6. Db = 1.20255(fbQb2/pL)

0.2 → Db(f,Q,pL)

Step 7. Vb = 4Qb(12/Db)2/ → Vb(D,Q)

Step 8. Deq,f = 1.265(ab)0.6/(a+b)0.2 → D(a,b)

total static-pressure loss = sum of equivalent lengthpL (in.wt/100 ft)

10

Sample 18-8 1. PLD 2.Qmain & all Qbranch 3. hloss 0.04 in.wg. Of each diffuser

Find duct sizes and PL (in.wg.)

total static-pressure loss in ft = sum of duct length and equivalent length = Ltotal

total static-pressure loss in ft =pL (in.wt/100 ft)*Ltotal/100

Leq, diffuser = no of diffuser*pL (in.wt/100 ft)*100/pdiff

total static-pressure loss in ft = sum of Lduct + Leq

total static-pressure loss in ft =pL (in.wt/100 ft)*Ltotal/100

11

Frictional Losses in Rectangular Ducts (Ch.18, N. C. Harris, Modern AC practice)

Rectangular Ducts: Aspect ratio < 8:1Dhydralic = 1.265(ab)3/5/(a+b)1/5 by laboratory studies

[But W.F. Stoecker R&A: Deq,f = 1.3(ab)0.625/(a+b)0.25]

V (fpm) & Hw (in. of water)v (fps) = (2gha)

0.5

ha = f (L/D)(V2/2g)ha = velocity pressure head, ftaha = wHw/12ha = 62.4Hw/(12a)ha = 5.2Hw/a

V/60 = (2g* 5.2Hw/a)0.5

g = 32.2 ft/s2, a = 0.075 lb/ft3

V = 4005Hw0.5

Supply air: Qs = qS /1.23(ts – ti)qS (W), t (C)

cfm = qS /1.08(ts – ti)qS (Btu/h), t (F)

12

Prob:

Static-Pressure Losses in Duct Fitting

13

Assume: low noise – specify V = 1300 fpm

14

IJ:100 cfm

HI:700 cfm

GH:1300 cfm

FG:1900 cfm

AF:2200 cfm

1. Select V = 1300 fpm2. Given Qs = 3200 cfm at main duct and Qs or all duct sections3. Find ff = (Qs, V)

CD:200 cfmCE:200 cfm

BC:400 cfm

AB:1000 cfm

A:3200 cfm

15

Use ff = 0.1 in. water per 100 ftin duct size calculation by equal-friction method

Fig. 18-15 Friction-loss chart for low-velocity straight ducts.

V = 1300 fpmQs = 3200 cfmff = 0.1 in. water per 100 ftof straight duct

Dh = 21.5 inA 3200 cfm, 22”, 1300 fpm

AF 2200 cfm, 18”, 1200 fpm

FG 1900 cfm, 17”, 1100 fpm

GH 1300 cfm, 15”, 1000 fpm

HI 700 cfm, 12” , 900 fpm

IJ 100 cfm, 12” , 550 fpm

16

Fig. 18-16 Rectangular equivalents of round ducts.

Dh = 21.5 inAssume b = 12”Dh = 1.265(ab)3/5/(a+b)1/5

Main duct = 34” x 12”

AF 2200 cfm, 18”, 1200 fpm24” x 12”

FG 1900 cfm, 17”, 1100 fpm21” x 12”

GH 1300 cfm, 15”, 1000 fpm17” x 12”

HJ 700 cfm, 12” , 900 fpm12” x 12”

IJ 100 cfm, 12” , 550 fpm6” x 6”

17

Duct Calculator- Imperial Unit Duct Calculator- SI Unit

18

Duct Calculator- Imperial Unit

19

Duct Calculator- SI Unit

20

Duct CalculatorMain duct section A

V = 1300 fpmQs = 3200 cfmff = 0.115 in. water per 100 ft of straight duct

Round Dh = 21 in Assume b = 12”(ceiling space limit)Main duct A = 33” x 12”

Qs = 3200 cfm

ff = 0.115 in. wtper 100 ft

Dh = 21 in

Select b = 12”, a = 33“Rectangular duct a x b = 33” x 12”

21

Duct CalculatorBranch duct section AF

ff = 0.115 in. wt /100 ftAF: Qs = 2200 cfmV = 1300 fpm

Round Dh = 18 in Assume b = 12”(ceiling space limit)duct AF = 24” x 12”

AF 2200 cfm

ff = 0.115 in. wt per 100 ft

DAF = 18 in

Select b = 12”, a = 24“Rectangular duct a x b = 24” x 12”

22

Use ff = 0.1 in. water per 100 ft in duct size calculation by equal-friction method

3-4

1. Select V = 1300 fpm2. Given Qs = 3200 cfm at main duct and Qs or all duct sections3. Find ff = 0.1 in. water per 100 ft of straight duct4. Select ff = 0.1 in. w/100 ft5. Find Dh from (Qs, ff)6. Specify b, find a from Dh = 1.265(ab)3/5/(a+b)1/5

7. Check V = Q/(ab) gradually decrease ad duct size

5 612

23

duct size calculation uing equal-friction method by calculations

12 3

45 67

24

The total static pressure

ff = 0.028

L/W= 4

ff = 0.04

10% for friction loss by dampers included → htot = 0.283 in.wg

25

Prob: Size SAD & RAD, fan static presureVSAD = 1500 fpmVRAD = 1300 fpm20% OAQsupply = 3800 cfm

Sizing Return-Air Ducts

Qreturn = 0.8*3800 = 3040 cfm = 800+1280+960

QOA = 0.2*3800 = 760 cfm

26

12 3

45 67

3-45 6

12

27

Min static pressure = SPsupply– (-SPreturn)

SPmin= 0.223– (-0.215) = 0.438 in.wg

SpsupplySpreturn

28

Min static pressure = SPsupply– (-SPreturn)

SPmin= 0.219– (-0.261) = 0.480 in.wg

29

The losses of coils, filters, and dampers must be added to determine the total fan static pressure.

Since the grille loss at J is only 0.05 in.wg, an excessive pressure difference exists across this return grille which will cause unwanted noise.A damper in stalled at J would have to be nearly closed to dissipate the static pressure and it would further increase the problem of noise.The best procedure is 1. to eliminate return-air locations close to fan or2. to minimize the return-duct losses by using larger duct. The latter choice adds to cost, of course.