Calculate Size of Staircase Pressurization Fan for Highrise Building
May 25, 2025 Leave a comment
Calculate Size of Staircase Pressurization Fan having following Details
- Building Height is 49 Meter
- No of Staircase Door is 14 No’s
- No of Fire Escape Door is 1 No at ground Floor
- Staircase Door is 0.9 meter width and 2 meter height
- Door is single Leaf and opening at Staircase (Pressurization) Side
- Air Velocity across door is 0.75 m/sec
Calculation:
(1) Air volume required when All doors are closed:
- Design Pressure differential Level as per building height is as under
|
Pressure Level |
|||
|
Building Height (meter) |
Fire Pressure (Pa) |
Wind Stack Effect (Pa) |
Design Pressure (Pa) |
|
0 |
8.5 |
8 |
25 |
|
5 |
8.5 |
8 |
25 |
|
25 |
8.5 |
10.5 |
25 |
|
50 |
8.5 |
13 |
50 |
|
100 |
8.5 |
19.5 |
50 |
|
150 |
8.5 |
29.5 |
50 |
- Air volume required when all doors are closed (Q1) = 0.827 x AE x P(1/n)
- Where AE = Leakage Area from the space (m2)
- P =Pressure Differential
- n=Leakage Factor
- As per above Table considering Air Pressure differential (P) = 50 Pa
- As per following Table Single Leaf Doors in Frame Opening into Pressurized Space =0.01 m2
|
Type of Door |
Leakage Area (m2) |
|
Single Leaf Doors in Frame Opening into Pressurized Space |
0.01 |
|
Single Leaf Doors in Frame Opening Outwards |
0.02 |
|
Double Leaf Doors with or without Central Rebate |
0.03 |
|
Lift Door |
0.06 |
- Here No of Staircase Door are 14 No’s
- AE=Total Leakage Area = 0.01 x14 = 0.14 No’s
- n=Leakage factor for Door is 2 as per following Table
|
Leakage Factor |
n |
|
Leakages area like Door |
2 |
|
Leakages small area like window crack |
1.6 |
- Air volume required when all doors are closed (Q1) = 0.827 x AE x P(1/n)
- Air volume required when all doors are closed (Q1) = 0.827 x 0.14x 50(1/2)
- Air volume required when all doors are closed (Q1) =0.82 m3 / sec
- It is assumed that there is other leakage, which are not calculated above is 50%.
- Air volume required when all doors are closed (Q1) =0.82 X50%
- Air volume required when all doors are closed (Q1) =1.23 m3/sec
(2) Air volume required when doors are opened:
- Area of Staircase Door = Length x width = 2 x 0.9
- Area of Staircase Door (A)= 1.8 m2
- Air Velocity across door (V) is 0.75 m/s =148 fpm
- It is assumed that Minimum Number of Opened Doors =Escape Door + 10% of remaining Doors.
- Minimum Number of Opened Doors = 1 + (14×10%) = 1+2
- Opened Door Area = Escape Door Area + 50% of Remanning Door area
- Opened Door Area = (1×1.8) + ((2×1.8) x50%) =1.8 +1.8 =3.6 m2
- Opened Door Area (A) =3.6 m2
- Air volume required when doors are opened (Q2) = A x V
- Air volume required when doors are opened (Q2) = 3.6 x 0.75
- Air volume required when doors are opened (Q2) = 2.7 m3/sec
(3) Total Air Supplied by the fan:
- Total Air Supplied = Air volume when doors are Closed+ Air volume when doors are opened
- Total Air Supplied =Q1+Q2
- Total Air Supplied = 1.23 + 2.7 m3/sec
- Total Air Supplied = 3.93 m3/sec
- Total Air Supplied = 3.93 x 2113
- Total Air Supplied = 8300 CFM
- Total Air Supplied per Floor = 8300/ 14 = 593 CFM
Conclusion:
- Capacity of Staircase Pressurization Fan = 8300 CFM





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