Transcript
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MECHANICAL VENTILATIONMECHANICAL VENTILATION
Compiled by
Mohd Rodzi IsmailSchool 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• Indoor air is withdrawn and replaced by fresh air continuously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 at approximately 21% of air volumeremoval of CO2control of humidity – between 30 & 70% RH is acceptable for human comfortprevention of heat concentrations from machinery, lighting and peopleprevention of condensationdispersal of concentrations of bacteriadilution and disposal of contaminants such as smoke, dust gases and body odoursprovisions of freshness – an optimum air velocity lies between 0.15 and 0.5 ms-1
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VENTILATION REQUIREMENTS
Control of ventilation rates - normally based on recommendations by authorities 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 “air changes per hour”
Air supply rate x nos. occupantsRoom volume
Example 1A private office of 30 m3 volume designed for 2 people
air changes per hour86.223043
=x
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MECHANICAL VENTILATION
An alternative to the unreliable natural systemsComponents involved:
FanFiltersDuctworkFire dampersDiffusers
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Table 1.0 - Fresh air supply rates (BS 5720)
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Fans
Provide the motive for air movement (imparting static energy or pressure and kinetic energy or velocity)It’s capacity for air movement depends on
TypeSizeShapeNumber of bladesSpeed
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Basic law of fan capabilities (at a constant air density):1. Volume of air varies in direct proportion to
the fan speed, i.e.
where,• Q = volume of air (m3/s)• N = fan impeller (rpm)
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2
1
2
NN
=
10
2. Pressure of, or resistance to, air movement is proportional to fan speed squared, i.e.
where,• P = pressure (Pa)
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1
2
)()(
NN
PP
=
11
3. Air and impeller power is proportional to fan speed cubed, i.e.
where,• W = power (W or kW)
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1
2
)()(
NN
WW
=
12
Example 2A fan of 2kW power discharges 4 m3/s with impellers rotating at 1000 rpm to produce a pressure of 250 Pa. If the fan impeller speed increases to 1250 rpm, calculate Q, P and W.
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1
2
1
2
NN
=1.10001250
42 =
Qtherefore, Q2 = 5 m3/s
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22
1
2
)()(
NN
PP
=2.2
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)1000()1250(
250=
Ptherefore, P2 = 390 Pa
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1
2
)()(
NN
WW
=3.3
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)1000()1250(
2=
Wtherefore, W2 = 3.9 kW
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As fans are not totally efficient, the following formula may be applied to determine the percentage
1100x
(W)powerAbsorbedvolumeairxpressurefanTotalEfficiency =
So, for the previous example,
%501
100x3900
5x903Efficiency ==
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Types of fan1. Cross-flow or tangential2. Propeller3. Axial flow4. 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-cooled motor in greasy, hot & corrosive gas situations
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Types of axial flow fans
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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 fansWall 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. Dry2. Viscous3. Electrostatic4. Activated carbon
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Dry filters
Roll filter Disposable element filter
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Viscous filters
Viscous filter
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Electrostatic filters
Electrostatic filter
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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 into building fabric
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Table 3.0 - Ductwork data
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Duct conversionFor 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
baabd+
=2
2.03)(265.1 ⎥⎦
⎤⎢⎣
⎡+
=ba
abxd
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Example 3 (duct conversion)A 450 mm diameter duct converted to rectangular profile of aspect ratio 2 : 1 (a = 2b).
baabd+
=2
For equal velocity of flow:
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222450
2 bbb
bbbxbx
==+
=
44503 xb = Therefore, b = 337.5 mm and a = 2b = 675 mm
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2.03)(265.1 ⎥⎦
⎤⎢⎣
⎡+
=ba
abxd
2.03
2)2(265.1450 ⎥⎦
⎤⎢⎣
⎡+
=bb
bxbx
For equal volume of flow:
2.032
3)2(265.1450 ⎥⎦
⎤⎢⎣
⎡=
bbx
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 airflow
patterns
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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 systemsMechanical extract/natural supplyMechanical supply/natural supplyCombined mechanical extract & supply
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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 supply
Plenum ventilation system
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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 appropriate
to each section or branch of ductworkEqual friction method
• the air velocity in the main duct is selected and the size and friction determined from a design chart. The same frictional resistance is used for all other sections 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 hourTime in seconds
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GivenRoom volume = 480 m3
Air changes per hour = 6
Therefore
smxQ /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 Table 4.0)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 Ø• Duct B = 320 mm Ø
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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 or Pa per unit length of ductwork
From the design chart
Duct A = 0.65 Pa x 5 m effective duct length = 3.25 PaDuct B = 1.00 Pa x 10 m effective duct length = 10.00 Pa
Total = 13.25 Pa
Therefore, the fan rating or specification is 0.8 m3/s at 13.25 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/sSelected air velocity in duct B = 3 m/sQ, 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 chartDuct A and B are both coincidentally 420 mm Ø
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From duct design chart (Velocity reduction method)
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Friction in duct A = 1.00 Pa x 5 m = 5.0 PaFriction in duct B = 0.26 Pa x 10 m = 2.6 Pa
Total = 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/sCalculated airflow through duct A = 0.8 m3/sCalculated airflow through duct B = 0.4 m3/s
From the chart:Duct A at 0.8 m3/s = 450 Ø with a frictional resistance of 0.65 Pa/mDuct B (using the same friction) at 0.4 m3/s = 350 Øwith an air velocity of approximately 4.2 m/sThe 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 changese.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 Ø
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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 = Q x time seconds
Room volume
= 0.1 x 3600180
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). Building Services Handbook. Oxford: Elsevier.
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QUIZ
Name 5 purposes of ventilationWhat is “coanda effect”?
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