Load Calculation Presentation

8/13/2019 Load Calculation Presentation

1/32

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 theRequirements For The Degree Of B.Sc. In Mechanical Engineering.

Supervisor: The Students:Dr. Ramiz Alkhaldi Amin Udeh (10611326)

Salem Hassoun (10615008)

Omar Sabanah (10644946)Mohammad jawabri (10717105)

May - 2011


2/32

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.


3/32

1) HVAC SYSTEM :

Hot water heating system :

Hot water heating system consists of:

Boiler.

Piping net work.

Expansion tank.

Circulation pump.


4/32

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 lossthrough a variety of building envelope components and situations.

1- Walls2- Roofs3- Windows4- Doors5- Basement Walls Basement Floors6- Infiltration Ventilation


5/32

The Heat Loss Equation:

Qs)cond = U* A* ( Ti - To ). Q s)vent = 1.2 V vent *(T i-To)

Qs = Q s)cond + Q s)vent ......... Q boiler = (Q s+Q w)*1.1

Uext = 1.24 W/m 2.oCU int = 2.53 W/m 2.oC

Ti = 22 oCT o = 5.7 oC

Uwind = 5 W/m 2.oC... U door = 3.6 W/m 2.oC

Uceiling = 0.88 W/m 2.oC

Where:

U: overall heat transfer coefficientA: areaTi: inside design temperatureTo: outside design temperature


6/32

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


7/32

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


8/32

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)


9/32

For people :

Qs=q s*n*CLF

qL=qL*n

where:

Qs,QL: sensible and latent heat gain

qs,qL: sensible and latent gains per person

n: number of people



10/32

For lighting :

Qs=W*CLF

Where:

Qs: net heat gain from lighting

W:lighting capacity: (watts)


11/32

For equipments :

Qs=q s*CLF

QL=qL

Qs,QL: sensible and latent heat gain.



12/32

4 th Floor3 rd Floor2nd Floor1st FloorFloor No.

34.32630.932.7HeatingLoad (ton)

60.2485659.8CoolingLoad (ton)

Results of heating and cooling load for each floor:


13/32

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 highindoor 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 smellsand excessive moisture, introduce outside air, to keep interior building air circulating, and to preventstagnation of the interior air.

Method of calculating ventilation: Its calculated from tables depending on the number of people or on the area of the roam in general.

Definition of exhaust air: Its the air which withdrawn from the room to the atmosphere to make complete cycle in which weallow fresh air to enter the room and withdraw the exhaust air to the atmosphere.

Method of calculating exhaust air: Its 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.


14/32

2) POTABLE WATER :

The pipes conveying water to water closets shall be of sufficient sizeto supply the water at a rate required for adequate flushing withoutunduly 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 sewershall have a separate connection.


15/32

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 leftcolumn.

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.


16/32

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.


17/32

Drainage system sizing :

Fixture unit values presently recommended for assignment to various kinds of plumbing fixtures whichdischarged 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 loadingeffects produced by the discharge of many different kinds of plumping fixtures commonly installed in

buildings.


18/32

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.


19/32

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 theresearch 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 Palestiniansystem 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 oxygensource.

Sizing :

Each pipe of landing valve is 2 , each pipe of cabinet is 1 and the rest of pipes is 4.


20/32

5) MEDICAL GASES :

Medical gases are very important in hospitals, the main application of it inside the surgery operations roomsand in patient rooms to help the patients.

There is many medical gases are used in hospitals, like :

Oxygen (O 2) Medical Air (MA) Medical Vacuum (MV) Nitrous Oxide (N 2O) Nitrogen (N 2) Instrumental Air (IA)

Carbon Dioxide (CO 2) Waste Anesthesia Gas Disposal (WAGD or EVAC)


21/32

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)


22/32

Pipe Sizing :

Flow Rate (considering diversity), the flow rate inside the pipes arecalculated 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 calculatedand multiplied by the diversity factor because not all the outlets are used inthe 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 sizethat was selected in the previous step, the friction loss are calculated for each pipeof 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 thetotal length by 1.5 and then the friction loss inside the elbows are including inthe calculations.

From table 6-4, increasing in the pipe diameter will decrease the friction lossand that useful for the compressor power and keep money.


23/32

WBDGPMSensible

capacity(BT

UH)

Total

capacity

(BTUH)

Entering

wet bulb

temp ( oF)

DBTAirflowrate

Fanspeed

modelFan coilno.

4.029.954394654976763761816HDC18

4Rows

1

6) SELECTION :

Example of Fan coil unit selection:


24/32

MODELLWTAmbient

temp.(o

F)

CAPGPMKWWPD

APS 205-

2S

5085229.9551.82076.25

Chiller selection :


25/32

.

Maximum

pressure (PSIG)

Maximum

temp ( oF)

KWDescriptionModel

135358500High 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) = 430 + 70 = 500 Kw


26/32

CFMNeck size in mmTotal pressureOutlet velocity

FPM

410225*2250.25 2000

Diffuser selection:


27/32


28/32

.

PUMPMODEL

Flow(GPM)

Pressure(Psi)

CastingMaterial

Ports(Inlet*Discharge)

Horsepower(HP)

MotorType

4K192MT325082304S2-1/2*1- 1/2 ANSI 10215FRElectric

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 buildingFriction losses (F.L) = (16*3.28*22.5 + 4.5*3.28*17.2 + 23*3.28*12.8 + 6.5*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) = 5PsiResidual Pressure ( Fixture pressure) = 8 PsiPump head = 40+85 = 43PsiPump flow rate =190 GPM (Using fixture unit (880F.U))


29/32

.

PUMP

MODEL

Flow

(GPM)

Pressure

(Psi)

Casting

Material

Ports

(Inlet*Discharge)

Horsepower

(HP)

Motor

Type3K392B

T05213304S1-1/2*1 NPT 1/256FR

Electric1PH

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 buildingFriction losses (F.L) = (16*3.28*40 + 4.5*3.28*31 + 23*3.28*22.5 + 6.5*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 PsiBack Pressure(B.P) = 4*5= 20PsiResidual Pressure ( Fixture pressure) = 10 PsiPump head = 140+20+10=170PsiPump flow rate =50 GPM (Using fixture unit 120F.U)


30/32

.

Fire duty

Usgpm

Range PsigRange (m)Pump

Model

Pump Item

No.

Diesel

EngineKW/rpm

Motor

KW/rpm

1000193.2-210.2133-145100-315K451A000170/2900185/2900

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 buildingFriction losses (F.L) = (4.95*9*3.28 +4.73*38*3.28 +15*3.5*3.28+4.95*69+3.28+ 18.9*12*3.28)/100Multiplying by 1.5 to compensate the losses in fittings:-F.L=1.5*27.8=42 PsiBack Pressure(B.P)=5PsiResidual Pressure( Fire Fighting)=100 PsiPump head = 42+100-5=137PsiPump flow rate =1000 GPMTank size =1000*90 = 90000Gallon = 90000*3.7/1000 = 333 m 3


31/32

Boiler chimney :

The assumption of the boiler chimney:

H = 15 m, CV=39000 , = 80%, g= 1.1, v = 5m/sPb=101.32, Ra= 287, Ta=25+273=298, Tg=250+273=523.

mf = Qb/ CV* = 500/39000*0.8 = .016kg/smg = 25.2*m f = .403 kg \sAc = m g / g v = 0.0733 m 2

Ac = / 4 * D 2 - D = 30.5cm =35 cmQ flow= m g/ g =.403 / 1.1= 0.366 m 3/s

---- p/ L = 0.5 pa/m

p= 0.5*(15*1.5) = 11.25 pa

But

P boiler = (Pb*g*H)/Ra * ((1/Ta)-(1/Tg)) = 71.5 paP boiler > p= 0.5*(15*1.5) = 11.25 pa

So, no need for fan.


32/32

Finished

Thank you for your attention

Documents

Load Calculation Presentation