Design of an Air Conditioning System for M-2 Hall

SESSION 2003PROJECT ADVISOR

PROF. DR. ARSHAD HUSSAIN QURESHIPROJECT COORDINATOR MR. MUHAMMAD ASLAM

GROUP MEMBERSMUHAMMED SHOAIB JAMIL 2003-UET-ME-RCET-02 ADEEL AKHTAR CH. 2003-UET-ME-RCET-27 SAJJAD AHMED 2003-UET-ME-RCET-28 SAJJAD QAISER MEHMOOD 2003-UET-ME-RCET-13

DEPARTMENT OF MECHANICAL ENGINEERINGRACHNA COLLEGE OF ENGINEERING & TECHNOLOGY, GUJRANWALA

UNIVERSITY OF ENGINEERING & TECHNOLOGY, LAHORE

Design of an Air Conditioning System for M-2 Hall(USING CANAL WATER) This report is submitted to Department of Mechanical Engineering at Rachna College of Engineering & Technology, Gujranwala For the partial Fulfillment of the requirements for the

BACHELORS DEGREE

In

MECHANICAL ENGINEERING

INTERNAL EXAMINER: NAME: PROF. DR. ARSHAD HUSSAIN QURESHI

SIGN: _______________

EXTERNAL EXAMINER: NAME: ________________________________

SIGN: _______________

DEPARTMENT OF MECHANICAL ENGINEERINGRACHNA COLLEGE OF ENGINEERING & TECHNOLOGY, GUJRANWALA

UNIVERSITY OF ENGINEERING & TECHNOLOGY, LAHORE

DedicationThis thesis report is dedicated to Our beloved parents who brought up us for this stage of learning And to the

Engineering Professionin recognition of the pivotal role played by refrigeration and air conditioning engineering in the analysis and design of the engineering systems

AcknowledgementsWe are graciously thankful to Prof. Dr. Arshad Hussain Qureshi for his valuable suggestions and supervision. His uniqueness and expertise in the area of Heat & Mass Transfer, Fluid Mechanics, and Thermal Engineering etc that really proved to be a source of continuous learning and inspiration for us, causing to work us hard. We wish to thank also our project coordinator Mr. Muhammad Aslam for his excellent guidance to complete this project before time. We are indebted to many of our lecturers, class fellows also who directly or indirectly helped us in preparing this thesis report, in particular, the present form of it. Various parts of the manuscripts were copied and distributed to many concerned persons in the college and outside, they returned with a lot of suggestions and improvements. Discussions with professors and engineers were of great help to us. We want to mention particularly Prof. Dr. Riaz Ahmed Mirza, Prof. Dr. Yunas Jamal, and Associate Prof. Mr. Asif Aslam, Mr. Ali Jibran (Former Lecturer in Rachna College), Mr. Qasim Ali Ranjha (Lecturer), Mr. Waqar Ahmed Qureshi (Former Lecturer in College) and many others who helped us in completion this thesis report with their substantial improvements. Our very special cordial thank goes to Mr. Sheikh Naseem Ahmed, founder of Siddique Sadiq Trust Hospital Gujranwala, who permitted us to visit the hospital central air conditioning system for pursuing the basics of HVACR principles. We also like to thank HVAC Engr. Muhammad Qayyum, PITB Lahore; Engr. Muhammad Riaz of Rupafil Limited Shiekhupura, who helped us with their polite and cooperative manners and provides us technical information & permission to use HVAC software. We also want to thank our non-academic friends like, Muhammad Akram Antal, Rana Abdul Jabbar, Abdul Haq (computer expert), and many others who helped us in various aspects. Furthermore, we wish to thank Mr. Akbar of Lalazar Printers, UET Lahore for his effective cooperation and great care in printing very efficiently all the crucial sections of this thesis report.

Group Members25 May, 2007

PAC-105

ContentsAcknowledgement

Unit 1 Introduction to PAC-105--Introduction ............ 1.1

Unit 2 Cooling Load Calculation and Psychrometrics--Introduction .......... 2.1 Cooling Load Calculation of M-2 Hall. 2.2 Pschyrometric Analysis of PAC-105 Air Conditioning System 2.1 2.2 2.4

Unit 3 Design of Air-Handling Unit and Ducting System--Introduction .......... 3.1 Function of Air-Handling Unit and Types ............ 3.2 Design of PAC-105 AHU ....................................... 3.3 Design of PAC-105 Cooling Coil of AHU ........ 3.4 Arrangement of Chilled Water Circuits for PAC-105 Cooling Coil ... 3.5 Design of Ducting System for Hall M-2 ................. 3.6 Indoor air quality (IAQ) and Selection of Air Filter for PAC-105 AHU .... 3.7 Fan Selection for PAC-105 AHU ............................ 3.8 Suggestions for PAC-105 AHU Fabrication ........ 3.9 Design of Condensate Drain Pan of the PAC-105 AHU 3.10 Results and Comments.. 3.1 3.2 3.2 3.3 3.12 3.14 3.26 3.26 3.29 3.31 3.34

Unit 4 Design of PAC-105 Refrigeration System-- Introduction .......... 4.1 The PAC-105 A Vapor Compression Refrigeration System 4.2 Refrigerant and its Selection Criteria for PAC-105 Refrigeration System .. 4.3 Graphical Evaluation of the PAC-105 Refrigeration 4.4 Comparison between PAC-105 and Conventional Air-Cooled Chilling System... 4.5 Designing and Selection of the PAC-105 Compressor 4.6 Results and Comments .. 4.1 4.2 4.2 4.3 4.6 4.8 4.20

Unit 5 Design of Water Chiller of PAC-105--Introduction ...................................... 5.1 Chiller and Its Types ........................................ 5.2 Flooded Liquid Chiller of PAC-105 ........... 5.3 Design of PAC-105 Flooded Liquid Chiller ... 5.4 Comments on PAC-105 Flooded Liquid Cooler design .......................... 5.5 Suggestion for PAC-105 Flooded Liquid Cooler ................ 5.1 5.2 5.4 5.5 5.10 5.10

Unit 6 Design of Water Cooled Open-type Condenser

i

PAC-105-- Introduction ..................................... 6.1 Condensers .................................... 6.2 Classification of Condensers .................... 6.3 Water-Cooled Open-type Condenser of PAC-105 ....... 6.4 Design of PAC-105 Water-Cooled Open-type Condenser .... 6.5 Suggestions for Construction and Installation of PAC-105 Condenser . 6.6 Results and Comments .................. 6.1 6.2 6.2 6.4 6.4 6.10 6.13

Unit 7 Design of PAC-105 Hydronic System-- Introduction ..................................... 7.1 Piping System of the PAC-105 .... 7.2 Centrifugal Pump Selection Criteria for PAC-105 .. 7.3 Determination of the Pump Head of PAC-105 .. 7.4 Estimation of Centrifugal Pump Power of the PAC-105 . 7.5 Results and Comments ......... 7.1 7.2 7.3 7.4 7.8 7.10

Unit 8 Design of PAC-105 Control System--Introduction ...................................... 8.1 Control System of PAC-105 ............................ 8.2 Hydronic and Refrigeration System Control ............................................... 8.3 Supply Air System Control...................................................... 8.1 8.2 8.2 8.4

Unit 9 Commissioning and Maintenance of PAC-105 Air Conditioning System--Introduction 9.1 Fundamental of Commissioning ...... 9.2 Commissioning Check Lists of PAC-105 9.3 Maintenance and Repair of PAC-105 .. 9.4 PAC-105 Maintenance and Repairing Schedule 9.1 9.2 9.3 9.9 9.9 A.1

-- Appendix A

Table A1 Abbreviations Table A2 Conversion of Units from I-P to SI -- Appendix B ... B.1 Table B1: Table of Thermo physical Properties of Air at Atmospheric Pressure Table B2: Thermophysical Properties of Saturated Water' Figure B1: R134a Pressure-Enthalpy Chart (S.I. units) Table B3: Properties of Refrigerant-134a Table B4: Local Loss Coefficients, Elbows Table B5: Loss Coefficients, converging junctions Table B6: Loss Coefficients, Transitions (Convergent Flow) Table B7: Loss Coefficients, Diverging Junctions Table B8: Circular Equivalents of Rectangular Ducts for Equal Friction and Capacity --References

ii

PAC-105

Unit 1

Introduction to PAC-105

Introductiont is well known fact that energy is the soul of human activities directly or indirectly. This is why; the measuring rod of a countrys development is the per capita use of energy. In this respect we are perhaps at the tail end and are using only few hundreds of energy units per annum per head as against about seven thousand in Europe and about ten thousand in USA. We have to go a long way to attain some measures of respectability in this respect. During last few years the situation has deteriorated and we have to resort to loadshedding which caused a lot of inconvenience to domestic consumers and industrialists as well. Although we are feeling well now economically, but still we are standing at the lowest rung of the ladder micro and macroecnomicaly as compared to the developed nations. This lack of energy can impart a serious setback to economic growth of country and can make our progress shaky.

I

1.1

Unit 1: Introduction to PAC-105Pakistan has availability of various kinds of energy resources like oil, coal, natural gas, etc more and more reservoirs of these energy resources are being explored. These resources are created by the natural process deep down in the earth after thousands of years and are described as Non-renewable fuel recourses. However large these resources may be, they are being consumed at high rate and are more liable to be depleted within few decades. It will seriously disrupt the normal existence of our energy demanding and energy oriented territory. This problem of energy crises is world wide accepted and is a challenge for engineers and scientists. It can be coped by introducing energy audits and better energy management techniques in Pakistan. At the local edge, solution seems to be main three fold, i.e. Discovery of renewable energy resources Energy recovery techniques / energy from waste Energy management plans etc

As for as the first option is concerned, due efforts are being made by the government to use the wind power and solar energy across the coastal areas, deserts and nationwide respectively. Although, they have limited application but the trend is uncourageous. The second option is also being utilized by installation of incineration plants in various cities like Lahore, Karachi etc As for as last option of energy saving is concerned this is seem to be more realistic now a days that Prime Minister Shaukat Aziz has launched Load-shedding Management Plan LMP on the first of May this year. In an interview he said that by LMP we could save 500 MW electricity each month. This is clear interpretation of energy management plans that how we can save the energy effectively in this regards? Our suggestion to solve this energy crisis and make our local resources of energy more reliable and long-lasting is An efficient use of every kind of resources (both renewable and Non-renewable) This option is of utmost importance. It will impart the independence and self reliance in energy sector. This suggestion can be implemented by the engineers by designing and introducing the local energy-efficient systems. This energy-efficient system might be an energy saving system (related to energy consumption) or a system which utilizes the energy resources (related to energy generation) more effectively. This suggestion can be applied to any energy resource like solar energy, wind energy, hydroelectric power and fossil fuels, but especially to the renewable energy resources. For example flowing canal or river water can be used to drive a water wheel or small water turbine unit, causing generation of mechanical work, which in turn may rotate an auxiliary generator for electricity generation. This electricity can be supplied to nearby population or even to a small industry in accordance with its generation capability. By careful observation of our surrounding, we can explore numerous ways to use the energy resources in our local vicinity in more efficient manner. Using above mentioned idea of designing an energy-efficient system to utilize a local energy resource. We utilized canal water as follows: Our institute, Rachna College of engineering and technology Gujranwala (constituent college of UET Lahore) has a nearby canal about 70 meters away from its main campus building. During summer season especially in months of July-September the weather becomes more severe, thus creating problem for all (students, teachers, and managerial staff etc) to continue their normal activities. It becomes more difficult for Engineering drawing students performing their drawing work in hall M-2, located upstairs on the first floor of admin block. Although the hall is well furnished with fans, ventilators, window pans provided with curtains to reduce solar radiation, white washed with a light color, but due to hot climate every thing become insufficient

1.2

Unit 1: Introduction to PAC-105and demanding for an air conditioning system for better temperature and humidity control inside the drawing hall. Because we belong to the same institute and had faced the same problem of hot climate and uncomfortable working environment during our Drawing hours. The solution of problem was to Air condition the hall simply. But the running cost of this summer Air conditioning system was a major factor to ponder. We started thinking on the problem. Another consideration in our mind was the availability of low temperature (of order 16 o C) canal water closer to campus. Then, we conceived an idea that Is there any way to condition this hall at lower running cost using this low temperature canal water flowing idler? The answer was YES, and was explored by the study of vapor compression cycle that is an approximation to reversed Carnot cycle. In vapor compression cycle the purpose of compressor is to compress the vapor refrigerant up to such an extent so that the refrigerant may easily reject its heat of condensation to some cooling medium like air or water. Higher is the temperature of cooling medium, higher will be the compression needed to reject heat and vice versa. Thus more compression work needed for given capacity of refrigeration machine and temperature limits in the former case. That means a large size compressor will be needed. Hence resulting in more energy consumption and higher running cost accordingly. Using this idea we utilized the low temperature canal water as a heat sink and designed a complete air conditioning system for summer Comfortable Air Conditioning (CAC) of hall M-2 and operating at lower running cost. The detailed discussion is provided in unit 4 of this thesis report. There it is mathematically proved that, how much energy can be saved using this suggested way of Air conditioning for hall M-2? The aspect of energy efficiency is shown by too high COP (Coefficient of Performance) of the refrigeration machine. Thereof it is: 1st, final year Project of design of an Air Conditioning system at RCET which was conceived and initiated in year 2005 Therefore, it was patented as PAC-105. Throughout the thesis report, this system will be called by its patent name i.e. PAC-105 and most of the equipments also will be discussed headed by this name i.e. patent name until specified or mentioned. Because it is an energy conservation project, therefore a concept of energy saving has been strictly followed during design of each and every component of this air conditioning system (referred to as PAC-105 shortly), rendering it more uniqueness. Another feature of this system is that it has greater capability to cope with seasonal variations in climatic temperatures. For example during peak loading hours like in months of June-July when outside temperature rises to even 40oC - 45oC of locality, conventional systems (air cooled systems / systems other than PAC-105) will be running at their extreme upper loading condition. This severe climatic change may lead the conventional system to malfunctioning or tripping. But PAC-105 will be running at its normal pace because such sever climatic change can never rise the temperature of bulk cooling water of canal only by few degrees that will be negligible for the system performance. It shows that offset between the performance curves of PAC-105 and conventional refrigeration system will increase at worst climatic temperatures range. Thus it can be inferred that higher climatic temperatures will be conducive towards the better performance of PAC-105 refrigeration system, which is quite impossible in case of any conventional air cooled systems.

1.3

Unit 1: Introduction to PAC-105It is important to realize here that in each unit of this thesis report, the information is given purely related to the PAC-105 equipments otherwise unless specified or mentioned. All the illustrations regarding PAC-105 air conditioning system are constructed or produced in the Computer Aided Drafting and Design, CADD with own self which is a clear interpretation of the hardware equipments. There may be a little variation in the drawings and dimension during the fabrication of the hardware equipments of the PAC-105 air conditioning system. It is also important to mention here that we also employed the few engineering software in the determination of the properties of the air and water, which are provided to us from our friends working in the fields of HVAC&R. We are much pleasure to mention here that about most of the text material which is added in each unit concerning PAC-105 is not yet copied or transferred massively, except major concepts; so we are adding learning concepts, which we learnt during study of the course. At the start of the each unit a short introduction is given related to the concerned unit, which has key information about the unit that what approach or idea is behind it? Note: An objection by some of our friends and colleagues on this system was that our suggested system has its limited application, because it is based on the utilization of natural water resource. Our answer was always the same and was as follows, 1. Our objective and suggestion is to use every resource of energy in an efficient way. In our case it was canal water and we used it. For other applications and locations any other available source of energy can be utilized Also, 2. It is a sort of Case study, and Case study is always related to some specific area of research Result: Finally it is suggested that, for all systems having vapor compression cycle (VCC) based refrigeration systems and nearby natural water source it is recommended that they must be designed in a way to avail the opportunity to reduce their energy consumption and making it more economical. If possible, during site selection for a refrigeration unit like cold storage units, ice plants, industrial refrigeration systems, or even a commercial air conditioning system employing VCC, due importance should be given to availability of low temperature water resource.

1.4

PAC-105

Unit 2

Cooling Load Calculation and Psychrometrics

Introduction

I

n this unit we are discussing two basic portions of HVAC system design procedure. In the first portion we will discuss the Cooling load calculation of M-2 Hall and in the second portion, the Psychrometrics analysis of PAC-105. In the first portion of this unit, we will calculate the cooling load of M-2 hall, which is the amount of heat that must be removed from the Hall, to maintain comfortable level of temperature and humidity of the air. In the cooling load calculation, we shall calculate the heat gain into the Hall due to all possible factors. At the end of the unit there is a cooling load table for M-2 Hall, in which all the calculations are shown in the form of a table. In Psychrometrics analysis portion, we shall evaluate the total volume of supply air, which must be delivered to the M-2 hall at design temperature through ducting in order to maintain the indoor design conditions. The design indoor and outdoor conditions for the M-2 hall will be set to be an optimum value, which is a perfect balance between energy and the occupants satisfaction.

2.1

Unit 2: Cooling Load Calculation and Psychrometrics

2.1 Cooling Load Calculation of Hall M-2he air inside a building receives heat from a number of sources during the cooling season. If the temperature and humidity of the air are to be maintained at a comfortable level, this heat must be removed. The amount of heat that must be removed is called the cooling load. The cooling load must be determined because it is the basis for selection of the proper size air conditioning equipment and distribution system. It is also used to analyze energy use and conservation. For determination of cooling load of a building it is very crucial to identify its construction type. Therefore before proceeding, first of all we shall try to explore the construction type of hall M-2.The architectural plane is shown in figure 2.1. The walls are 9 in. common brick type with in plaster on both the sides. There are 18 windows on the South Wall and 10 ventilators and 2 Doors on the North wall (as shown in architectural plan). So the net conduction area of these walls will be calculated by subtracting the areas of the windows, ventilators and doors from the total area of walls. Also the South and East Walls are exterior walls and North and West walls are interior walls. The entire building exterior is covered with light color weather sheet (I.e. white washed).GROUND FLOOR PLAN1'-8"WIN. 6'-10.5" WIN. 6'-10.5" WIN. 6'-10.5" WIN. 6'-10.5" WIN. 6'-10.5" WIN. 6'-10.5" WIN. 6'-10.5" WIN. 6'-10.5"

T

6'-101"

2

M-2 HALL ADMIN BLOCK, FIRST FLOOR 78'X18'4'-9"VENT. 4'-9" VENT. 4'-9" VENT. 4'-9" VENT. 4'-9" VENT. 4'-9" VENT. 4'-9" VENT. 4'-9"

4'-2"

2'-9"

4'-2" 8'-0" 4'-2"

STREET 8'-0"4'-2"

TOILETS

OTHER BUILBING, CLASSROOMS

Figure 2.1 Architectural Plan of Hall M-2 (First Floor, Admin Block RCET Gujranwala Campus) There are a few different, acceptable procedures for calculating cooling loads that take into account the phenomena we have discussed. All of them are more accurate than past methods and are often required in state energy codes and standards. These methods often lead to use of smaller equipment and sometimes result in less energy use. The cooling load calculation procedure that we will be using is called the CLF/CLTD method. This procedure is relatively easy to understand and use. The CLTD values for roof and wall constructions and the values of U, LM, CLTD and all of other technical parameters are available from respective Tables in ASHRAE Fundamentals Handbook 2005 OR principles of air conditioning by ADWARD G. PIITTA .All the walls used for the M-2 Hall are of medium construction, so the value of U for walls is 0.45 and that for roof is 0.13 and for floor is 0.26 The CLF/CLTD method can be carried out manually or by using a computer. In actual practice the use of computers and cooling load calculation software are very common to use now a days. The

2.2

Unit 2: Cooling Load Calculation and Psychrometricscomplete summary of load calculation for hall M-2 is given table 2.1

Table 2.1 Cooling Load CalculationRoom: M-2 Hall Admin Block, First Floor Location: Rachna College of Engineering and Technology, Gujranwala, Pakistan DB F Out Door 108 Design Conditions Room 77 WB F 56 50 Area ft2 1.04 1.04 RH% Daily Range: 50-55 F Day: July/ August Latitude: 32 North W gr/lb Average:52.5 F Time:12 PM

Conduction Dir.

Color U South North

CLTD, F Gross Net Table 310.5 14 95 Corr. 37.5 26

SCL BTU/hr 12109.5 2568.8 0 0 11480.738 5169.825 2957.5125 8041.95 11955.06 5475.6 0 1551.42

Glass South East

Wall

D D West North

Roof/Ceiling D Floor Partition North Door

0.45 0.45 D D D 0.26 1.04

586.5 24 43.5 207 32 55.5 0.45 103.5 0.45 744.63 10 0.13 1404 40 1404 15 57.375 26

63.5 24 65.5

Glass

Solar South North

Dir. no no

Sh. SHGF Area SC 60 103.5 1 0.58 44 152.4 1 0.86

SLF 3601.8 5765.87 0 0 1 CLF 2557.5 5115 LCL BTU/hr 15300 15300 0 0 0 0 88174.975 15300

600 Lights W x 3.41x 1.25 BF x 250 x 10 W x 3.41x 1 Fan CLF 255 People SHG x 60 nx 1 CLF 255 LHG x 60 n Equipment *Infiltration 1.1x 0.68x

CFMx CFMx

TC gr/lb Subtotal

SA Duct gain

2.3

Unit 2: Cooling Load Calculation and PsychrometricsSA Duct leakage SA Fan Gain(draw thru) Room Load SA Fan Gain (Blow thru) **Ventilation 1.1x 0.68x RA Duct gain RA Fan Gain CFMx CFMx TC gr/lb Cooling 88174.975 15300 Total CL, BTU/hr 103474.98

Cooling Coil Load Pump gain Refrigeration load

88174.975

15300

103474.98

103474.98 8.70Tons/ 30.65 kW

* Infiltration will not be considered for Hall M-2, because during normal operation of system hallwill be pressurized slightly above the atmospheric, thus causing no infiltration.

Note: For simplicity additional heat gains like return duct heat gain, fan/ Blower heat gain,ventilation heat gain etc are not considered here. (The reason will be explained at the end of this unit)

2.2 Psychrometrics Analysis of the PAC-105 Air Conditioning SystemRoom temperature = 77 F =25 o C Relative humidity = RH =50% From psychrometrics chart, at 77 F and 50% RH Specific humidity or humidity ratio = w = 60 grains / lb of d.a Room sensible cooling load (RSCL) = 26.16 KW =89181.8 BTU/ hr Room latent cooling load (RLCL) = 4.486 KW =15293.18 BTU/ hr Room total cooling load (RTCL) = RSCL+ RLCL= 30.65 KW= 8.707 TRo o

2.2.1 Coil Entering and Leaving Conditionso Temperature of air entering the coil = 26 o C =78.8 F o Temperature of air leaving the coil = 13.33 o C = 56 F Coil by pass factor

BPF =Where

Tc Tb Tc Ta

Ta = The DB temperature of the air entering the coil Tb= The DB temperature of the air leaving the coil Tc = The mean effective temperature of the coil surface

2.4

Unit 2: Cooling Load Calculation and PsychrometricsNow, Room sensible heat factor = RSHF =

RSCL = 0.8536 0.85 RTCL Using value of room sensible heat factor (RSHF), draw a line called coil process line on psychrometrics chart as shown in the figure 2.2. This coil process line intersects the curved line onchart with complete saturation at 11.1 o C . This point of intersection is called the coil or apparatus dew point (ADP). This is the temperature that must be maintained to cool and dehumidify the air according to desired conditions of entry and leaving from cooling coil. As

The temperature of water entering the coil = 6 o C The temperature of water leaving the coil = 14 o C Tw,i +Tw,o 6+14 = = 10 o C Therefore the mean temperature of water = 2 2 The actual (ADP) required for designed cooling and dehumidification is found to be 11.1 o C . But most of coil surface will be at mean temperature of water. Therefore, Let we select mean temperature of water as (ADP). Thus putting values in above relation we get, T -T 10-13.3 BPF= c b = =0.206 Tc -Ta 10-26 It shows that the coil configuration is such that 20.6 % quantity of air will be by passed.

2.2.2 Determination of Supply Air QuantityIn order to find the quantity of air required to maintain desired room conditions using relation 2.12 for sensible heat of air . qsen = m. c a pd + wc ps T = ma c paT .

(

)

All the parameters in above relation have been defined in unit 2.the modified form of relation isqsen = 1.1 CFM (TR TS )

The above relation relates all parameters in British system of units. So,CFM = qsen 88174 = = 3817 * 1.1(TR TS ) 1.1(77 56)

In SI units using density of air as 1.1614 kg/m3 at 26 o C , the corresponding mass flow rate of air is 2.0937 kg/sec. *this is the maximum value of the supply air CFM, which must be maintained during the peak load season, so in the unit 3 for the duct design purposes we shall take a value of 3790CFM, which is an optimum value of the supply air so the fan/blower will attain this value in the season automatically.

2.2.3 Condensate Formation during Air Conditioning ProcessNow from psychrometrics chart in figure 2.2, at leaving conditions of air the relative humidity of supply air is w = 54 lb/ kg of d.a S Thus mass of water vapor condensed = m. = m. ( wR wS ) w a = 2.1963 (60-54) = 13.1778 grains /sec=13.1778/7000 =1.8825lb/sec=3.0735kg/hr

2.5

ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURER

55R

60

BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY Copyright 1992

OD90

AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.

5000

1.00.8

SEA LEVEL0.60. 5

50

.028

60

1.0

--2000 -1000

850 15..026

2.0

SENSIBLE HEAT TOTAL HEAT0.40.3

Qs Qt

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4.0 8.0 - -48.0 .0 -2. 0

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HUMIDITY RATIO - POUNDS MOISTURE PER POUND DRY AIR

PY

EN

TH AL

SA TU

15

40 3535

DRY BULB TEMPERATURE - F

100

105

110

115

10

15

20

25

ENTHALPY - BTU PER POUND OF DRY AIR

120

35

40

45

50

55

60

65

70

75

80

85

90

95

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3000

4585 WE T.024

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ENTHALPY HUMIDITY RATIO

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75

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HUMIDITY

30

Unit 2: Cooling Load Calculation and PsychrometricsFor calculating maximum condensate formation in PAC-105 air conditioning process can be calculated by mixing of 5% outdoor air quantity in the return air from the M-2 hall. So if the outdoor air which is taken inside the AHU is at the DB= 44 0C, RH65, w =38.81 g/kg then the condensate flow rate is. = ma ( wo ws ) = 2.1964 ( 38.81 7.637 )

= 68.468g/sec = 4.108 kg /min also 0.903GPMIn this case condensate may be significant problem, so therefore a proper drain pan should be carefully designed to collect this condensate, in order to minimize or mitigate the bacterial and microorganisms growth etc.

2.8

PAC-105

Unit 3

DDC C o n tro l P anel

Design of AirHandling Unit and Ducting SystemIntroduction

ne of the most important function of an Air-conditioning system is to reduce the temperature of an air stream. It is accomplished by the use of air handling units, which mainly contain an air cooling coil. Various types of AHUs are used depending upon the applications. This unit is concerned with the design of an air handling unit for PAC-105. An air-handling unit consists of cooling coil, air filters and blower for treatment of air. First of all we shall design the most important component of AHU i.e. chilled-water cooling coil. Convection correlations developed by Zhukauskas for bundle of tubes in cross flow will be used to find the air-side coefficient. The water-side convection coefficient will be found by using Dittus-Boelter and colburn correlation. Using these coefficients and appropriate fouling factors, finally the required heat transfer area will be found for cooling coil. After arranging the tubes in rows, specific paths (technically called circuits) of chilled water via coil will be planned. Secondly, other important component called fan will be selected. Its selection is based on the head loss of air stream in ducting system. Therefore before fan selection we have to design the ducting system for M-2 hall by using Cost Optimization-T Method, supply ducting system will be designed and then finally we shall design an air handling unit for PAC-105.

O

3.1

Unit 3: Design of Air Handling Unit and Ducting System

3.1 Function of Air-Handling Unit and Typen air-handling unit (AHU) is the primary equipment in an air system of a central hydronic system; it conditions the air and distributes it to conditioned space. In an AHU, the required amounts of outdoor air and recirculating air are often mixed and conditioned. The temperature of the discharge air is then maintained within predetermined limits by means of control systems. After that, the conditioned supply air is provided with motive force and is distributed to various conditioned spaces through ductwork and space diffusion devices. Many air-handling units are modular so that they have the flexibility to add components as required. An AHU basically consists of an outdoor air intake and mixing box section, a fan section including a supply fan and a fan motor, coil section with water cooling coil, a filter section and a control section. A return or relief fan, a heating coil, a precooling coil, and a humidifier may also be included depending on the application. Supply volume flow rates of AHUs vary from 2000 to 63,000 cfm (945 to 29,730 l/s). Whether a return fan or a relief fan should be added to an air system depends on the construction and operating characteristics of the air system and the total pressure loss of the return system. A heating coil is mainly used in the air-handling unit that serves the perimeter zone or for morning warm-up in the heating season. Humidifiers are employed for processing air conditioning and health care facilities where space humidity must he controlled. Air handling units used commercially may fall into any of the following categories

A

Horizontal or Vertical Unit Draw-Through Unit or Blow-Through Unit Outdoor Air (Or Makeup Air) AHU or Mixing AHU Rooftop AHU or Indoor AHU Factory-Fabricated AHU or Field-Built AHU, Custom-Built or Standard Fabrication Single-Zone AHU or Multizone AHU

3.2 Design of PAC-105 AHUBefore going into details of design and selection of components for PAC-105 AHU, first of all we have to specify its location. The suggested location for AHU is Along the eastern wall of hall M2 on ground floor. The location of PAC-105 AHU is shown in figure 7.1, it would be an optimum selection of location for air handling unit, because this place assures the easy supply and delivery of chilled water from/to the refrigeration system located at canal bank. Also this location provides better outdoor area, easy installation and maintenance, supply & return ducts connections and minimum condensate flow problems. After selecting an adequate location, the next step is to select an appropriate type of air handling unit. Because hall M2 is on the first floor, therefore the supply and return ducts from/to the PAC-105 AHU will run upward/downward direction respectively. For such a location it is better to use a vertical type AHU. As for as the physical sequence of fan and coil in AHU is concerned, we shall select a draw-through configuration for blower and coil. In blow-through units air stream suffers excessive pressure drop while passing through the cooling coil, thus require larger fan and thus more energy consumption. Also in blow-through units the air stream is more liable to carry condensate or mist droplets with it. The schematic of PAC-105 AHU is shown in figure 3.1. The design of PAC-105 air handling unit consists of following steps. Design of cooling coil Design of ducting system Selection of an air filter

3.1

Unit 3: Design of Air Handling Unit and Ducting System Selection of blower/fan

Supply Air Duct Flexible Connection b/w Fan and Supply Duct Return Air Duct Cooling Coil Saddle for fan Guide Vanes for AirDDC Control Panel

Supply Air Fan (Direct Drive )

Air Filter, HEPA

Air Mixing Box Fresh Air Dampers

Foundation

Coil Section Condensate Pan

Exchaust Air Dampers(Optional)

Figure 3.1 the Schematic of PAC-105 Air Handling Unit

3.3 Design of PAC-105 AHUs Cooling CoilKnown parametersAmount of heat rejected by coil=Cooling load of hall M-2 = 30.65 Kw (neglecting return duct heat gain) Mass flow rate of air =2.077 kg/sec Temperature of air entering cooling coil=26 o C Temperature of air leaving the cooling coil=12 o C Temperature of water entering the cooling coil=6 o C Temperature of water leaving the cooling coil=14 o C .

Assumptions1. First of all we shall select proper tubing for coil. Let we use copper tubing of type K with outer diameter as 12.7mm and inner diameter as 10.2 mm. 2. Usually cooling coils consists of an array of tubing arranged in a special way. Let the cooling coil of PAC-105, consists of an array or bank of tubes arranged in staggered

3.2

Unit 3: Design of Air Handling Unit and Ducting Systemarrangement, With 30 tubes in each row and 10 circuits of chilled water.

Properties1. All the properties of air will be evaluated from table B-1 at air mean temperature, Ta , m= T a ,m Ta ,i +Ta ,o 26+12 o o = = 19 C 20 C 2 2

2. All the properties of water will be evaluated from table B-2 at water mean temperature,

Tw,m= T w,m Tw,i +Tw,o 6+14 o = = 10 C 2 2

Design ProcedureUsing equation, which relates the heat transfer of any heat exchanging device with its surface area and temperature,Q = UAT lm

Where, Q = Total heat transfer rate Kw

U = Over-all heat transfer coefficient W/m2 .K A= Total heat transfer area m2 Tlm = Log-mean temperature difference K The relation for finding value of U is of form[a],

R f ,o R f ,i ln D d 1 1 1 = + + + + UA hi Ai ho Ao Ao Ai 2 kl Where, l is required length of a single tube in array, k is conductivity of copper = 400 W / m.k , d and D are inner and outer diameters of tubing. R f ,i , R f ,o are the fouling factors of inner and outersides of tubing, and there values are taken as, R f ,i =0.0002 m2.K/W (for chilled water side, [a])

( )

R f ,o =0.0005 m2.K/W (For air side), accounts for conductive resistance of the copper tubing. 2 kl Because the outer surface of tubing is exposed to higher temperature, Therefore, U will be based on outer surface area of tubing Now modified form of above expression for tube array is as under, D ln d 1 1 D 1 D = + + R f ,i + R f ,o + D U o hi d ho d 2k All parameters in above relation have been defined except, hi = Chilled water side heat transfer coefficient W/m 2 .K ho = Airside heat transfer coefficient W/m2.K Now we shall find these parameters one by one.

The term

ln

D d

3.3

Unit 3: Design of Air Handling Unit and Ducting System Determination of Air-side Heat Transfer CoefficientAs convection heat transfer coefficient is very strong function of velocity of fluid, therefore first we shall find the velocity of air flowing over the cooling coil. According to . ma = a AF Va continuity equation, for air flowing over cooling coil Where, . ma = Mass flow rate of air kg/sec a = density of air kg/m3 AF = Face area of cooling coil m2 Va = face velocity of air m/sec From table B-1 a = 1.2034 kg/m3 at 20C. In above relation only two parameters ma & a are known. In order to find the face velocity of air, the face area of coil must be known. Let the length of coil is 1m.Now the next step is to find the height of coil. As we have already assumed that coil consists of successive no. of rows, with 30 tubes in each row. Assuming staggered arrangement of tubes in each row, as shown in [d].From geometry of staggered tubes array the height of each row can be correlated as, Height of each Row = H = ( NT D)+( NT 1) A 1 Where, NT = No. of tubes in each rowD = Outer dia. of tube m A = gap b/w successive tubes m 1 Now the next step is to choose an appropriate value of parameter A1 , let we take.

A1 =21mm=0.021mThus Height of each row = H = ( NT D )+( NT 1) A1 Height of each row = H = (300.0127)+( 301)0.021 H = 0.990m

Therefore the face area of coil will be AF = length(l ) Height ( H ) 1 0.990 = 0.990m 2 . Now, using above relation ( ma = a AF Va ) to find the velocity of air . ma 2.077 = = 1.755m / sec Va = a AF 1.20340.990 Now, for assumed configuration of tubes array the transversal pitch (center to center distance of tubes in a row, in vertical direction) will be, Transversal pitch ST = D + A1 = 0.0127 + 0.021 = 0.0337 = 33.7 mm = 0.0337 mm The next step is to specify the longitudinal pitch (center to center distance b/w successive rows of tubes, in horizontal direction) Let[d],

3.4

Unit 3: Design of Air Handling Unit and Ducting SystemLongitudinal pitch S L = 2 D = A1 + D S L =212.7=25.4 mm=0.0254m

2 S Hence, Diagonal pitch = S D = S L2 + T = 30 mm 2 Now we have completely described the configuration of tubes array, the next step is to find the value of convection heat transfer for the assume geometry of tubes array, using Zhukauskas relation [d], for average Nusselt number. Nu Of tubes 0.25 m ( Pr )0.36 Pr Pr max s All of the parameters used in above relation stand for their usual notations, now from [d] S For Staggered tubes with T = 1.35 < 2 SL 1 ST 5 C = 0.35 m = 0.6 S = 0.37 & L The relation for Reynoldss no based on max. Velocity of air will be as under, D V = a max ReD, a max a = viscosity of air at mean temp. of air = 181.6 107 Pa.sec S +D = 23.2 mm Now, because S D = 30 mm ? T 2 So, Vmax (Max. velocity of air) occurs at A1 and is: Nu = CReD, S 33.7 Vmax = T Va = 1.755 = 2.816m / sec S D 33.712.7 T

Thus,

V D 1.20342.8160.0127 = a max = = 2369.9 2370 ReD, a max 181.6107

And, from table B-2

Pr At mean temp. Of 20 o C (293K) = Pr293 = 0.7085And Prs = prandtle no. at surface temperature of tubing = Pr = 0.713 283K Putting all values in relation for Nusselt no, 0.7085 0.25 Nu = 0.37 (2370)0.6 (0.7085)0.36 ( ) 0.711 Nu = 34.577 h D Nu = a ka k ha = a Nu D Where, ka = Thermal conductivity of air at mean temp = 0.026w/m.k

3.5

Unit 3: Design of Air Handling Unit and Ducting System0.026 33.54 = 69.5 0.0127 ha = 70.788W/m 2 .K =

Determination of Water-side Heat Transfer CoefficientAs So,. Q. = mwC p T. mw =

C p T

Q.

=

30.75103 = 0.91Kg/s 4.18103(146 )Kg/s

. mw = Total mass flow rate of water required in

Assuming 10 no. of circuits for cooling coil, then mass flow rate of water through each circuit Will be, . w. 0.91 mw = w = = 0.091Kg / s 10 10per tube

Now, Where, So,

Re,w =

4 m. w

d w

per tube

w = viscosity of water at mean temp. of air Tw,m = 1.31 103 Pa.s

40.091 Re,w = = 8671 0.01021.31103 As Re,w > Rc (i.e. 2000), so flow is turbulent Therefore, Dittus-Boelter relation [g] using for turbulent flow between 0.6>Pr>100 0.8 Nu = 0.023 ( Re,w ) ( Pr )0.4 More recent information by Gnielinski [n] suggest that better results for turbulent flow in smooth tubes may be obtained from the following relations:

Nu = 0.0214 ( Re 0.8 100 ) Pr 0.4

for 0.5