BASIC AIR CONDITIONING

BASIC AIR CONDITIONING
SYSTEM DESIGN

INIRODUCTION

Air conditioning system design is a broad and sometimes complex field with many Aspect to consider and many options available for selection of the final system for a particular project.Air conditioning system is design is more than just a technical exercise and in order to Design an air conditioning system we need to understand what air conditioning is, what The clients needs are, what constraints there are on the design and what the architect trying to achieve. That is, we need to know the technical aspects of air conditioning system As well as the effect of the design of the other members of the project team. The design process involves a mix of technical interpersonal and management skills. This paper covers a typical air conditioning system design process.
So how do we go about designing air conditioning system? first of all we need to understand. What air conditioning is.

THE TECHNICAL ASPECT

Air conditioning is the treatment of the environment to achieve a set of required conditioning the field of building services it usually relates to air providing comforts For the building occupants but could also cover other situation such as fumigation and specialist storage environments. this presentation will deal only with treatment of air for Comforts conditions. The environmental condition, which is controlled in a given design, may include:
Ø Dry bulb temperature
Ø Moisture contents of the air (humidity, both relative and absolutes)
Ø Air movements
Ø Air quality

The need for control of these variables and there control points will be based on the Clients requirements. For example, if a factory owner wants to provide some limited Cooling for their factory .then dry bulb temperature control over a fairly wide range would probably suffice. However if a pharmaceutical manufacture wants to provide suitable condition for theirs manufacturing process, then we would needs to controls all the above variables to close tolerance. Air quality is control by filtration system, and an air movement is controlled by the distribution system. These aspects of air conditioning system design are worth separate Discussion and there own paper .we will look at the control of temperature and moisture control only. A useful tool to help visualize the control of temperature and moisture control is the Psychometric chart. Atmospheric air is a mixture of air and water vapour.The psychometric chart is a graph showing the various concentration of water vapor in the air and the associated temperatures, densities and energy contents. The task of the air conditioning system is to control the temperature and moisture Contents of the air by one or more of the following process.

Ø Heating
Ø Cooling
Ø Dehumidification
Ø Humidification


The Metabolic Rate

The rate at which body produces heat is called the metabolic rate. The heat produce by a
normal healthy person while sleeping is called the basal metabolic rate which is of the order of 60W.The maximum value may be 10 times much as this for a person engaged in sustained hard work.
The temperature of the body remains comparatively constant at about 36.9 C(98.4)for tissues or the skin and abt 37.2 C for the deep tissues or the core. it is found that the body temperature in the morning after sleep is about 0.5C less than its temperature in the afternoon. A value of 40.5C (104.9F) is considered serious and 43.5C (110F) is certainly fatal..
Human comfort is influenced by physiological factor determined by the rate of heat
generation within the body and the rate of heat dissipation to the environment


OUTSIDE DESIGN CONDITION:

It is observed that there is kinds of sinusoidal relationship the air dry bulb temperature and the sun time. For example, in the month of June in a certain locality where the sun rise is at about 5 am. and the sunset at about 7pm.the time of minimum temperature falls at about 4 pm.,i.e.,whith laps of about 12 hours.
As regards relative humidity, it is seen that it reaches a minimum value in the afternoon.
Since the mean daily maximum dry bulb temperature occurs between 1 pm., it is reasonable to assume that the minimum relative humidity would occur during the same period.

BRIEF HISTORY OF REFRIGERATION

The method of production of cold by mechanical process is quite recent. Long back in 1748, William coolen of Glasgow University produced refrigeration by creating partial Vacuum over ethyl ether, but he could not implement his experience in practice. The first development took place in 1834 when Perkins proposed a hand-operated compressor machine working on ether. Then in1851 came Gorrie, s air refrigeration machine, and in1856 Linde developed a machine working on ammonia.
The pace of development was slow in the beginning when steam engine was the only prime mover known to run the compressors. With the advent of electric motor and Consequent higher speed of the compressor, the scope of applications of refrigeration widened.the pace of development was considerably quickened in 1920 decade when due pont put in the market a family of new working substances, the fluoro-chloro derivates of Methane ,ethane ,etc.popularly known as chloro fluorocarbon or CFCs—under the name of Freons, since it has been found that chlorine atoms in Freons are responsible for the depletion of ozone layer in the upper atmosphere. Water vapors absorption machine by Carre.These developments account for the major commercial and industrial application in the field of refrigeration.
A phenomenon called Peltier effect was discovered in 1834 which is still not commercialized. Advances in cryogenics, a field of very low temperature refrigeration, were registered with the liquefaction of oxygen by pictet in 1877.Dewar made the famous Dewar flask in 1898 to store liquids at cryogenic temperatures. Then followed the liquefaction of other permanent gases including helium in 1908 by Ones which led to the discovery of the phenomenon of superconductivity. Finally in 1926,Giaque and Debye independently proposed adiabatic demagnetization of a paramagnetic salt to reach temperatures near absolute zero Two of the most common refrigeration application ,viz.,a window-type conditioner and a domestic refrigerator ,have been described in the following pages.

ROOM AIR CONDITIONER

Flowing figure shows schematic diagram of a typical window –type room air conditioner, which works according to the principle described below:
Consider that a room is maintained at constant temperature of 25C.In the air conditioner, the air from the room is drawn by a fan is made to pass over a cooling coil, the surface of which is maintained, say, at a temperature of 10C.After passing over the coil, the air is cooled (for example, 15C) before being supplied to the room. After picking up the room heat, the air is again return to the cooling coil at 25C.
Now, in the cooling coil, a liquid working substance called a refrigerant, such as CHCIF2 (monochloro-difluoro methane), also called Freon 22 by trade name,

Schematic diagram of a Room air conditioner

Or simply refrigerant 22 (R22) ,enter at a temperature of ,say,5C and evaporates, thus absorbing its latent heat of vaporization from the room air. This equipment in which the refrigerant evaporates is called an evaporator.
After evaporation, the refrigerant becomes vapour.To enable it to condense back and to release the heat –which it has absorbed from the room while passing through the vapor-its pressure is raised by a compressor. Following this the high pressure vapors enter the condenser. In the condensers, the outside atmospheric air, say, at a temperature of 45C in summer, is circulating by a fan. After picking up the latent heat of condensation from the condensing refrigerant, air is let out in to the environment ,say, at temperature of 55C.The condensation of refrigerant may occure,for example, at temperature of 60C
After condensation, the high pressure liquid refrigerant is reduced to the low pressure of the evaporator by passing it through pressure reducing device called the expansion device, and thus the cycle of operation is completed. A partition wall separates the high temperature side of the condenser from the low temperature side of the evaporator
The principle of working of large air conditioning plants is also the same, except that condenser is water cooled instead of being air cooled .

Unit of Refrigeration capacity:
The standard unit of refrigeration in vogue is ton refrigeration or simply ton denoted By the symbol TR.It is equivalent to the production of cold at the rate at which heat is to be removed from one US tone of water at 32 F to freeze it to ice at 32 F in one dayor 24 hours .Thus
1TR=1×2,0001b×144Btu/1b
24 hrs
=12,000Btu/hrs=200Btu/min
where the latent heat of fusion of ice has been taken as 144 Btu/1B.The term one ton refrigeration is a carry over from the time ice was used for cooling. In general 1TR always means 12,000Btu of heat removal per hours, irrespective of the working substance used and the operating condition, viz, temperature of refrigeration and heat rejuction.This unit of refrigeration is currently in used in the USA, the UK and India. in many countries, the standard MKS unit of kcal/hr is used

It can be sent that
1TR=12,000 BTU/hr
=12000 =3,024kcal/hrs
3.968
=50kcal/min =50 kcal/min
Also, since 1Btu =1.055kj,the conversion of ton in to equivalent SI unit is
1TR=12,000×1.055=12,600 kj/hours
=211 kj/min=3.5167K

Design for air conditioning of a 4 m high-story office building located at 30N latitude, the plan of which is shown in fig the following data are given
Fig.

Plan of building for example
Plaster on inside =11/4cm

Outside wall construction =20cm concrete block
Partition wall construction=33cm brick
Roof construction =20cm RCC slab with 4cm asbestos cement board
Floor construction =20cm concrete
Densities, brick =2000kg/m3
Concrete =1900kg/m3
Plaster = 1885kg/m3
Asbestos board =520kg/m3
Fenestration =2m×11/2m glass
(Weather stripped loose fit) U=5.9wm-2 k-1
Doors =11/2m×2m wood panels
U=0.63 Wm-2 k-1
Outdoors-design condition=43C DBT,27C WBT
Indoor-design condition =25C DBT, 50% RH
Daily range =31C to 43C =12C
Occupancy =100
Light =15,000W florescent
Assumed by pass factor of coil: 0.15
Find the room sensible and latent heat loads, and also the grand total heat load.
Solution
Thermal conductivities from table 18.1
k glass =0.78 Wm-1 k-1
k concrete =1.73 Wm-1 k-1
k brick =1.32 Wm-1 k-1
k plaster =8.65 Wm-1 k-1
k asbestos =0.154Wm-1 k-1

Assumed film coefficients
ƒ=23 Wm-2 k-1
ƒi=7 Wm-2 k-1
Outside wall
1/U=1/23+0.1/1.32+0.2/1.73+1/7+0.0125
U=7 Wm-2 k-1
Partition wall
1/U=1/7+0.33/1.32+1/7+2 (0.0125)/8.65

U=1.86 Wm-2 k-1
Roof
1/U=1/23+0.2/9+0.04/0.154+0.0125/8.65+1/7
U=2.13 Wm-2 k-1
Floor
1/U=1/7+0.2/9
Area and volume of space
A= (27) (17) =459m2
V= (459) (4) =1836m2

Ventilation rate of office
QV/person=0.28 cmm (from Table 16.2)
Qv=0.28(100) =28cmm

Number of air changing of ventilation air
(28)(60)/1836=0.92(satisfactory)
Mass of wall per unit area
Outside wall: 0.2(1900) + (2000) +0.0125(1885)
=604 kg/m2


Partition wall: 0.33 (2000) + 2(0.0125) (1885) =707 kg/m2

Roof: 0.2(1900) + 0.04 (520) =401 kg/m2
Correction for equivalent temperature differential
For daily range of 12C=12-11.1 =0.45C
2
For (t-ti)of 18C = 18-8.3=9.7C
Total correction= -0.45+9.7=9.25C
Equivalent temperature differential in C, from Table 18.9 and 18.10 and
Incorporating correction:
2 p.m. 3 p.m. 4 p.m 5 p.m 6 p.m 7 p.m
West wall 14.4 14.8 15.2 16.5 17.5
North wall 9.6 10.2 9.6 11.3 11.7
South wall 13.1 14.7 16.0 17.4 17.8
Roof (exposed) 24.0 25.8 28.0 29.7 30.5 30.2
__


Rates of solar gain through glass on June 21 W/m2 from Table 17.8(d)
2pm 3pm 4pm 5pm
West glass 309 451 508 492
North glass 44 44 51 91
South glass 47 44 38 32
_____________________________________________________________
Door area = 11/2×2=3m2
Glass area
West glass=4(2×11/2) =12m2
North glass=2×11/2 =3m2
South glass =2(2×11/2) =6m2
Outside wall areas
West wall = (27) (4)-12=96m2
North wall= (10) (4)-3-3=34m2
South wall = (17) (4)-3-6=59m2
Partition wall area
East wall= (27) (4)-3=105m2
North wall= (7) (4) =28m2
Estimated time of maximum cooling load:
From the above calculation, it is obvious that the major components of the variable
Cooling loads are solar and transmission heat gains through the west wall and glass
And the roof. Of these glass and roof loads are the predominant loads. The roof load
Is maximum at 6 pm.when the equivalent temperature differential is 30.5C.The solar
gain through the west glass has a maximum value of 508 w/m2 at 4 pm.Thus the time
Of maximum loads is most likely to be near 5pm. Heat transfer through floor:
Assume a temperature difference of 2.5 C across the floor wind pressure
Assume a wind velocity of kmp, we have
Δp=0.00047(15) =0.11cm H2O
Infiltration rate for window, from table 18.11 for 0.11 wind pressure
=2.5m3/h/m crack
Length of crack for 7 window=7<2(2+11/2)>=49m
SHL=75W/person
LHL=55W/person
Other assumption
Only 10%of the supply duct outside the condition space
No return duct outside the conditioned space
Fan horsepower,5 per cent of RSH
The detail of cooling loads calculation are given on the calculation sheet in table
Calculation sheet fir cooling load Estimation
Space used for office
Size 27×17=459m2×4=1836m2
Estimate for 5 pm LOCAL TIME SUN TIME
HOURSE OF OPERATION DAY TIME
Conditions DB WB %RH DP h, kj/kg kg/kg
Outdoors 43 27 29 21.3 85.0 0.016
ROOM 25 18 50 15.7 50.85 0.01
Difference 18 34.15 0.006
OUTDOOR AIR
100 PEOPLE×0.28cmm/PERSON=28cmm
VENTILATION cmm=28

SWINGING
REVOLVING____________PEOPLE ×___________cmm/PERSON=cmm
DOORS
OPEN 3 DOORS×1.9813cmm/DOOR=17.8cmm
DOORS
EXAUST _____________________________________________=cmm
FAN
CRACK 49m×2.5/60 cmm/m=2.0cm
INFILTRATION cm

LOAD CALCULATIONS
ITEM AREA OR SUN GAIN OR FACTOR W
QUANTITY TEMP.DIFF.OR
HUMIDITY
DIFF.
SENSIBLE HEAT
SOLAR GAIN-GLASS
EAST GLASS -m2 ___ ___ ___
WEST GLASS 12m2 492 __ 5,900
NORTH GLASS 3m2 91 __ 270
SOUTH GLASS 6m2 32 __ 190
SKY LIGHT -m2 __ __ __

SOLAR TRANSMITION GAIN-WALLS AND ROOF
EAST WALL -m2 __ ___ ___
WEST WALL 96m2 16.5 3.5 5,540
NORTH WALL 34m2 11.3 3.5 1,345
SOUTH WALL 59m2 17.4 3.5 3,590
ROOF-SUN 459m2 29.7 2.13 29.035
ROOF-SHADED -m2 __ __ __

TRANSMITION GAIN- OTHERS
DOORS 9m2 18 0.63 100
ALL GLASS (12+3+6) m2 18 5.9 2,230
PARTITION (108+28) m2 15.5 1.86 3,930
CEILING -m2 __ __ __
FLOOR 459 2.5 6.05 6,940
INFILTRATION 19.8cmm 18 20.4 7,270

INTERNAL HEAT GAIN
PEOPLE 100 __ 75 7,500
POWER __ __ __ __
LIGHT 15,000 __ 1.25 18,750
APPLIANCES __ __ __ __
ADDITIONAL __ __ __ __

_________________
SUB TATAL 92,690
STORAGE (Neglected) __ __ __ __
SAFETY
FACTOR 5% 4,635
ROOM SENSIBLE HEAT 103,090
SUPPLY DUCT
SUPPLY DUCT
HEAT GAIN 0.5%+LEAKAGE O.5%+Fan 5% 5,560
HP
OUTDOOR AIR
BY PASSED 28cmm 18C 20.4×0.15 1540
EFECTIVE ROOM SENSIBLE HEAT 104,425
LATENT HEAT
INFILTRATION 19.8cmm 0.006 50,000 5,940
PEOPLE 100 --- 55 5,500
STEAM -- -- -- --
APPLIANCES -- -- -- --
ADDITIONAL -- -- -- --
VAPOR TRANS -- -- -- --
_________________________
SUB TOTAL 11,440
SAFETY FACTOR 5% 570
ROOM LATENT HEAT 12,010
SUPLY DUCT
LEAKAGE LOSS 0.5% 60
OUTDOOR AIR
BY PASSED 28 0.006 50,000×0.15 1,260
EFFECTIVE ROOM LATENT HEAT 13,330
EFFECTIVE ROOM TOTAL HEAT 117,755

OUTDOOR AIR TOTAL HEAT (on equipment)
SENSIBLE 28cmm 18 20.4× (1-0.15) 8740
LATENT 28cmm 0.006 50,000× (1-0.15) 7140
RETURN 0%+RETURN DUCT 0%PUMP %+DEUH. %
DUCT LEAKAGE GAIN PIPE
HEAT GUN GAI


GRAND TOTAL HEAT

133,635(38TR)
Note
Many designers do not like in to account the filtration loads separately. It is consider to be taken care of by ventilation air if the ventilation cmm is designer is greater than filtration cmm.one such simplified loads estimation calculation sheet fir the ground, first and third floor of television studio building, without considering infiltration loads. note that in such a case, there is actually no infiltration as the room is under positive pressure. There is, however, infiltration which is equivalent to exhaust of the room air.