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1. 0
MECHANICAL VENTILATION
Compiled by
Mohd Rodzi Ismail
School of Housing Building & Planning
2. 1
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• Indoor air is withdrawn and replaced by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
3. 2
The importance of ventilation – to maintain air purity, i.e.:
preservation of O2 content – this should be maintained at
approximately 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
4. 3
VENTILATION
REQUIREMENTS
Control of ventilation rates - normally
based on recommendations by
authorities or code of practice.
e.g. BS 5720
6. 5
Conversion from “m3/hour per person” to “air
changes per hour”
Air supply rate x nos. occupants
Room volume
Example 1
A private office of 30 m3 volume designed for 2
people
43
x 2 = 2.86 air changes per hour
30
7. 6
MECHANICAL
VENTILATION
An alternative to the unreliable natural
systems
Components involved:
Fan
Filters
Ductwork
Fire dampers
Diffusers
9. 8
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
Type
Size
Shape
Number of blades
Speed
10. 9
Basic law of fan capabilities (at a
constant air density):
Volume of air varies in direct proportion to
1.
the fan speed, i.e.
Q2 N 2
=
Q1 N1
where,
• Q = volume of air (m3/s)
• N = fan impeller (rpm)
11. 10
Pressure of, or resistance to, air
2.
movement is proportional to fan speed
squared, i.e.
P2 ( N 2 ) 2
=
P ( N1 ) 2
1
where,
• P = pressure (Pa)
12. 11
Air and impeller power is proportional to
3.
fan speed cubed, i.e.
W2 ( N 2 )3
=
W1 ( N1 ) 3
where,
• W = power (W or kW)
13. 12
Example 2
A 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.
15. 14
As fans are not totally efficient, the following formula
may be applied to determine the percentage
Total fan pressure x air volume 100
Efficiency = x
Absorbed power (W) 1
So, for the previous example,
390 x 5 100
Efficiency = = 50%
x
3900 1
16. 15
Types of fan
Cross-flow or tangential
1.
Propeller
2.
Axial flow
3.
Centrifugal
4.
17. 16
● Cross-flow or tangential fan
Tangential or cross-flow fan
44. 43
Ductwork
Circular, square or rectangular cross-sections
More efficient, less
frictional resistance
to airflow
Convenience,
more easily
fitted into
building fabric
Circular & rectangular ductwork
46. 45
Duct conversion
For equal velocity of flow
2ab
d=
a+b
For equal volume of flow
0.2
⎡ (ab) ⎤ 3
d = 1.265 x ⎢ ⎥
a+b⎦
⎣
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
47. 46
Example 3 (duct conversion)
A 450 mm diameter duct converted to rectangular
profile of aspect ratio 2 : 1 (a = 2b).
For equal velocity of flow:
2 x 2b x b 4b 2 4b
2ab
d= 450 = = =
a+b 2b + b 3b 3
3 x 450
b= Therefore, b = 337.5 mm and a = 2b = 675 mm
4
48. 47
0.2
⎡ (ab) ⎤
3
For equal volume of flow: d = 1.265 x ⎢ ⎥
a+b⎦
⎣
3 0.2
⎡ (2b x b) ⎤
450 = 1.265 x ⎢ ⎥
⎣ 2b + b ⎦
0.2
⎡ (2b ) ⎤
23
450 = 1.265 x ⎢ ⎥
⎣ 3b ⎦
From this, b = 292 mm and a = 2b = 584 mm
49. 48
Duct conversion – using conversion chart (simpler
but less accurate)
Circular to rectangular ductwork conversion chart
56. 55
“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
64. 63
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 ductwork
Equal 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
66. 65
Example 4 (ventilation design calculation)
Q, air volume flow rate (m3/s) = Room volume x air changes per hour
Time in seconds
67. 66
Given
Room volume = 480 m3
Air changes per hour = 6
Therefore
480 x 6
Q= = 0 .8 m 3 / s
3600
68. 67
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
70. 69
450
320
From the design chart:
A
• Duct A = 450 mm Ø
• Duct B = 320 mm Ø
B
71. 70
From duct design
chart (equal
velocity method)
72. 71
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 Pa
Duct 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.
73. 72
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 Ø
74. 73
From duct design
chart (Velocity
reduction method)
75. 74
Friction in duct A = 1.00 Pa x 5 m = 5.0 Pa
Friction 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.
76. 75
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 Ø with a frictional
resistance of 0.65 Pa/m
Duct B (using the same friction) at 0.4 m3/s = 350 Ø
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
77. 76
From duct design
chart (Equal friction
method)
78. 77
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 Ø
80. 79
From
Q = Room volume x air changes per hour
Time in seconds
and,
Air changes per hour = Q x time seconds
Room volume
= 0.1 x 3600
180
Thus, 2 changes per hour would be provided
81. 80
REFERENCES
Greeno, R.(1997). Building Services,
Technology and Design. Essex:
Longman.
Hall, F. & Greeno, R. (2005). Building
Services Handbook. Oxford: Elsevier.
82. 81
QUIZ
Name 5 purposes of ventilation
What is “coanda effect”?