Data: 1. pipe line diagram 2.Q main & all Q branch (m 3 /s or cfm) Topic6-4 Duct Calculations by Chart Step 1. Select air velocity: V main (m/s or fpm) [cost, noise, install, mainten] building; V main 4 – 12 m/s (800 – 2400 fpm), V brance 2.5 – 10 m/s (500 – 2000 fpm) Industrial; V main 12–25 m/s (2400 – 5000 fpm), V brance 10–20 m/s (2000 – 4000 fpm) Step 2. Main duct Graph p L (Q,V,D) → Known (Q,V) -- find p L ,D Step 3. Branch ducts Graph p L (Q,V,D) → Known (Q,p L ) – find D,V Step 4. Conversion round - ab ducts Graph D(a,b) → Select b(D) – find a
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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
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
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)
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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.
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)
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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.
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”
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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
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duct size calculation uing equal-friction method by calculations
12 3
45 67
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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
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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
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12 3
45 67
3-45 6
12
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Min static pressure = SPsupply– (-SPreturn)
SPmin= 0.223– (-0.215) = 0.438 in.wg
SpsupplySpreturn
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Min static pressure = SPsupply– (-SPreturn)
SPmin= 0.219– (-0.261) = 0.480 in.wg
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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.