Mechanical Smoke Ventilation Calculations
For …………………………………
Mechanical Ventilation system for ………………………
Mf
=
Q g L2
p Cp T0
0.58 p
1/3
2 +
1+
0.22
Z +2
L
WHERE :Q
p
Cp
g
T0
L
Z
=
=
=
=
=
=
=
Heat Flux
= 5000
Density of smoke
1.2 Kg/m3
Specific Heat
1.01 Kj/KgK
Gravity Accel.
9.81m/sec2
KW
FOR Shoping Araes
=
=
=
=
=
=
=
=
=
=
=
Abs.Temp - Amb.Temp.
303 K
Length of void edge past which gases spill
Width of void
3 m
Hight of rise of thermal plume above void edge
H - d2 = 4.0 - 0.8
3.2 m
Emprical hight of virtual source below void edge
0.3 H = 0.3 x 4
1.2 m
THEN :-
Mf
=
0.58 p
Mf
=
5193.05
Q g L2
p Cp T0
1/3
3
2 +
1.2
1+
0.22
Z + 2 1.2
3
2/3
2/3
Mechanical Smoke Ventilation Calculations
For …………………………………
Mechanical Ventilation system for ………………………
# Assumptions:Fire Area
=
3x3
=
Smoke Plume High (Y)
=
Total height (H)
=
Smoke layer thickness (db) = H - Y
=
Floor Area
=
9
2.5
3.5
1
sq.m
M
M
M
sq.m
# Determining the heat release
Q = q x A …………………………(1)
where q =
93 KW/sq.M (for vehicles)
Refference :- BS7436 tables with aid of NFBA 204M tables .
From equation (1)
Qs
=
837
HEAT RELEASE =
93 KW/M2 FOR FURNATURE
310 KW/M2 FOR CELLULOSICS
(‫)ورقكققارتونلققلتعبئه واققلتغليفو م نتجاتقققطنية‬
260 KW/M2 FOR VIHICLES(car park)
kw
# Determining smoke mass flow rate Ms
Ms = 0.188 PY 3/2 ………………… (2)
where P = Fire Perimeter
=
Y = Clear height of smoke plume =
From equation (2)
Ms = 0.188 x12 x 2.53/2
Ms =
8.9 kg/s.
12 M
2.5 M
# Determining plume temperature rise :
Qs
dT
=
………….(3)
Ms x Cps
dT
=
93 K
And
Ts = To + DT …………………….(4)
Where :
Ts = Smoke temp.
To =Ambient temp. =
303 K
Ts
=
396 K
# Determining the required air flow rate :
V
=
Ms Ts
To ps
……………….(5)
Where :V = Required air volumetric flow rate
ps = Smoke density = 1.2 Kg/m3
V =
=
9.7 M3/Sec.
20,584 CFM
~~=
21,000 CFM
# Determining duct size and fan static pressure :V = A X v ………………………….(8)
Where v = volumetric flow rate for one fan =
OR.
v = volumetric flow rate for one fan =
A = duct cross sec. Area
V = desired air velocity in the duct =
21,000 cfm
9.7 M3/Sec.
6 M/Sec.
A =
1.62 M2
Static pressure ( in.wg) = 0.5 + duct friction loss
where duct friction loss = 0.5 in.wg
Total Static pressure ( in.wg) = 1 in.wg.
Conclusion :
System Discription :
Exhaust Smoke Fans :
Nunmers:
2 Nos. Fans/Zone
Capacity:
10,500 cfm (each)
Type
: Tubilar ducted
Location
: ………………………………………….
# Determining fresh air intake area :
No. of entrances and doors =
Average door size (m)
=
=
Total fresh air intake area
=
V
= AXv
2
2.2
4.4
8.8
input the sugested fans
numbers here .
If u put it as one it gives
the total flow required .
doors
(width) ,
m2
m2
EDIT THE ACTUAL NO.
AS PER ARCH.
DRAWINGS .
2 (height)
………………………….(8)
Where V = The required fresh air supply
=
21,000 CFM
=
9.7 M3/Sec.
Assuming average air velocity through doors to be 2m/s ( to allow door opening for
occupants evacuation )
in case if the enterances area
A=
4.86 m2
is less than the required
The required area is less than doors and enterance areas
area , so, the fresh air fans
SO, Enterance area is sufficient as fresh air supply openings .
are required .
Fresh Air Fans:
Fresh Air Fans capacity shall be half of exhaust air capacity depending on the doors and
park ramp opening to deliver adequate fresh air amount for the other half of the floor area .
Fresh Air properties shall be :
Nunmers:
2 Nos. Fans
Capacity:
5,250 cfm (each)
Type
: Tubilar ducted
Location
: ………………………………………….
Total Static pressure ( in.wg) = 1 in.wg.
GENERAL
- All fans used in smoke management and exhaust system ( exhaust and fresh air fans )
shall be capable to withstand 200 oC temp.for 30 min.
- All fans shall be interconnected with fire allarm panel for automatic operation also, shall be
provided with ON /OFF switch near enterances for manual operation by occupants.
Mechanical Ventilation Calculations
For Car Park ( Basement )
inter the floor area here
1-Mechanical Ventilation system
inter the floor hight here
Location
Floor Area
Floor Hiegh
Floor Volume
:
:
:
=
690 Sq.M
3.5 M
2415 Cu.M
follow the regulation for the
area application ,
for basement 4times/hr.
No. of air change per hour
=
6 times
Total air volume
=
=
14490 Cu.M
512014.1 Cu.Ft
Total flow rate (cfm)
=
8533.569 cfm
<=>
9000 cfm
Fan selection
Fan Capacity
No. of fans
=
=
1800 cfm
5 nos. fans
CONCLUSION
System Discription :
Exhaust Smoke Fans :
Nunmers:
5 Nos. Fans
Capacity:
1,800 cfm (each)
Type
: Tubilar ducted
Location
: ………………………………………….
according to plane (or the
posibility to install many
nos. of fans to reduce the
cap. Of the fan .
Stair Case Pressurization For Internal Stairs .
No. Of Floors
=
No. Of Doors
=
Qe
Gr. +
(Gr.)
Floors +
+
Roof =
(Typical Floors)
2 Floors
+
(Roof )
=
0 Doors
= KAe (P)1/2
Where :
Ae
= 0.003 x
(( 2.2 + 0.8 ) x 2 ) x
0 =
0 M2
K = 0.839
P = 50 Pascal.
Qe = 0.839 x Ae x (50)1/2
Qe
=
0
M3/Sec.
=
0
CFM
Incase One Door Opened :
Qo
=
VA
0.6
=
0.75 x 2.2 x 0.8
0.6
=
2.2
M3/Sec.
=
4659.6
M3/Sec.