Hvac Project

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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http://www.energysavers.gov/your_home/water_heating/index.cfm/mytopic=12860

http://www.energysavers.gov/your_home/water_heating/index.cfm/mytopic=12860

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


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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


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


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°)


(W/m².C) (m²) (C°) (W)

Floor 1.68 15 0 0

Windows 5.7 2.2 11 138

Doors 3 2 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

http://en.wikipedia.org/wiki/Thermal_expansion

http://en.wikipedia.org/wiki/Water_heating

http://en.wikipedia.org/wiki/Water_tank

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

http://www.engineeringtoolbox.com/water-specific-volume-weight-d_661.html


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


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:






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


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)


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:



39




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.


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°)


(W/m².C) (m²) (C°) (W)

External walls 0.84 18 17 257



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°)


(W/m².C) (m²) (C°) (W)

Floor 1.68 15 0 0

Windows 3.5 2.2 11 85

Doors 3 2 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 = 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 = 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


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.


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.


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


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



http://en.wikipedia.org/wiki/Thermal_expansion



http://en.wikipedia.org/wiki/Water_tank

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.


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:


72





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.


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


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


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






KHRQ22M20T 3 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


] x (1000L

m3 )/ ρdiesel

Burner capacity= 6 [ Kghr ] x (1000

L

m3 )/850=7Lh



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


=7Lh

x 4368hryr

x 1.5$L=45,864

$yr

Chiller Operating Cost:

Electrical chiller capacity = 555 kW.



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



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 ]



hryr


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:




hryr


kWh ]

84

Pump operating cost = (Pump capacity [ Kwunit ] x number of pump) x (operating period [

hryr


=(4[ Kwunit ] x 2unit )x 1764

hryr

x 0.175[ $kWh ]=2,470

$yr

DHW Pump Operating Cost




hryr




hryr



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]



hryr


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 ]



hryr


kWh ]Pump operating cost = (Burner capacity [ Kw


¿) 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


] x (1000L

m3 )/ ρdiesel

87

Where ρ (diesel) = 850 kg/m³.

Burner capacity= 30[ Kghr ] x (1000

L

m3 )/850=35.3Lh



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


¿35.3( Lhr )x 1764

hryr

x1.5$L=93,404

$yr

\

Boiler Operating Cost (DHW):

DHW boiler capacity is 50 KW


] x (1000L

m3 )/ ρdiesel

Burner capacity= 6 [ Kghr ] x (1000

L

m3 )/850=7Lh



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


=7Lh

x1680hryr

x1.5$L=17,640

$yr

88

Chiller Operating Cost:

Electrical chiller capacity = 432 kW.



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



hryr


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 ]



hryr


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 :




hryr




hryr


=(3 [ Kwunit ] x 2unit ) x1764

hryr

x 0.175[ $kWh ]=1,852

$yr

DHW Pump Operating Cost




hryr




hryr



hryr

x0.175 [ $kWh ]=1,176

$yr

90

Burner (Heating) Operating Cost




hryr


kWh ]Pump operating cost = (Burner capacity[ Kw


¿) x (unit energy

cost[ $kWh ])

=(0.6 KW ) x1764hryr

x 0.175[ $kWh ]=185

$yr

Burner (DHW) Operating Cost




hryr


kWh ]Pump operating cost = (Burner capacity [ Kw


¿) 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.


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;



95



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



130% 35.7100% 25.870% 15.950% 10.8


4=22.05 KW

Now, to calculate the electric consumption for the 33.6 ton unit;





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



130% 38.1100% 27.670% 17.150% 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


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



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

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