# Mechanical System For AL- Rehan Hospital - PowerPoint PPT Presentation

Mechanical System For AL- Rehan Hospital

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Mechanical System For AL- Rehan Hospital

## Mechanical System For AL- Rehan Hospital

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##### Presentation Transcript

2. Objective • The objective of this project is to design Conventional (HVAC) System (Fan coil units with chiller and Boiler ) for AL-Rehan Hospital. • In addition with all recommended mechanical systems that should contain like (potable water, drainage, medical gases and fire fighting) systems.

3. HVAC • HVAC (heating, Ventilation, and Air conditioning) is the technology of indoor and automotive environmental comfort. HVAC system design is a major sub discipline of mechanical engineering, based on the principles of thermodynamics, fluid mechanics, and heat transfer. Also the mechanical systems are to be introduced in our research including potable, drainage, firefighting and medical gases.

4. Building Description • Location Country: Palestine . City: Ramallah Region: Dahyet Al-Rehan. Elevation: 800 m above sea level. Latitude: 32 N. Wind’s speed in Ramallah is above 5 m/s.

5. Al-Rehan Hospital consist of tenth floor ,four Basement Floor , Ground Floor and Five Up Floor and each Floor has approximately 1500 m^2

6. Inside and Outside Design Conditions

7. Overall Heat Transfer coefficient,Uoverall Overall heat transfer coefficient depends on the construction of the unit. To find the overall heat transfer coefficient , the construction was taking in consideration because Uoverall control with the quantity of losses by wall , ceiling , ground , windows and doors. • Uoverall is given by: U= 1/Rtot R tot = Ri+ R + Ro R= ∑ x/K

8. External walls construction Ri = 0.12 m2.C\W Ro= 0.03 m2.C\W Rtot= Ri+Rw+Ro Rtot=0.12+1.0082+0.03 =1.18814 (m^2).C/W U=0.8634 W/(m^2).C

9. Internal walls construction Ri = 0.12 m2.C\W Ro= 0.03 m2.C\W Rtot= Ri+Rw+Ro Rtot=0.12+0.17221+0.03 =0.41221 (m^2).C/W U=2.4259 W/(m^2).C

10. Ceiling Construction Ri = 0.12 m2.C\W Ro= 0.03 m2.C\W Rtot= Ri+Rw+Ro Rtot=0.12+0.17221+0.03 =0.41221 (m^2).C/W U=2.4259 W/(m^2).C

11. Windows and Doors U overall for windows and doors taken directly from energy efficient building code as follow: • Windows double glasses with aluminum material type, wind speed more than 5 m/s Uwindow = 3.5 W/m2.C. • Doors with wood material type without storm door , wind velocity more than 5 m/s Udoor= 2.4 W/m2.C

12. Summary The value of the overall heat transfer coefficient for each element construction

13. Heating Load Calculation PROCEDURES • Select inside design condition Temperature, relative humidity(Tin, Φin). • Select outside design condition Temperature, relative humidity(Tout, Φout). • Select unconditioned temperature(Tun). • Find over all heat transfer coefficient Uo for wall, ceiling, floor, door, windows, below grade. • Find area of wall, ceiling, floor, door, windows, below grade. • Find Qs conduction. • Find V inf , V vent . • Find Qs, QL vent, inf. • Find Q domestic hot water. • Find Q boiler.

14. EQUATIONS • Q = U* A* ( Ti - To ) • Vvent= n * value of ventilation • Vinf= (ACH * inside volume *1000) /3600 • Qs)vent , inf= 1.2 Vvent,inf*(Ti-To) • Ql)vent , inf = 3 Vvent,inf*(Wi-Wo). • Qw= (Mw *cp*(Th – Tc ))/∆t Unconditioned Temperature In summer Tun = Ti+2/3*( To - Ti ) Unconditioned Temperature In winter Tun = Ti+0.5*( Ti - To )

15. Sample Calculation • Single Room ( 1 ) Specification Area of outside wall = 14 m² Area of unconditioned wall = 32.2 m² Area of window is = 1.8 m² Ceiling area = 18.6 m²

16. Conduction heat gain Qs = U*A*∆T Q.ext = 0.8634*14*17.3 = 209.11548 Watt. Q. un = 2.4259*32.2*8.65 = 675.686 Watt. Q cel. = 0.8379*18.6*17.3 = 278.01 Watt. Qs.cond = Q.ext + Q.un + Q.cilling Qs.cond= 1268.5 Watt. • Ventilation and Infiltration heat gain Q.s.ven =1.2*5*18.6 *17.3 = 1930.068 Watt • Q.tot = 1268.5 + 1930 = 3199.18 Watt

18. Boiler Selection • Boiler specification • Domestic hot water load 145.03 KW • Heating Load 834.4 KW • Boiler Capacity = 1077.37 KW • Annual fuel consumption of diesel is 154.86 cubic meter per year .

19. From Obrien boilers Company Catalogue we chose • REX K120 F with capacity of 1200 KW

20. Pump Specification

21. Pump Selection • Required pump for boiler have the following specification • pump head 149.75PSI . • pump flow rate 20.14 L/s. • From WILO catalog we chose NL 50/315-45-2-12-50Hz series .

22. Expansion Tank • From Wessels Company Catalogue depending on pump flow rate(20.17 L/s) (319.7 GPM )we select TXA 1400

23. Coaling Load Calculation • EQUATIONS • For Ceiling: Q=U*A*(CLTD)corr (CLTD)corr =(CLTD + LM) K + (25.5 – Ti )+ (To – 29.4) CLTD: cooling load factor K:color factorK=1 dark color K=0.5 light color • For Walls: Q=U*A*(CLTD)corr (CLTD)corr =(CLTD + LM) K + (25.5 – Ti )+ (To – 29.4) K=1 dark color K=0.83 medium color K=0.5 light color

24. For Glass • Heat transmitted through glass Q=A*(SHG)*(SC)*(CLF)SHG: solar heat gainSC: shading coefficientCLF: cooling load factor • Convection heat gain Q=U*A*(CLTD)corr • For people Qs=qs*n*CLF qL=qL*nQs,QL: sensible and latent heat gainqs,qL: sensible and latent gains per person n: number of people CLF: cooling load factor • For lighting Qs=W*CLF W:lighting capacity: (watts) • For equipments Qs=qs*CLF QL=qL

25. Sample Calculations For Single Room (1) • Area of outside wall = 14 m² • Area of unconditioned wall = 32.2 m² • Area of window is = 1.8 m² • Ceiling area = 18.6 m² Sensible and latent heat gain for one person from table (A-16). • Qs = 71.5 W • QL= 57 W Sensible heat gain • Qs in, wall = U x A x (CLTD) correct = 90.58 W Transmission heat gain (window) = 51.86 W Convection heat gain = U*A*CLTD.corr = 63.63 W

26. Convection heat gain • U*A*CLTDcorr = 63.63 W Load from ventilation and infiltration • Qs vent/inf= 1.2*Vvent* ∆t = 1.2*75.8*7.4 = 669.6 W • QL vent = 3*Vvent*(Wi-Wo) = 3*75.8*6.5 =1478.7 W Heat load due to people • Qs = qs*n*CLF = 70*1*0.84 = 62.3 W • Ql = ql*n = 44*1 = 44 W Heat load due to lighting • Qs/l = Aroom*CLF= 279 * 0.85 = 237.15 W Heat load due to equipment • Qs = 522 W • QL = 0 W Total sensible load ∑Qs = 1826.87 W Total latent load = ∑Ql = 1522.7 W Total load = 3349.3 W

27. Chiller Selection • Building Load = 1270.06 kW OR 362.87 T.R • From PETRA Company we select APSa 385 – 3S AC1 50 Hz. • with 385 Ton. Refrigeration cooling load capacity .

28. Pump Selection • Friction loss = 66.17 PSI • Fitting loss = 33.08 PSI • Head loss = 50.49 PSI. • Pump head = 149.75 PSI • Pump Flow rate = 50.64 L/s • From WILO catalog we select SCP 150/580HA series.

29. Pressure Tank Selection From Wessels Company Catalogue depending on Pump flow rate (19.44 L/s) (308 GPM) the suitable pressure tank FXA-1200

30. Fan Coil Unit For Single Room (1) • Cooling Load = 3.35 KW • V cir. = 465.17 L/s • V c.f.m = 1023.34 CFM • From Petra Catalogue we select DCC 20 H/C 4Rows Model

31. Pipe Sizing ( FCU 1) • Supply Pipe sizing • m = (Qs +Ql.) / (4180*6) m = (1826.6 +1552.7) / (4180*6) = 0.135 L/s • Pressure drop assumption 400 pa/m. • Preferred size at operating condition is 0.75 in

32. Duct Sizing ( FCU 1) • Qt = 3.35 kw • V circulation air = Qt / 1.2 * (Tcir. – Ti) • V circulation = 3.35 / (1.2* 10) = 0.465 m3/ s • Pressure drop = 0.55 pa / m • velocity = 4.1m / s • A = 0.11 m2 • D = 0.385 m • High = 0.3 m • Width = 0.425 m

33. Fresh Air Duct For single room ( 1 ) • V.vent. = ( L/S/ Area ) * Area • V.vent. = 5 * 18.6 = 93 L/s • Pressure drop = 0.55 pa / m • velocity = 3m / s • A = 0.0314 m2 • D = 0.2 m • High = 0.2 m • Width = 0.175 m

34. Fresh Air Fan Selection Fan specification • Flow rate = 1173.5 L/s • Static Pressure = 13.75 pa • From Rosenberg RoVentcatalog we select DHAD 355-4 series .

35. Exhaust Air Duct • V = ( A.C.H *Vin* 1000) / 3600 • V = ( 10 * 3.5*3.1*1000 ) / 3600 • V = 30 L/s/path. • Pressure drop = 0.55 pa / m • velocity = 2m / s • D = 0.136 m • High = 0.2 m • Width = 0.75 m

36. Exhaust Fan Selection Fan specification • Flow rate = 210( L/s ). • Static Pressure = 13.75 pa • From Rosenberg RoVentcatalog we select DHAD 355-4 series .