Transcript Air conditioning project #Heating load calculations. #Hot water

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:
Dr. Ramiz Alkhaldi
The Students:
Amin Udeh
(10611326)
Salem Hassoun
(10615008)
Omar Sabanah
(10644946)
Mohammad jawabri (10717105)
May - 2011
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 m2.
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:
Qs)cond = U* A* ( Ti - To )………. Qs)vent = 1.2 Vvent*(Ti-To)
Qs = Qs)cond + Qs)vent ……………......... Qboiler = (Qs+Qw)*1.1
Uext = 1.24 W/m2.oC……………………Uint = 2.53 W/m2.oC
Ti = 22 oC…………………………………………To = 5.7 oC
Uwind = 5 W/m2.oC……………………... Udoor = 3.6 W/m2.oC
Uceiling = 0.88 W/m2.oC
Where:
U: overall heat transfer coefficient
A: area
Ti: inside design temperature
To: 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 :
Qs=qs*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
CLF: cooling load factor
For lighting :
Qs=W*CLF
Where:
Qs: net heat gain from lighting
W:lighting capacity: (watts)
For equipments :
Qs=qs*CLF
QL=qL
Qs,QL: sensible and latent heat gain.
CLF: cooling load factor
Results of heating and cooling load for each floor:
Floor No.
1st Floor
2nd Floor
3rd Floor
4th Floor
Heating
Load (ton)
32.7
30.9
26
34.3
Cooling
Load (ton)
59.8
56
48
60.2
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.
6) SELECTION :
Example of Fan coil unit selection:
Fan coil
no.
1
model
DC18
4Rows
Fan
speed
H
Air
flow
rate
1816
DBT
76
Entering
Total
Sensible
wet bulb
capacity capacity(BT
temp (oF)
(BTUH)
UH)
63
49767
39465
GPM WBD
9.954
4.02
Chiller selection :
WPD
KW
GPM
CAP
Ambient
LWT
MODEL
50
APS 205-
temp.(oF)
6.25
207
551.8
229.9
85
2S
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
Model
CHS
Description
High capacity
KW
500
Maximum
Maximum
temp (oF)
pressure (PSIG)
358
135
Diffuser selection:
Outlet velocity
Total pressure
Neck size in mm
CFM
0.25”
225*225
410
FPM
2000
Exhaust fan selection:
External state pressure (Inch water)
1
RPM
0.75
KW
RPM
0.5
KW
RPM
CFM MODEL
0.25
KW
RPM
K
W
797
5.306
754
4.9
709
4.5
663
4.1
1010
05
0
1000
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*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) = 5Psi
Residual Pressure ( Fixture pressure) = 8 Psi
.
Pump head = 40+85 = 43Psi
Pump flow rate =190 GPM (Using fixture unit (880F.U))
Motor
Type
215FR
Electric
3PH
Horsepower
Ports
Casting Pressure Flow
PUMP
(HP)
(Inlet*Discharge) Material
(Psi)
(GPM) MODEL
10
2-1/2*1-1/2” ANSI
304S
82
250
4K192MT3
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*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 Psi
Back Pressure(B.P) = 4*5= 20Psi
.
Residual Pressure ( Fixture pressure) = 10 Psi
Pump head = 140+20+10=170Psi
Pump flow rate =50 GPM (Using fixture unit 120F.U)
Motor
Type
56FR
Electric
1PH
Horsepower
(HP)
1/2
Ports
(Inlet*Discharge)
1-1/2*1” NPT
Casting
Material
304S
Pressure Flow
PUMP
(Psi)
(GPM) MODEL
13
52
3K392B
T0
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*3.28 +4.73*38*3.28 +15*3.5*3.28+4.95*69+3.28+ 18.9*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 = 42+100-5=137Psi
Pump flow rate =1000 GPM
Tank size =1000*90 = 90000Gallon = 90000*3.7/1000 = 333 m3
Motor
KW/rpm
185/2900
Diesel
Engine
KW/rpm
170/2900
Pump Item
No.
Pump
Model
K451A000
100-315
Range (m) Range Psig Fire duty
Usgpm
133-145
193.2-210.2
1000
Boiler chimney :
The assumption of the boiler chimney:
H = 15 m, CV=39000, ή= 80%, ρg= 1.1, v = 5m/s
Pb=101.32, Ra= 287, Ta=25+273=298, Tg=250+273=523.
mf = Qb/ CV*ή = 500/39000*0.8 = .016kg/s
mg = 25.2*mf = .403 kg \s
Ac = mg / ρg v = 0.0733 m2
Ac = π/4 * D2 - D = 30.5cm =35 cm
Q flow= mg/ ρg =.403 / 1.1= 0.366 m3/s
----∆p/ L = 0.5 pa/m
∆p= 0.5*(15*1.5) = 11.25 pa
But
∆Pboiler = (Pb*g*H)/Ra * ((1/Ta)-(1/Tg)) = 71.5 pa
∆Pboiler > ∆p= 0.5*(15*1.5) = 11.25 pa
So, no need for fan.
Finished
Thank you for your attention