This document outlines calculations for a mechanical smoke ventilation system for a basement floor with a store. It determines the heat release rate and smoke mass flow rate based on the fire area and assumptions. It then calculates the required air flow rate, duct size, and fan specifications. The system will include two exhaust fans of 18,500 CFM each and two fresh air intake fans of 10,500 CFM each, all located on the roof. The ducts will be sized to accommodate the required air flow rates. The entrance areas are determined to provide sufficient fresh air intake.

1. ‫شرح‬‫مروحة‬‫الدخان‬
Mechanical Smoke Ventilation Calculations
For Basement Floor ( Store)
Mechanical Ventilation system for basement floor
#Assumptions:-
Fire Area = 3 x 3 = 9 sq.m
Smoke Plume High (.Y) = 2.5 M
Total height (H) = 3.5 M
Smoke layer thickness (db) = H - Y = 1 M
Floor Area = 1050 sq.m
#Determining the heat release
Q = q x A…………………………(1)
where q = 390 KW/sq.M
Refference :- BS7436 tables with aid of NFBA 204M tables.
From equation(1)
Qs = 3510 kw
#Determining smoke mass flow rate Ms
Ms = 0.188 PY 3/2…………………(2)
where P = Fire Perimeter = 12 M
Y = Clear height of smoke plume = 2.5 M
From equation(2)
Ms = 0.188 x12 x 2.53/2
Ms = 8.9 kg/s.
#Determining plume temperature rise:
dT = Qs…………(.3)
Ms x Cps
dT = 390 K
And
Ts = To + DT……………………(.4)
Where:

2. Ts = Smoke temp.
To =Ambient temp. = 303 K
Ts = 693 K
#Determining the required air flow rate:
V = Ms Ts………………(.5)
To ps
Where:-
V = Required air volumetric flow rate
ps = Smoke density = 1.2 Kg/m3
V = 17.0 M3/Sec.
=310,14CFM ~~= 37,000 CFM
#Determining duct size and fan static pressure:-
V = A X v…………………………(.8)
Where v = volumetric flow rate for one fan = 37,000 cfm
OR. v = volumetric flow rate for one fan = 17.0 M3/Sec.
A = duct cross sec. Area
V = desired air velocity in the duct = 6 M/Sec.
A = 2.83 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 Description:
Exhaust Smoke Fans:
Nunmers: 2 Nos. Fans/Zone
Capacity: 18,500 cfm (each)
Type : Tubilar ducted
Location : Roof Top
Duct Size = 12.33 Sq.ft = 60" X 22"
Duct Size = 1.15 Sq.M = 152 X 56 CM
#Determining fresh air intake area:

3. No. of entrances and doors = 2 doors
Average door size (m) = 2.2 (width) , 2 (height)
=4.4m2
Total fresh air intake area = 8.8 m2
V = A X v…………………………(.8)
Where V = The required fresh air supply
=1CFM
=-M3/Sec.
Assuming average air velocity through doors to be 2m/s ( to allow door opening for
occupants evacuation)
A = - m2
The required area is less than doors and enterance areas
SO, Enterance area is sufficient as fresh air supply openings.
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 No. Fan
Capacity: 10,500 cfm
Type : Tubilar ducted
Location : Roof Top
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.‫شرح‬
‫مروحة‬‫الدخان‬
Mechanical Smoke Ventilation Calculations
For Basement Floor ( Store)

4. Mechanical Ventilation system for basement floor
#Assumptions:-
Fire Area = 3 x 3 = 9 sq.m
Smoke Plume High (.Y) = 2.5 M
Total height (H) = 3.5 M
Smoke layer thickness (db) = H - Y = 1 M
Floor Area = 1050 sq.m
#Determining the heat release
Q = q x A…………………………(1)
where q = 390 KW/sq.M
Refference :- BS7436 tables with aid of NFBA 204M tables.
From equation(1)
Qs = 3510 kw
#Determining smoke mass flow rate Ms
Ms = 0.188 PY 3/2…………………(2)
where P = Fire Perimeter = 12 M
Y = Clear height of smoke plume = 2.5 M
From equation(2)
Ms = 0.188 x12 x 2.53/2
Ms = 8.9 kg/s.
#Determining plume temperature rise:
dT = Qs…………(.3)
Ms x Cps
dT = 390 K
And
Ts = To + DT……………………(.4)
Where:
Ts = Smoke temp.
To =Ambient temp. = 303 K
Ts = 693 K
#Determining the required air flow rate:

5. V = Ms Ts………………(.5)
To ps
Where:-
V = Required air volumetric flow rate
ps = Smoke density = 1.2 Kg/m3
V = 17.0 M3/Sec.
=310,14CFM ~~= 37,000 CFM
#Determining duct size and fan static pressure:-
V = A X v…………………………(.8)
Where v = volumetric flow rate for one fan = 37,000 cfm
OR. v = volumetric flow rate for one fan = 17.0 M3/Sec.
A = duct cross sec. Area
V = desired air velocity in the duct = 6 M/Sec.
A = 2.83 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 Description:
Exhaust Smoke Fans:
Nunmers: 2 Nos. Fans/Zone
Capacity: 18,500 cfm (each)
Type : Tubilar ducted
Location : Roof Top
Duct Size = 12.33 Sq.ft = 60" X 22"
Duct Size = 1.15 Sq.M = 152 X 56 CM
#Determining fresh air intake area:
No. of entrances and doors = 2 doors
Average door size (m) = 2.2 (width) , 2 (height)
=4.4m2
Total fresh air intake area = 8.8 m2

6. V = A X v…………………………(.8)
Where V = The required fresh air supply
=1CFM
=-M3/Sec.
Assuming average air velocity through doors to be 2m/s ( to allow door opening for
occupants evacuation)
A = - m2
The required area is less than doors and enterance areas
SO, Enterance area is sufficient as fresh air supply openings.
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 No. Fan
Capacity: 10,500 cfm
Type : Tubilar ducted
Location : Roof Top
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.