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MECHANICAL VENTILATIONMECHANICAL VENTILATION

Compiled by

Mohd Rodzi I smail

School of Housing Building & Planning


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INTRODUCTION

Definition

the process of changing air in an

enclosed space Indoor air is withdrawn and replaced byfresh air continuously

From clean external source


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The importance of ventilation to maintain air purity, i.e.:

preservation of O2 content this should be maintained atapproximately 21% of air volume

removal of CO2 control of humidity between 30 & 70% RH is acceptable for

human comfort prevention of heat concentrations from machinery, lighting

and people

prevention of condensation

dispersal of concentrations of bacteria dilution and disposal of contaminants such as smoke, dust

gases and body odours

provisions of freshness an optimum air velocity lies

between 0.15 and 0.5 ms-1


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VENTILATIONREQUIREMENTS

Control of ventilation rates - normallybased on recommendations byauthorities or code of practice.

e.g. BS 5720


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Table 2.0 - Air changes rates (BS 5720)


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Conversion from m3

/hour per person to airchanges per hour

Air supply rate x nos. occupants

Room volume

Example 1

A private office of 30 m3 volume designed for 2people

air changes per hour86.22

30

43=x


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MECHANICALVENTILATION

An alternative to the unreliable naturalsystems

Components involved:

Fan Filters

Ductwork

Fire dampers Diffusers


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Table 1.0 - Fresh air supply rates (BS 5720)


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Basic law of fan capabilities (at aconstant air density):

1. Volume of air varies in direct proportion tothe fan speed, i.e.

where, Q= volume of air (m3/s) N= fan impeller (rpm)

1

2

1

2

N

N

Q

Q=


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2. Pressure of, or resistance to, airmovement is proportional to fan speed

squared, i.e.

where,

P= pressure (Pa)

2

1

2

2

1

2

)(

)(

N

N

P

P

=


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3. Air and impeller power is proportional tofan speed cubed, i.e.

where,

W= power (W or kW)

31

3

2

1

2

)(

)(

N

N

W

W=


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1

2

1

2

N

N

Q

Q=1.

1000

1250

4

2 =Q

therefore, Q2 = 5 m3/s

2

1

22

1

2

)()(

N

N

P

P =2. 2

2

2

)1000()1250(

250=P therefore, P2 = 390 Pa

3

1

3

2

1

2

)()(

NN

WW =3. 3

3

2

)1000()1250(

2=W therefore, W2 = 3.9 kW


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As fans are not totally efficient, the following formula

may be applied to determine the percentage

1

100x

(W)powerAbsorbed

volumeairxpressurefanTotalEfficiency =

So, for the previous example,

%501

100x3900

5x903Efficiency ==


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Types of fan1. Cross-flow or tangential

2. Propeller

3. Axial flow

4. Centrifugal


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Cross-flow or tangential fan

Tangential or cross-flow fan


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Tangential flow fan


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How tangential flow fans work


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Propeller fan

Wall mounted propeller fanFree standing propeller fan


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Types of propeller fans


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Axial flow fan

Axial flow fan Bifurcated axial flow fan

To protect the fan-cooledmotor in greasy, hot &

corrosive gas situations


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Heavy duty Counter rotating


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Bifurcated axial-flow fan


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Centrifugal fan

Centrifugal fan


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Air in

Air out


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Centrifugal fan impellers


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Centrifugal fans

Wall type


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HVAC duty centrifugal fan Industrial duty

centrifugal fan

Tubular

centrifugal fan


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Filters

Four categories of filters1. Dry

2. Viscous

3. Electrostatic4. Activated carbon


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Dry filters

Roll filter Disposable element filter


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35


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38


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Electrostatic filters

Electrostatic filter

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40


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Activated carbon filters

Commercial cooker hood

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HEPA filters

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Ductwork

Circular, square or rectangular cross-sections

Circular & rectangular ductwork

More efficient, less

frictional resistance

to airflow

Convenience,

more easily

fitted intobuilding fabric

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Table 3.0 - Ductwork data

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Duct conversion

For equal velocity of flow

For equal volume of flow

where

d= diameter of circular duct (mm) a= longest side of rectangular duct (mm) b= shortest side of rectangular duct (mm) 0.2= fifth root

ba

abd

+=

2

2.03

)(265.1

+=

ba

abxd

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Example 3 (duct conversion)

A 450 mm diameter duct converted to rectangularprofile of aspect ratio 2 : 1 (a = 2b).

ba

abd

+=

2

For equal velocity of flow:

3

4

3

4

2

22450

2b

b

b

bb

bxbx==

+=

4

4503xb = Therefore, b= 337.5 mm and a= 2b= 675 mm

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2.03

)(265.1

+= baabxd

2.03

2

)2(265.1450

+

=

bb

bxbx

For equal volume of flow:

2.032

3

)2(265.1450

=

b

bx

From this, b= 292 mm and a= 2b= 584 mm

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Duct conversion using conversion chart (simpler

but less accurate)

Circular to rectangular ductwork conversion chart

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Noise control

Sound attenuation

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Table 4.0 - Recommended maximum ducted air velocities

and resistance for accepted levels of noise

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Volume &

direction control

Air movement control

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Fire dampers

Fire dampers

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Diffusers

Grills &

diffusers

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Diffusers

airflowpatterns

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Coanda effect created by restricted air and pressure at the adjacent

surface due to limited access for air to replace the entrained air above

the plume

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Suspended ceilings as plenum chambers

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SYSTEMS

Mechanical ventilation systems

Mechanical extract/natural supply

Mechanical supply/natural supply

Combined mechanical extract &supply

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Mechanical extract/natural


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Mechanical extract/natural

supply

Extract ventilation to a commercial kitchen

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Extract ventilation to a lecture theatre

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Application of shunt ducts to a block of flats

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Mechanical supply/natural


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Mechanical supply/natural

supply

Plenum ventilation

system

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Combined mechanical


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Combined mechanical

extract & supply

Combined mechanical extract and supply

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VENTILATION DESIGN

Three methods of designing ductwork and fan:

Equal velocity method the designer selects the same air velocity for use

through out the system

Velocity reduction method

the designer selects variable velocities appropriateto each section or branch of ductwork

Equal friction method

the air velocity in the main duct is selected and thesize and friction determined from a design chart. The

same frictional resistance is used for all othersections of ductwork

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Duct design chart

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Example 4 (ventilation design calculation)

Q, air volume flow rate (m3/s) = Room volume x air changes per hour

Time in seconds

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Given

Room volume = 480 m3

Air changes per hour = 6

Therefore

smx

Q /8.03600

6480 3==

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Equal velocity method

Air velocity throughout the system (duct A &

duct B) = 5 m/s (selected based on Table4.0)

Q, the quantity of air = 0.4 m3/s is equally

extracted through grille Duct A will convey 0.8 m3/s; duct B will

convey 0.4 m3/s

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(0.8 m3/s)

0.4 m3/s 0.4 m3/s

(0.4 m3/s)

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320

450

A

B

From the design chart: Duct A = 450 mm

Duct B = 320 mm

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From duct design

chart (equal

velocity method)

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The fan rating relates to the frictional resistance obtained

in N/m2

orPaper unit length of ductwork

From the design chart

Duct A = 0.65 Pax 5 m effective duct length = 3.25 PaDuct B = 1.00 Pax 10 m effective duct length = 10.00 Pa

Total = 13.25 Pa

Therefore, the fan rating or specification is 0.8 m3/s at13.25 Pa

Effective duct length the actual length plus additional allowances for bends, offsets, dampers, etc.

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Velocity reduction method

Selected air velocity in duct A = 6 m/s

Selected air velocity in duct B = 3 m/s Q, the quantity of air = 0.4 m3/s is equally extracted

through grille

Duct A will convey 0.8 m3/s; duct B will convey 0.4

m3/s

From the design chart

Duct A and B are both coincidentally 420 mm

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From duct design

chart (Velocity

reduction method)

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Friction in duct A = 1.00 Pax 5 m = 5.0 Pa

Friction in duct B = 0.26 Pax 10 m = 2.6 PaTotal = 7.6 Pa

Therefore, the fan rating or specification is 0.8 m3/s at 7.6

Pa

Effective duct length the actual length plus additional allowances for bends, offsets, dampers, etc.

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Equal friction method

Selected air velocity through duct A = 5 m/s

Calculated airflow through duct A = 0.8 m3

/s Calculated airflow through duct B = 0.4 m3/s

From the chart:

Duct A at 0.8 m3/s = 450 with a frictionalresistance of 0.65 Pa/m

Duct B (using the same friction) at 0.4 m3/s = 350 with an air velocity of approximately 4.2 m/s

The fan rating is 0.8 m3/s at 0.65 Pa/m x 15 m =9.75 Pa

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From duct design

chart (Equal friction

method)

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Determination of sufficient air changes

e.g.: Library (max. velocity of 2.5 m/s with a max.

resistance of 0.4 Pa/m length) from Table 4.0

From the chart:

Maximum air discharged, Q = 0.1 m3/s

Duct size = 225 mm

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Duct design chart

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From

Q = Room volume x air changes per hourTime in seconds

and,

Air changes per hour = Qx time secondsRoom volume

= 0.1 x 3600

180

Thus, 2 changes per hour would be provided

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REFERENCES

Greeno, R.(1997). Building Services,

Technology and Design. Essex:Longman.

Hall, F. & Greeno, R. (2005). BuildingServices Handbook. Oxford: Elsevier.

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QUIZ

Name 5 purposes of ventilation

What is coanda effect?

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