Mechanical Systems For Al-Zakah Hospital Graduation Project 2 Mechanical Systems For Al-Zakah Hospital Students: Mohammed Sawalhi (10840740) Mohammed Younis (10824906) Mohammed Hannon (10822899) Nedal Al-Masri (10823665) Aya Nairat (10821417) Supervisor: Dr. Iyad Assaf

Introduction: In our project, we are going to Design the following Mechanical systems for Al Zakah hospital in Palestine-Tulkarm which consists of 7 floors two of them under the ground : 1- HVAC System “Heat, Ventilation & Air conditioning” 2- Plumbing System 3- Fire Fighting System 4- Medical Gases System.

Building’s Description Country : Palestine / West bank . City: Tulkarm. Street Al-Montaza street . Latitude: 31.9 Building face sits at south orientation. The wind speed is greater than 5 m/s

Heating 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: Floor No. Heating Load (Kw) (TON Ref.) (CFM) Basement 2 27.38 7.823 3129.14 Basement 1 Ground Floor 24.81 7.088 2835.42 Second Floor 23.46 6.704 2681.71 Third Floor 29.48 8.423 3369.14 Fourth Floor 70.85 20.244 8097.14 Fifth Floor 39.48 11.280 4512 Total heating load 209.14 59.753 23901.71

Boiler selection Boiler capacity = (Q building + Q domestic hot water) * 1.1 = (209.137 + 232) * 1.1 = 485.25 KW we select boiler model: FM500 that gives power = 500 KW

Cooling Equations: Qs| transmitted = A * SHG * SC * CLF . From glass Qs = U * A * CLTD corc. For Wall And Ceiling. CLTD corr = ( CLTD + LM ) K + ( Tin -25.5 ) + ( To – 29.4 ). Qs| transmitted = A * SHG * SC * CLF . From glass Qs| convection = U * A * ( CLTD ) correction. From glass Qs| vent = 1.2 *A * ( To – T in ) . QL = 3 * A ( W o –W in ) Qs| people = qs * n * CLF QL| people = ql * n Qs| lighting = W * CLF Qs| equipment = qs * CLF Ql| equipment = ql

Cooling Results Floor No. Cooling Load (Kw) (TON Ref.) Cooling Load (CFM) Basement 2 65.68 18.76 7504 Basement 1 Ground Floor 85.93 24.55 9820 Second Floor 57.77 16.5 6600 Third Floor 74.70 21.34 8536 Fourth Floor 141.36 40.39 16156 Fifth Floor 69.36 19.82 7928 Total Cooling load 468.81 133.95 43580

Chiller Selection Q chiller = Q cooling for building * 1.1 = 515.6891 KW = 147.34 Ton Refr. From the Petra catalogue according to the following data: Ambient temperature = 95 (Fo) Leaving chiller water temperature = 45 (Fo) Q chiller = 147.34 Ton Ref. Frequency = 50Hz

Air Handling Units Selection From Petra catalogue:

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

Fans selection:

Pumps selection: Chiller Pump 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 Pump 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 Sizing

Steel Pipes Sizing

Plumbing Results Potable water Sizing “Steel Pipes”: As an Example for 3rd Floor:

Potable Pump Head (ΔP) pump = (ΔP) friction + fitting ± (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 Flow rate |Pump = 8.94 L/s

Drainage System: At each number of fixtures we find the diameter size from tables. Main Results : Horizontal branch diameter: 4" Riser diameter: 4" Building drain diameter: 4" Building drain slope: 1%

Fire 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 head (ΔP) pump = (ΔP) friction + fitting ± (ΔP) head + (ΔP) flow 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 Jockey Pump Flow rate = (5 – 10) g.p.m (ΔP)Jockey pump = (ΔP) pump + (10 PSI* 6.8) = 500.6175 kpa

FM-200 system: Mass of clean agent: M = (V/S) [C / (100 – C)] Leakage rate: Leakage rate = 0.608 * Pc (8.7) > Pc = g * Ho * (rm – ra) Results :

Volume of FM-200 Cylinder: Weight = (ρ * g * Volume) Total mass * g = (ρ * g * Volume) Volume = (Total mass/ ρ) = (398.682/100) = 3.98 m3

Medical Gases Results As an Example for 3rd floor:

Medical 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