Post on 10-Apr-2015
Chapter 1 BASE CASE DESCRIBTION
1.1 IntroductionEnergy saving in buildings is a very important topic that is being studied locally
and globally to reduce the cost of energy consumption, and different electrical and
mechanical equipment that are used for different purposes. The energy saving in
buildings depends on several factors including: the comfort, the durability, the cost, and
the efficiency. These factors must be studied carefully for each building that is chosen as
a base case to take the best options that are available ones. From two factors of them, the
cost and efficiency, we create the comfort conditions for buildings, and have durability
for along time.
This type of study is chosen because its importance in our daily life. Most of
engineers are dealing with air conditioning from many years ago until these days, and try
to make successive improvements on it.
In Mechanical Engineering field, there is a combination of factors and methods
that are discussed to save energy using the best choices and decisions for constructing
and founding mechanical system with more efficient in energy, durability, and the cost.
One of the mechanical energy fields is air conditioning or HVAC engineering; it
is an applied engineering science looking for specifying the techniques to get a medium
with comfort conditions for human, no matter what is the case of surrounded air.
However, the climate of outdoor conditions is very important role to determine the
comfort indoor conditions, by increasing or decreasing the humidity and temperature for
indoor.
HVAC science also concerns on the quality of indoor air, by ventilation
techniques to obtain healthy conditions, by taking the air inside that is polluted from
microbial and smoking outside and bring a fresh air to inside, to get a comfort and
healthy conditions for humans, and to produce an excellent products like medicines as an
example.
A case study for studying energy saving in buildings is a hospital in Palestine that
is located in Bethlehem city called Caritas Hospital. The hospital is chosen because the
heath sector is considered as an important sector in any country, the development of
healthy sector is a proof of the growth and progress of the country. Thus, the idea of
saving energy in hospitals was found to provide comfortable conditions for doctors,
patients, employees, and the visitors. In addition, there is diversity in energy facilities that
number of facilities like the emergency centers that are open 24 hours, sterilization
rooms, laundering rooms, and cooking (kitchens). Also, some rooms needed 100% fresh
air or high ventilation rates like surgical rooms and sterilization rooms to remove the
microbial contamination.
To manage the energy saving in hospitals, there are several steps under
consideration: Every hospital must identify one person to manage the hospital energy
efficiency, and observing the rate of using energy and water in hospital. In addition to,
controlling on air handling units, HVAC system, and lighting.
1.2 Base Case Study
Caritas hospital which was selected as case study in this project contains four
levels, basement level, ground level, first level, and second level. Ground level and first
level are chosen to study because those levels contain the main activities in the hospital.
This hospital belongs to nursing faculty in Bethlehem city, thus there are exam
rooms, school section and classrooms for the students, and there is a section for patients.
The ground level consists of :
(See plan in the Appendix).
1. Five rooms of exams
2. Two rooms for triage purposes
3. Two rooms for feeding and changing for babies.
4. Water cycle (bathroom), the next room is for janitor
5. Store room and another bathroom.
6. Control area room.
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7. Audiology booth room.
8. Social worker room.
9. Personal use room.
10. Accountant room.
11. Nurse room.
12. Reception.
13. Waiting and play kids area.
14. Minor entrance and two corridors.
15. Emergency treatment room.
16. Two rooms for dirty and clean linen (laundry)
17. Dirty wash room.
18. Blood receiving room.
19. Two labs of Chemistry and Micro-biology.
20. Lab store room.
21. Laboratory office room.
22. Two rooms for laboratory use.
23. Blood storage room.
24. X-ray room.
25. Ultrasound room.
26. Dark room.
27. Visitor's water cycle.
28. Staff water cycle.
29. Equipment room.
30. Another bathroom.
31. Two rooms of dirty utility.
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32. Playroom
33. Physiotherapy.
34. 11 bedrooms, each room contain 5 beds.
35. Two rooms for treatment discharge.
36. Two isolation rooms.
37. Corridor for main entrance.
38. Two rooms for tea kitchen.
The area of the this level is 1300 m²
First level consists of :
(See plan in the Appendix).
1. 9 bedrooms.
2. 3 classrooms. (school section)
3. 3 school offices.(school section)
4. Laundry room.
5. Lobby.
6. Living room + kitchen.
7. Dirty utility.
8. Janitor room.
9. Waiting. (school section)
10. 3 doctors rooms. (school section)
11. Library. (school section)
12. Two bathrooms. (school section)
13. Doctor waiting room (school section).
14. Dirty utility. (School section).
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15. 3 premature rooms. (Premature section).
16. Ultra sound+ social worker. (Premature section).
17. Premature patient waiting. . (Premature section).
18. Doctors' offices+ conference room. (Premature section).
19. Water cycle. . (Premature section).
The area of the first level is 1210 m².
1.3 Climates for the region of case study
Caritas hospital is located in Bethlehem city in Palestine, thus the climate in
Palestine is affected by Mediterranean Sea climate, dry summer and short, cool, rainy
water.
The climate of Bethlehem is pleasant in the most of the time in the year; winter
extends three months from mid December to mid March and may be severe, during the
rest of year the climate is temperate. However, the hottest months are July and August.
The summer is eased by breezes that are coming from Mediterranean Sea. In addition, in
the summer season the climate is hot in daytime and fairly cool at night. The sunshine in
summer is thirteen hours during a day, but in winter the sunshine is seven hours during a
day. [1]
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1.4 Overview of Solar Water heating systems.
Solar energy and the application of solar energy are increasing because the solar
radiation is parental source of energy and it is easy to get and collect and it has variety of
application. This energy reaches the earth on the form of radiation.
Solar radiation is renewable energy that comes from the sun as a result of a
nuclear fusion that takes place in the sun.
The radiation reaches to the earth on the shape of waves, about half of these waves are
invisible (short-waves) these waves have an electromagnetic energy .The temperature of
solar radiation reaches to 5800K.
Solar water heaters also called solar domestic hot water systems can be a cost-
effective way to generate hot water for the homes. They can be used in any climate, and
the fuel they use is sun shine and it is free.
“Solar water heating systems include storage tanks and solar collectors. There are
two types of solar water heating systems: active, which have circulating pumps and
controls, and passive, which don't.” [2]
“Most solar water heaters require a well-insulated storage tank. Solar storage
tanks have an additional outlet and inlet connected to and from the collector. In two-tank
systems, the solar water heater preheats water before it enters the conventional water
heater. In one-tank systems, the back-up heater is combined with the solar storage in one
tank.” [2]
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Chapter 2
HVAC Load Analysis.
2.1 Introduction
The main aim of the (HVAC) systems is to create comfort condition for the
people inside spaces, and these conditions can be achieved by controlling the
(temperature, humidity and solar gains) in the interest space. Hospitals are one of the
most important spaces that must be conditioned in order to provide the best service for all
patients.
The most important thing that must be available in hospitals is the air
conditioning which include cooling, heating and filtration of the space, nowadays it
becomes one of the most important thing that must be provided in any hospitals specially
when we are talking about surgery rooms or intensive care rooms where the climate in
these rooms must be kept steady on a specific comfort condition.
In our case we are to design an air conditioning system for (Bethlehem) hospital,
we need to find how much power must be used to accomplish this task, heating and
cooling loads must be calculated, since they help in the selection of the air conditioning
equipment needed for this project. The calculations of the heating and cooling load are
based on the climate information that is available about Jerusalem region since there are
nearly the same climate between Bethlehem and Jerusalem. When calculating the
heating needed for the project the main load that was taken into perspective was the
transmission load, since this load takes a large space when considering cooling in winter.
The hospital buildings have a proper design shape and there service that must be
provided. Technical systems must be designed and adjusted to meet the requirements and
needs of each individual environment. Most countries have regulations that outline how
these requirements may be fulfilled, through proper design and operation of technical
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systems and the building it self. This is mainly accomplished by regulation for thermal
insulation, ventilation, lighting, and indoor temperature and humidity levels.
2.2 Comfort Temperature in medical sectors
Indoor temperature in hospitals is usually (1-4) C° higher than other natural
building. A typical temperature for patient’s rooms is (22-23) C°. This minimum value
should be maintained during the colder part of the year.
When temperature discomfort is identified, it is usually more efficient to localize
the source of discomfort and treat it, by controlling the temperature increasing or
decreasing. This can be done, by covering cold walls, erecting screens against cold drafts
from windows, window frames and badly positioned air vents, and minimizing insulation
by installing sun shades.
2.3 Indoor Air Humidity
Indoor air should be at the middle neither too dry nor high humid (which cause
perspiration and increase the risk of fungal growth). The comfort rang covers relative
humidity of 50-60%, at temperature normal for hospitals (22-24 °C), see fig. (2.1) for
comfort condition in Psychometric chart.
Humidification (due to operating cost), and especially dehumidification (due to
initial cost) are expensive. For this reason it is often acceptable to allow humidity levels
to fall below the supposed comfort level. Strict applied in rooms where conditions are
more critical, i.e. in operating theatres, intensive-care rooms, etc.
2.4 Indoor Air Quality
Good indoor air quality may be defined as air that is free of pollutants that cause
irritation, discomfort or ill health to occupants, or premature degradation of the building
materials, paintings, and furnishings or equipment. Thermal conditions and relative
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humidity also impact the perception of air quality in addition to their effects on thermal
comfort.
Focus on indoor air quality issues increased as reduced ventilation energy-saving
strategies, and consequently increased pollution levels, were introduced. A poor indoor
environment can manifest itself as a sick building in which some occupants experience
mild illness symptoms during periods of occupancy. More serious pollutant problems
may result in long-term or permanent ill-health effects.
Figure (2.1): Comfort condition point on the Psychometric Chart. [3]
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Comfort condition point
2.5 Heating Load Analysis
Space indicates either a volume or a site without a partition or a partitioned room
or group of rooms. A room is an enclosed or partitioned space that is usually treated as a
single load. A conditioned room often has an individual control system. A zone is a
space, or several rooms, or units of space having some sort of coincident loads or similar
operating characteristics. A zone may or may not be an enclosed space, or it may consist
of many partitioned rooms. It could be a conditioned space or a space that is not air
conditioned. A conditioned zone is always equipped with an individual control system. A
control zone is the basic unit of control. To determine heating load for a specific building
the following factors must be considered. [3]
2.5.1 Transmission Load
Heat loss or heat gain due to a temperature difference across a building element
The heat transferred through walls, ceilings, roof, windows glass, floors and doors is all
sensible heat transfer, referred to as transmission heat loss. [4]
2.5.2 Ventilation
Ventilation is the building service most associated with controlling the indoor air
quality to provide a healthy and comfortable indoor environment. In large buildings
ventilation is normally supplied through mechanical systems, but in smaller buildings
such as single-family homes it is principally supplied by leakage through the building
envelope (i.e., infiltration).
All structures have some air leakage or infiltration. Heat loss--because the cold
dry outdoor air must be heated to inside design temperature and moisture must be add to
the designed humidity. So the heat losses include Sensible and Latent Heat. [4]
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2.5.3 Infiltration
Infiltration is the process of air flowing in (or out) of leaks in the building
envelope, thereby providing ventilation in an uncontrolled manner. All buildings are
subject to infiltration, but it is more important in smaller buildings. In larger buildings
there is less surface area to leak for a given amount of building volume, so the same
leakage matters less. More important, the pressures in larger buildings are usually
dominated by the mechanical ventilation system and the leaks in the building envelope
have only a secondary impact on the ventilation rate. However, infiltration in larger
buildings may affect thermal comfort, control, and system balance.
2.6 Cooling Load Analysis
Cooling Load: is a rate at which energy must be removed from a space to
maintain the temperature and humidity at the design values. The cooling load will
generally differ from the heat gain because the radiation from the inside surface of walls
and interior objects as well as the solar radiation coming directly into the space through
opening doesn’t heat the air within the space directly.
There are two types of cooling load:
2.6.1 External Cooling Loads
These loads are formed because of heat gains in the conditioned space from
external sources through the building envelope or building shell and the partition walls.
Sources of external loads include the following:
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1. Heat gain entering from the exterior walls and roofs
2. Solar heat gain transmitted through the windows.
3. Conductive heat gain coming through the windows.
4. Heat gain entering from the partition walls and interior doors
5. Infiltration of outdoor air into the conditioned space
2.6.2 Internal Cooling Loads
These loads are formed by the release of sensible and latent heat from the heat
sources inside the conditioned space. These sources contribute internal cooling loads:
1. People
2. Electric lights
3. Equipment and appliances
If moisture transfers from the building structures and the furnishings are excluded,
only infiltrated air, occupants, equipment, and appliances have both sensible and latent
cooling loads. The remaining components have only sensible cooling loads. All sensible
heat gains entering the conditioned space represent radiative heat and convective.
2.7 Heat Transfer Coefficient Calculation
Heat transfer coefficient or overall heat transfer coefficient is the reciprocal of the
overall R-value. R-value is the thermal resistance for constructed materials that contain
(walls, windows, doors, etc)
U= 1Rtotal
W
m2 .˚ C (2.1)
Rtotal=X1
K1
+X2
K2
+X3
K3
+…+Ro+ Rim2 . ˚ C
W (2.2)
Where:
X: Thickness of wall (m).
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K: Thermal conductivity of the wall. W
m. ˚ C
Ro : Thermal Resistivity of the outside m2 . ˚CW
Ri : Thermal Resistivity of the inside m2 . ˚CW
2.8 Heating load Calculation
2.8.1 Transmission Load [3]
q=U × A × ¿ (2.3)
U: Overall Heat transfer Coefficient. (W/m2. ºC)A: Area of plane. (m2)To: Outdoor Temperature. (ºC)Ti: Indoor Temperature. (ºC)
2.8.2 Ventilation [3]
qs=1.23 ×V v ×(T o−T ¿) (2.4)
q l=3000 ×V v ×(ωo−ω¿) (2.5)
qS: sensible heat gain from ventilation.qL : latent heat gain from ventilation.Vv: volume flow rate of outside air. (L/s)wo: humidity ratio of outdoor temperature. (kgwater/kgair).wi: humidity ratio of indoor temperature. (kgwater/kgair).
2.8.3 Infiltration [3]
qs=1.23 ×V i×(T o−T ¿) (2.6)
q l=3000 ×V i ×(ωo−ω¿) (2.7)
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2.9 Cooling load
2.9.1 Transmission Load
q=U × A ×(T o−T ¿) (2.8)
U: Overall Heat transfer Coefficient. (W/m2. ºC)A: Area of plane. (m2)To: Outdoor Temperature. (ºC)Ti: Indoor Temperature. (ºC)
2.9.2 Ventilation
qs=1.23 ×V v ×(T o−T ¿) (2.9)
q l=3000 ×V v ×(ωo−ω¿) (2.10)
qS: sensible heat gain from ventilation.qL : latent heat gain from ventilation.Vv: volume flow rate of outside air. (L/s)wo: humidity ratio of outdoor temperature. (kgwater/kgair).wi: humidity ratio of indoor temperature. (kgwater/kgair).
2.9.3 Infiltration
qs=1.23×V i×(T o−T ¿) (2.11)
q l=3000 ×V i ×(ωo−ω¿) (2.12)
2.9.4 External walls heat Gain
q=U × A × CLTDadj (2.13)
CLTD: Cooling Load Temperature Load. U: Overall Heat transfer Coefficient. (W/m2. ºC)A: Area of plane. (m2)
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2.9.5 Glass Heating Gain
q=SHGF × A × SC xCLF (2.14)
SHGF: Solar Heat Gain Factor. (W/ m2)SC: Shading coefficient.A: Area of window. (m2).CLF: Cooling load factor
2.9.6 Lighting Heat Gain
q=Lighting Intensity [ W
m2 ]× Area [m2] (2.15)
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Figure (2.2): Lighting Intensity for any kind of lamps.
2.10 Sample calculation
2.10.1 Overall heat transfer coefficientBy taking the heat transfer coefficient of external walls as an example.
Figure (2.3): Construction of the external wall
Table (2.1): The thickness, thermal conductivity and thermal resistance for each constructed material of external wall.
External Wall Thickness(m) K(W/m.˚C) R(m2.˚C/W)Face Stone 0.05 1.7 0.03Concrete 0.2 1.75 0.11Air gap 0.02 0.28 0.07
Insulator 0.02 0.045 0.44Hollow bricks 0.1 0.9 0.11Cement plaster 0.02 1.2 0.02
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Ro 0.03
Ri 0.15
0.97 U= 1.03
Substituting in equation (2.1):
Rtot=0.051.7
+ 0.21.75
+ 0.020.28
+ 0.020.045
+ 0.10.9
+ 0.021.2
+0.03+0.15=0.97 m2 .CW
U= 10.97
=1.03W
m2 .℃
Figure (2.4): Construction of the Concrete Partition
Table (2.2): The thickness, thermal conductivity and thermal resistance for each constructed material of Concrete Partition.
Concrete partitions Thickness(m
)K(W/m.˚C)
R(m.˚C/W)
Cement plaster 0.02 0.72 0.03
Concrete 0.2 1.75 0.11
Cement plaster 0.02 0.72 0.03
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0.17 U= 5.89
Figure (2.5): Construction of the Exposed Roof.
Table (2.3): The thickness, thermal conductivity and thermal resistance for each constructed material of Roof.
Exposed RoofThickness(m
)K(W/m.˚C)
R(m.˚C/W)
Asphalt water proving 0.02 0.7 0.03
Concrete baking 0.08 1.75 0.05
Insulator 0.01 0.05 0.22
Concrete 0.06 1.75 0.03
Hollow bricks and concrete ribs 0.18 0.95 0.19
Cement plaster 0.02 0.72 0.03
Ro 0.03
Ri 0.14
0.72 U= 1.39
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Figure (2.6): Construction of the Roof.
Table (2.4): The thickness, thermal conductivity and thermal resistance for each constructed material of Exposed Roof.
Roof(other floor)Thickness(m
) K(W/m.˚C) R(m.˚C/W)
Ceramic( karmica) 0.01 1.05 0.01
Cement mortar 0.025 1.75 0.01
Sand and gravel 0.2 0.7 0.29
Concrete baking 0.08 1.75 0.05
Hollow bricks and concerts baking 0.2 0.95 0.21
Cement plaster 0.02 0.72 0.03
0.59 U= 1.68
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2.10.2 Heating Load Sample Calculation
Design conditions:
To=5 °C
Tin =22 °C
ωo =0.0160 Kg moisture/kg dry air
ωin = 0.0080 Kg moisture/kg dry air
This calculation for EXAM ROOM.5 in ground level in the base case
2.10.2.1 Transmission Load
q=U × A ×(T o−T ¿)
q External walls = 1.03 [W
m2 .℃] * 18 [m2 ] * 17 [˚C] = 315 W
q Concrete partitions = 5.89 [W
m2 .℃] * 8.3 [m2 ] * 0 [˚C] = 0 W
q Roof = 1.68 [W
m2 .℃] *15 [m2 ] * 0 [˚C] = 0 W
q Windows = 5.7 [W
m2 .℃] *2.2 [m2 ] * 17 [˚C] = 213 W
U (Windows Single glass )=5.7[ W
m2 .℃] [5]
q Door = 3 [W
m2 .℃] * 2 [m2 ] * 0 [˚C] = 0 W
q Total = 528 W
2.10.2.2 Ventilation
qs=1.23 ×V v ×(T o−T ¿)
q l=3000 ×V v ×(ωo−ω¿)
V v=[ No. of person xCFM
Person] [CFM] (2.16)
V v=CFM ×0.472 (L/s) (2.17)
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Where0.472 is a conversion factor¿ ft3
min¿ L
s
V v = 10 [ persons ] X 10 [ CFM
Person ] = 100 CFM
V v=47.2 L/s
qs = 1.23 x47.2 [L/ s] x17 [˚C] = 987 W
qL = 3000 x 47.2 [L/ s]x (0.016-0.0080) [Kg moisture/kg dry air] = 1133 W
qventilation = 987 + 1133 = 2120 W
2.10.2.3 Infiltration
qs=1.23 ×V i×(T o−T ¿)
q l=3000 ×V i ×(ωo−ω¿)
V i= (Volume of the room)*(Air change/hr)*(1000/3600) (L/s) (2.18)
Δω= ωO-ωin humidity difference (2.19)
Substituting in equation (2.17):
V i =105[m¿¿3 ]¿x ( 10003600
) = 29 L/s
qs =1.23 x 29 [Ls¿x 17 [˚C] = 606 W
qL=3000 x 29[ Ls¿x (0.016-0.0080) [Kg moisture/kg dry air] = 696 W
qInf = 606 + 696 = 1302 W
Q total = 528 + 2120 + 1302 = 3950 W = 3403 Kcal/hr
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Table (2.5): Sample Calculation for heating load on Excel Sheet.
Floor # : Ground ωo= 0.016 To = 5 (C°)
room : EXAM RM.5 ωin= 0.008 Tin = 22 (C°)
Specifications U A (To-Tin) Q
(W/m².C) (m²) (C°) (W)
External walls 1.03 18 17 315
Concrete partition 1 5.89 8.3 0 0
Concrete partition 2 5.89 18 0 0
Roof 1.68 15 0 0
Floor 1.68 15 0 0
Windows 5.70 2.2 17 213
Doors 3.00 2 0 0
528
Q ventilation (W) # of person cfm/person V QL/s W
10 10 47.2 2120
Q infiltration(W) Volume Air changes V Qm³ /hr L/s W
52.5 2 29 1310
Grand total (W) 3950
Grand total (Kcal/hr) 3403
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2.10.3 Cooling Load Sample Calculation
Design conditions:
To =33 °CTin =22 °Cωo =0.0028 Kg moisture/kg dry airωin =0.0082 Kg moisture/kg dry air
2.10.3.1 Transmission Loadq=U × A × ¿
qwindows = 5.7 [W
m2 .℃] x 2.2 [m2] x 11[˚C ] = 138 W
- External walls
q=U × A × CLTDadj East ……… q = 1.03[ W
m2 .℃] x 18 [m2] x 22.5 [˚C ] = 417 W
N/ Shaded ……... q = 1.03[ W
m2 .℃] x 8 [m2] x 13.5 [˚C ] = 111W
q total = 528 W- Glassq=SHGF × A × SC ×CLF
North …………. q = 120¿ ]* 2.2 [m2] * 0.95 = 251 W23
2.10.3.2 Internal Load
- Light
q =25[ W
m2 ] xArea [m2]
q = 25 x15=375 W
- Occupantsq = (# of Persons) x (q sensible+ q latent) (2.20) q= (10) x (60 + 40) = 1000 W- MachinesHeat gain due to the machines = (q sensible+ q latent) = 305 W
2.10.3.3 Ventilation
qs=1.23 ×V v ×(T o−T ¿) q l=3000 ×V v ×(ωo−ω¿)
V v=[ No. of person xCFM
Person] (CFM)
V v=CFM x 0.472 (L/s)V v=10 [ person ] x10 [ CFM
Person]=100 CFM
Where0.472 is a conversion factor¿ ft3
min¿ L
s
V v = 100¿ L/sqs =1.23x 47.2 [L/s] x 11[˚C ] = 638 WqL =3000 x 47.2 [L/ s] x (0.0082-0.0028) [Kg water/kg air]= 765 Wqventilation = 638 +765 = 1403 W24
2.10.3.4 Infiltration
qs=1.23 ×V i×(T o−T ¿) q l=3000 ×V i ×(ωo−ω¿) V i= (Volume of the room)*(Air change/hr)*(1000/3600) (L/s)Δω= ωO-ωin humidity differenceV i =105 [m3]*(1000/3600) = 29 L/sqs =1.23x29 [L/s] x 11[˚C ] = 395 WqL=3000*29 [L/s] *(0.0082*0.0028) [Kg water/kg air]= 473 WqInf = 395+473 = 867 Wq total = 138 + 528 + 251 + 375+1000 + 305 + 1403 + 867 = 4867 Wq total = 16608 (Btu/hr)q total = 1.4(ton ref.)q sensible,total = 3230 Wq total sensible=mC p(Troom−T supply) (2.21)
m=q total sensible
C p(Troom−T supply)
Flow rate∈CFM=m x 1765.83 Kgs
m: Mass flow rate of air in kg/s.Cp: Specific heat capacity of air Cp=1004 J/kg.Kρair =1.169 Kg/m³
m = ( 32301004 x (22−13)) = 0.357 Kg/s
Supply air flow rate = (0.357) [Kg/s] * (1765.83) = 631 CFM25
For the supply of air inside the room the sensible ventilation load is subtracted from the total sensible load, since it consumes on the fan of the fan coil then the new supply air flow rate is: q total sensible=mC p(Troom−T supply)
m=q total sensible
C p(Troom−T supply)
m = ( 25921004 x (22−13)) = 0.287 Kg/s
Supply air flow rate = (0.287) [Kg/s] * (1765.83) = 507 CFM
Table (2.6): Sample Calculation for Cooling Load on Excel Sheet.
Floor # : Ground ωo= 0.0028To = 33
(C°)
room : EXAM RM.5 ωin= 0.0082Tin = 22
(C°)
Specifications U A (To-Tin) Q
(W/m².C) (m²) (C°) (W)
Floor 1.68 15 0 0
Windows 5.7 2.2 11 138
Doors 3 2 0 0
Concrete partition 1 5.89 8.3 0 0
Concrete partition 2 5.89 18 0 0
Sum (W) 138
External walls U A CLTD Q
(W/m².C) (m²) (W)
E 1.03 18 22.5 417
S 1.03 0 19 0
W 1.03 0 27 0
N/ Shaded 1.03 8 12 111
Sum (W) 528
Glass SHGF A SC Q
(m²) (W)
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E 690 0 0.95 0N 120 2.2 0.95 251
S 350 0 0.95 0
W 690 0 0.95 0
Sum (W) 251
Exposed roof U A CLTD Q
(W/m².C) (m²) (W)0 0 15 0
Intesity Area
Light 25 W/m² 15
q (W) 375
# of q/person Q
Occupants person W/person (W)
10q/sensible 60
1000q/latent 40
Kcal/hr Q (W)q sensible 262 305
Machines q latent 0 0
sum 305# of person cfm/person V Q
Q ventilation (W) L/s W
10 10 47 1403
Volume Air changes V Q
Q infiltration(W) m³ /hr L/s W
52.5 2 29 867TOTAL (W) 4867
TOTAL (Btu/hr) 16680
TOTAL (ton ref.) 1.38
TOTAL q/sensible 2592
Supply air flow rate Kg/s 0.287TOTAL (CFM) 507
2.11 Domestic Hot Water Load Calculation
Domestic hot water is used in several fields in the hospitals, in laundry to clean the
clothes, and in the kitchen to clean the dishes and preparing the food, in water cycles or
in baths for cleaning and washing, and may be used in washing the cars, and for the
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machines that are used for hot drinks like coffee and tea, etc. and there are many
applications for the domestic hot water.
The temperature of domestic hot water that is reaching to the desired place often
ranges 50-60 ºC. But for kitchens the temperature of water is usually 65 ºC according to
the hygienic standards. In addition to, for kindergartens the temperature of water is
between 40 and 50 ºC that is for safety reasons.
Basic elements of solar water system are flat plate collector, storage tank, pump,
connecting pipes and valves.
Water circulation in the storage tank and collector can be either by natural
circulation or by forced circulation where a pump is activated by means of temperature
differential controller.
Water heating systems can also classify as:
1. Direct circulation systems: where the drawn water is the same, which circulates
through the collectors.
2. Indirect circulation systems: the fluid that removes heat from collector is different
than that drawn water. A heat exchanger is added where the heat is exchanged
(transferred) from the collector loop or fluid with the water withdrawn from the
system.
The evaluation of hot water consumption is based on the fact that the occupants
and visitors need no more than cleaning and lavatory use.
2.11.1 Calculation of Hot Water Consumption
For Medical and educational center staff , the average daily consumption of hot
water assumed as 35 L/Person.
The daily hot water demand can be calculated by using the following equation
The daily hot water demand = Average personal demand × # person
For 260 occupants
28
The daily hot water demand = 35 × 260 =9100 L/day
2.11.2 Domestic Hot Water Capacity :
As the heating load of building varies during the year, depending on the
temperature outside and the building will be under service all over the year season.
The heat load of the system can be calculated by the following equation
QD. H .W=mD. H .W x Cpw x ∆ T (2.22)
Where:
mD. H .W : Hot water flow rate (kg/s)
ΔT: temperature difference between required temperature hot water supply &
inlet water temperature (oC).
CP : specific heat of water = 4.184 kJ/kg.K
Q : heat load of system (kW).
The mass flow rate of water is calculated by the following equation
mD. H .W =ρwater×V (2.23)
ρ at 40 °C = 992 (kg/m³)
Where
ρ: water density (kg/m³).
V : volume flow rate of hot water (m³/s).
mD. H .W :hot water flow rate (kg/s).
For 12 hour operation per day:
⇒
[9100 x10−3 ] [ m3
day ]x [ 992 ] [ kgm3 ].
[12 ][ hrday ] x [3600][ s
hr]
=0.209kgs
From weather data Palestinian information, the January has the lowest ambient
temperature which will be our design temperature for cold water supply.
Tin = 5 °C
29
Tout = 60 °C
⇒ Q = 0.209 × 4.184× (60-5) = 48.09 kW (Without heat losses)
2.11.3 Heat Losses Estimation
Piping system (steel pipes, Schedule- 40). Pipes system insulated using VEEDO
FLEX insulator with low thermal conductivity k= 0.027 W/m.K.
Total pipe length = 50 m
Heat losses in insulated Piping System = 30 W/ m. [6]
Pipe heat losses = 50 × 30 = 1.5 kW
Q tot = heat load + piping losses
Q tot = 48.09 + 1.5 = 49.59 kW
2.12 System and Equipment Selections
Table (2.7): The Heating Cooling load for both two levels in Kilowatts for base case.
Level Heating Load (KW) Cooling Load (KW)
Ground 194 301
First 185 254
Total 379 555
2.12.1 Boiler Selection for Heating LoadTotal Heating load for first and ground level =379 KW
Piping system (steel pipes, schedule 40) with insulation of type VEEDO FLEX
insulator with low thermal conductivity K=0.027 W/m.K.
Total pipe length 70 m.
From (ASHRAE handbook, HVAC Application) [6]
Heat losses in insulated piping system =30 W/m
Pipelosses=70 x301000
=2.1 Kw
30
QTotal=Heating load+Pipinglosses
QTotal=379+2.1=381.1 Kw
Using Dedietrich boiler Catalog.
we select the boiler that has capacity from 300 to 390 Kw
Of type GT 408
Table (2.8): Technical data for heating boiler from Catalog Appendix[C ]
Length L mm 1505Flue nozzle Φ out side A mm 250Flow and returntapping Φ outside B inches 2'' 1/2D mm 235E mm 1427Water contact liters 366Water resistance for ∆t =15 K mbar 17 Diameter mm 530Combustion Chamber Width mm 638 Length mm 1183 Volume m^3 0.31Flue ways volume m^3 0.563Flue gas weight Oil kg/h 650
Natural gas kg/h 700
Combustion Chamber resistance mbar 1.1Section number 8Weight kg 1470
2.12.2 Burner Selection
The selection of Burner was from De Dietrich FUEL OIL BURNERS Products
Catalog.
Table (2.9): Technical data for burner from Catalog Appendix[C]
MODEL M 42-1SNumber of stages 2Burner output kW 185*/285-515Fuel oil flow rate kg/h 15,6/24,1-43,5Preset output kW 280*/450Can be fitted to boilers GT 309,408 GTE 507
31
Recommended nozzle for these boilers Max. absorbed power W 1100Motor power W 650Power supply 230~mono/50HzNoise level at 1m dB (A) ≈ 69Net weight kg/h 51
2.12.3 Heating System Pump Selection
Hot water Pump for heating systemThe total heating load = 379 kW.
mheating boiler=Total loadC p x ∆ T
ΔT = TS - Tr =15 ºC
Where
TS = supply hot water temperature = 70ºC
Tr = return water temperature = 55º C
m= 3794.18 × 15
=6 Kg /sec
Q= m
ρwater
= 6980
=¿0.006122 m3/s
Set the velocity of water =2.5 m/s
Area=Water flow rate( m3
s)
Velocityms
(2.24)
A=π d2
4=Q
V =
0.0061222.5
=2.45 x10−3 m2
d=0.056 m we take d=2.5' '
32
To select the hot water pump the flow rate and the head must be known.The flow rate was calculated and it is equal → mD. H .W= 6 Kg/sec.
Calculating the head friction losses
To find the friction factor from Moody chart the Reynolds number and relative
roughness must be known.
At flow rate equals to 6 L/s and diameter equal 56 mm
V = 2.5 m/s
The kinematic viscosity of water is 0.45 x10−6 m²/s. [7]
ℜ=VDρµ
=VDv
=2.5 x0.056
0.45 x10−6=3.1 x 105
For commercial steel the roughness (Є) =0.0006 m. [7]
ЄD
=0.00060.056
=0.0107
From Moody chart at Re =3.1 x105 and ЄD
=0.0107, the friction factor is 0.041
hp = hf +hm + ΔZ ( 2.25)
Where:
hp = pump head required in m.
ΔZ = elevation between the boiler and the highest point in the system .
hf = head friction losses in m.
hm = head minor losses in m.
h f=fxlxV 2
2 xgxD=0.041 x 67 x2.52
2 x9.81 x0.056=15.62 m
Head minor losses.
33
The equivalent length of the pipes within fittings, elbows and valves in this building
about 70 m.
hm=fx lequ . xV 2
2 xgxD=0.041 x14.5 x2.52
2 x9.81 x 0.056=3.38 m
hp = 15.62+3.38+ 8( Maximum building height) = 27 m
The selection of the Pump was from LOWARA Products Catalog, the pump
selection at total head = 27 m and flow rate =22m3
h, the model is
(SH 25 – SH 32 series) ( 32-200/30)
Operating characteristics at 2900 rpm 50 Hz , 2 Poles. Appendix (C)
2.12.4 Expansion Tank Selection for HW BoilerAn expansion tank or expansion vessel is a
small tank used in closed water heating systems and
domestic hot water systems to absorb excess water
pressure, which can be caused by thermal expansion
as water is heated.
Required volume of closed expansion tanks can be
expressed as
V exp=2V w [ ( v1
v0)−1
( pa
p0)−( pa
p1) ] (2.26)
34
Where:
Vw = Volume of water in the system (gallon, liter).
v0 = Specific volume of water at initial (cold) temperature ( m3/kg).
v1 = Specific volume of water at operating (hot) temperature (m3/kg).
pa = Atmospheric pressure (psia).
p0 = System initial pressure - Cold pressure (psia).
p1 = System operating pressure - Hot pressure (psia).
Volume of water in the boiler = 366 L (From the catalog)
Volume of water in the heating coil = 200 L
Volume of water in the piping = Total length of pipe × area
¿260 xπ4
x ( 0.019052 )+50 xπ4
x ( 0.05082 )+10 xπ4
x (0.06352)=207 L
Volume of water in the supply tank = 2000 L
Total volume of water in the system = 2773 Liter.
For this project:
To = 5° C , T1 = 75° C
Vw = 2773 L
v0 = 0.001 m3/kg, v1 = 0.001026 m3/kg
pa = 101.3 kPa, p0 = 80 kPa, p1 = 700 kPa
V exp=2 x 2773 x [ 0.0010260.001
−1
101.380
−101 .3700
] = 129 L
From the Bell and Gosset Catalog in the Appendix [C ] ITT Industries , the
Model # is HFT-90V, thus the volume of the expansion tank is 166.1 L
35
2.12.5 Chiller Selection
Total Cooling load for first and ground level=555 kW = 159 TR
Using Petra Catalog the type of chiller is PSC 165 .
Table (2.10): Technical data for Chiller from Catalog Appendix[C ]
2.12.6 Chilled Water Pump Selection
The total cooling load =555 kW.
m=Total Load kwC p x∆ T
Where:
ΔT = TS - Tr =15
T s=8 ˚ C
T r=23 ˚ C
m= 5554.18 x 15
=8.852 Kg /s
At T avg=8+23
2=15.5 ˚ C ρwater=998
Kg
m3
Q= mρwater
=8.852998
=0.008869m3
s
Set the velocity of water =2.5m/s
36
Area=Water flow rate( m3
s)
Velocityms
A=0.0088692.5
=3.55 x 10−3 m2
D= 0.0672 m , D= 3’’
To select Chilled Water pump flow rate and head must be known.
hp = hf +hm + ΔZ
Where:
hp = pump head required in m.
ΔZ = elevation between the boiler and the highest point in the system .
hf = head friction losses in m.
hm = head minor losses in m.
Calculating the head friction losses
At flow rate equal 8.852 L/s and diameter equal 67.20 mm
V = 2.5m/s
The kinematic viscosity of water is 1 .005×10−6 m²/s. [7]
ℜ=VDρµ
=VDv
=2.5 x0.0672
1.005 x 10−6=167164
For commercial steel the roughness (Є) = 0.0006 m. [7]
ЄD
=0.00893
From Moody chart at Re=167 164 and Є/D =0.00893, the friction factor is 0.038.
h f=fxlxV 2
2 xgxD= 0.038 x67 x 2.52
2 x 9.81 x 0.0672=12 m
37
Head Minor losses.
The equivalent length of the pipes within fittings , elbows and valves in this building a
bout 14.5 m .
hm=fx Lequ xV 2
2 xgxD=0.038 x 14.5 x2.52
2 x9.81 x 0.0672=2.6 m
hp = 12+2.6 +8 = 22.6 m
The selection of the Pump was from LOWARA Products Catalog, the pump
selection at total head = 22.6 m and flow rate = 31.87m3
h, the model is (SH40series)
(40-160/40 ) ,Operating characteristics at 2900 rpm 50 Hz , 2 Poles. Appendix(C)
2.12.7 Boiler Selection for DHW
The capacity of boiler that used in Domestic hot water is 50 Kw. From De Dietrich Thermique S. A. Niederbronn, FRANCE .The type selected is DTG 120-10
Table (2.11): Technical data for D.H.W Boiler from Catalog Appendix[C ]
38
2.12.8 Pump selection for DHW
The total load for domestic hot water = 50 kW.
m=Total Load kwC p x∆ T
Where:
ΔT = TSupply – Tfeed =55
T supply=60 ˚C
T feed=5 ˚C ( The worst temperature in winter)
m= 504.18 x 55
=0.2175 Kg /s
At T avg=5+60
2=32.5 ˚ C ρwater=995
Kg
m3
Q= mρwater
=0.2175995
=2.186 x10−4 m3
s
Set the velocity of water =1.5 m/s Appendix (A)
Area=Water flow rate( m3
s)
Velocityms
A=2.186 x 10−4
1.5=1.4572 x10−4 m2
D= 2√ 4 x Aπ
= 2√ 4 x1.4572 x10−4
π=0.01362 m
D=0.536 ' ' .<<<<<< D=34
' '
To select DHW pump flow rate and head must be known.
hp = hf +hm + ΔZ
Where:
hp = pump head required in m.
ΔZ = elevation between the boiler and the highest point in the system .
39
hf = head friction losses in m.
hm = head minor losses in m.
Calculating the head friction losses
At flow rate equal 0.2186 L/s and diameter equal 13.62 mm
V = 1.5 m/s
The kinematic viscosity of water is 0.075 x10−6 m²/s. [7]
ℜ=VDρµ
=VDv
=1.5 x 0.01362
0.075 x 10−6=272 400
For commercial steel the roughness (Є) = 0.0006 m. [7]
ЄD
=0.044053
From Moody chart at Re=272 400 and Є/D =0.044053, the friction factor is 0.067.
h f=fxlxV 2
2 xgxD= 0.067 x 62 x 1.52
2 x 9.81 x 0.01362=34.98 m
Head Minor losses:
The equivalent length of the pipes within fittings , elbows and valves in this building a
bout 12 m .
hm=fx Lequ xV 2
2 xgxD= 0.067 x12 x1.52
2x 9.81 x 0.01362=6.77 m
hp = 34.98+6.77+ 8 = 49.75 m.
The selection of the Pump was from LOWARA Products Catalog, the pump
selection at total head = 49.75 m and flow rate = 0.786m3
h, the model is (SH25 -
SH 32 series) (32-200/40 ) ,Operating characteristics at 2900 rpm 50
Hz, 2 Poles. Appendix (C)
2.12.9 Expansion Tank Selection for DHW
The boiler water content is 19 L
40
The pipes water content is about 500 L
Total volume of water in the system = 519 L
V exp . tank=1
20xV syst=
120
x 519=26 L
The expansion tank volume is 28.7 L [See appendix].
The expansion tank is from ITT Company Model HFT-60. Figure (2.7) : Expansion Tank
Table (2.12): Technical data for D.H.W Expansion Tank from Catalog Appendix[C]
2.12.10 Storage Tank Selection for DHW
The daily hot water demand = Average personal demand × Number of persons
For 260 occupants,
The daily hot water demand = 35 × 260 = 9100 L/ day
It is recommended by the designer to design the
volume of storage tank assuming that 70% of hot water in
the tank is usable;
storage tank capacity = 9100/0.7 = 13000 L
Storage tank capacity = 13 m3
2.12.11 Burner Selection for DHW
With boiler capacity 50 KW from De Dietrich
catalogue the burner selected is
41
M 200/1S Module of capacity range is from 38 -71 KW Appendix (C)
2.12.11 Fan Coil Selection
Fan coil unit (FCU) is one of different HVAC systems, it form as closed-loop that
containing cooling and heating coils and fan ,It can be used to control the temperature in
the space where it is installed only . It is controlled either by a manual on/off switch or by
thermostat.
FCU could be used in residential, commercial, and industrial buildings. The most
important characteristic of fan coil system is that it does not have any ductwork.
Figure (2.8): Schematic diagram for the cycle of the fan coil.
FCU work by receiving hot or cold water from a central plant by pipes and add or
removes heat from air in the limited space using fan to move the air through the coil.
The selected fan coils is selected from Petra Catalogue See Appendix (C)
2.12.12 Air Distribution and Duct Selection
The supply air should in the right temperature, humidity and in the right quantity
so that when it is mixed with the room air, the resultant room air condition falls within
the comfort condition.
42
-The correct amount of air (m3/s or CFM) that passes in each section or branch of duct
must be known and can be calculated from room sensible heat gain
-Noise level.
-Pressure drop in dampers, outlets, coils…etc. are obtainable from manufacturer
catalogues.
-Pressure drop in straight ducts and fittings is calculated or selected based on
recommended velocities.
Components of a duct system
a) Straight sections.
b) Fittings (Bends, Branches … etc).
c) Dampers.
d) Terminal units and air outlets as grills, diffusers, register …
etc).
e) Heating coil.
f) Filters.
Methods of duct sizing:
(1) Velocity method (simple, not accurate)
Volume flow rate in each branch and main duct has been given.
Velocity of air in each branch and main duct has been selected from recommended
velocities.
(2) Equal pressure drop (or equal friction method, accurate enough, widely used)
This method gives better results. It reduces size of duct and cost and is suitable for
completer system.
(3) Static region method (for balancing of branches).
(4) T-method (for computer simulation)
The method used is
Equal friction method
Main duct recommendation velocity is from 8 m/s.
43
Branches recommendation velocity is from 6 m/s
Steps of using this method
1- Determine the air quantity required for each zone to cover the cooling load (since the
cooling load is more critical than the heating load. So the duct design must depend on the
cooling load).
2- Locating the supply diffusers on the plane.
3-take the main duct velocity 7 m/sec (From velocity recommended for the buildings).
And by using the ductilator;
We can determine the remaining unknown values:
Pressure Friction (Pa/m)
Rectangular Duct Dimensions (Width X Depth)
2.12.13 Grill Selection
Supply Grill
The grills selected is from Anemostat Catalogue
The selected grills are shown in plans of duct distribution for the project
See Appendix ( C)
44
Supply grill
Return air
All return grills are located in the corridors and we added door grills for each
room in order to allow return air to pass through these door grills to return grills exists in
corridors. The grills selected is from Anemostat Catalogue See Appendix ( C)
CHPTER THREE: IMPROVED CASE
3.1 Introduction
The improvements in one side of energy consumption, the other sides are
affecting with those improvements, such as selecting techniques to decrease the
infiltrated air in doors and windows by using insulation in windows, thus will reduce the
cooling and heating loads in the buildings, and by specific kinds of walls and using the
insulation will reduce the heating and cooling loads, too. In addition to, the kind of
glazing (single or double glazing) that is using the double glass is more efficient than
single for decreasing the loss in energy; thus, give us a comfort condition with low
energy consumption.
45
Return grill
Those improvements are done by providing procedures for improving such as
lighting system by using control system on lights in the buildings, which no need to keep
the lights switching on if there is no body in the space in the building and using the lights
with low energy consumption per unit time, and using a specific types of lights that give
us a lower effect on the cooling load and more efficient in lighting, also more efficient in
consuming electrical power from the source. Furthermore, using a specific kind of
shading such as internal shading or external shading like roller blades is also affect on the
load.
3.2 Improved factors that affect the load
3.2.1 Overall Heat Transfer Coefficient Calculations
The U -value or overall heat transfer coefficient is one of the factors that affect in
heating and cooling modes. The U-value represents the value of transferring heat of any
kind of fixtures in the space per unit area in such a specific temperature difference, and it
is the inverse of the total resistances of all the components in the fixture such as walls,
thus when the total thermal resistance of a specified wall increases, then the value of
overall heat transfer coefficient decreases. Thus, the thermal insulation is used in both
base case and improvement, but in the improving the thickness of the insulation is
increased from 2cm to 3cm, that is increasing the value of the total thermal resistance of
the external walls and exposed roofs.
In windows, the base case the type of the windows is chosen as single glass and
its overall heat transfer coefficient is 5.7 W/m².K, and in the double glass with 6mm
thickness the value of U is 3.5 W/m².K. The data below showed the values for base case
and improvement for U-values.
Base Case External walls (U-Value = 1.03 W/m².K.)Insulation thickness, 2 cm (Extruded Polystyrene)
Exposed Roof (U-Value = 1.39 W/m².K.)Insulation thickness, 1 cm (Extruded Polystyrene)
46
Windows (U-Value = 5.7 W/m².K.)Single glass, clear glass.
Improved CaseExternal walls (U-Value = 0.84 W/m².K.)Insulation thickness, 3 cm (Extruded Polystyrene)
Exposed Roof (U-Value = 1.06 W/m².K.)Insulation thickness, 2 cm (Extruded Polystyrene)
Windows (U-Value = 3.5 W/m².K.)Single glass, clear glass.
Table 3.1: U_ Value for different walls and windows.
Base Case Improvement CaseU-factor W/m2.K U-factor W/m2.K
External wall 1.03 0.84Internal walls 5.89 5.89Exposed Roof 1.39 1.06
floor 1.68 1.68Glass of windows 5.7 3.5
Figure (3.1): cross section shows the main construction of the wall.
47
3.2.2 Shading Coefficient
The shading coefficient is defined as the ratio of solar heat gain of a glazing
assembly of specific construction and shading devices at summer design solar intensity
and outdoor and indoor temperatures, to the solar heat gain of a reference glass at the
same solar intensity and outdoor and indoor temperatures.
The shading coefficient SC is an indication of the characteristics of a glazing and
the associated shading devices. The data below shows the values for base case and
improvement for SC value. [4]
Base Case
Shading coefficient With no shading 0.95
Improved Case
Shading coefficient Dark roller shads with double glass regular shade 0.6.3.2.3 LightingFor electric lights installed inside the conditioned space, such as Light fixtures
hung below the ceiling, the sensible heat gain released from the electric lights, the
emitting element, and light fixtures is equal to the sensible heat released to the
conditioned Space, both depend mainly on the criteria of illumination and the type and
efficiency of electric lights.
Base Case Incandescent light bulb ( 60 W )
Intensity 25W
m2
Improved Case Compact fluorescent ( 15 W )
48
Figure (3.2): Compact Florescent.
There are two types of lights that are used in the project:
1. Incandescent – that are used in the base case, they have relatively short lives
(typically 1000 to 2000 hours of use) and are the lowest efficient of common light
sources. In fact, only about 15 percent of the energy they use as light – the rest
becomes heat.
2. Compact Fluorescent Lamps – that are used in the improved case , this type is
similar in operation to standard fluorescent lamps but it is manufactured to
produce colors similar to incandescent lamps. They are available in a range of
types and sizes to meet most applications including down lighting, ambience, task
and general space lighting. Compact Fluorescent Lamps are about four times as
efficient as incandescent and last up to 10 times longer. CFL’s combinations that
replace incandescent in standard fixtures are substantially more expensive than
their incandescent counterparts.[8]
49
Figure (3.3): The comparison between base and improved cases for the previous three
factors.
50
Lighting SC U-V
Comparison between Base Case and Improved Case
Base CaseImproved Case
3.3 Improved Case Load Calculations
3.3.1 Heating Load Calculations
After improving the three factors(U-values, SC and lighting) , we took as a sample
calculation as base case EXAM RM.5 to compare between heating load and cooling load
Table ( 3.2) Heating load calculation for EXAM RM.5 room in ground level
Floor # :GROUND ωo= 0.016 To = 5 (C°)
room : EXAM RM.5 ωin= 0.008 Tin = 22 (C°)
Specifications U A (To-Tin) Q
(W/m².C) (m²) (C°) (W)
External walls 0.84 18 17 257
Concrete partition 1 5.89 8.3 0 0
Concrete partition 2 5.89 18 0 0
Roof 1.68 15 0 0
Floor 1.68 15 0 0
Windows 3.50 2.2 17 131
Doors 3.00 2 0 0
388
Q ventilation (W) # of person CFM/person
V Q
L/s W
10 10 47 2120
Q infiltration(W) Volume Air changes V Q m³ /hr L/s W
52.5 2 29 1310
Grand total (W) 3818
Grand total (Kcal/hr) 3283
3.3.2 Cooling Load Calculations
51
Table (3.3): Cooling load calculation for EXAM RM.5 room in ground level
Floor # : Ground ωo= 0.0028To = 33
(C°)
room : EXAM RM.5 ωin= 0.0082Tin = 22
(C°)
Specifications U A (To-Tin) Q
(W/m².C) (m²) (C°) (W)
Floor 1.68 15 0 0
Windows 3.5 2.2 11 85
Doors 3 2 0 0
Concrete partition 1 5.89 8.3 0 0
Concrete partition 2 5.89 18 0 0
Sum (W) 85
External walls U A CLTD Q
(W/m².C) (m²) (W)
E 0.84 18 22.5 340
S 0.84 0 20.5 0
W 0.84 0 28.5 0
N/ Shaded 0.84 8 13.5 91
Sum (W) 431
Glass SHGF A SC Q
(m²) (W)
E 690 0 0.6 0
N 120 2.2 0.6 158
S 350 0 0.6 0
W 690 0 0.6 0
Sum (W) 158
Exposed roof U A CLTD Q
(W/m².C) (m²) (W) 0
q lamps Fu Fb CLF
Light 225 1.2 1 1
q (W) 270
# of q/person Q
Occupants person W/person (W)
10q/sensible 60
1000 q/latent 40
Kcal/hr Q (W) q sensible 262 305
Machines q latent 0 0
sum 305
52
# of person CFM/person V Q
Q ventilation (W) L/s W
10 10 47 1403
Volume Air changes V Q
Q infiltration(W) m³ /hr L/s W
52.5 2 29 867TOTAL (W) 4474
TOTAL (Btu/hr) 15266
TOTAL (ton ref.) 1.27
TOTAL q/sensible 2198
Mass flow rate m 0.243
TOTAL (CFM) 430
Table (3.4) Total heating and cooling load in base and improved case
Level Heating Load (Kw) Cooling Load (Kw)
Ground 169 237
First 157 195
Total 326 432
For Calculating the Percentage saving Energy
% Saving Energy=Total load∈base case – Total load improved caseTotal load∈base case
% Saving Energy=Total load∈base case – Total load improved caseTotal load∈base case
For the percentage saving energy in cooling load is
53
% Saving Energy=555 – 432555
=¿ 22.16 %
For the percentage saving energy in heating load is
% Saving Energy=379 – 326379
=¿ 13.98 %
Table (3.5) The percentage saving energy in cooling load .
SpaceBase Case Improved Case
Saving Energy % SavingTotal Load Total Load
KW KW KWGround Floor 301 237 64 21.26
First Floor 254 195 59 23.23Total 555 432 123 22.16
Table (3.6) The percentage saving energy in heating load .
SpaceBase Case Improved Case
Saving Energy % SavingTotal Load Total Load
KW KW KWGround Floor 194 169 25 12.89
First Floor 185 157 28 15.14Total 379 326 53 13.98
Base Case Improved Case0
50
100
150
200
250
300
350
400
Heati
ng Lo
ad K
w
54
Figure (3.4 ) : The percentage saving energy in Heating load .
Base Case Improved Case0
50
100
150
200
250
300
350
400
450
500
550
600
Cool
ing
Load
Kw
Figure (3.5 ) : The percentage saving energy in Cooling load .
3. 4 Solar water heating system
Solar domestic water system is used the sun to provide hot water to the
buildings to use in different fields. A solar system is installed in a building that has
conventional heating system by oil or gas boiler. Solar water heating systems are a good
technology that works well even in cold climates.
3.4.1 Types of solar collectors
A solar collector is a kind of heat exchanger than transform the solar radiant
energy into heat. In the solar collector, the energy transfer is from distant source of
radiation energy to fluid. Collectors are divided into three types.
55
a) Flat plate collector.
b) Concentrating collectors.
c) Evacuated tube collectors.
And each type divided into other individual types, in this chapter we will focus on the flat
plate and evacuated tube collectors which will be used in our project.
56
Figure (3.6) : Solar collectors DHW system
3.4.2 Flat plate solar collectors system design
Flat-plate collectors are the most common solar collectors for use in solar water-
heating systems.
A flat-plate collector consists basically of an insulated metal box with a glass or
plastic cover (the glazing) and a dark-colored absorber plate. Solar radiation is absorbed
by the absorber plate and transferred to a fluid
57
Figure (3.7) :Flat plate solar collectors
that circulates through the collector in tubes. In an air- based collector the
circulating fluid is air, whereas in a liquid-based collector it is usually water.
Flat-plate collectors heat the circulating fluid to a temperature considerably less than that
of the boiling point of water and are best suited to applications where the demand
temperature is 30-70°C (86-158°F) and/or for applications that require heat during the
winter months.
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
1 2 3 4 5 6 7 8 9 10 11 12
MONTH
AC
CU
MU
LA
TIV
E S
OL
AR
IRR
AD
IAN
CE
(W
.H/M
2)
Figure (3.8 ) Monthly average daily solar radiation for Bethlehem regions. [10]
58
Flat collectors can be mounted in a variety of ways, depending on the type of building,
application, and size of collector. Options include mounting on a roof, in the roof it self,
or free standing.
3.4.3 Evacuated tube solar collectors system design
A type of solar collector that can achieve high temperatures, in the range 170°F
(77°C) to 350°F (177°C) and can, under the right set of circumstances, work very
efficiently. Evacuated-tube collectors are, however, quite expensive, with unit area costs
typically about twice that of flat-plate collector. They are well-suited to commercial and
industrial heating applications and also for cooling applications (by regenerating
refrigeration cycles). They can also be an effective alternative to flat-plate collectors for
domestic space heating, especially in regions where it is often cloudy. For domestic hot
water heating, flat-plate collectors tend to offer a cheaper and more reliable option.
Figure (3.9): Evacuated tube solar collectors
An evacuated-tube collector consists of parallel rows of glass tubes connected to
a header pipe. Each tube has the air removed from it to eliminate heat loss through
convection and radiation. Evacuated-tube collectors fall into two main groups.
59
3.4.4 Comparison between flat plat and evacuated collectors
By the same procedures and calculations that were used in base case, the water
heating load calculations and the heat losses estimation calculated for the improved case.
Q tot = heat load + piping losses
Q tot = 48.09 + 1.5 = 49.59 kW
After calculating the water heating load, this equals to the useful heat from the
collectors, the collectors was selected from AMCOR COMPANY and the effective
absorber area for single glazing collector = 1.32 m2.
Bethlehem city is located at 31.5 o latitude Northern hemispheres and 35 o
longitude and the following condition must be satisfied for water heating system design:
Collectors must be oriented to the south.
Collector's inclination angle calculated as follows.
Latitude + 15 o (National energy research center, Amman, Jordan)
= 31.5+ 15 = 46.5 o to make use of solar radiation at winter seasons, but for easier
designed it was taken as 45 o.
Where:
η col : collector efficiency.
I : solar radiation falling on the tilted collector (W/ m2 ).
Q u : useful heat (W).
A : collector area (m2).
In May month, by the following procedure the collectors area could be found:
A=Qtot
η × I
The value I= 900 W/m2 is the average solar radiation value falling on the tilted collector.
[5]
A= 495900.45 × 900
=122. 44 m 2
60
Number of collectors needed = A
Aabsorber
=122. 441.32
=92.7
Number of collectors = 93
In order to determine the distance between the collectors to avoid shading effect,
the following equation are used:
D=L cos β+(L sin β cos α )
tan Φ
L : Collector length (m).
β : Collector slope angle
α : Azimuth angle of the sun.
Φ : Solar altitude angle.
D=1.90 cos 45+(1.9 sin 45 cos25)
tan30
D = 3.45 m
Values of sun’s position taken from “guidelines for energy efficient
building design”[5]
61
Evacuated Tube Solar system
By the same procedures and calculations that were used in base case, the water
heating load calculations and the heat losses estimation calculated for the improved case.
Q tot = heat load + piping losses
Q tot = 48.09 + 1.5 = 49.59 kW
3.9.1 System Design
After calculating the water heating load, this equals to the useful heat from the
collectors, the collectors was selected from AMCOR COMPANY and the effective
absorber area for single glazing collector = 2.5 m2.
Bethlehem city is located at 31.5 o latitude Northern hemispheres and 35 o
longitude and the following condition must be satisfied for water heating system design:
Collectors must be oriented to the south.
Collector's inclination angle calculated as follows.
Latitude + 15 o (National energy research center, Amman, Jordan)
= 31.5+ 15 = 46.5 o to make use of solar radiation at winter seasons, but for easier
designed it was taken as 45 o.
The average annual efficiency of the collector was assumed to be η col = 56 %
A=Qtot
η × I
Where:
η col : collector efficiency.
I : solar radiation falling on the tilted collector (W/ m2 ).
Q u : useful heat (W).
A : collector area (m2).
62
In May month, by the following procedure the collectors area could be found:
A=Qtot
η × I
A= 495900.56 × 900
=98.39 m 2
Number of collectors needed = A
2.5=
98.392.5
= 39.35
Number of collectors = 40
The value 900 W/m2 is the average solar radiation value falling on the tilted collector. [5]
F-Chart is a program that is developed by (national energy research center, solar
water heater, Jordan, 2003). The energy supplied by the collector to the total consumed
load called the F-factor, which can be calculated using the following equation.
F=1.029Y-0.065X-0.245Y²+0.0018X²+0.0215Y³…………
Where the factors X,Y are defined as:
Χ=[ FR U L×(F 'R /F R )×(T ref −T a)×Δt×AC / L ]×[ (11.6+1. 18 TW+3.86 T m−2 .32 T a) ]Y=F R ( τα )n×(F 'R /FR )×(τα )av / (τα )n×HT×N×AC/ L
Where:
63
F 'R / FR=1+ [ ( AC×F R×U L )/ (m×c p )×(1/ηHX )]−1
(τα )av / (τα )n=(0 . 096−0 .94 )
T a : Ambient temperature, C°
T ref : Reference temperature derived experimentally (100 C°)
AC : Effective collector area, m²
HT : Solar irradiance (J/m²)
FR U L : Factor of collector efficiency slope
FR (τα )n : Factor of intersection of collector efficiency curve with y-axis
The value of F-factor equals to zero (when total consumed load covered by the
auxiliary heating), and one (when total load covered by solar collectors).
The economical calculations are necessary to decide the optimum number of
collectors to provide fuel and money saves along a utilizing time.
64
Figure (3.11): The number of flat collectors using F-chart program
Area: solar collector area, (m²).
Tank area: storage tank surface area, (m²).
U tank: storage tank heat transfer coefficient, (W/m²).
HX eff: heat exchanger efficiency.
τα Efficiency: average τα to normal τα when solar radiation is perpendicular to
collector surface (recommended value 0.96).
Two: required collector’s outlet water temperature, (C°).
Storage capacity: amount of water to be stored for each square meter of collectors area,
(litters/m²).
Slope: collectors slope, (degrees).
65
CP : Specific heat capacity of the fluid to be heated (4186 J/kg.C°. for water),
Daily water consumption: quantity of required hot water, (kg/day).
Figure(3.12) The number of evacuated collectors using F-chart program
From the previous F-chart figures we see that the total utilizing of water heating is
39% of all months in the year , The other 61% of hot water will depend on the auxiliary
system such as boiler
The evacuated tube will give a bout 48 % utilizing of water heating is of all
months in the year The other 52 % of hot water will depend on the auxiliary system such
as boiler .
66
3.5 System and Equipment Selections of the improved case
3. 5.1 Boiler Selection for heating load
Total Heating load for first and ground level =326 KW
Piping system (steel pipes, schedule 40) with insulation of type VEEDO FLEX
insulator with low thermal conductivity K=0.027 W/m.K.
Total pipe length 70 m.
From (ASHRAE handbook , HVAC Application )
Heat losses in insulated piping system =30 W/m
Pipelosses=70 x301000
=2.1 Kw
Qtotal=Heatingload+Pipin glosses
Qtotal=379+2.1=328.1 Kw
Using Dedietrich boiler Catalog.
we select the boiler that has capacity 360 Kw
Of type GT 339A
See Appendix (C)
3. 5.2 Burner Selection
The selection of Burner was from De Dietrich FUEL OIL BURNERS Products of type M 32-9 S [ ] See Appendix (C)
3. 5.3 Heating System Pump Selection
Hot water Pump for heating systemAs the same procedure in base case
67
The total heating load = 379 kW.
m=Total loadC p x ∆T
ΔT = TS - Tr =15 ºC
Where
TS = supply hot water temperature = 70ºC
Tr = return water temperature = 55º C
m= 3264.18 × 15
=5.2 Kg /sec
Q= m
ρwater
= 5.2980
=¿0.005305 m3/s
Set the velocity of water =2.5 m/s
Area=Water flow rate( m3
s)
Velocityms
A=π d2
4=Q
V =
0.0053052.5
=2.122 x10−3
D= 2√ 4 x Aπ
= 2√ 4 x2.122 x10−3
π=0.0519 m
D=2 ' '
To select the hot water pump the flow rate and the head must be known.
The flow rate was calculated and it is equal → m¿
w = 5.2 Kg/sec.
Calculating the head friction losses
To find the friction factor from Moody chart the Reynolds number and relative
roughness must be known.
At flow rate equals to 5.2 L/s and diameter equal 51.9 mm
68
V = 2.5 m/s
The kinematic viscosity of water is 0.45 x10−6 m²/s.
ℜ=VDρµ
=VDv
=2.5 x0.0519
0.45 x10−6=2.88 x 105
For commercial steel the roughness (Є) =0.0006 m.
ЄD
=0.00060.0519
=0.0115
From Moody chart at Re =2.88 x105 and ЄD
= 0.0115, the friction factor is 0.040
h f=fxlxV 2
2 xgxD= 0.040 x67 x 2.52
2 x 9.81 x 0.0519=16.44 m
Head minor losses.
The equivalent length of the pipes within fittings, elbows and valves in this building
about 70 m.
hm=fx lequ . xV 2
2 xgxD=0.040 x14.5 x2.52
2 x 9.81 x 0.0519=3.56 m
hp = 16.44+3.56 + 8( Maximum building height) = 28 m
The selection of the Pump was from LOWARA Products Catalog, the pump
selection at total head = 28 m and flow rate =19m3
h, the model is (SH25 – 32Series)
Operating characteristics at 2900 rpm 50 Hz , 2 Poles. See Appendix (C)
3.5.4 Expansion Tank Selection
69
An expansion tank or expansion vessel is a small tank used in closed water
heating systems and domestic hot water systems to absorb excess water pressure, which
can be caused by thermal expansion as water is heated.
Required volume of closed expansion tanks can be expressed as
V exp=2V w [ ( v1
v0)−1
( pa
p0)−( pa
p1) ] (3.1)
Where:
Vw = Volume of water in the system (gallon, liter).
v0 = Specific volume of water at initial (cold) temperature ( m3/kg).
v1 = Specific volume of water at operating (hot) temperature (m3/kg).
pa = Atmospheric pressure (psia).
p0 = System initial pressure - Cold pressure (psia).
p1 = System operating pressure - Hot pressure (psia).
Volume of water in the boiler =350 L (From the catalog)
Volume of water in the heating coil = 200 L
Volume of water in the piping = Total length of pipe × area
Volume of water in the supply tank =1500 L
Total volume of water in the system = 2257
Liter.
For this project:
To = 5° C , T1 = 75° C
Vw = 2773 L
v0 = 0.001 m3/kg, v1 = 0.001026 m3/kg
pa = 101.3 kPa, p0 = 80 kPa, p1 = 700 kPa
70
V exp=2 x 2257 x [ 0.0010260.001
−1
101.380
−101 .3700
] = 105 L
From the Bell and Gosset Catalog in the Appendix [C ] ITT Industries , the Model #
is HFT-60V, thus the volume of the expansion tank is 121.1 L See Appendix (C)
Figure (3.14) Technical data for expansion tank.
3.5.5 Chiller Selection
Total Cooling load for first and ground level=432 kW = 123.5 TR
Using Petra Catalog the type of chiller is PSC 125.
Figure (3.15): Technical Data for chiller selected from Petra catalog. (See Appendix C)
LWT: Leaving water Temperature = 5˚C.
CAP: Total Cooling Capacity (TR)
3.5.6 Cooling System Pump Selection
The total cooling load = 432 kW.
m=Total Load (kw )
C p x ∆ T (3.2)
71
Where:
ΔT = TS - Tr =15
T s=8 ˚ C
T r=23 ˚ C
m= 4324.18 x 15
=5.98 Kg /s
At T avg=8+23
2=15.5 ˚ C ,
- Then the density of water is: ρwater=998Kg
m3
- The flow rate of water :
Q= mρwater
=5.98998
=0.005999m3
s
- Set the velocity of water =2.5m/s.
Area=Water flow rate( m3
s)
Velocityms
A=0.0059992.5
=2.396 x10−3 m2
D= 2√ 4 x Aπ
= 2√ 4 x2.396 x 10−3
π=0.05552 m
D=2.5 ' '
- To select Chilled Water pump flow rate and head must be known.
hp = hf +hm + ΔZ
Where:
hp = pump head required in m.
72
ΔZ = elevation between the boiler and the highest point in the system .
hf = head friction losses in m.
hm = head minor losses in m.
Calculating the head friction losses
At flow rate equal 5.98L/s and diameter equal 55.25 mm
V = 2.5m/s
The kinematic viscosity of water is 1 .005×10−6 m²/s.
ℜ=VDρµ
=VDv
=2.5 x0.05552
1.005 x 10−6=137 453
For commercial steel the roughness (Є) = 0.0006 m.
ЄD
=0.01080
From Moody chart at Re=137 453and Є/D =0.01080, the friction factor is 0.039.
h f=fxlxV 2
2 xgxD= 0.039 x67 x 2.52
2 x 9.81 x 0.05552=14.99 m
Head minor losses.
The equivalent length of the pipes within fittings, elbows and valves in this building
about 14.5 m.
hm=fx Lequ xV 2
2 xgxD=0.039 x 14.5 x2.52
2 x9.81 x 0.05552=3.24 m
hp = 14.99+3.24 + 8 = 26.23 m
The selection of the Pump was from LOWARA Products Catalog, the pump
selection at total head = 26.23 m and flow rate =Q=21.6m3
h, the pump selected is (SH25-
SH32 series) (32 -160/15) , Operating characteristics at 2900 rpm 50 Hz , 2 Poles.
73
3.5.7 Boiler Selection for DHW
The boiler used to provide hot water in winter is the same as the base case boiler
of 50 KW. The capacity of boiler that used in Domestic hot water is 50 Kw. From De-
Dietrich Thermique S. A. Niederbronn, FRANCE .The type selected is DTG 120-10.
The burner selection is same as base case.
3.5.8 Pump selection for DHW
The selection of the Pump was from LOWARA Products Catalog, the pump
selection at total head = 49.75 m and flow rate = 0.786 m3/hr, the model is
(SH25 - SH 32 series) (32-200/40 ) ,Operating characteristics at 2900 rpm 50
Hz, 2 Poles.
3.5.9 Expansion Tank Selection for DHW
The expansion tank selected for DHW is the same as Base case expansion tank
3.5.10 Storage Tank Selection for DHW
The daily hot water demand = Average personal demand × Number of persons
For 260 occupants,
- The daily hot water demand = 35 × 260 = 9100 L/ day
It is recommended by the designer to design the volume of storage tank assuming
that 70% of hot water in the tank is usable;
storage tank capacity = 9100/0.7 = 13000 L
Storage tank capacity = 13 m3
Assume that L = 3D = 6R
Where:
L = storage tank length. (m).
D = storage tank base diameter (m).
R = Storage tank base radius (m).
Storage tank volume can be estimated using the following equation:
74
V=π R2 × L=6 π × R3
Where:
V = storage tank volume (m3).
R = (V
6 π) 1/3 = 0.884 m.
D = 1.767 m.
L = 5.304 m.
Storage tank surfaces area:
Storage tank bottom and top area = 2 π × r2 = 4.91 m2
Storage tank side surface area = 2 r π L = 12 π r2 = 29. 46 m2
Storage tank total surface area = 4.91+ 29. 46 = 34.37 m2
Storage Tank Insulation
Equation can be used to determine the vertical tank insulation factor (W/m2)
(ƒQ/AӨ)
1R
= fQAθ
1( tavg−ta )
Where:
R: thermal resistivity of insulation, m2· K / W.
f : specified fraction of stored energy that can be
lost in time θ.
Q : stored energy, (J).
A : exposed surface area of storage unit, (m2 ).
θ : given time period, (s).
tavg: average temperature in storage unit, (°C)
ta : ambient temperature surrounding storage unit during season when it will be
heated, (°C).
75
Where:
t avg = 55 °C
t a = 15 °C
R = 3.66 (m2· K/w)
1/R = 0.273 (w/ m2· K)
(fQ/AӨ) = 10.9
But R = L / k
Where:
L : insulation thickness (m).
k : thermal conductivity (W/m.K)
(From Palestinian energy efficient building code, page 112). [5]
For extruded polystyrene as insulator;
k = 0.028 (w/m.K)
L = k × R = 0.028 × 3.66 = 0.1025 m = 10.25 cm.
U = thermal transmittance (W/ m2· K) = 1/R
U = 0.273 W/ m2· K
3.5.11 Solar Collectors Selection
From the calculation in section 3.3.4, the collector is selected is evacuated tube
with # of collectors that are 40 collectors. From the Catalog NASTEP COMPANY, the
collection area of the collector is 2.62 m2, See Appendix (C) for more details.
3.5.12 Exhaust Fan Selection
For the exhaust Fan that is selected by calculating the total flow rate of the supply
air in (CFM) for the bathrooms and kitchens, and dirty utilities, the total flow rate is 5400
CFM. Thus, it is needed to provide the system with 3 exhaust fans from the catalog of
SOLER & PALAU Company, the model of those 3 fans is (CBM-270/200- ½), for
more information See Appendix (C).
3.6 Variable Refrigerant Volume (VRV) HVAC system
3.6.1 What is VRV?
76
A VRV uses the same principle as every airconditioner;A compressor compresses
the refrigerant (gas phase) to high pressure / high temperature, Then gas condensate in
the condenser and it give the heat to the outside air, then lower the pressure by use of an
expansion valve to get low pressure / low temperature liquid, This liquid passes then to
the indoor where it takes on heat by evaporating to gas again, And the cycle restarts.
The difference with a simple split is that, with VRV, there is an expansion valve
in the indoors , every indoor can be controlled individually (of course this means another
control algorithm than split, Separate rooms with different heat loads and different
required room temperatures
Figure (3.16): The distribution of the indoor units of VRV system.
The presence of people in a room will increase heat production in the room, so the
temperature will start to rise. When the temperature sensor feels this it will react by
increasing the opening of the expansion valve. This will increase the refrigerant flow
through the indoor unit and so increase his capacity.
77
outdoor unit
many indoor units
Figure (3.17): The distribution of air of VRV system.
When the refrigerant flow through an indoor unit is increased, this effect the
pressure the main line which will go down. The compressor will react to this by
increasing his rotation speed and so increasing refrigerant flow.
Figure (3.18): The principle of the outdoor units of VRV system.
VRV systems represent the only capital project applications that genuinely
operate on the direct expansion principle. VRV systems enable a single outdoor unit to
operate in conjunction with as many as 30 indoor units, each of which can be controlled
individually. The major difference between VRV and conventional split (SRA)
applications is that in the latter, every indoor unit must be piped directly to the outdoor
78
LiquidGas
PINVST1ST2INV
INV
INV
ST2
ST1ST1 INVST1ST20%
100%
100%
Compressor capacity control
load PCB
55 capacity steps
10%
unit. The latest VRV systems however, link indoor and outdoor units in much the same
way that computers are connected to a network.
3.6.2 Equipment used in VRV system
The main equipment used in VRV system is the outdoor unit, indoor unit, copper
pipes and control system.
Figure (3.19): VRV system including Outdoor, Indoor units and piping system.
In a VRV system, the heat absorbed from a room is exhausted directly to the
outside air without the intervention of any other medium. The system is therefore, the
most energy efficient today for capital project applications. The drawing shows clearly,
the impact made by a VRV system on a typical building.
79
Figure (3.20): Indoor unit used in VRV system.
Figure(3.21 ) : refnet joint
3.6.3 VRV Equipment Selection
All equipment the selected in VRV system taken from Daiken Company
Material ListTable (3.7): the devices that are used in VRV system in base case [See Appendix D]
Model Qty DescriptionRXYQ42P7W1BA 1 Heat pump VRV III P COMPACT
RXYQ44P8W1BA 1 Heat pump VRV III P COMPACT
FXSQ125P7VEB 1 S - Concealed ceiling mounted
FXSQ32P7VEB 2 S - Concealed ceiling mounted
FXSQ40P7VEB 1 S - Concealed ceiling mounted
FXSQ50P7VEB 30 S - Concealed ceiling mounted
FXSQ63P7VEB 2 S - Concealed ceiling mounted
FXSQ80P7VEB 4 S - Concealed ceiling mounted
KHRQ22M20T 3 REFNET branch piping kit
KHRQ22M29T 3 REFNET branch piping kit
KHRQ22M64T 12 REFNET branch piping kit
KHRQ22M75T 20 REFNET branch piping kit
BRC1D52 40 Remote Controller
BHFQ22P1517 2 Outdoor unit multi connection piping kit for 3 outdoors
80
Chapter 4Economic Analysis
4.1 Base Case Cost Analysis
In determining the base case cost; the effected equipments and elements only will
be compared because the other equipments and elements are constant by size and cost
and will not change in the total cost of the project.
4.1.1 Fixed cost analysis for base case :
Table below shows the details of fixed cost of the equipment and devices.
Table (4.1) Fixed cost base analysis: this is shown in table below.
EquipmentManufacturer
(Company) Features Quantity Unit Price $Total Price
US ($ )Boiler (heating)
Capacity De-detritch 381 KW (327601 kcal/hr) 1 No. 9100 9,100Expansion tank
(heating) Volume ITT 234 L 1 No. 260 260
Chiller Capacity Petra 555 KW(159 TR) 1 No. 180,000 180,000
H.W.Pump Lowara H =27 m , Q= 22 m³⁄h 1 No. 198000 198,000
C.W. Pump Lowara H=22.6 ,31.87 m³⁄h 2 No. 2250 4,500Boiler (DHW)
Capacity De-detritch 50 KW (42992 kcal/hr) 2 No. 2000 4,000
DHW Pump Lowara H=49.75 m , Q= 0.786 m³⁄h 1 No. 2200 2,200Expansion tank
(DHW) ITT 28.7 L 1 No. 50 50
Fan Coils price Petra 0
DC6 4 KW 20 No. 300 6,000DC8 5 KW 5 No. 360 1,800
DC10 6 KW 15 No. 450 6,750DC12 8 KW 9 No. 570 5,130DC14 9 KW 8 No. 650 5,200DC18 10 KW 1 No. 720 720DC20 11 KW 2 No. 800 1,600DC24 14 KW 2 No. 1000 2,000DC30 19 KW 2 No. 1360 2,720
Storage tank price (DHW) - 9 m³ 1 No. 1200 1,200
Insulated Air Ducts - 650 m² 650 No. 25 16,250Non- insulated Air
Duct - 630 m² (Area) 630 No. 16 10,080
81
Ceiling Mounted Diffuser (Square) - 220 m2 100 22,000
Ceiling Return Grill - 150 m2 10 1,500
Fuel Tank (Heating and DHW), Weekly
consumption
Fitz-Sommons Systems 4000L 1 No. 2560 2,560
Fuel Pipes - 16 mm copper fuel pipes 10 m 8 80External Walls
Insulation - 2 cm thickness 1340 m2 1 1,340Exposed roof
Insulation - 1 cm thickness 1200 m2 0.75 900
Glass price - single U = 5.7 200 m2 8.5 1,700
Exhaust FanSOLER & PALAU Q=5400CFM 3 No. 3000 9,000
Lighting lamps cost Incandescent 800 No. 1 800
Total of Mechanical Bill ( Us $ ) 497,440
4.1.2 Annual Operating Cost for base case:
Boiler Operating Cost (Heating):
Heating boiler capacity 381 KW
Burner capacity= Fuel oil flow rate [Kghr
] x (1000L
m3 )/ ρdiesel
Where ρ (diesel) = 850 kg/m³.
Burner capacity= 35[ Kghr ] x (1000
L
m3 )/850=41.176Lh
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Where 21[ weekyr ] represents the period of the winter season and it is equivalent to 5
months in the year.
Diesel price= 6 NIS
L= 1.5
$L
Boiler operating cost = (burner capacity) × (operating period) × (diesel price)
41.176Lh
x 1764hryr
x1.5$L=108,952
$yr
82
Boiler Operating Cost (DHW):
DHW boiler capacity is 50 KW
Burner capacity= Fuel oil flow rate [Kghr
] x (1000L
m3 )/ ρdiesel
Burner capacity= 6 [ Kghr ] x (1000
L
m3 )/850=7Lh
Operating period=12 [ hrday ]x 7[ day
week ]x 52[ weekyr ]=4368
hryr
Where 52[ weekyr ] represents all weeks in the year that the boiler used to supply hot water
in 12 hr per day.
Diesel price= 6 NIS
L= 1.5
$L
Boiler operating cost = (burner capacity) × (operating period) × (diesel price)
=7Lh
x 4368hryr
x 1.5$L=45,864
$yr
Chiller Operating Cost:
Electrical chiller capacity = 555 kW.
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric ( KWh ) price=0.7NIS
KWh=0.175
$KWh
Chiller Operating cost = (Chiller capacity) x (operating period) x (unit energy cost)
= 555 [ kW ] x1764 [ hryr ] x 0.175[ $
KWh ]=171,329$yr
Fan Coils Operating Cost :
Average electrical Fan coil capacity = 0.7 KW [see Appendix C ]
These values refer to theory chapter as in catalog, see Appendix
83
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]F.Cs operating cost = (F.C capacity x number of F.C[ Kw
unit ]) x (operating period[ $kWh ]) x
(unit energy cost[ $kWh ])
= (0.7 [ Kwunit ] x64 unit ) x1764
hryr
x 0.175[ $kWh ]=13,688
$yr
Pump Operating Cost ( heating) :
Average electrical pump capacity = 3 KW [see Appendix C ]
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]Pump operating cost = (Pump capacity [ Kw
unit ] x number of pump) x (operating period [
hryr
¿) x (unit energy cost[ $kWh ])
=(3 [ Kwunit ] x 2unit ) x1764
hryr
x 0.175[ $kWh ]=1,852
$yr
Chilled water Pump Operating Cost:
Average electrical pump capacity = 4 KW [see Appendix C ]
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]
84
Pump operating cost = (Pump capacity [ Kwunit ] x number of pump) x (operating period [
hryr
¿) x (unit energy cost[ $kWh ])
=(4[ Kwunit ] x 2unit )x 1764
hryr
x 0.175[ $kWh ]=2,470
$yr
DHW Pump Operating Cost
Average electrical pump capacity = 4 KW [see Appendix C ]
Operating period=12 [ hrday ]x 7[ day
week ]x 52[ weekyr ]=4368
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]Pump operating cost = (Pump capacity [ Kw
unit ] x number of pump) x (operating period [
hryr
¿) x (unit energy cost[ $kWh ])
=(4[ Kwunit ] x 1unit )x 4368
hryr
x 0.175[ $kWh ]=3,058
$yr
Burner (Heating) Operating Cost
Average electrical pump capacity = 1.1 KW ( at max. consumption) [see Appendix C]
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]Pump operating cost = (Burner capacity[ Kw
unit ]) x (operating period [hryr
¿) x (unit energy
cost[ $kWh ])
=(1.1 KW ) x 1764hryr
x0.175 [ $kWh ]=340
$yr
85
Burner (DHW) Operating Cost
Average electrical pump capacity = 0.215 KW ( at max. consumption) [see Appendix C ]
Operating period=12 [ hrday ]x 7[ day
week ]x 52[ weekyr ]=4368
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]Pump operating cost = (Burner capacity [ Kw
unit ]) x (operating period [hryr
¿) x (unit energy
cost [ $kWh ])
=(0.215 KW ) x4368hryr
x 0.175[ $kWh ]=165
$yr
4.2 Improved Case Cost Analysis:
4.2.1 Fixed cost analysis for improved case:
Table below shows the details of fixed cost of the equipment and devices for the
improved case.
Table (4.2) Fixed Cost analysis: using the table below.
Equipment Features Quantity Unit Price $Total Price
US ($ )Boiler (heating)
Capacity 326KW(860 Kcal/hr) 1 No. 8400 8,400Expansion tank
(heating) Volume 121 L 1 No. 135 135Chiller Capacity 432 KW(124 TR) 1 No. 110,000 110,000
H.W.PumpH =27 m , Q= 22
m³⁄h 2 No. 1800 3,600
C.W. PumpH=22.6 ,Q=31.87
m³⁄h 2 No. 2200 4,400
Boiler (DHW) Capacity50 KW (42992
kcal/hr) 1 No. 2000 2,000DHW Pump H=49.75 m , Q= 1 No. 2200 2,200
86
0.786 m³⁄hExpansion tank (DHW) 28.7 L (Volume) 1 No. 50 50
Fan Coils price No. 0DC6 4 KW 17 No. 300 5,100DC8 5 KW 5 No. 360 1,800DC10 6 KW 13 No. 450 5,850DC12 8 KW 8 No. 570 4,560DC14 9 KW 7 No. 650 4,550DC18 10 KW 1 No. 720 720DC20 11 KW 2 No. 800 1,600DC24 14 KW 1 No. 1000 1,000DC30 19 KW 1 No. 1360 1,360
Storage tank price (DHW) 9 m3 1 No. 1200 1,200
Insulated Air Ducts 500 m² (Area) 500 m2 25 12,500Non- insulated Air Duct 630 m²(Area) 630 m2 17 10,710
Ceiling Mounted Diffuser (Square) 200 No. 100 20,000
Ceiling Return Grill 130 No. 90 11,700Fuel Tank (Heating and
DHW), Weekly consumption 1000galon(3785 L) 1 No. 2400 2,400
Fuel Pipes16 mm copper fuel
pipes 10 m 8 80External Walls
Insulation 3 cm thickness 1340 m2 1.5 2,010Exposed roof Insulation 2 cm thickness 1200 m2 1 1,200
Glass price Double U = 3.5 200 m2 18 3,600Solar Collectors
( evacuated ) 40 No. 400 16,000Exhaust Fan CBM/270/270 3 No. 3000 9,000
Lighting lamps cost Fluorescent 800 No. 13 10,400
Total of Mechanical Bill ( Us $ ) 258,125
4.2.2 Annual Operating Cost for Improved case:
Boiler Operating Cost (Heating):
Heating boiler capacity 328.1 KW
Burner capacity= Fuel oil flow rate [Kghr
] x (1000L
m3 )/ ρdiesel
87
Where ρ (diesel) = 850 kg/m³.
Burner capacity= 30[ Kghr ] x (1000
L
m3 )/850=35.3Lh
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Where 21[ weekyr ] represents the period of the winter season and it is equivalent to 5
months in the year.
Diesel price= 6 NIS
L= 1.5
$L
Boiler operating cost = (burner capacity) × (operating period) × (diesel price)
¿35.3( Lhr )x 1764
hryr
x1.5$L=93,404
$yr
\
Boiler Operating Cost (DHW):
DHW boiler capacity is 50 KW
Burner capacity= Fuel oil flow rate [Kghr
] x (1000L
m3 )/ ρdiesel
Burner capacity= 6 [ Kghr ] x (1000
L
m3 )/850=7Lh
Operating period=12 [ hrday ]x 7[ day
week ]x 20[ weekyr ]=1680
hryr
Where 20[ weekyr ] represents the weeks in winter that the boiler used to supply hot water
in 12 hr per day.
Diesel price= 6 NIS
L= 1.5
$L
Boiler operating cost = (burner capacity) × (operating period) × (diesel price)
=7Lh
x1680hryr
x1.5$L=17,640
$yr
88
Chiller Operating Cost:
Electrical chiller capacity = 432 kW.
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric ( KWh ) price=0.7NIS
KWh=0.175
$KWh
Chiller Operating cost = (Chiller capacity) x (operating period) x (unit energy cost)
= 432 [ kW ] x1764 [ hryr ] x 0.175[ $
KWh ]=133,358$yr
Fan Coils Operating Cost :
Average electrical Fan coil capacity = 0.7 KW.
These values refer to theory chapter as in catalog, see Appendix
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]F.Cs operating cost = (F.C capacity x number of F.C [ Kw
unit ]) x (operating period [ $kWh ])
x (unit energy cost [ $kWh ])
= (0.7 [ Kwunit ] x55 unit) x1764
hryr
x0.175 [ $kWh ]=11,885
$yr
Pump Operating Cost ( heating ) :
Average electrical pump capacity = 2.3 KW [see Appendix C ]
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]89
Pump operating cost = (Pump capacity [ Kwunit ] x number of pump) x (operating period [
hryr
¿) x (unit energy cost [ $kWh ])
=(2.3 [ Kwunit ]x 2unit ) x1764
hryr
x 0.175[ $kWh ]=1,420
$yr
Chilled water Pump Operating Cost :
Average electrical pump capacity = 3 KW [see Appendix C ]
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]Pump operating cost = (Pump capacity [ Kw
unit ] x number of pump) x (operating period [
hryr
¿) x (unit energy cost [ $kWh ])
=(3 [ Kwunit ] x 2unit ) x1764
hryr
x 0.175[ $kWh ]=1,852
$yr
DHW Pump Operating Cost
Average electrical pump capacity = 4 KW [see Appendix C ]
Operating period=12 [ hrday ]x 7[ day
week ]x 20[ weekyr ]=1680
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]Pump operating cost = (Pump capacity [ Kw
unit ] x number of pump) x (operating period [
hryr
¿) x (unit energy cost [ $kWh ])
=(4[ Kwunit ] x 1unit )x 1680
hryr
x0.175 [ $kWh ]=1,176
$yr
90
Burner (Heating) Operating Cost
Average electrical pump capacity = 0.6 KW ( at max. consumption) [see Appendix C ]
Operating period=12 [ hrday ]x 7[ day
week ]x 21[ weekyr ]=1764
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]Pump operating cost = (Burner capacity[ Kw
unit ]) x (operating period [hryr
¿) x (unit energy
cost[ $kWh ])
=(0.6 KW ) x1764hryr
x 0.175[ $kWh ]=185
$yr
Burner (DHW) Operating Cost
Average electrical pump capacity = 0.215 KW ( at max. consumption) [see Appendix C ]
Operating period=12 [ hrday ]x 7[ day
week ]x 52[ weekyr ]=4368
hryr
Electric (kWh) price = 0.7 [ NISkWh ] = 0.175 [ $
kWh ]Pump operating cost = (Burner capacity [ Kw
unit ]) x (operating period [hryr
¿) x (unit energy
cost [ $kWh ])
=(0.215 KW ) x4368hryr
x 0.175[ $kWh ]=165
$yr
Lighting Operating Cost
Using Fluorescent Lamps
Operating period=20 [ hrday ] x7 [ day
week ] x52[ weekyr ]=7280
hryr
91
Electric (kW) price = 0.7 [ NISkW ] = 0.175 [ $
kW ]Operating Electrical Cost =
Watt used x Energy rate xOperating period1000
12000 x 0.175 x72801000
=15,288$yr
Table(4.3) Summary of the operating cost in base case.
Base Case
Equipment Operating Cost ($/yr)
Boiler Operating Cost (Heating) 109,292Boiler Operating Cost (DHW) 46,029Chiller Operating Cost 171,329Fan Coils Operating Cost 13,690 Pump Operating Cost (heating ) 1,852Chilled water Pump Operating Cost 2,470DHW Pump Operating Cost 3,058
Lighting Operating Cost 61,152
Total 408,872$
Table (4.4) Summary of the operating cost in the improved case.
Improved Case
Equipment Operating Cost ($/yr)
Boiler Operating Cost (Heating) 93,589Boiler Operating Cost (DHW) 17,805Chiller Operating Cost 133,358Fan Coils Operating Cost 5,094Pump Operating Cost ( heating ) 1,420Chilled water Pump Operating Cost 1,852DHW Pump Operating Cost 1,176
Lighting Operating Cost 15,288
Total 269,582$
92
4.3 Total Cost Calculation
Fixed charge Rate
FCR=i+ i
(1+i )n−1+ Insurance+Taxes
n=40 year (lifeof project) , i =12 %
FCR=0.12+ 0.12
(1+0.12 )40−1+0.02+0.01=15.14 %
For Base case
Total AnnualCost=[ FCR x Capital Cost ]+Operating Cost
Total AnnualCost=[ 0.1514 x 497,440 ]+408,872=484,184$yr
For Improved case
Total AnnualCost=[ FCR x Capital Cost ]+Operating Cost
Total AnnualCost=[ 0.1514 x 258,125 ]+269,582=308,662$yr
Total saving cost=Total costbase−Total cost improved
Total saving cost=484,184$yr
−308,662$yr
=175,522$yr
% Saving Cost=Total costbase−Total cost improved
Totalcostbase
% Saving Cost=484,184−308,662484,184
=36.25 %
4.4 Solar system cost analysis :
Solar collectors are used to heat domestic hot water which saves fuel used for this
job especially in summer. However these collectors aren’t used in the base case.
Collector price = 400$/unit × 40 collector = 16000 $
93
Expansion tank price = 50$
Storage tank price = 1360 $
Pipes price = 13 $/m × 2 × 25 m = 650 $
Total price = 16000 + 50 + 1360 + 650 = 18,060 $
Cost of fuel saved by inserting solar system:
Heat gain by collector = 50 kW × 40 (collectors) = 2000 kW
Period of heating = 12(hr/day) × 7 (day/wk) × 21 (wk/yr) = 1764 hr/year
High heating value of diesel = 36693 KJ/Kg
Amount of diesel early = 2000 (kJ/s) × 3600 (s/hr) × 1764 (hr/yr) / 36693
= 346137 kg/year
= 346137 (kg/yr) × 1000(L/m³) / 850(kg/m³)
= 407220 L/year
Cost of diesel yearly = 407220 × (6/4) $/L
= 610,830 $/ year
Repair cost (Pay back period) = Solar System Cost / cost of fuel saved
= 18,060 / 610,830
= 0.029 YEAR
4.5 Case study between chiller system and VRV III
The case study will be for outdoor units only for both systems:
4.5.1 VRV POWER INPUT:
94
According to cooling capacity tables in VRV system, we found that the VRV
system work most time in the partial load.
We will calculate the electrical power consumption depend in four values of the
combination ratio (partial time) for VRV that taken from cooling capacities, and then take
the average value for them as follows: (indoor temperature=23 ⁰C, outdoor tem. =39⁰ C);
For RXYQ20P=20hp=20 ton=70 KW.
Combination ratio Power input KW (compressor +outdoor unit fan)
130% 15.2100% 1170% 6.8450% 4.61
Average power input= (15.2+11+6.84+4.61)
4= 9.4 KW
Now, to calculate the electric consumption for the20 ton unit;
Cost=energy*price (NIS)
Energy=power*time (K.W.H)
If we assume that the system will work for 12 hours a day, and seven days a week
and 30 days in month, and price for KWH=0.7 NIS;
Energy=power*time=9.4 KW *12*30=3384 K.W.H (FOR 30 DAYS)
Cost=energy*price=3384 KWH*0.7NIS/KWH=2368.8 NIS
For RXYQ32P=32hp=35.6 ton=124.6 KW.
Combination ratio Power input KW (compressor +outdoor unit fan)
130% 28.7100% 20.870% 12.950% 8.68
Average power input=(28.7+20.8+12.9+8.68)
4=17.77 KW
Now, to calculate the electric consumption for the 35.6 ton unit;
Cost=energy*price (NIS)
Energy=power*time (K.W.H)
95
If we assume that the system will work for 12 hours a day, and seven days a week
and 30 days in month, and price for KWH=0.7 NIS;
Energy=power*time=17.77 KW *12*30=6397.2 K.W.H (FOR 30 DAYS)
Cost=energy*price=6397.2 KWH*0.7NIS/KWH=4478 NIS
For RXYQ42P=42hp=33.6 ton=152.6 KW.
Combination ratio Power input KW (compressor +outdoor unit fan)
130% 35.7100% 25.870% 15.950% 10.8
Average power input=(35.7+25.8+15.9+10.8)
4=22.05 KW
Now, to calculate the electric consumption for the 33.6 ton unit;
Cost=energy*price (NIS)
Energy=power*time (K.W.H)
If we assume that the system will work for 12 hours a day, and seven days a week
and 30 days in month, and price for KWH=0.7 NIS;
Energy=power*time=22.05 KW *12*30=7938 K.W.H (FOR 30 DAYS)
Cost=energy*price=7938 KWH*0.7 NIS/KWH=5556 NIS
For RXYQ44P=44hp=35.2 ton=123.2 KW.
Combination ratio Power input KW (compressor +outdoor unit fan)
130% 38.1100% 27.670% 17.150% 11.5
Average power input=(38.1+27.6+17.1+11.5)
4=23.6 KW
96
If we assume that the system will work for 12 hours a day, and five days a week
and 30 days in month, and price for KWH=0.7 NIS;
Energy=power*time=23.6 KW *12*30 = 8496 K.W.H (FOR 30 DAYS)
Cost=energy*price= 8496 KWH*0.7NIS/KWH=5947 NIS
WE HAVE IN THE PROJECTS THIS NUMBER OF OUTDOOR UNITS:
RXYQ20P7W1B: ONE UNIT
RXYQ32P7W1B: TWO UNITS
RXYQ42P7W1B: ONE UNIT
RXYQ44P7W1B: ONE UNIT
SO WE CONCLUDE THAT THE ELECTRICAL CONSUPTION FOR ALL THE
UNITS WILL BE:
If we assume that the system will work for 12 hours a day, and seven days a week and 30
days in month, and price for KWH=0.7 NIS;
TOTAL CONSUPTION IN (NIS) =2369+4478*2+5556+5947=22,828 NIS
month
1 $ = 4 NIS
Then
TOTAL CONSUPTION IN ($) =
2 2,828NIS
monthx
1 $4 NIS
x12 month1 year
=86,484$
year
4.5.2 Chiller System Calculation Electrical Power:
As we know from all catalogues of most types of Chiller systems, each one ton
cooling need 8 Ampere to work= 1.76 KW, and all know that the chiller system work
with just ON-OFF compressors (no capacity steps, no combination ratio), so these types
can work only with 1.76 KWton
or BY ZERO.
As for our project, 200 hp=160 ton of cooling need (160*1.76 KW) =281.6 KW.
As before, we calculate now the cost by;
Energy =power*time=281.6 KW*12*30 =101,376 K.W.H (FOR 30 DAYS)
Cost=energy*price=101,376 KWh* 0.7 NIS/KHW= 70,963 NIS
97
70,963NIS
monthx
1 $4 NIS
x12 month
1 year=212,889
$year
So we conclude that the owner will save
¿Chiller OperatingCost−VRV Operating Cost
¿212,889$
year−86,484
$year
=126,400$
year
CHAPTER FIVE: CONCLUSION AND RECOMMENDATIONS
98
5.1 CONCLUSION
In this project, the main goals that will be considered in the beginning of the
project will be achieved successfully in spite of performing them partially. The
improvement that is taking place in this project is performed by saving in the cost in each
fixed cost and annual cost.
Saving energy in the building was achieved by reducing the U-value of the
external walls and the roof, this reducing was by inserting Polyethylene layer in the
external walls construction and increasing hollow bricks thickness. Also the U-value of
the glass was changed by choosing shaded glass with different thickness. The solar
collectors save energy used for heating domestic hot water; this energy was consumed by
the boiler in the base case, the type of collectors that is chosen evacuated tube collectors,
since it is more efficient in maitainting on the temperature of water and it is more
efficient in collecting the solar energy with area that is less than flat plate collector.
In shading coefficient factor, the base case is without shading, and in improved
case with roller blades shading, and this is minimized the load, and thus minimizing in
the cost of the boiler. In lighting system, it is used Incandescent type in base case, and in
the improved case, it is used florescent type of lights. The florescent type of lighting is
more efficient because the life of it is more than the incandescent.
In VRV system, the annual operating cost is less than the annual operating cost in
fan coil system, and also in fixed cost, and the problems of VRV system is less then
VWV system or fan coil system.
Saving energy appears clearly in chapters of improved case and economic
analysis. And the results will be shown in figures as a comparison between the cases, and
it is noted the saving cost is about 175,000 $ and that is good amount of saved money .
99
5.2 Recommendations:
100
Base Case Improved Case Saving
Heating Load 381 326 55
Cooling Load 555 432 123
50
150
250
350
450
550
Comparison between Cooling and Heating Load with Saving
Load
KW
Fixed Cost Operating Cost Total Annual Cost
Base Case 497440 408872 484184
Improved Case 258125 269582 308662
Saving 239315 139290 175522
50000
150000
250000
350000
450000
550000
Coparison between Base case and Improved case
$/ye
ar
It is recommended that many factors should be taken in consideration in any
project that will be performed and designed, these factors are:
1) The effect of insulation
2) the effect of lighting,
3) The effect of shading coefficient in HVAC system and its effects on the load.
Finally: choosing the system that has good features from any other systems in
the side of fixed cost, operating cost, and choosing techniques to minimize the
load.
101
102
103