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Design calculation sheetProject no: Date: Sheet no.: 1 2 Computed by:
Subject: ACC Checked by:
Stairwell Pressurisation with some doors opened. Summer Approved by:
Fire Escape on : 4 FloorsStair Doors: 4Escape Door : 1Building Height : 15.6 m
25 Pa
1-
Q =
= 0.030 Double Leaf Doors with or without Central Rebate= 0.120 4 Stair Doors
P = Pressure Differential = 25 Pa n = Leakage Factor = 2
0.827 x 0.12 x 25 : 0.50
Although the effective leakage area is known, there are other leaks we are not aware of.To account for these leaks, Q is increased by 50%, hence:
0.50 +50% 0.74
2-
Door Area = 2.1 x 0.9 = 1.89Air Velocity Across Door = 2 m/s = 394 fpm
Minimum Number of Opened Doors = main escape doors + 10% of remaining straiwell doors= 1 + 1
Open door area = Area of escape doors + 50% of the area of the other doors= 1.89 + 0.945= 2.84
5.673- Total Air Supplied by the fan :
6.4 = 13562 cfm3391 cfm per floor
Conditions :
Pressure Level :
Air volume required when all doors are closed : Q1= 0.827 x AE x P1/n
Volume of Air Required (m3)
AE = Leakage Area from the space (m2)(m2) for(m2) for
Q1 = Q1 = m3 / sec
Q1 = Q1 = m3 / s
Air volume required when doors are opened :
m2
m2
Q2 = A x V => Q2 = m3 / s
QT = Q1 + Q2 => QT = m3 / s
1/2
of
Design calculation sheetProject no: Date: Sheet no.: 2 2 Computed by:
Subject: Checked by:
Stairwell Pressurisation with some doors opened. Summer Approved by:
6.4 = 13562 cfm = 6400 l/s
at 2500 fpm => f = 0.5 "/100 ft
Duct Length = 120 m
Friction Loss through duct: = (Duct Length x 1.25 x 3.28 x f)
Friction Loss = 2.46 "
Total Static Pressure = (PD at Register + Friction Loss + Volume Damper)*30%= 0.20 + 2.46 + 0.2= 3.80 "= 9.7 mm= 946 Pa
Total Static Pressure on Fa = 946 + 25
Total Fan Static Pressure = 971 Pa = 3.9 " = 99.0 mm
5.7
Area of Pressure Relief A = 5.70.827 x 25
A = 1.37
Since we are using two fans:
Air flow per Fan = 6781 cfm = 3200 lps@ 2500 fpm Φ = 600 mm
Total Static Pressure on Fan = 633 + 25
Total Static Pressure on Fan = 658 Pa = 2.6 "
QT = m3 / sec
Relief Damper Sizing :
Air to be releived = Q2 = m3/sec
m2
½
of
main escape doors + 10% of remaining straiwell doors
Design calculation sheetProject no: Date: Sheet no.: 1 2 Computed by:
Subject: LAB Checked by:
Stairwell Pressurisation with some doors opened. Summer Approved by:
Fire Escape on : 3 FloorsStair Doors: 3Escape Door : 1Building Height : 11.4 m
25 Pa
1-
Q =
= 0.030 Double Leaf Doors with or without Central Rebate= 0.090 3 Stair Doors
P = Pressure Differential = 25 Pa n = Leakage Factor = 2
0.827 x 0.09 x 25 : 0.37
Although the effective leakage area is known, there are other leaks we are not aware of.To account for these leaks, Q is increased by 50%, hence:
0.37 +50% 0.56
2-
Door Area = 2.1 x 0.9 = 1.89Air Velocity Across Door = 2 m/s = 394 fpm
Minimum Number of Opened Doors = main escape doors + 10% of remaining straiwell doors= 1 + 1
Open door area = Area of escape doors + 50% of the area of the other doors= 1.89 + 0.945= 2.84
5.673- Total Air Supplied by the fan :
6.2 = 13138 cfm4380 cfm per floor
Conditions :
Pressure Level :
Air volume required when all doors are closed : Q1= 0.827 x AE x P1/n
Volume of Air Required (m3)
AE = Leakage Area from the space (m2)(m2) for(m2) for
Q1 = Q1 = m3 / sec
Q1 = Q1 = m3 / s
Air volume required when doors are opened :
m2
m2
Q2 = A x V => Q2 = m3 / s
QT = Q1 + Q2 => QT = m3 / s
1/2
of
Design calculation sheetProject no: Date: Sheet no.: 2 2 Computed by:
Subject: Checked by:
Stairwell Pressurisation with some doors opened. Summer Approved by:
6.2 = 13138 cfm = 6200 l/s
at 2500 fpm => f = 0.5 "/100 ft
Duct Length = 120 m
Friction Loss through duct: = (Duct Length x 1.25 x 3.28 x f)
Friction Loss = 2.46 "
Total Static Pressure = (PD at Register + Friction Loss + Volume Damper)*30%= 0.20 + 2.46 + 0.2= 3.80 "= 9.7 mm= 946 Pa
Total Static Pressure on Fa = 946 + 25
Total Fan Static Pressure = 971 Pa = 3.9 " = 99.0 mm
5.7
Area of Pressure Relief A = 5.70.827 x 25
A = 1.37
Since we are using two fans:
Air flow per Fan = 6569 cfm = 3100 lps@ 2500 fpm Φ = 600 mm
Total Static Pressure on Fan = 633 + 25
Total Static Pressure on Fan = 658 Pa = 2.6 "
QT = m3 / sec
Relief Damper Sizing :
Air to be releived = Q2 = m3/sec
m2
½
of
main escape doors + 10% of remaining straiwell doors
Design calculation sheetProject no: Date: Sheet no.: 1 2 Computed by:
Subject: ONC Checked by:
Stairwell Pressurisation with some doors opened. Summer Approved by:
Fire Escape on : 3 FloorsStair Doors: 3Escape Door : 1Building Height : 11.4 m
25 Pa
1-
Q =
= 0.030 Double Leaf Doors with or without Central Rebate= 0.090 3 Stair Doors
P = Pressure Differential = 25 Pa n = Leakage Factor = 2
0.827 x 0.09 x 25 : 0.37
Although the effective leakage area is known, there are other leaks we are not aware of.To account for these leaks, Q is increased by 50%, hence:
0.37 +50% 0.56
2-
Door Area = 2.1 x 0.9 = 1.89Air Velocity Across Door = 2 m/s = 394 fpm
Minimum Number of Opened Doors = main escape doors + 10% of remaining straiwell doors= 1 + 1
Open door area = Area of escape doors + 50% of the area of the other doors= 1.89 + 0.945= 2.84
5.673- Total Air Supplied by the fan :
6.2 = 13138 cfm4380 cfm per floor
Conditions :
Pressure Level :
Air volume required when all doors are closed : Q1= 0.827 x AE x P1/n
Volume of Air Required (m3)
AE = Leakage Area from the space (m2)(m2) for(m2) for
Q1 = Q1 = m3 / sec
Q1 = Q1 = m3 / s
Air volume required when doors are opened :
m2
m2
Q2 = A x V => Q2 = m3 / s
QT = Q1 + Q2 => QT = m3 / s
1/2
of
Design calculation sheetProject no: Date: Sheet no.: 2 2 Computed by:
Subject: Checked by:
Stairwell Pressurisation with some doors opened. Summer Approved by:
6.2 = 13138 cfm = 6200 l/s
at 2500 fpm => f = 0.5 "/100 ft
Duct Length = 120 m
Friction Loss through duct: = (Duct Length x 1.25 x 3.28 x f)
Friction Loss = 2.46 "
Total Static Pressure = (PD at Register + Friction Loss + Volume Damper)*30%= 0.20 + 2.46 + 0.2= 3.80 "= 9.7 mm= 946 Pa
Total Static Pressure on Fa = 946 + 25
Total Fan Static Pressure = 971 Pa = 3.9 " = 99.0 mm
5.7
Area of Pressure Relief A = 5.70.827 x 25
A = 1.37
Since we are using two fans:
Air flow per Fan = 6569 cfm = 3100 lps@ 2500 fpm Φ = 600 mm
Total Static Pressure on Fan = 633 + 25
Total Static Pressure on Fan = 658 Pa = 2.6 "
QT = m3 / sec
Relief Damper Sizing :
Air to be releived = Q2 = m3/sec
m2
½
of
main escape doors + 10% of remaining straiwell doors
Design calculation sheetProject no: Date: Sheet no.: 1 1 Computed by:
Subject: Checked by:
Approved by:
6.4 m3/s = 13562 cfm = 6400 lpsRoof
At 2500 fpm : f = 0.50 ''/100 ft.700 x 4006400 lps Grille Sizing based on 1000 fpm
Eight face velocity
600 x 400 10 grilles => 640 lps/grille5760 lps 1356 cfm/grille
SevenP.D = 3.8 '' = 971 Pa
550 x 4005120 lps Fan Selected : Woods
SixModel : 50JM/20/2/6/22
500 x 4004480 lps 380/3/50 3.3 kW
Five
450 x 4003840 lps
Four
400 x 4003200 lps
Three
400 x 3502560 lps
NTV
350 x 3501920 lps
NTV
350 x 2501280 lps
Mezzanine
350 x 150640 lps
Ground
SR 500x150640 lps Typical
of
Data
Page 14
Building Height Fire Pressure Wind Stack Effect Design Pressure
(m) (Pa) (Pa) (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
Size
2 m x 800 mm 5.6 0.01
Single Leaf Doors in Frame Opening Outwards 3 m x 800 mm 5.6 0.02
Double Leaf Doors with or without Central Rebate 2 m x 1.6 m 9.2 0.03
Lift Door 2 m High x 2 m Wide 8.0 0.06
Double Leaf Doors with or without Central Rebate
7.2.3.9 Stair Pressurization.
7.2.3.9.1
Smokeproof enclosures using stair pressurization shall use an approved engineered system with a design
pressure difference across the barrier of not less than 0.05 in. water column (12.5 Pa) in sprinklered
buildings, or 0.10 in. water column (25 Pa) in nonsprinklered buildings, and shall be capable of
maintaining these pressure differences under likely conditions of stack effect or wind. The pressure
difference across doors shall exceed that which allows the door to begin to be opened by a force
of 30 lbf (133 N) in accordance with 7.2.1.4.5.
7.2.1.4.5
The forces required to fully open any door manually in a means of egress shall not exceed 15 lbf (67 N) to release
the latch, 30 lbf (133 N) to set the door in motion, and 15 lbf (67 N) to open the door to the minimum required width.
Opening forces for interior side-hinged or pivoted-swinging doors without closers shall not exceed 5 lbf (22 N).
These forces shall be applied at the latch stile.
Crack Length (m)
Leakage Area (m2)
Single Leaf Doors in Frame Opening into Pressurized Space
Design calculation sheetProject no: Date: Sheet no.: Computed by:
Subject: Checked by:
Approved by:
of