This document summarizes the design of an air conditioning system for Al-Zayan Mall in Alexandria, Egypt. The mall has 3 basement floors, 4 floors of retail space, and 6 floors of administrative offices. The design involves calculating the cooling load using the HAP program, selecting appropriate air handling units and fan coil units from manufacturer catalogs based on the load results, and designing ductwork using the equal friction method. Key components of the chilled water system are also described, including water chillers, cooling towers, pumps, and the water distribution system. Duct material, thickness, insulation, installation and testing are discussed.

1. Al-Zayan Mall Air Conditioning System
Supervised By:
Prof. Dr. Samy Morsy EL-Sherbiny
Graduation Project

2. Introduction

3. 1-Brief about the mall
• The aim of the project is to design an air conditioning system for a mall.
• This mall calls El Zayan mall.
• This mall locates in Semouha, Alexandria.
• Area of mall =5500 𝑚2.
• All surrounding buildings are residential buildings.

4. Building Construction
Building
Construction
3 Floors as
Basements 4 Floors as
malls
6 Floors as
administrative
Surface roof

5. Basement Construction
3 Floors as
basements
First
basement
(upper)
Second
basement
Third
basement
(lower)

6. Basements are used for car barking

7. Part of the third basement contains the chillers and
pumps for HVAC system.

8. 4 Floors
as mall
Ground
floor
First
floor
Second
Floor
Third
Floor

9. Six floors of administrative offices.

10. Surface roof

11. 2-Project main design procedures
• 3 floors (basements)
Ventilation by using supply and suction Fans through galvanized
steel ducts.
• 4 Floors (Mall)
Each floor is conditioned by Air Handling Units (AHUs) for open
regions and ducted Fan Coil Units (FCUs) for closed regions
(spaces) .

12. Cooling Load calculations
Cooling load calculations is important because the HVAC equipment needs to be the
right size to do it’s job .
The selection of (HVAC) system components and equipment should always be based on
accurate determination of building heating and cooling loads .
The consequences for choosing the wrong-sized system can result:
Noisy operation, inability to keep people comfortable and a failure to maintain proper moisture
control.

13. Required Data for Cooling Load Calculations:
 Building location and orientation (architectural Plans)
 Building construction (architectural plans)
 Outdoor design conditions
 Indoor design conditions
 Occupancy schedule
 Lighting
 Equipment schedules
 Infiltration / ventilation

14. Data for AL-ZAYAN mall
 Building location and orientation
Smouha – Alex - Egypt
 Outdoor design conditions
DBT = 37oc WBT =28 oc
 Indoor design conditions
DBT= 24 oc RH = 50 %
 Occupancy schedule

15. Data for AL-ZAYAN mall
 Lighting
(20 Watts/ square meter)
 Equipment schedules
 Infiltration / ventilation
( 20% fresh air)

16. Components of cooling load:

17. Methods for estimating cooling loads:
•Manual calculations
•Using HAP program
 manual calculations:
we calculate:
 Transmission heat gain
 Solar heat gain
 People load Q total= ∑Q
 Equipment load cooling load = Q total × F.O.S
 Light load
 Ventilation load
and any other loads.

18. the results from HAP and manual calculations were so close to
each other for single zone in third floor so,
we used HAP calculations in this project for the whole mall.
 HAP calculations:
at first we enter some input data like: weather, wall and roof construction,
equipment ,light and people schedule and then we select the type of system
used.

19. Results from HAP:
ground floor

20. Air Conditioning System
Classification of air conditioning systems
HVAC Systems
Chilled
water
System
Direct
Expansion
System
(DX System)

21. Direct Expansion System
The evaporator absorbs heat from the air supply by expanding
refrigerant, which then flows to a compressor . The compressed
refrigerant moves to a condenser, releasing the heat absorbed from the
air supply .

23. Chilled Water System
The chiller basically removes heat from the water. It is used as a
refrigerant to remove heat from the building. The chilled water
circulates through a chilled water loop and through coil located
in air handlers .

24. Why We Use Chilled Water System?
1- Due to a large thermal load of the place .
Available Full Load
Required System
7 TR
Window / Mini
Split
7 : 15 TR
Concealed
15 : 70 TR
Central Split Unit
>70 TR
Chiller

27. Air Handling Units :

28. AHU Components :

29. Fan Coil Units :
Ducted Fan Coil Unit
Operating Characteristics :
A one coil unit installed in a 2‐pipe system shall be capable of
providing cooling as determined by the operating mode of the
central water supply system.

30. Selection of air equipment devices
1) AHU selection
 Using CARRIER AHU catalogue:
Selection data required:
1) Supply air flow rate (L/S) from hap
2) Air velocity inside AHU (m/s) assume v=2.5 m/s
3) Cooling capacity (KW) from hap

31. - Enter the catalogue with supply air flow rate (L/S) and air velocity
inside AHU to determine AHU unit size as shown:
unit size as a function of air volume

32. model number nomenclature
- After determining AHU unit size , you can select AHU
components from the catalogue as shown:

33.  mixing damper
mixing damper specifications

34.  Cooling coil
cooling coil specifications

35.  Fan section
fan specifications

36. fan performance

37.  panel filter section
 bag filter section
pressure drop verses face velocity for bag filter and panel filter

38. 2) FCU selection
 Using saiver FCU catalogue:
Selection data required:
1) Cooling capacity (W) or (Btu/hr) from hap
2) Sensible cooling capacity (W) or (Btu/hr) from hap
3) Air flow rate (L/S) or (Cfm) from hap
4) Air inlet temperature
5) Chilled water inlet temperature

39. - Enter the catalogue with this information to determine the unit
model and its specifications as shown:

40. FCU specifications

41. DUCT DESIGN

42. Air Duct
 Ducts : pipes or passages are used in HVAC to deliver and remove air
For example
• Supply air duct
• Return air duct
• Exhaust duct
 Shape of Duct
• Round duct
• Oval duct
• Squared duct
• Rectangular duct

43. Rectangular Duct
 Aspect ratio of the duct : ratio of larger to smaller dimension of cross
section.
• Aspect ratio = a/b
• Recommended Aspect ratio= 1-4
Lower aspect ratio
• Less material due to lesser perimeter.
• Lesser cost of installation.
• Lesser insulation cost.
• Lesser running cost due to lesser pressure drop

44. Duct Sizing
 The factors which affected on design of ducts:
• CFM
• False ceiling height
• Velocity “V”
• Friction “f”
 Steps of design:
1. Load estimation (TR ,CFM).
2. Air outlets design (CFM for each grill).
3. Design drawing single line from ACU to air outlets “simple, symmetrical”.
4. Calculate CFM in each duct section.
5. Use the Duct Sizer program.
6. Constant friction.
7. Each section by “CFM, f” we get sizing of duct.

45. Equal Friction Method
 This method is based on principle that friction rate per unit length in entire
duct system would be maintained constant.
 This can be used by any engineer with knowledge of friction chart.
 Recommended friction factor is taken as 0.8 to 1.2 pa/m with duct
velocity not exceeding 8-12 m/s and noise level not exceeding 48 db
 Equal Friction Method Tools
• Friction chart.
• Duct sizer program

46. Duct Material
 Aluminum (K=205W/m K)
o Advantage: Fire resistant, Rust resistant, Bacteria resistant.
o Disadvantage: Expensive.
 Galvanized steel (K=50.2 W/m K)
o Advantage: Fire resistant, Rust resistant, Cheap
o Disadvantage: might rust over the days giving rise to bacterial growth.
 Fabric (textile).
o Advantage: flexibility, mounting, even air distribution and no insulation required.
o Disadvantage: could be damaged in case of fire.

47. Duct thickness
 The factors which affected on thickness of ducts:
• Pressure inside the duct.
• Dimensions of cross section .
• Type of duct material.
 Recommended thickness.
Larger size Required thickness
Up to 12” 0.6 mm
14” : 30” 0.7 mm
32” : 40” 0.8 mm
40’’-60’’ 1 mm
More than 60’’ 1.2 mm

48. Duct Insulation
 Heat insulation is outside the duct to decrease heat transfer and prevent
water vapor condensation on the outer duct surface by using fiber
glass(k=.04 W/m K).
• In door duct : Insulation density=12 𝑘𝑔/𝑚3,Insulation thickness=1”
• outdoor duct: Insulation density=24 𝑘𝑔/𝑚3,Insulation thickness=4”
 Sound insulation is inside duct to damp sound level by using arm flex.
• Insulation thickness= 1-3 cm
 Insulation area
A=(W+H)*L
fiber glass arm flex

49. Duct installation
 Types of Joint Used In Duct depend on the pressure inside the duct
• S slip Joint: small size (up to 600mm).
• Drive slip: medium size(up to 1000mm).
• Double S slip joint: large size (up to 1500 mm).
• Flange joint: very large size (more than 1500mm).
 Silicon is used in all joints to avoid leakage
 Duct supporters
HAGER SCHEDULE
Longest Dimension
of Duct
Rod Diameter Maximum Spacing
300- 600 mm 8 mm 3000 mm
600-1300 mm 10 mm 2500 mm
Over 1300mm 15 mm 2500 mm

50. Duct Test
 Smoke Test
1. All air outlets must be closed except one in the end of the duct.
2. Smoke equipment :to provide smoke inside the duct.
3. Checking all joints for any leakage.
4. Use silicon to fix any joint with small leakage.
5. Replace any joint with high leakage.
 Light Test
1. fall darkness.
2. Using very strong torch(1000 Watt) or more than one.
3. Test to be done for separate straight duct.
4. Checking all joints for any leakage.
5. Use silicon to fix any joint with small leakage.
6. Replace any joint with high leakage.

51. Duct Test
 Smoke Test
1. All air outlets must be closed except one in the end of the duct.
2. Smoke equipment :to provide smoke inside the duct.
3. Checking all joints for any leakage.
4. Use silicon to fix any joint with small leakage.
5. Replace any joint with high leakage.
 Light Test
1. fall darkness.
2. Using very strong torch(1000 Watt) or more than one.
3. Test to be done for separate straight duct.
4. Checking all joints for any leakage.
5. Use silicon to fix any joint with small leakage.
6. Replace any joint with high leakage.

52. SOURCES
 ASHRAE – Fundamentals
 SMACNA – HVAC Systems Duct Design

53. Water cycle components
1- water cycle components.
2- cooling tower.
3- pump head.
Contents

54. Chilled Water system components
The conventional chilled-water system consists of combinations of the following primary
components:
1-Water chillers.
2- Load terminals (chilled-water cooling coils in comfort-cooling applications) Cooling
towers in water-cooled systems.
3-Chilled and condensate pumps.
4-Chilled and condenser water distribution systems that include piping, an expansion
valve, control valves , check valves, strainers ,..

55. 1-chillers
A- The vapor-compression cycle chiller.
In the vapor-compression refrigeration cycle, refrigerant enters the evaporator in the form of a cool,
low-pressure mixture of liquid and vapor (A). Heat is transferred from the relatively warm air or
water to the refrigerant, causing the liquid refrigerant to boil. The resulting vapor (B) is then drawn
from the evaporator by the compressor, which increases the pressure and temperature of the
refrigerant vapor. The hot, high-pressure refrigerant vapor (C) leaving the compressor enters the
condenser, where heat is transferred to ambient air or water at a lower temperature. Inside the
condenser, the refrigerant vapor condenses into a liquid. This liquid refrigerant (D) then flows to the
expansion device, which creates a pressure drop that reduces the pressure of the refrigerant to that
of the evaporator.. The cool mixture of liquid and vapor refrigerant (A) travels to the evaporator to
repeat the cycle.
Refrigerant type is HFC-134a

56. Main components:
1-Compressor.
2-Condenser.
3-Evaporator.
4-Expansion valve.
1-Compressor.
The compressor function is to increase the pressure and temperature
of the refrigerant . The most common energy source to drive the
compressor is an electric motor. Water chillers using the vapor-
compression refrigeration cycle vary by the type of compressor used.
Reciprocating, rotary and centrifugal compressors are common types
of compressors used in vapor-compression water chillers.
We used a centrifugal compressor for these reasons:
1-discharge flow is relatively free of pulsation.
2-efficient performance over a wide range of pressure capacity.
3-less contamination.
4- small , quiet and cheap.

57. 2-Condenser
Types: water cooled, air cooled and evaporative.
We used a shell and tube water cooled condenser for these reasons:
1-high thermal conductivity of water
2-high specific heat of water
3- less flow if required to condensate the same heat as that in case of air
cooled condenser.
4-high heat transfer coefficient.
5- high cop.
Note: the water in the tubes and the
refrigerant in the shell.

58. 3- the evaporator
Types:
the flooded shell and tube and the direct expansion evaporators (DX).
We used the flooded shell and tube where the water flows in the tube
and the refrigerant in the shell
Advantages :
1- the configuration gives a large surface area in a small volume.
2- available with numerous passes.
3- available in large capacities ranging from 10 TR to thousands of TR.
4- easily cleaned.
5- high efficiency.

59. 4- the expansion valve
A valve through which refrigerant under pressure is allowed to
expand to a lower pressure and a greater volume .it also
controls the capacity of the refrigerant with load .we used
thermal expansion valve as it’s cheap and needs less
maintenance.

60. B-Absorption refrigeration cycle chiller
It is in reality the same refrigeration process discussed previously except the
compressor has been replaced with an absorber, generator, pump and recuperative heat
exchanger.
We used lithium bromide/water. The refrigerant (water) in the evaporator migrates to the
lower-pressure absorber where it is “soaked up” by a solution of lithium-bromide. While
mixed with the lithium-bromide the vapor condenses and releases the heat of
vaporization picked up in the evaporator. This heat is transferred to condenser water
and rejected out the cooling tower. The lithium-bromide and refrigerant solution (weak
solution) are pumped to a heat exchanger (generator) where the refrigerant is boiled off
and the lithium-bromide (strong solution) returns to the evaporator.

61. Direct-fired Absorption Chiller
The direct-fired absorption chiller includes an
integral burner, rather than relying on an external
heat source. fuel used to fire the burner is natural
gas. Typical COPs for direct-fired, double-effect
chillers are 0.9 to 1.1

62. Cooling tower
the conversion of liquid water to a gaseous phase requires the latent heat of
vaporization. Cooling towers use the internal heat from water to vaporize the
water in an adiabatic saturation process. A cooling tower’s purpose is to
expose as much water surface area to air as possible to promote the
evaporation of the water.
Types: spray , forced draft and induced draft.
We used the induced daft cooling tower as it has larger capacity than spray
and less circulation than forced draft.

63. Chilled water pump head
The pump used is single stage centrifugal pump for these reasons:
1-it has less friction losses and minimum wear .
2-there is almost no noise .
3-it has a magnetic coupling which breaks up on high load eliminating the risk of
damaging the motor.
How to calculate the pump head?
Pressure drop= friction losses+ chiller pressure drop + AHU or FCU pressure drop
Chiller & AHU or FCU pressure drop : from catalogue.
Hf =(h𝑓 ∗ 𝐿𝑡𝑜𝑡𝑎𝑙)/100
hf : from friction chart
𝐿𝑡𝑜𝑡𝑎𝑙 = Lpipe + Leq
Leq : from table

64. Valves losses:

65. Fittings losses:

66. Design of chilled water's pipeline

67. Design recommendation:-
- length of pipeline should be minimum length and using
minimum number of fittings like elbow, tee,… to decrease
cost of construction and to decrease the pressure drop of
pipeline grid to min value .
- For diameter 2" or less check for velocity doesn’t
increase than 1.2 m/s.
-For diameter 2.5" or more check for pressure drop doesn’t
increase than 400 pa/m (0.040 m h d)

68. • To get the minimum allowable diameter we should use
this chart (for steel pipes)
• pipe material is seamless black steel (schedule 40)

69. •we can determine pipe diameter by different
methods:-
• friction chart.
• friction table.
• pipe sizer program .
• pipe flow wizard program .

70. Piping design by pipe flow wizard program
• For example {space (S206)}(second floor)
• flow rate = 1.764 L/s -Start with (1/2") then check
velocity

71. • Last assumption was (2") and finally it was right
after check on velocity because diameter equal 2"

72. Pipe sizing for the 1stfloor (lower )

73. Flow control methods (system balance)
-Common reversed return

74. Direct return method (by using balancing
valve)

76. Chillers selection:
In our project there are two types of chillers :
1-Mechanical chiller
2-Absorption chiller

77. 1-Mechanical chiller selection :

78. From HAP results:
-Ground floor load=285,5 TR
-First floor load=268 TR
-Second floor load=344 TR
-Third floor load=305 TR
-Total load=1202,2 TR
-Chiller load =f.o.s *total load
=1.08 *1202,5=1300 TR
-Each Chiller load = 650 TR (2275 KW)

79. From carrier catalogue :
-we enter table with 2275 KW capacity

81. Table important output data :
-model=19XR-5R514QELEH52
-input power=416.6 kw
-cop=5.487
-operating weight=10.5 TN
-Dimensions (L*W*H)=4.46*2.054*2.207 m

82. 2-Absorption chiller

83. We enter lithium promide table with (2275
kw) capacity:

84. Selection output data:
-model=16 JL (98bar)
-cooling capacity=2321 kw
- operating weight=22 TN
-length=6,124 m
-width=2,004 m
-Height=3,464 m

85. Cooling tower output data
-Inlet temperature= 40 c
-Outlet temperature=32 c
-Flow rate=165,8 l/s (597m^3/h)
-Pressure drop=93 kpa
-The two cooling towers are the same

86. Pump selection:
-From calculations :
-Flow rate=246 l/s
-Head=7,6 m
-In GRUNDFOS company site
-From soft were program
-AS following steps:

87. Step one :

88. Step two :

89. Step three :

90. Step four :

93. Selection results :
-Product name= 2*HS 350-250-498
-Product no= 99129670
-Max operating pressure= 16 bar
-flow rate=885,6 m^3/h
-Head=7,6 m
-Energy consumption=57107 (kwh/year)
-Energy costs =8566,06 (EURP/year)
-Motor no=5712607301576
-Motor type= 250MS

95. Heat balance:

96. From calculations:
-Input power=416,16 kw
-Qevap=2445,3 kw
-Input power+Qevap=2861,3kw
-Qcond=3084,8 kw
So from balance :
-Qcond almost eual Qevap+ input power

97. Ventilation system
.
it’s changing air in any space to provide high indoor air quality.
Ventilation is required for several reasons

98. Types of ventilation:
1- Natural ventilation 2- Mechanical ventilation

99. Amount of exhaust air
CFMEXHAUST =
𝑉 ∗ 𝐴𝐶𝐻
1.7
CFM amount of exhaust air draft from space in cfm.
V volume of space = Area * Height ( 𝑚3
)
ACH Air change per hour , depend on application of space.

100. Mechanical
Ventilation
Kitchen
Ventilation
Basement
Ventilation
Bathroom
Ventilation

101. Kitchen Ventilation
The ventilation spot at the point of harmful gases
generation using hoods, That to avoid the spread of
gases in the space.
Q = A*V
A Area of hood
V Velocity = 0.5 m/s
=100 fpm
Hood types:
- Single Hood
- Double Hood

102. Bathroom Ventilation
For bathroom not used Fresh air fan and instead of them by install Louver
in door.

103. Basement Ventilation
There are two type of basement ventilation system:
1- Traditional system
depend on exhaust fans draft air by air ducts from spaces.
Consist of:-
- Exhaust fans
- Air duct
- Grills
- Supply fans

104. 2- Induction fan (jet fan) system
depend on air motion inside basement by jet fans to reach
exhaust air at draft point ( exhaust fans ) , basement should be
have many draft point distributed inside
Consist of:-
- Jet ( induction ) fans
- Heat & smoke detectors
- CO detectors
- Control panel connect of all detectors & fans

105. For Traditional system:-
There are 2 exhaust fans at one side of mall remove exhaust air and1
fresh fan from another side to balance system
- Due to heavy exhaust of cars we need to divide amount of L/S in to:-
1- 65% of exhaust air from high level
2- 35% of exhaust air from low level (40 cm from F.F.L)
- Amount of Fresh air (L/S) = (0.85 – 0.90) of exhaust air amount.
Amount of Exhaust air
(L/S)
Amount of Fresh air
(L/S)
26880
24192
upper basement
26622
23960
2nd basement
18060
16254
lower basement

106. Hourse power of fan:
Fan hp =
𝐶𝐹𝑀 ∗ 𝑆𝑡𝑎𝑡𝑖𝑐 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛.𝑤𝑔
6356 ∗ 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

107. Thank You