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Ventilation

10 Duct Design

Vladimír Zmrhal (room no. 814)

http://users.fs.cvut.cz/~zmrhavla/index.htmDpt. Of

Environmental

Engineering

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Laminar and turbulent flows

Reynolds number

� laminar flow Re ≤ 2300

� transitional flow 2300 < Re < 10000

� fully turbulent flow Re > 10 000

νair… kinematic viscosity [m2/s] = 14,5.10-6 [m2/s]

Rewd

ν=

2

3

Flow characteristics

n…exponent f(Re)

1/

max

1

nw y

w r

= −

2

1/

max2

1

11 2

s

s

S

n

s

S

V w S

w wdSr

yw w ydy

r r

π

ππ

=

=

= −

∫

∫ max

0,817sw

w=

Laminar and turbulent flows

4

Bernoulli equation (energy)

Pressures in the duct

Pressure losses

2 2

1 1 1 2 2 2 1,22 2

zp h g w p h g w pρ ρρ ρ+ + = + + + ∆

2

2c d

wp p p p ρ= + = +

2 2

1,2 1 1 2 2 1 22 2

z c cp p w p w p pρ ρ ∆ = + − + = −

3

5

� friction

� local pressure losses

2 2

1,2 .2 2

mt

z

pp

l w wp R l Z

dλ ρ ζ ρ

∆∆

∆ = + = +∑����������

22

1,22

z

l wp kV

dλ ζ ρ ∆ = + =

∑

Pressure losses

6

Laminar flow

Turbulent flow

Colebrook (1939)

ε/d – relative roughness

Smolik (1959) for ε = 0,15

64

Reλ =

0,125 0,11

0,0812

Re dλ =

1 / 2,512log

3,71 Re

dελ λ

= − +

Friction losses

4

7

Turbulent flow

for smooth pipes and duct (plastic)

5000 < Re 80 0004

0,3164

Reλ =

Friction losses

8

Material εεεε (mm)Galvanized steel 0,15

Concrete duct – smooth surface 0,5

Concrete duct – rough surface 1,0 – 3,0

Smooth brass, copper 0,015

Hose pipe 0,6 - 3

Plastic pipe 0,007

Roghness height of the conduit wall surfaces

Friction losses

5

9

Hydraulic diameter

Rectangular ducts

4 4 2

2( )h

A ab abd

O a b a b= = =

+ +

λ λ= dC

1,1 0,1b

Ca

= −

Friction losses

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Moody diagram

6

11

Local pressure losses are caused by the fluid flow through the duct

fittings:

� which change the direction of the flow (elbows, bands, etc.)

� affect the flow in the straight duct with constant cross-section

(valves, stopcocks, filters etc.).

� ζ … local loss coefficient (experiments - see Idelchik 1986)

� Borda loss prediction

Local pressure losses

2

2m d

wp pζ ζ ρ∆ = =∑ ∑

12

109876543210

Local pressure losses

7

13

109876543210

0,08

1,11a

bζ

− =

Local pressure losses

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Duct design

Methods

� velocity method

� equal-friction method

� static regain method

8

15

Velocity method

Duct design procedure:

1) Find the main line

Rule no. 1: the main line is the maximum pressure loss line

(longest line, most segment line (?))

2) Air flow rate V (m3/h) in duct sections is known

3) Selection of the air velocity in the duct w

Rule no. 2: Air velocity increase towards the fan

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Air velocity w (m/s)

Main section Side section

Ventilation and low-pressure air-

conditioningrecomend. max. recomend. max.

- residential buildings 3,5 - 5 6 3 5

- public buildings 5 – 7 8 3 – 4,5 6,5

- industry 6 - 9 11 4 - 5 9

High-pressure air-conditioning 8 - 12 15 - 20 8 - 10 18

Velocity method

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4) duct area A (m2) → diameter d or a x b

→ nominal diameter dN or aN x bN

Rule no. 3: Duct sizes: 80, 100, 125, 140, 160, 180, 200, 250, 315,

355, 400, 450, 500, 560, 630, 710, 800, 900, 1000, 1120, 1250,

1400, 1600, 1800, 2000

4Vd

wπ=

Velocity method

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5) dN → real velocity wreal

6) calculation of dynamic pressure pd

7) Reynolds number → friction coefficient λ8) local loss coefficients ζ9) pressure loss of the duct section ∆pz,i

2

4real

N

Vw

dπ=

2

,

, 2i i

z i

s i

l wp

dλ ζ ρ

∆ = +

∑

Velocity method

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Rule no. 4: Balancing

10) total pressure loss is the sum of the duct sections pressure

losses

,ext z ip p∆ = ∆∑

, , , ,z F z E z G z Ip p p p∆ + ∆ = ∆ + ∆

Velocity method

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, , , , ,z z A z B z D z G z Ip p p p p p∆ = ∆ + ∆ + ∆ + ∆ + ∆

1 2 3 4 5cV V V V V V= + + + +

Velocity method

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Example

Example 1: Dimension the air duct system. Use the velocity method.

air velocity w = 6 - 10 m/s,

air density ρ = 1,2 kg/m3,

kinematic viscosity ν = 14,5.10-6 m2/s.

V1 = 9 000 m3/s

V2 = 1 440 m3/s

V3 = 2 160 m3/s

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Example

Line l V V wcalc Dcalc DN wreal pd Re l R.l ΣζΣζΣζΣζ Z ∆∆∆∆pel ∆∆∆∆pz

- m m3/h m3/s m/s mm mm m/s Pa - - Pa - Pa Pa Pa

0,41 19

0,96 0

0,46 0

2,04 0

TOTAL XX

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Equal-Friction Method

Duct design procedure:

1) selection of pressure loss per unit length R = 0,8 – 4 Pa/m

2) local pressure losses → friction in straight duct with equivalent

length

3) duct section pressure loss

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2

wR

dλ ρ=

ζλ ρ ζ ρλ

= ⇒ =2 2

2 2e

e

l w wl d

d

( )z ep R l l∆ = +

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Friction chart

Choice:

R = 1 Pa/m

Air flow rate:

1 000 m3/h

diameter D:

280 mm

Velocity :

w = 4,5 m/s

Equal-Friction Method

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Static Regain Method

� for uniform air supply

� constant static pressure before the

branch

Principles

� cross section reduction after

branches to change the dynamic

pressure

� decreasing of dynamic pressure

balances the pressure losses in the

duct section

−∆ = −3 2 3 2z d dp p p

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Assumptions:

� V = const.

� b = const.

� i = n, n - 1, …. 1

� calculation of dimension a

λ −− −

−

−= + 11 1

1

11 i

i i i

i

ia a l

d i

Static Regain Method

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Deduction:

kdeλ ρ ρ ρ= −2 2 2

1 1 2 1

1 2 2 2

l w w w

d

−∆ = −2 1 2 1z d dp p p

= =1 21 2

1 2

,V V

w wa b a b

λ = −2 2 2

1 1 2 1

2 2 2 2 2 2

1 1 2 1

l V V V

d a b a b a b

Static Regain Method

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⇒

←λ = −

2 2 2

1 1 2 1

2 2 2

1 1 2 1

l V V V

d a a a= =1 2, 2V V V V

( ) ( )λ λ −

− − −

− −⋅ ⋅ ⋅+ = ⇒ + =2 22 2 2 2 2 2 2

1 1

2 2 2 2 2 2

1 1 1 2 1 1 1

1 11 1 2 i

i i i i

i il V V V l i

d a a a d a a a

( )22 2 21 11 1 1

1 1

11 1 1i i

i i i i i

i i

l ia i a i a a l

d d i

λλ − −− − −

− −

−+ − = ⇒ = +

Static Regain Method

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Duct systems

Shapes

� rectangular

� round

� flexible duct

Materials

� steel galvanized

� aluminium

� plastic PVC

� textile

� ALP

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Duct leakage rate

where Sv … duct surface [m2]

0,67

vV m p S= ∆

Duct systems

Class Charakteristics of the leakage path

m [m3/s perm2]

A 0,027 . 10-3

B 0,009 . 10-3

C 0,003 . 10-3

D 0,001 . 10-3

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Purpose

� condensation risk

� heat losses/gains

Thickness of TI

� indoor 45 – 60 mm

� outdoor 80 – 100 mm (with sheet covering)

Thermal insulation

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Thank you for your

attention