Mechanical Smoke Ventilation CalculationsFor typical offices floors

Mechanical Ventilation system for typical offices floors# Assumptions:-Fire Area = 3 x 3 = 9 sq.mSmoke Plume High (Y) = 2.5 MTotal height (H) = 3.5 MSmoke layer thickness (db) = H - Y = 1 MFloor Area = 1050 sq.m

# Determining the heat release

Q = q x A …………………………(1)where q = 93 KW/sq.M Refference :- BS7436 tables with aid of NFBA 204M tables .From equation (1)

Qs = 837 kw

# Determining smoke mass flow rate MsMs = 0.188 PY 3/2 ………………… (2)where P = Fire Perimeter = 12 M Y = Clear height of smoke plume = 2.5 MFrom equation (2)

Ms = 8.9 kg/s.

# Determining plume temperature rise :

dT =Qs

………….(3)Ms x Cps

dT = 93 KAnd 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

……………….(5)

Where :-V = Required air volumetric flow rate

Ms = 0.188 x12 x 2.53/2

To ps

ps = Smoke density = 1.2 Kg/m3

V = 9.7

= 20,584 CFM ~~= 21,000 CFM

# Determining duct size and fan static pressure :-

Where v = volumetric flow rate for one fan = 21,000 cfm

OR. v = volumetric flow rate for one fan = 9.7 A = duct cross sec. Area V = desired air velocity in the duct = 6

A = 1.62 Static pressure ( in.wg) = 0.5 + duct friction losswhere duct friction loss = 0.5 in.wgTotal Static pressure ( in.wg) = 1 in.wg.

Conclusion :System Description :Exhaust Smoke Fans :Nunmers: 2 Nos. Fans/ZoneCapacity: 10,500 cfm (each)Type : Tubilar ducted Location : Roof Top

Duct Size = 7.00 Sq.ft = 60" X 22"Duct Size = 0.65 Sq.M = 152 X 56 CM

Fresh Air Fans:Fresh Air Fans capacity shall be same as exhaust air capacity .

Fresh Air properties shall be :Nunmers: 2 No. FanCapacity: 10,500 cfm Type : Tubilar ducted Location : Roof TopTotal Static pressure ( in.wg) = 1 in.wg.

GENERAL

- All fans used in smoke management and exhaust system ( exhaust and fresh air fans )

- 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.

M3/Sec.

V = A X v ………………………….(8)

M3/Sec.

M/Sec.

M2

shall be capable to withstand 200 oC temp.for 30 min.

input the sugested fansnumbers here . If u put it as one it gives the total flow required .

in case if the enterances area is less than the required area , so, the fresh air fans are required .

Mechanical Smoke Ventilation CalculationsFor Basement Floor ( Store)

Mechanical Ventilation system for basement floor# Assumptions:-Fire Area = 3 x 3 = 9 sq.mSmoke Plume High (Y) = 2.5 MTotal height (H) = 3.5 MSmoke layer thickness (db) = H - Y = 1 MFloor 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 MsMs = 0.188 PY 3/2 ………………… (2)where P = Fire Perimeter = 12 M Y = Clear height of smoke plume = 2.5 MFrom equation (2)

Ms = 8.9 kg/s.

# Determining plume temperature rise :

dT =Qs

………….(3)Ms x Cps

dT = 390 KAnd Ts = To + DT …………………….(4)Where :Ts = Smoke temp. To =Ambient temp. = 303 K

Ts = 693 K

# Determining the required air flow rate :

V = Ms Ts

……………….(5)

Where :-V = Required air volumetric flow rate

V = 17.0

= 36,014 CFM ~~= 37,000 CFM

Ms = 0.188 x12 x 2.53/2

To ps

ps = Smoke density = 1.2 Kg/m3

M3/Sec.

# Determining duct size and fan static pressure :-

Where v = volumetric flow rate for one fan = 37,000 cfm

OR. v = volumetric flow rate for one fan = 17.0 A = duct cross sec. Area V = desired air velocity in the duct = 6

A = 2.83 Static pressure ( in.wg) = 0.5 + duct friction losswhere duct friction loss = 0.5 in.wgTotal Static pressure ( in.wg) = 1 in.wg.

Conclusion :System Description :Exhaust Smoke Fans :Nunmers: 2 Nos. Fans/ZoneCapacity: 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 doorsAverage door size (m) = 2.2 (width) , 2 (height)

= 4.4

Total fresh air intake area = 8.8

Where V = The required fresh air supply = 6 CFM

= - Assuming average air velocity through doors to be 2m/s ( to allow door opening for occupants evacuation )

A = - The required area is less than doors and enterance areasSO, 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 andpark ramp opening to deliver adequate fresh air amount for the other half of the floor area .

Fresh Air properties shall be :Nunmers: 2 No. FanCapacity: 10,500 cfm Type : Tubilar ducted Location : Roof TopTotal Static pressure ( in.wg) = 1 in.wg.

GENERAL

- All fans used in smoke management and exhaust system ( exhaust and fresh air fans )

V = A X v ………………………….(8)

M3/Sec.

M/Sec.

M2

m2

m2

V = A X v ………………………….(8)

M3/Sec.

m2

shall be capable to withstand 200 oC temp.for 30 min.

input the sugested fansnumbers here . If u put it as one it gives the total flow required .

EDIT THE ACTUAL NO. AS PER ARCH. DRAWINGS .

in case if the enterances area is less than the required area , so, the fresh air fans are required .

- 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.

Left Well Pressurization fan For elevators .

No. Of Stops = Gr. + 14 Floors + Service Floor = 16 Floors

No. Of elevators = 2 Nos.

No. Of Elev.Doors = 2 (Gr.) + 30 (Typical Floors) = 32 Doors

Door Size :-W = 0.8 MH = 2.0 m

Where :

Ae = 0.003 x ( ( 2.0 + 0.8 ) x 2 ) + 2.0 x 32 = 0.73

K = 0.839 P = 50 Pascal.

Qe = 4.3 = 152.8 CFM

In Case One Door Opened :

Where :- A = Door Area

Qe = KAe (P)1/2

Qe = 0.839 x Ae x (50)1/2

M3/Sec.

just inter the no. of floors

don’t touch it formated cell

don’t touch it formated cell

0.73 M2

Mechanical Ventilation CalculationsFor Car Park ( Basement )

1-Mechanical Ventilation system

Location :Floor Area : 8000 Sq.MFloor Hiegh : 2.8 M Floor Volume = 22400 Cu.M

No. of air change per hour = 4 times

Total air volume = 89600 Cu.M = 3166078 Cu.Ft

Total flow rate (cfm) = 52767.96 cfm <=> 52800 cfm

Fan selection

Fan Capacity = 13200 cfmNo. of fans = 4 nos. fans

CONCLUSIONSystem Discription :Exhaust Smoke Fans :Nunmers: 4 Nos. FansCapacity: 13,200 cfm (each)Type : Tubilar ducted Location : ………………………………………….

follow the regulation for the area application ,for basement 4times/hr.

inter the floor area here

inter the floor hight here

according to plane (or the posibility to install many nos. of fans to reduce the cap. Of the fan .

Mechanical Smoke Ventilation CalculationsFor …………………………………

Mechanical Ventilation system for ………………………

M f = 0.58 p Q g L2 1/3

2 + 1 + 0.22 Z + 2

L

WHERE :-

Q = Heat Flux = 5000 KW FOR Shoping Araes p = Density of smoke

=Cp = Specific Heat

= 1.01 Kj/KgKg = Gravity Accel.

=

= Abs.Temp - Amb.Temp. = 303 K

L = Length of void edge past which gases spill = Width of void = 3 m

Z = 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 :-

M f = 0.58 p 1/3

2 + 1.2 1 + 0.22 Z + 2 1.2

3

M f = 5193.05

3

p Cp T0

1.2 Kg/m3

9.81m/sec2

T 0

Q g L2

p Cp T0

2/3

2/3

Stair Case Pressurization For Internal Stairs .

No. Of Floors = Gr. + 6 Floors + Service Floor = 8 Floors

No. Of Doors = 1 (Gr.) + 6 Typical Floors + Service Flo = 8 Doors

Door Size :-W = 0.8 MH = 2.0 m

Where :Ae = 0.003 x ( ( 2.0 + 0.8 ) x 2 ) + 2.0 x 8 = 0.18

K = 0.839 P = 50 Pascal.

Qe = 11.9 = 418.8 CFM

In Case One Door Opened :

Where :- A = Door Area V = Air Velocity = 0.75 m/sec.

Area Factor = 0.6

Qo =V A

=0.75 x 2.0 x 0.8

= 2.00 Af 0.6

= 4236 CFM

Conclusion :System Description :Left Well Pressurization fan For elevators .Qty. : 1 Nos. Fan/stair caseCapacity : 4,300 cfm (each)Type : Tube Axial Fan Location : stair case Roof

1- Provide one no. fresh air pressurized fan located on each stair case roof of 4,300 CFM & 1.5 In.wg.

Qe = KAe (P)1/2

Qe = 0.839 x Ae x (50)1/2

M3/Sec.

Af =

M3/Sec.

2- Provide deffrential pressure sensor at the midlle way of stair case high and interfacewith fan control panel .

0.18 M2

MECHANICAL DEPARTMENT.

DESCRIPTION VALUE

NUMBER OF CARS 64.00

HIGHT,H 10.00

AREA OF PARKING,A 3700.00

AREA OF PARKING,AF 39826.80

TIME OF CAR OPERATION, Ө 120.00

MAX CO2 LEVEL REQUIRED 25.00

OPERATION CARS AT PEAK 40.00

NUMBER OF OPERATION CARS AT PEAK, N 25.60

EMISSION RATE , E 700.00

GENERATION RATE G 0.45

2.48

F 18.14

C 0.0001363

VENTELATION RATE 0.30

REQUIRED AIR CHANGE /HR 1.78

AIR FLOW RATE 11819

CAR PARKING VENTELATION CALCULATION *

NORMALIZE VALUE OF GENERATION RATE ,GO

STATIC PRESSURE 2.00

FAN POWER 7.88

* 1999 ASHRAE Application-A 12 Page ( 12.9 )

D:\MECH\ABD ELRAHIM\CURRENT PROJECTS-2008\05-08-1--- AL FAYHA\MECH\MECH-06

H:\MECHANICAL DEPT\05-08-1--- AL FAYHA\FAIHAA BASEMENT& GF\AR-02A_02B_02C BASEMENT FLR PLAN.dwg

DATE : 6 JUNE 2008

PROJECT : xxxx HOSP.

UNIT I/O

CAR INPUT

FT INPUT

M² INPUT

FT² OUTPUT

SEC RECOMMENDED

PPM RECOMMENDED

% RECOMMENDED

CAR OUTPUT

GAL/HR AVERAGE

g/hr.ft² OUTPUT

g/hr.ft² RECOMMENDED

- OUTPUT

CFM/FT²-S RECOMMENDED

OUTPUT

ACH OUTPUT

CFM OUTPUT

CAR PARKING VENTELATION CALCULATION *

CFM/FT²

IN" W INPUT

HP OUTPUT

CALCULATED BY :

ENG / M.ABD EL RAHIM

H:\MECHANICAL DEPT\05-08-1--- AL FAYHA\FAIHAA BASEMENT& GF\AR-02A_02B_02C BASEMENT FLR PLAN.dwg

VEVTILATION CALCULATION FOR THE TRANSFORMER ROOM

PROJECT :

1- ASSUMPTION - Ambient Temperature = 115 F - Max. Allowed Temp. Rise = 10.8 F - Sub- Station Ref. No. 1 - Number Of Transformers = 1 Nos. - Transformer Rateing = 1000 KVA - Number Of Transformers = 0 Nos. - Transformer Rateing = 1000 KVA - Power Factor = 0.8 - Transformer Eff. = 0.98

2- CALCULATIONHeat Loss / Transformer = ( 1 x 1000 + 0 x 1000 )x 0.8 x ( 1 - 0.98 ) = 16 KW = 16000 Watt

Total Heat Dissipation = = 16000 x 3.413 = 54608 Btu/hr = 1.08 x CFM x Temp. Rise = 1.08 x CFM x 10.8

Air Quantity Required for ventilationThen :-

The Air Quantity = 4,682 CFM

VENTILATION SYSTEM DISCRIPTION

Number of fans = 1Fan Capacity / Each = 4,682 = CFMFan rating KW/Each = 1.11 KWFan st. Pressure = 1 IN.WG

GENERATOR ROOM VENTILATION CALCULATION

Diesel Generator capacity = 750 KVA

Ventilating air (cu.m/min.) = H

+ Engine Combustion AirD x SH x Dt

Where :- H = Total Heat Radiation in (KW) (for engine + generator) H = 125.5 KW (Ref. To specs. Sheet) V = Ventilating air (cu.m/min.) D = Density of air (kg/m3) = 1.07 At 54 deg. C SH = Specific heat of air ( kw.min/Kg.C) = 0.017 d T = Temp. Rise inside engine room ( deg.C) = 10 deg.C ECA = Engine Combustion Air (Exhaust gas flow ) = 120 cu.m/min (Ref. To specs. Sheet)

7,182 cu.m/hr.

Radiator fan air flow = 17.6 cu.m/sec. (Ref. To specs. Sheet) = 1056 cu.m/min = 37,256 cu.ft/min.

THEN :-

Ventilating air ( V ) =125.5

+ 1201.07 x 0.017 x 10

V = 810 cu.m/min

V = 28,564 cu.ft/min.

INTAKE LOUVER SIZE

- Consider Air Velocity = 1000 ft/min.Total Air Quantity = 28,564 cu.ft/min. Louver Area = 29 Sq.ft

2.7 Sq.m

Important Notes :-

1-

2- Cool air shoul always be available for the engine air cleaner (air filter ).

3- For best ventilation results , air should flow first across the generator then to both sides of the engine .

4-

clean , cool , dry air circulates around the switchgear , flows through the rear of the generator , across the engine , and discharges through the radiator

Inlets located at the end of the room will provide adequate ventilation only to the engine nearast the inlet.

5- Air flow restriction should be 0.5 inch W.G

For best ventilation results , air should flow first across the generator then to both sides of the engine .

clean , cool , dry air circulates around the switchgear , flows through the rear of the generator , across

Inlets located at the end of the room will provide adequate ventilation only to the engine nearast the