Graduation Project 2 Mechanical Systems For Al-Zakah Hospital Students:

Graduation Project 2Graduation Project 2

Mechanical Systems For Al-Zakah HospitalMechanical Systems For Al-Zakah Hospital

Students:Students: Mohammed Sawalhi (10840740)Mohammed Sawalhi (10840740) Mohammed Younis (10824906)Mohammed Younis (10824906) Mohammed Hannon (10822899)Mohammed Hannon (10822899) Nedal Al-Masri (10823665)Nedal Al-Masri (10823665) Aya Nairat (10821417)Aya Nairat (10821417)

Supervisor:Supervisor:Dr. Iyad AssafDr. Iyad Assaf

Introduction:Introduction:In our project, we are going to Design the following In our project, we are going to Design the following

Mechanical systems for Al Zakah hospital in Palestine-Mechanical systems for Al Zakah hospital in Palestine-Tulkarm which consists of 7 floors two of them under Tulkarm which consists of 7 floors two of them under the ground :the ground :

1- HVAC System “Heat, Ventilation & Air conditioning”1- HVAC System “Heat, Ventilation & Air conditioning”

2- Plumbing System2- Plumbing System

3- Fire Fighting System3- Fire Fighting System

4- Medical Gases System.4- Medical Gases System.

Building’s DescriptionBuilding’s DescriptionCountry : Palestine / West bank .Country : Palestine / West bank .

City: Tulkarm. City: Tulkarm.

Street Al-Montaza street .Street Al-Montaza street .

Latitude: 31.9 Latitude: 31.9

Building face sits at south orientation.Building face sits at south orientation.

The wind speed is greater than 5 m/s The wind speed is greater than 5 m/s

Heating load EquationsHeating load Equations

Qs,cond = U A (Ti – To) .

Q s,vent = 1.2 Vvent (Ti – To).

Q L,vent = 3 Vvent (Wi- Wo).

Qtotal = Qs,cond + Qs,vent +Ql,vent

Heating Results:Heating Results:Floor

No.Heating Load

(Kw)Heating Load

(TON Ref.)Heating Load

(CFM)

Basement 227.3827.387.8237.8233129.143129.14

Basement 1000000

Ground Floor24.8124.817.0887.0882835.422835.42

Second Floor23.4623.466.7046.7042681.712681.71

Third Floor29.4829.488.4238.4233369.143369.14

Fourth Floor70.8570.8520.24420.2448097.148097.14

Fifth Floor39.4839.4811.28011.28045124512

Total heating load209.14209.1459.75359.75323901.7123901.71

Boiler selectionBoiler selection Boiler capacity = (Q building + Q domestic hot water) * 1.1 Boiler capacity = (Q building + Q domestic hot water) * 1.1

= (209.137 + 232) * 1.1 = 485.25 = (209.137 + 232) * 1.1 = 485.25 KWKW we select boiler model: FM500 that gives power = 500 we select boiler model: FM500 that gives power = 500 KW KW

Cooling Equations:Cooling Equations: Qs = U * A * CLTD Qs = U * A * CLTD corc. corc. For Wall And Ceiling.For Wall And Ceiling.

CLTD CLTD corr corr = ( CLTD + LM ) K + ( T= ( CLTD + LM ) K + ( Tinin -25.5 ) + ( T -25.5 ) + ( Too – 29.4 ). – 29.4 ).

QQs|s| transmittedtransmitted = A * SHG * SC * CLF . From glass = A * SHG * SC * CLF . From glass

QQs| s| convectionconvection = U * A * ( CLTD ) = U * A * ( CLTD ) correction. correction. From glassFrom glass

QQss|| vent vent = 1.2 *A * ( T= 1.2 *A * ( To o – T – T inin ) . ) .

QQL L = 3 * A ( W = 3 * A ( W oo –W –W inin ) )

QQs| s| peoplepeople = q = qs s * n * CLF* n * CLF

QQLL| | peoplepeople = ql= ql * n* n

QQs| s| lightinglighting = W * CLF = W * CLF

QQs| s| equipmentequipment = q = qs s * CLF* CLF

QQl| l| equipmentequipment = ql = ql

Cooling ResultsCooling ResultsFloor

No.Cooling Load

(Kw)Cooling Load

(TON Ref.)Cooling Load

(CFM)

Basement 265.6865.6818.7618.7675047504

Basement 1000000

Ground Floor85.9385.9324.5524.5598209820

Second Floor57.7757.7716.516.566006600

Third Floor74.7074.7021.3421.3485368536

Fourth Floor141.36141.3640.3940.391615616156

Fifth Floor69.3669.3619.8219.8279287928

Total Cooling load468.81468.81133.95133.954358043580

Chiller SelectionChiller Selection Q chiller = Q Q chiller = Q cooling for building cooling for building * 1.1 = 515.6891 * 1.1 = 515.6891 KWKW = 147.34 = 147.34 Ton RefrTon Refr.. From the Petra catalogue according to the following data: From the Petra catalogue according to the following data: Ambient temperature = 95 Ambient temperature = 95 ((Fo) )

Leaving chiller water temperature = 45 Leaving chiller water temperature = 45 ((Fo) )

Q Q chillerchiller = 147.34 = 147.34 Ton Ref.Ton Ref.

Frequency = 50Frequency = 50Hz Hz

Air Handling Units SelectionAir Handling Units SelectionFrom Petra catalogue:From Petra catalogue:

Fan coil selection:Fan coil selection:From Petra catalogue As example: Doctor Room1 at ground floorFrom Petra catalogue As example: Doctor Room1 at ground floor

Fans selection:Fans selection:

Pumps selection: Pumps selection: Chiller PumpChiller Pump

From Salsmon catalogues: From Salsmon catalogues:

m pump = 20.56 (L/s ) & ΔH = 7.8m >> ΔP = 76518 Pa

L total= L to the farthest diffuser * 2 * 1.5 = 204 m

(ΔP/L) = 375 (Pa/m) Where 200 < (ΔP/L) < 550

Model No. 80-160 from Salsmon catalogues

Boiler PumpBoiler Pump

From Salsmon catalogues:From Salsmon catalogues:

m pump = (Qs)/(Cp * ΔT) = 3.75 (L/s)

m pump = 3.75 (L/s) , ΔH = 7.1m >> ΔP = 69651 Pa

L total = 204 m

(ΔP/L) = 341.1 (Pa/m) Where 200 < (ΔP/L) < 550

Model No. 40-160 from Salmson catalogues

Duct SizingDuct Sizing

Steel Pipes SizingSteel Pipes Sizing

Plumbing ResultsPotable water Sizing “Steel Pipes”:Potable water Sizing “Steel Pipes”:

As an Example for 3As an Example for 3rdrd Floor: Floor:

Potable Pump HeadPotable Pump Head

(ΔP) pump = (ΔP) (ΔP) pump = (ΔP) friction + fitting friction + fitting ± (ΔP) head + (ΔP) flow ± (ΔP) head + (ΔP) flow

Where:

- (ΔP) friction + fitting = 1.8 * (ΔP) friction

> (ΔP) friction = (ΔP/L) AB * LAB + (ΔP/L) BC * LBC + (ΔP/L) CD * LCD

- (ΔP) friction + fitting = 12.726 kpa - (ΔP) head = ρ * g * H = -29.4 Kpa

- (ΔP) flow = 15*6.8 = 102 Kpa

(ΔP) pump = 85.32 Kpa(ΔP) pump = 85.32 Kpa

Flow rate |Flow rate |Pump Pump = 8.94 L/s= 8.94 L/s

Drainage System:

At each number of fixtures we find the diameter size from tables.At each number of fixtures we find the diameter size from tables.

Main Results :Main Results : Horizontal branch diameter: 4" Horizontal branch diameter: 4"

Riser diameter: 4" Riser diameter: 4"

Building drain diameter: 4" Building drain diameter: 4"

Building drain slope: 1% Building drain slope: 1%

Fire Fighting SystemFire Fighting System

.In this project, we used 1 landing valve and 1 cabinet for each floor:

Cabinet: Residual pressure = 65 PSI

Size = 1½" Landing valve: Residual pressure = 100 PSI

Size = 2½" Stand pipe diameter = 4" Pump flow rate = 500 g.p.m Tank volume = ((Flow rate * 3.8 * t) / (1000))

= ((500*3.8*120)/ (1000))

= 228 m3

Fire Fighting Pump headFire Fighting Pump head(ΔP) pump = (ΔP) (ΔP) pump = (ΔP) friction + fitting friction + fitting ± (ΔP) head + (ΔP) flow ± (ΔP) head + (ΔP) flow

Where:Where: - (ΔP) friction + fitting = 1.5 * (ΔP) friction

> (ΔP) friction = (ΔP/L) AB * LAB + (ΔP/L) BC * LBC + (ΔP/L) CD * LCD…

- (ΔP) friction + fitting = 33.289548 kpa - (ΔP) head = ρ * g * H = 1000* 9.81* 3 = -29 Kpa

- (ΔP) flow = 100*6.8 Kpa = 680 kpa

(ΔP) pump = 684.289 Kpa(ΔP) pump = 684.289 Kpa

Jockey Pump Jockey Pump Flow rate = (5 – 10) g.p.m

(ΔP)Jockey pump = (ΔP) pump + (10 PSI* 6.8) = 500.6175 kpa

FM-200 system:FM-200 system:

Mass of clean agent: Mass of clean agent:

M = (V/S) [C / (100 – C)] Leakage rate: Leakage rate:

Leakage rate = 0.608 * Pc (8.7)

> Pc = g * Ho * (rm – ra) Results :Results :

Weight = (ρ * g * Volume)

Total mass * g = (ρ * g * Volume)

Volume = (Total mass/ ρ) = (398.682/100) = 3.98 m3

Volume of FM-200 Cylinder:Volume of FM-200 Cylinder:

Medical GasesMedical GasesResultsResults As an Example for 3rd floor:

Medical GasesMedical Gases

Volume of cylinders for each gas:

1 -Oxygen gas : 54.5 SCFM >> 24.77 (L/s) >> 0.02477 (m3/s)

2 -Medical vacuum :

92.5 SCFM >> 42 (L/s) >> 0.042 (m3/s)

3 -Medical air : 7 SCFM >> 3.1 (L/s) >> 0.0031 (m3/s)

Thank You