Presentation on theme: "AN-NAJAH NATIONAL UNIVERSITY FACULTY OF ENGINEERING DEPARTEMENT OF MECHANICAL ENGINEERING MECHANICAL SYSTEMS OF AL-OYOON HOSPITAL-NABLUS Graduation Project."— Presentation transcript:
AN-NAJAH NATIONAL UNIVERSITY FACULTY OF ENGINEERING DEPARTEMENT OF MECHANICAL ENGINEERING MECHANICAL SYSTEMS OF AL-OYOON HOSPITAL-NABLUS Graduation Project Submitted In Partial Fulfillment Of the Requirements For The Degree Of B.Sc. In Mechanical Engineering. Supervisor: The Students: Dr. Ramiz Alkhaldi Amin Udeh ( ) Salem Hassoun ( ) Omar Sabanah ( ) Mohammad jawabri ( ) May
INTRODUCTION. Al-oyoon hospital, north of Nablus, located at the top of the north mountain at Nablus-Aseera street. The hospital has four departments A,B,C and D. In this project, only the mechanical systems of department A and C are calculated, A and C department has four stores, each one has an area of more than 2000 m 2. In this project, there is many mechanical systems was designed, like: HVAC systems (heating, cooling, ventilation, fresh air, exhaust are systems). Potable water system (hot water supply and return, cold water supply). Drainage system, Venting system, Fire fighting system, Medical gases system.
1) HVAC SYSTEM : Hot water heating system : Hot water heating system consists of: Boiler. Piping net work. Expansion tank. Circulation pump.
General procedure for calculating total heat load: Select inside design condition (Temperature, relative humidity). Select outside design condition (Temperature, relative humidity). Select unconditioned temperature. 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 total and Q boiler. The heating load calculation begins with the determination of heat loss through a variety of building envelope components and situations. 1- Walls 2- Roofs 3- Windows 4- Doors 5- Basement Walls Basement Floors 6- Infiltration Ventilation
The Heat Loss Equation: Q s ) cond = U* A* ( Ti - To )………. Q s ) vent = 1.2 V vent *(T i -T o ) Q s = Q s ) cond + Q s ) vent …………… Q boiler = (Q s +Q w )*1.1 U ext = 1.24 W/m 2. o C……………………U int = 2.53 W/m 2. o C T i = 22 o C…………………………………………T o = 5.7 o C U wind = 5 W/m 2. o C……………………... U door = 3.6 W/m 2. o C U ceiling = 0.88 W/m 2. o C Where: U: overall heat transfer coefficient A: area T i : inside design temperature T o : outside design temperature
Cooling Load Calculation: For ceiling : Q=U*A*(CLTD) corr Where: (CLTD) corr = (CLTD + LM) K + (25.5 – Ti )+ (To – 29.4) Where : CLTD: cooling load factor K:color factor: K=1 dark color K=0.5 light color
For walls : Q=U*A*(CLTD) corr Where: (CLTD) corr =(CLTD + LM) K + (25.5 – Ti )+ (To – 29.4) Where : CLTD: cooling load factor K:color factor: K=1 dark color K=0.83 medium color K=0.5 light color
For glass : Heat transmitted through glass Q=A*(SHG)*(SC)*(CLF) SHG: solar heat gain SC: shading coefficient CLF: cooling load factor Convection heat gain: Q=U*A*(CLTD) corr (CLTD) corr = (CLTD)+(25.5 – Ti )+ (To – 29.4)
For people : Q s =q s *n*CLF q L =q L *n where: Q s,Q L : sensible and latent heat gain q s,q L : sensible and latent gains per person n: number of people CLF: cooling load factor
For lighting : Q s =W*CLF Where: Q s : net heat gain from lighting W:lighting capacity: (watts)
For equipments : Q s =q s *CLF Q L =q L Q s,Q L : sensible and latent heat gain. CLF: cooling load factor
4 th Floor3 rd Floor2 nd Floor1 st FloorFloor No Heating Load (ton) Cooling Load (ton) Results of heating and cooling load for each floor:
Ventilation and exhaust air: Definition of ventilation: Ventilating (the V in HVAC) is the process of "changing" or replacing air in any space to provide high indoor air quality (i.e. to control temperature, replenish oxygen, or remove moisture, odors, smoke, heat, dust, airborne bacteria, and carbon dioxide). Ventilation is used to remove unpleasant smells and excessive moisture, introduce outside air, to keep interior building air circulating, and to prevent stagnation of the interior air. Method of calculating ventilation: It’s calculated from tables depending on the number of people or on the area of the roam in general. Definition of exhaust air: It’s the air which withdrawn from the room to the atmosphere to make complete cycle in which we allow fresh air to enter the room and withdraw the exhaust air to the atmosphere. Method of calculating exhaust air: It’s equal to the fresh air rate in general but we can make it larger than fresh air if we need positive pressure or smaller than fresh air if we want negative pressure.
2) POTABLE WATER : The pipes conveying water to water closets shall be of sufficient size to supply the water at a rate required for adequate flushing without unduly reducing the pressure at other fixtures Separate sewer connections. Every building intended for human habitation or occupancy on premises abutting on a street in which there is a public sewer shall have a separate connection.
How to size pipe based on flow rates. Identify the gpm requirement of the furthest head from the zone valve. For systems with only one zone use the head furthest from the main line point of connection (POC). On a Friction Loss Chart for the type of pipe selected, find the gpm amount from 1st item in the far left column. In that row, move right across the chart until a velocity of less than 5 feet per second is reached. Move up that column to find the minimum pipe size necessary to carry the flow to this head. Add the gpm requirement of the last and the next to the last head together to size the next pipe. Find the total gpm in the 1st column of the Friction Loss Chart and repeat steps 3 and 4. Continue this process until you reach the zone valve or main line POC. Select the largest of the pipe sizes for the entire zone.
3) DRINAGE AND VENT SYSTEMS : In modern plumbing, a drain-waste-vent is a system that removes sewage and grey water from a building and vents the gases produced by said waste. Waste is produced at fixtures such as toilets, sinks and showers, and exits the fixtures through a trap, a dipped section of pipe that always contains water. All fixtures must contain traps to prevent gases from backing up into the house. Through traps, all fixtures are connected to waste lines, which in turn take the waste to a soil stack, or soil vent pipe, which extends from the building drain at its lowest point up to and out of the roof. Waste is removed from the building through the building drain and taken to a sewage line, which leads to a septic system or a public sewer. Cesspits are generally prohibited in developed areas.
Drainage system sizing : Fixture unit values presently recommended for assignment to various kinds of plumbing fixtures which discharged into sanitary drainage systems are stated in table10-1. They are provides as a means for computing sizes of soil, waste, and vent piping bases upon the loading effects produced by the discharge of many different kinds of plumping fixtures commonly installed in buildings.
Venting mechanisms : To prevent the problems of high pressure in a drain system, sewer pipes will usually vent via one of two mechanisms. Sizing of vent piping : Table 4-3 used for sizing vent stacks in accordance with drainage stack capacity loads.
4) FIRE FIGHTING SYSTEM : Fire protection is the prevention and reduction of the hazards associated with fire. If involves the study of the behavior compartment, suppression and investigation of fire and its related emergencies as will as the research and development production, testing and application of mitigating. The wet stand pipe system is designed in our project for its advantages on dry system and the Palestinian system in fire fighting. Fire dampers are installed in the supply of each fan coil to not allow fire to spread of fire and stop the oxygen source. Sizing : Each pipe of landing valve is 2 ½”, each pipe of cabinet is 1 ½” and the rest of pipes is 4”.
5) MEDICAL GASES : Medical gases are very important in hospitals, the main application of it inside the surgery operations rooms and in patient rooms to help the patients. There is many medical gases are used in hospitals, like : Oxygen (O2) Medical Air (MA) Medical Vacuum (MV) Nitrous Oxide (N2O) Nitrogen (N2) Instrumental Air (IA) Carbon Dioxide (CO2) Waste Anesthesia Gas Disposal (WAGD or EVAC)
Designing Medical Gas Systems : Estimating flow requirements Selecting equipment Pipe sizing Zone valves and alarms Electrical service Equipment space requirements Number of outlets Flow rate per outlet (depends on the specific gas and outlet type) Diversity factor (depends on the number and type of outlets)
Pipe Sizing : Flow Rate (considering diversity), the flow rate inside the pipes are calculated due to the diversity factor, in the pipes outlets the flow rate is (1), in the other lines the ones in outlets that the pipe supplies it are calculated and multiplied by the diversity factor because not all the outlets are used in the same time, then the pipe sizes are calculated from table 6-4. The friction loss in the pipes are shown in table 6-4, and it's depend on the size that was selected in the previous step, the friction loss are calculated for each pipe of the longest line, and this help us to now the medical gas compressor power, then calculate the equivalent length of the longest pipe line by multiplying the total length by 1.5 and then the friction loss inside the elbows are including in the calculations. From table 6-4, increasing in the pipe diameter will decrease the friction loss and that useful for the compressor power and keep money.
WBDGPM Sensible capacity(BT UH) Total capacity (BTUH) Entering wet bulb temp ( o F) DBT Air flow rate Fan speed modelFan coil no H DC18 4Rows 1 6) SELECTION : Example of Fan coil unit selection:
MODELLWT Ambient temp.( o F) CAPGPMKWWPD APS S Chiller selection :
. Maximum pressure (PSIG) Maximum temp ( o F) KWDescriptionModel High capacityCHS Boiler selection : Q(boiler) = Q(heating) + Q(domestic) Q(domestic) = m*Cp*∆T M = 59 GPM = 59*3.7/60 = 3.64 L/S Q(domestic) = m*Cp*∆T = 3.64*4.18*60/24 = 35 Kw/hour Assuming that hot water is used for 2 hours then: Q(domestic) = 35*2 = 70 Kw Q(boiler) = Q(heating) + Q(domestic) = = 500 Kw
MODELCFMExternal state pressure (Inch water) KWKW RPMKWRPMKWRPMKWRPM Exhaust fan selection:
. PUMP MODEL Flow (GPM) Pressure (Psi) Casting Material Ports (Inlet*Discharge) Horsepower (HP) Motor Type 4K192MT S2-1/2*1-1/2” ANSI10215FR Electric 3PH Potable water pump selection: Pump head = Friction losses(F.L)-Back Pressure(B.P)+Residual Pressure (Fixture pressure) Now to Find Friction losses(F.L): The longest distance and the largest residual pressure was taken in the top story of the building Friction losses (F.L) = (16*3.28* *3.28* *3.28* *3.28*10.8)/100 Multiplying by 1.5 to compensate the losses in fittings:- F.L = 1.5*26.5 = 40 Psi Back Pressure (B.P) = 5Psi Residual Pressure ( Fixture pressure) = 8 Psi Pump head = = 43Psi Pump flow rate =190 GPM (Using fixture unit (880F.U))
. PUMP MODEL Flow (GPM) Pressure (Psi) Casting Material Ports (Inlet*Discharge) Horsepower (HP) Motor Type 3K392B T S1-1/2*1” NPT1/256FR Electric 1PH Hot water pump selection : Pump head = Friction losses(F.L)+Back Pressure(B.P)+Residual Pressure (Fixture pressure) Now to Find Friction losses(F.L): The longest distance and the largest residual pressure was taken in the top story of the building Friction losses (F.L) = (16*3.28* *3.28* *3.28* *3.28*17.2)/100 Multiplying by 2 because there is a return line: F.L = 2*46 = 93 Psi Multiplying by 1.5 to compensate the losses in fittings: F.L=1.5*93=140 Psi Back Pressure(B.P) = 4*5= 20Psi Residual Pressure ( Fixture pressure) = 10 Psi Pump head = =170Psi Pump flow rate =50 GPM (Using fixture unit 120F.U)
. Fire duty Usgpm Range PsigRange (m) Pump Model Pump Item No. Diesel Engine KW/rpm Motor KW/rpm K451A000170/ /2900 Fire pump selection and tank size : Pump head = Friction losses (F.L)Back Pressure(B.P)+Residual Pressure Fire Fighting. Now to Find Friction losses (F.L): The longest distance was taken in the top story of the building Friction losses (F.L) = (4.95*9* *38* *3.5* * *12*3.28)/100 Multiplying by 1.5 to compensate the losses in fittings:- F.L=1.5*27.8=42 Psi Back Pressure(B.P)=5Psi Residual Pressure( Fire Fighting)=100 Psi Pump head = =137Psi Pump flow rate =1000 GPM Tank size =1000*90 = 90000Gallon = 90000*3.7/1000 = 333 m 3
Boiler chimney : The assumption of the boiler chimney: H = 15 m, CV=39000, ή= 80%, ρ g = 1.1, v = 5m/s Pb=101.32, Ra= 287, Ta=25+273=298, Tg= =523. m f = Qb/ CV*ή = 500/39000*0.8 =.016kg/s m g = 25.2*m f =.403 kg \s Ac = m g / ρ g v = m 2 Ac = π/4 * D 2 - D = 30.5cm =35 cm Q flow = m g / ρ g =.403 / 1.1= m 3 /s ---- ∆p/ L = 0.5 pa/m ∆p= 0.5*(15*1.5) = pa But ∆P boiler = (Pb*g*H)/Ra * ((1/Ta)-(1/Tg)) = 71.5 pa ∆P boiler > ∆p= 0.5*(15*1.5) = pa So, no need for fan.