design and simulation of air cooledconditioner

CHAPTER 1

INTRODUCTION

1.1 Project Introduction

Nowadays, there are many usage of air conditioning involve either in

commercial, industry and residential. The usage is worldwide and global, not only

certain part of earth. There are varies type and choice of air conditioning can be

found. The selection of air conditioning is depend on the application and type of

building or space.

The system involve in my research is Air Cooled system. Air cooled air

conditioner is a kind of air conditioning equipment that adopts air as the cold or heat

source to regulate the indoor air. So cooling tower, water pump, boiler and relevant

piping system, as well as many auxiliaries are not needed in this air conditioner,

making this ducted air conditioner simpler in structure, easier to install and maintain,

and more energy efficient

1.2 About the System

Air conditioning is a combined process that performs many functions

simultaneously. It conditions the air, provides heating and cooling, controls and

maintains the temperature, and humidity, ensures air movement, air cleanliness,

sound level, and pressure differential in a space within predetermined limits for the

comfort and health of the occupants. A cooling system is a part of a heating,

ventilation and air-conditioning (HVAC) system that provides space cooling. [1]

1

Direct expansion type or DX type is the air is directly cooled from the

refrigerant; therefore the cooling coil is filled with refrigerant. These cooling systems

are widely used in small to medium sized buildings. For larger and more complex

applications, a secondary cooling medium is used to deliver cooling to one or more

locations needing it. This is accomplished by utilizing the chiller to cool the water,

which in turn is pumped to the cooling coil(s). The heat flow path is from the space

to the chilled water to the refrigerant to the atmosphere. [1]

In direct expansion (DX) systems, the air is cooled with direct exchange of

heat with refrigerant passing through the tubes of the finned cooling coil. A basic DX

system comprises of a hermetic sealed or open compressor/s, evaporator (cooling

coil fabricated out of copper tubes and aluminum fins), a supply air blower, filter,

and a condenser. The term "expansion" refers to the method used to introduce the

refrigerant into the cooling coil. The liquid refrigerant passes through an expansion

device (usually a valve) just before entering the cooling coil (the evaporator). This

expansion device reduces the pressure and temperature of the refrigerant to the point

where it is colder than the air passing through the coil. [1]

In the split system, the condensing unit comprising of the condenser,

compressor and condenser fan with motor are located outside, while the indoor unit

consisting of the evaporator, evaporator fan with motor, expansion valve and air

filter is located inside the conditioned room. The indoor and outdoor units are

connected by refrigerant piping. Flexibility is the overriding advantage of a split

system. Because a split system is connected through a custom designed refrigerant

piping system, the engineer has a large variety of possible solutions available to meet

architectural and physical requirements particularly for buildings with indoor and/or

outdoor space constraints. [1]

1.3 Objective

To analyse the effect of evaporating temperature on cooling capacity.

To determine the evaporating temperature for Malaysia conditions.

2

1.4 Project overview

This project is about to design and analyze the effect of evaporating and

condensing temperature on system sizing and design. Its mean by using varies and

different evaporating temperature according to Malaysia conditions that can be found

in Malaysia Standard 1525 (MS1525) handbook. Then determine the effect of varies

temperature on system sizing and design. The process involve is to select and the

compressor, condenser, evaporator and capillary tube. After that, by using the varies

temperature for condensing and evaporating, determine the component sizing and

design for each temperature have tested. Collect the data, then make a graph and

analyze the result.

The software use for this project is :

1. Copeland Selection Software 6.6

2. Microsoft Excel 2007

3. Solvay Fluor

4. Autocad 2010

1.5 Scope of work

Design and select compressor.

Determine the evaporator and condenser capacity.

Design the condenser.

Design the evaporator.

Design and calculate the length of capillary tube.

Doing the analysis then evaluate the result.

Make a conclusion.

3

4

GANTT CHART

The gantt chart is for a period 37 weeks (Semester 6 and Semester 7)

1) Semester 6

JANUARY FEBRUARY MARCH APRILTOPIC / WEEK 1 2 3 4 5 6 7 8 9 10 11 12 13 14Registration of FYP subject FYP briefing Select & register title Discussion about the project Propose the objective Introduction Literature review Methadology Presentation preparing Presentation

2) Semester 7

5

MAY JUNE JULY AUGUST SEPTEMBER OCTOBER NOVEMBER

TOPIC / WEEK 15 16 1718 19 20

21 22 23

24 25 26 27

28 29 30

31 32 33

34 35 36 37

Research & learn how to design system

Design compressor

Design condenser & evaporator

Design capilary tube

Analysis

Conclusion

Preparing the report

Presentation preparing

Presentation

CHAPTER 2

LITERATURE REVIEW

2.1 Basic Concepts

Comfort Air-conditioning – a process of controlling the air temperature,

relative humidity, ventilation, air movement and air cleanliness of a given

space in order to provide the occupants with a comfortable indoor

temperature. [2]

Air-conditioning System – consists of a group of components or equipment

connected in series to control the environmental parameters. [2]

2.2 The Goal of Air Conditioning System Design

The goal of an air conditioning system design is to achieve a highly quality

system that functions effectively and is energy-efficient and cost-effective. The

following are essential for a system to function effectively: [2]

All design criteria are fulfilled, and the requirements of the owner and the

user are satisfied.

A good indoor air quality is provided.

The system is reliable and has adequate fire protection level (e.g. smoke

management).

2.3 Split-System Air Conditioners design standard

Provide complete a DX system for central station air conditioning. The system must consist of matching air-cooled condensing units, compressors, piping, controls, wiring, and other accessories, as well as the appurtenances necessary to provide a fully-automatic system. [3]

6

2.3.1 Materials

i. Condenser coils must be aluminum plate fins, mechanically bonded to

seamless copper tubes, circuited for subcooling.

ii. Provide openings for power and refrigerant connections.

iii. Provide a service access panel.

iv. Provide copper tubes, circuited for sub-cooling. Provide propeller fans

arranged for vertical discharge. Condenser fan motors must have

inherent protection, and must be permanently-lubricated and

resiliently-mounted. Fans must have safety guards. Provide controls

for cycling fans.

v. Compressors must be serviceable, hermetic compressors, with

external spring isolators and an automatically reversible oil pump.

a. Compressors must unload in steps, in response to suction

pressure, for partial load operation. Separate compressors from

condenser fans and coils.

b. Multiple compressor units must have stop-start fans and coils.

Compressor motors must have a part-winding start.

vi. Provide refrigerant piping between air-cooled condensing units and

air conditioning units. Refrigerant piping must be equipped with the

necessary auxiliary equipment, such as strainers, sight glasses, oil

traps, scale traps, and other devices, to make the system complete and

operable under fully-automatic control.

vii. Refrigeration piping must be ACR copper tubing made up with

wrought copper fittings, using silver solder and installed with a

nitrogen charge while soldering. Use the piping size recommended by

the manufacturer of the air conditioning unit and matching air-cooled

unit. Casings must be galvanized steel finished with baked enamel.

7

2.3.2 Quality Control

If this portion of the project includes commissioning, verify that insertions in

the project specifications have been made that refer to the commissioning procedures

in the commissioning specification section. Verify that the systems and equipment

identified in this section of the standards, and listed in the project specifications, do

not conflict with commissioning procedures for testing and training. [3]

2.4 Design Criteria for Scroll compressor

2.4.1 Scroll Compression Process

The diagram shown describes the scroll compression process. The two

components shown are mating involute scrolls. One scroll is fixed in place and the

other scroll orbits within this fixed scroll. One part that is not shown in this diagram

but is essential to the operation of the scroll is the anti-rotation coupling. This device

maintains a fixed angular relation of 180 degrees between the fixed and orbiting

scrolls. This fixed angular relation, coupled with the movement of the orbiting scroll,

is the basis for the formation of gas compression pockets. [4]

As shown here, the compression process involves three orbits of the orbiting

scroll. In the first orbit, the scrolls ingest and trap-off two pockets of suction gas.

During the second orbit, the two pockets of gas are compressed to an intermediate

pressure. In the final orbit, the two pockets reach discharge pressure and are

simultaneously opened to the discharge port. [4]

This simultaneous process of suction, intermediate compression, and

discharge leads to the smooth continuous compression process of the scroll

compressor. [4]

8

Figure 2.1 : The process of the scroll compressor. [4]

2.4.2 Compressor Handling

It is recommended that the plugs in the compressor line connections be left in

place until the compressor is set into the unit. This reduces the chance of

contaminants and moisture getting into the compressor especially if the compressor

is charged with the more hygroscopic POE oil. If the compressor has two lifting tabs,

both must be used for lifting. The discharge connection plug should be removed first

before pulling the suction connection plug to allow the dry air pressure inside the

compressor to escape. Pulling the plugs in this sequence prevents oil mist from

coating the suction tube making brazing difficult. The copper-coated steel suction

tube should be cleaned before brazing. No object (e.g. a swaging tool) should be

inserted deeper than 50 mm into the suction tube or it might damage the suction

screen. [5]

Figure 2.2 : Suction tube brazing. [5]

9

2.4.3 Lubrication and Oil Removal

The compressor is supplied with an initial oil charge. The standard oil charge

for use with refrigerants R407C and R134a is a polyolester (POE) lubricant

Copeland 3MAF (32 cSt). In the field the oil level could be topped up with ICI

Emkarate RL 32 CF or Mobil EAL Arctic 22 CC, if 3MAF is not available. In R22

applications Suniso 3GS is used. Suniso 3GS is compatible with Texaco WF 32 and

Fuchs KM. These oils may be used if an addition is required in the field. [5]

When a compressor is exchanged in the field it is possible that a major

portion of the oil from the replaced compressor may still be in the system. While this

may not affect the reliability of the replacement compressor, the extra oil will add to

rotor drag and increase power usage. To remove this excess oil an access valve has

been added to the lower shell of the compressor. The compressor should be run for

10 minutes, shut down and the access valve opened until oil is somewhere between

1/4 to 1/3 of the sight glass. This operation should be repeated at least twice to make

sure the proper oil level has been achieved. One disadvantage of POE is that it is far

more hygroscopic than mineral oil. Only brief exposure to ambient air is needed for

POE to absorb sufficient moisture to make it unacceptable for use in a refrigeration

system. Since POE holds moisture more readily than mineral oil it is more difficult to

remove it through the use of vacuum. Compressors supplied by Copeland contain oil

with a low moisture content, and this may rise during the system assembling process.

Therefore it is recommended that a properly sized filter-drier is installed in all POE

systems. [5]

Figure 2.3 : Absorption of moisture in ester oil in comparison to mineral oil in [ppm] by weight at 25°C and 50% relative humidity. h = hours. [5]

10

This will maintain the moisture level in the oil to less than 50 ppm. If oil is

charged into a system it is recommended to charge systems with POE containing no

more than 50 ppm moisture content. If the moisture content of the oil in a

refrigeration system reaches unacceptable high levels, corrosion and copper plating

may occur. The system should be evacuated down to 0.3 mbar or lower. If there is

uncertainty, as to the moisture content in the system, an oil sample should be taken

and tested for moisture. Sight glass/moisture indicators currently available can be

used with the HFC refrigerants and lubricants; however, the moisture indicator will

just show the moisture contents of the refrigerant. The actual moisture level of POE

would be higher than the sight glass specifies. This is a result of the high

hygroscopicity of the POE oil. Oil samples would have to be taken from the system

and analyzed to determine the actual moisture content of the lubricant. [5]

2.4.4 Accumulators

Due to the Copeland Scroll’s inherent ability to handle liquid refrigerant in

flooded start and defrost cycle operation, an accumulator is not be required for

durability in most systems, especially those systems designed with thermostatic

expansion valves. However, large volumes of liquid refrigerant which repeatedly

flood back to the compressor during normal off cycles or excessive liquid refrigerant

floodback during defrost or varying loads, no matter what the system charge is, can

dilute the oil. As a result, bearings are inadequately lubricated and wear may occur.

If an accumulator must be used, an oil return orifice size in the range 2 mm² is

recommended. A large-area protective screen no finer than 30 x 30 mesh (0.6 mm²

openings) is required to protect this small orifice from plugging with system debris.

Tests have shown that a small screen with a fine mesh can easily become plugged

causing oil starvation to the compressor bearings. [5]

11

Accumulators are a standard item in air to air heat pumps and are used even

when a thermostatic expansion valve is used to meter refrigerant in the heating mode.

During low ambient conditions the oil returning from the outdoor coil will be very

viscous and difficult to return through the accumulator if the expansion valve is

working properly by maintaining superheat. To prevent slow oil return it may be

possible to remove the accumulator from systems that use expansion valves in

heating. To determine if the accumulator can be removed a defrost test must be done

at an outdoor ambient of around 0 °C in a high humidity environment to ensure that

excessive liquid does not flood back to the compressor during reversing valve

operation, especially when coming out of defrost. Excessive flood back occurs when

the sump temperature drops below the safe operation line shown in Figure 2.4 for

more than 10 seconds. [5]

Figure 2.4 : Bottom shell temperature. [5]

2.4.5 Screens

The use of screens finer than 30 x 30 mesh (0,6 mm² openings) anywhere in

the system is not recommended. Field experience has shown that finer mesh screens

used to protect thermal expansion valves, capillary tubes, or accumulators can

become temporarily or permanently plugged with normal system debris and block

the flow of either oil or refrigerant to the compressor. Such blockage can result in

compressor failure. [5]

12

2.4.6 Crankcase Heaters

The crankcase heater must be mounted below the oil removal valve located

on the bottom shell. The crankcase heater must remain energized during compressor

off cycles. The initial start in the field is a very critical period for any compressor

because all load bearing surfaces are new and require a short break-in period to carry

high loads under adverse conditions. The crankcase heater must be turned on a

minimum of 12 hours prior to starting the compressor. This will prevent oil dilution

and bearing stress on initial start up. If it is not feasible to turn on the crankcase

heater 12 hours in advance of starting the compressor, then use one of the techniques

listed below to prevent possible flooded-start damage to the compressor: [5]

i. Direct a 500 watt heat lamp or other safe heat source (do not use torch) at

the lower shell of the compressor for approximately 30 minutes to boil off

any liquid refrigerant prior to starting; or

ii. Bump start the compressor by manually energizing the compressor

contactor for about one second. Wait five seconds and again manually

energize compressor for one second. Repeat this cycle several times until

the liquid in the shell has been boiled off and the compressor can be

safely started and run continuously.

Due to the Copeland Scroll’s inherent ability to handle liquid refrigerant in

flooded conditions, no crankcase heater is required when the system charge does not

exceed following values:

a) 7,7 kg for ZR 90 K3* ... ZR 19 M3*

b) 11,3 kg for ZR 250 KC*

c) 13,6 kg for ZR 310 KC*…ZR 380 KC*

A crankcase heater is needed to drive out excessive amounts of refrigerant

that have migrated into the shell during standstill periods and no accumulator is

piped to provide free liquid drainage during the off cycle as shown in Figure 2.5. For

correct mounting location of such a heater please see Figure 2.6. [5]

13

Figure 2.5 : Suction accumulator. [5]

Figure 2.6 : Crankcase heater location. [5]

2.4.7 Pumpdown

Recycling pumpdown for control of refrigerant migration may be used

instead of, or in conjunction with, a crankcase heater when the compressor is located

so that cold air blowing over the compressor makes the crankcase heater ineffective.

The Scroll compressor discharge check valve is designed for low leak back and will

allow the use of recycling pump down without the addition of an external check

valve. The low pressure control cut-in and cut-out settings have to be reviewed since

a relatively large volume of gas will re-expand from the high side of the compressor

into the low side on shut down. A one time pump down at the end of a run cycle is

not recommended since refrigerant can still migrate into the compressor after a long

shut down. If a one time pump down is used a crankcase heater must be installed. [5]

14

2.4.8 Minimum Run Time

There is no set answer to how often scroll compressors can be started and

stopped in an hour, since it is highly dependent on system configuration. There is no

minimum off time, because the scrolls start unloaded, even if the system has

unbalanced pressures. The most critical consideration is the minimum run time

required to return oil to the compressor after startup. This is easily determined since

these compressors are equipped with a sight glass. The minimum on time becomes

the time required for oil lost on compressor startup to return to the compressor sump

and restore a normal level in the sight glass. Cycling the compressor for a shorter

time than this, for instance to maintain very tight temperature control can result in

progressive loss of oil and damage to the compressor. [5]

2.4.9 Discharge Temperature Protection

A thermistor with a nominal response temperature of 140 °C is located in the

discharge port of the fixed scroll (Figure 2.7). Excessive discharge temperature will

cause the electronic protector module to trip. The discharge gas sensor is wired in

series with the motor thermistor chain. [5]

Figure 2.7 : Internal discharge temperature protection. [5]

15

2.5 Design Criteria Condensing Temperature

This is a key economic design decision. Lowering the design condensing

temperature lowers the energy consumption by the compressors. However, lowering

the design condensing temperature also increases the size and cost of the condenser,

raising installation costs. Thus, the decision is balance between initial construction

cost and operating costs. [6]

There are 3 design condensing temperatures typically used for ammonia

refrigeration systems. [6]

96.3ºF 185 psig (older standard design)

95ºF 180.7 psig (newer standard design)

90ºF 165.9 psig (enhanced design)

Table 2.1 : Design condensing temperature typically used. [6]

Using the older standard of 96.3ºF as a base, lowering the design condensing

temperature has the following effect on size, or base rating, of the condenser. [6]

96.3

95

90

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

Condenser size

Condensin

g tem

pera

ture

Chart 2.1 : How condensing temperature affects The size of the condenser. [6]

Thus, a 90ºF condenser is over 50% larger than a 96.3ºF condenser.

Condenser size, or capacity, is mostly achieved through a larger coil bundle – more

surface area of pipes. In general, a 90ºF condenser has 50% more tube area than a

96.3ºF condenser. [6]

2.5.1 Sizing Condenser

16

There are three factors influencing the sizing of a condenser: [6]

1) The total heat rejection required – the capacity of the system compressors.

2) The local climate conditions – the design wet bulb temperature

3) The design condensing temperature

The first two factors are a given and cannot be changed. The third is a design

decision between the owner and designer and the subject of this article. [6]

2.5.2 Total Heat Rejection

Total heat rejection is based on the total compressor capacity of the plant

including a factor for oil cooling and excluding any back-up machines. The equation

is: (BHP x 2545) + (TR x 12000) = Total Heat Rejection, BTU/hr [6]

2.5.3 Wet Bulb

Evaporative condensers work by evaporating water over the condenser’s tube

bundle –cooling the refrigerant gas inside. The temperature at which the water will

evaporate on any given day is called the wet bulb temperature. The wet bulb

temperature is a measure of the humidity in the air. If the humidity is high, the wet

bulb temperature is high. The dry bulb temperature, which is the temperature

measured by a thermometer, has significantly less effect on condenser performance.

Refrigeration system designers size condensers using the ASHRAE 1% wet bulb

temperature for a given location. In other words, 99% of the time the wet bulb

temperature will be below that value. Typical wet bulb values for the Northern

United States are between 72ºF and 80ºF. [6]

2.5.4 How Does a Larger Condenser Affect Energy Consumption

Consider a typical high stage compressor, operating at 20ºF suction and

discharging at various pressures. As the discharge pressure gets lower, the capacity

of the compressor increases, and the BHP/TR (energy input per ton refrigeration)

goes down. Therefore, in effect, it does more for less. The following two graphs

show the relationship between condensing pressure and compressor

performance/energy consumption for various head pressures. [6]

17

Graph 2.1 : RXF-101 Perfomance at various condensing pressures. [6]

Graph 2.2 : Compressor energy consumption vs condensing pressure. [6]

Clearly, the lower the condensing pressure, the lower the energy consumption

of the compressor. The difference between 95ºF (180.7 psig) and 90ºF (165.9 psig)

condensing is approximately 15% lower energy consumption. Thus, at peak

condensing conditions, the larger condenser saves 15% of the compressor energy

costs. However, in Northern climates, and only during daytime hours. Thus the non-

peak performance is also a significant factor. [6]

18

Note that Chart 2.2 shows a relatively linear relationship between condensing

pressure and energy consumption. The larger condenser will achieve consistently

lower head pressures throughout this period, due to its greater surface area. The 90ºF

condenser will typically have 15 psig lower condensing pressures than the 95ºF

condenser, maintaining 12% lower compressor energy cost during this time period as

well. [6]

2.6 Design Criteria Evaporators

In the evaporation process, concentration of a product is accomplished by

boiling out a solvent, generally water. The recovered end product should have an

optimum solids content consistent with desired product quality and operating

economics. It is a unit operation that is used extensively in processing foods,

chemicals, pharmaceuticals, fruit juices, dairy products, paper and pulp, and both

malt and grain beverages. Also it is a unit operation which, with the possible

exception of distillation, is the most energy intensive. [7]

2.6.1 Materials of Construction

The two parameters which control the selection of the material of

construction are corrosion and ease of cleaning. All evaporators for hygienic duties

must be capable of being frequently cleaned in place. In most cases, this means

rinsing the equipment with water, followed by washing with caustic and then acid

cleaning agents, and finally, a further rinsing with water. It is important, particularly

with dairy and meat products, that the evaporator is completely cleaned of all

deposits. The cleaning processes eliminate the use of carbon steel as the material of

construction. Most hygienic evaporators are manufactured in either 304 or 316

stainless steel. [7]

Corrosion is often a major problem with chemical duties and some hygienic

applications. A particular problem with evaporators is the range of concentration of

solids in the process fluid, since the corrosive component will be concentrated as it

passes through the evaporator. In some evaporators, the concentration range can be

as high as 50 to 1.

19

For example, waste water with a chloride content of 40ppm in the feed would

have 2000ppm in the product. While stainless steel would be acceptable for the

initial stages of evaporation, a more corrosion resistant material would be required

for the last one or two stages. [7]

Corrosion is also a major consideration in the selection of gasket materials.

This is particularly important with plate evaporators with elastomeric gaskets sealing

each plate. Many solvents such as chlorinated and aromatic compounds will severely

attack the gaskets. A less obvious form of attack is by nitric acid. This is important

since nitric acid can be present in some cleaning materials. While concentrations of

about 1% up to 140°F (60°C) can be accepted, it is best to eliminate nitric acid from

cleaning materials. Phosphoric and sulfamic acids are less aggressive to gaskets. [7]

Typical materials of construction for a number of evaporator applications are shown

below: [7]

Table 2.2 : Typical materials of construction. [7]

In some cases, the type of evaporator is controlled by the materials of

construction. For example a sulfuric acid evaporator, where the acid concentration

can reach 50%, would utilize graphite tubular heat exchangers and non-metallic

separators and piping. [7]

20

2.7 Effect of Evaporator Temperature

The effect of evaporator temperature on performance of the system is

obtained by keeping the condenser temperature (pressure) and compressor

displacement rate and clearance ratio fixed. To simplify the discussions, it is further

assumed that the refrigeration cycle is an SSS cycle. [8]

a) On Volumetric efficiency and refrigerant mass flow rate:

The volumetric of the compressor with clearance is given by:

For a given condensing temperature (or pressure), the pressure ratio rp

increases as the evaporator temperature (or evaporator pressure) decreases. Hence,

from the expression for clearance volumetric efficiency, it is obvious that the

volumetric efficiency decreases as evaporator temperature decreases. This is also

explained with the help of Graph 2.3, which shows the P-V diagram for different

evaporator pressures. As shown, as the evaporator pressure decreases, the volume of

refrigerant compressed decreases significantly, since the compressor displacement

remains same the clearance volumetric efficiency decreases as evaporator

temperature decreases. In fact, as explained in the earlier lecture, at a limiting

pressure ratio, the volumetric efficiency becomes zero. [8]

Graph 2.3 : P-V diagram for different evaporator pressures and a fixed condenser pressure. [8]

21

The mass flow rate of refrigerant m is given by:

As the evaporator temperature decreases the clearance volumetric efficiency

decreases and the specific volume of refrigerant at compressor inlet ve increases. As

a result of these two effects, the mass flow rate of refrigerant through the compressor

decreases rapidly as the evaporator temperature decreases as shown in Figure 2.8. [8]

Figure 2.8 : Effect of evaporator temperature on clearance volumetric efficiency and refrigerant mass flow rate. [8]

b) On refrigeration effect and refrigeration capacity:

A compressor alone cannot provide refrigeration capacity. By refrigeration

capacity of compressor what we mean is the capacity of a refrigeration system that

uses the compressor under discussion. Figure 12 shows the SSS cycle on P-h

diagram at different evaporator temperatures. It can be seen from the figure that the

refrigeration effect, qe (qe = h1-h4) increases marginally as the evaporator

temperature is increased. This is due to the shape of the saturation vapour curve on

P-h diagram. The effect of Te on refrigerant effect is also shown in Chart 2.2. [8]

22

The refrigeration capacity of the compressor Q is given by:

Chart 2.2 : Effect of evaporator temperature on refrigeration effect on P-h diagram. [8]

Since mass flow rate of refrigerant increases rapidly and refrigerant effect

also increases, though marginally with increase in evaporator temperature, the

refrigeration capacity increases sharply with increase in evaporator temperature as

shown in Chart 2.3. [8]

Chart 2.3 : Effect of evaporator temperature on refrigeration effect and refrigeration capacity. [8]

23

c) On work of compression and power requirement:

At a constant condenser temperature as evaporator temperature increases the

work of compression, Δh (= h-hc21) decreases as shown in Figure 12. This is due to

the divergent nature of isentropes in the superheated region. The work of

compression becomes zero when the evaporator temperature becomes equal to the

condenser temperature (T=Tec) as shown in Figure 14. [8]

The power input to the compressor is given by:

As discussed before, for a given clearance ratio and condenser temperature,

the volumetric efficiency and hence the mass flow rate becomes zero at a lower

limiting value of evaporator temperature (Te = Te,lim). Since the work of

compression becomes zero when the evaporator temperature equals the condenser

temperature, the power input to the compressor, which is a product of mass flow rate

and work of compression is zero at a low evaporator temperature (at which the mass

flow rate is zero). And the power input also becomes zero when evaporator

temperature equals condenser temperature (at which the work of compression

becomes zero). This implies that as evaporator temperature is increased from the

limiting value, the power curve increases from zero, reaches a peak and then

becomes zero as shown in Chart 2.4. [8]

Chart 2.4 : Effect of evaporator temperature on work of compression (Δhc) and power input to compressor (Wc). [8]

24

The variation of compressor power input with evaporator temperature has a

major practical significance. As a mentioned before, there is an evaporator

temperature at which the power reaches a maximum value. If the design evaporator

temperature of the refrigeration system is less than the evaporator temperature at

which the power is maximum, then the design power requirement is lower than the

peak power input. However, during the initial pull-down period, the initial

evaporator temperature may lie to the left of the power peak. [8]

Then as the system runs steadily the evaporator temperature reduces and the

power requirement passes through the peak point. If the motor is designed to suit the

design power input then the motor gets overloaded during every pull-down period as

the peak power is greater than the design power input. Selecting an oversized motor

to meet the power peak is not an energy efficient solution, as the motor will be

underutilized during the normal operation. [8]

One way of overcoming the problem is to throttle the suction gas during the

pull-down so that the refrigerant mass flow rate is reduced and the motor does not

pass through the power peak. In multi-cylinder compressors, some of the cylinders

can be unloaded during the pull-down so as to reduce the power requirement. [8]

d) On COP and volume flow rate per unit capacity:

The COP of the system is defined as:

As discussed before, as the evaporator temperature increases the refrigeration

effect, q increases marginally and the work of compression, Δhec reduces sharply.

As a result the COP of the system increases rapidly as the evaporator temperature

increases as shown in Figure 15. [8]

25

The volume flow rate per unit capacity, V is given by:

As evaporator temperature increases the specific volume of the refrigerant at

compressor inlet reduces rapidly and the refrigerant effect increases marginally. Due

to the combined effect of these two, the volume flow rate of refrigerant per unit

capacity reduces sharply with evaporator temperature as shown in Figure 2.9. This

implies that for a given refrigeration capacity, the required volumetric flow rate and

hence the size of the compressor becomes very large at very low evaporator

temperatures. [8]

Figure 2.9 : Effect of evaporator temperature on COP and volume flow rate per unit capacity (V)

2.8 R407C Characteristics

Compressors designed for operation with HFC refrigerants are factory

supplied with one of the approved oils and are suitably identified. They also can be

used with HCFC R22. Different from single component, azeotropic and near

azeotropic refrigerants, the zeotropic R407* blends are characterized by its relatively

large temperature glide. Therefore, certain influences on system design, service and

maintenance need to be considered. The composition of liquid and vapour of a

zeotropic blend is different when both liquid and vapour are present. Therefore it is

important that charging is carried out in the liquid phase only. [5]

26

If a leak occurs there could be a change in composition of the refrigerant

remaining in the system. Large changes in composition can result in decreasing

superheat at the expansion valve leading to liquid returning to the compressor. [5]

Another characteristic of significant composition change can be the

appearance of sudden “hunting” of the expansion valve. Tests by R 407C

manufacturers have shown that recharging with the original refrigerant can typically

restore satisfactory system operation simply. If it does not result in satisfactory

system operation, it is recommended to analyse the refrigerant composition in the

system prior to replacing the entire system charge. Working with zeotropic blends

requires an understanding of the effect of temperature glide shown in Figure 2.10 [5]

Figure 2.10 : Temperature Glide. [5]

Evaporation at constant pressure takes place with the temperature of the

refrigerant increasing from tex to to2 and condensation takes place with a falling

temperature from tc1 to tc2. Thus the terms “evaporating temperature” and

“condensing temperature” must be redefined. High glide causes considerable

temperature differences within the heat exchangers. Similarly, clear definitions for

“superheating” and “subcooling” are needed. Such new definitions are also necessary

in order to ensure accurate comparisons of performance against other azeotropic or

near azeotropic refrigerants.

27

Evaporating temperature is defined as the mean temperature (tom) between

the dew point temperature (to2) resulting from constant suction pressure (pv1) and

the temperature at which the refrigerant enters the evaporator (tex). Condensing

temperature is defined as the mean temperature (tcm) between the dew-point

temperature (tc1) resulting from constant discharge pressure (pv2) and the boiling-

point temperature (tc2) of the refrigerant. [5]

The superheating of the suction gas is then calculated as the difference in

temperature at the compressor inlet (tv1) and the dew-point temperature (to2) of the

refrigerant at suction pressure (pv1). It is essential that these definitions be followed

when adjusting the superheat setting of expansion valves. Liquid subcooling is

calculated as the difference between actual liquid temperature and the bubble point

temperature (tc2) of the refrigerant at discharge pressure (pv2). [5]

The definitions presented here are based on those specified by the Air-

Conditioning and Refrigeration Institute (ARI) as part of their Alternative

Refrigerants Evaluation Program (AREP). These definitions are used to provide

performance comparison with R 22. System designers normally use data based on the

dew point temperatures as specified in EN 12900. With the Copeland Selection

Software version 4 and above performance data sheets for both mid point and dew

point definitions are available. [5]

It is essential that the glide of R407 refrigerant blends are given careful

consideration when adjusting pressure controls. Furthermore, it is crucial to consider

the effect of pressure losses on glide when sizing heat exchangers. Pressure losses

effectively increase glide significantly in the system. Failure to consider this in heat

energy balance calculations will likely result in undersizing heat exchangers and

other system components. These effects are especially apparent when operating a

system near the limits of its application range. [5]

28

2.9 Malaysia Standard ( MS1525 )

2.9.1 ACMV System Equipment, Electrically Operated, Cooling Mode

These requirements apply to but are not limited to unitary (central) cooling

equipment (air-cooled, water-cooled and evaporatively-cooled) packaged terminal

airconditioners. [9]

ItemAir-cooled

Water-cooled

(water-source)

Dry-bulb Wet-bulb Inlet Outlet

Room air entering

equipment (oC)27 19 - -

Condenser ambient

(air-cooled) (oC)35 24 - -

Refrigerant-water

heat exchanger (oC)- - 29.4 35

NOTES:

1. Data in this table apply to the following types of equipment:

a. Central Air Conditioners Air Evaporatively and Water Cooled, ARI Std 210/240

b. Commercial/Industrial Unitary Air- Conditioning Equipment, ARI Std 340/360

2. Standard Ratings are also based on other standard rating conditions such as but not

limited to electrical conditions; cooling coil air quantity; requirements for separated

(split) assemblies; and minimum external static conditioned-air flow resistance, as

provided in the applicable standards.

Table 2.3 : Standard rating temperatures – cooling. [9]

29

2.9.2 ACMV System Equipment/Component – Heat-Operated (absorption), Cooling Mode

In the heat-operated (absorption) system equipment/components, pumps

included in the package for circulating refrigerant and absorber fluids in the

refrigeration cycle are included in determining the COP of the

equipment/components. [9]

For heat-operated cooling equipment /component, the heat energy input

should be limited to: [9]

a) solar energy;

b) recovered energy from other processes, and

c) natural gas or others (non electric).

Standard rating conditionsHeat source

Direct fired(Gas, oil)

Indirect fired(Steam, hot water)

Airconditioner

Entering conditioned airEntering condenser air

Units Temperatures Temperatures

⁰C

⁰C26.7 DB, 19.4 WB35.0 DB, 23.9 WB

--

Water chillers

Leaving chilled waterFouling factor

Entering chilled waterEntering condenser

Fouling factorLeaving condenser water

Condenser water flow rate

⁰Cm2 K/W

⁰C

⁰Cm2 K/W

⁰Cl/min

7.2-

Per mfg. Spec23.9

-35.0

6.70.00009

12.229.4

0.00018-

Per mfg spec

1. Per ANSI Standard Z21.40.1-1994 and Addenda for Gas-fired absorption summer air-conditioning appliances.2. Per ARI Standard 560-92 for Absorption water-chilling packages.

Table 2.4 : Standard rating conditions – cooling, heat-operated. [9]

30

CHAPTER 3

METHODOLOGY

To design and simulation of air cooled conditioning, there is varies method

can be choose to done this project. This design and simulation is use the software :

a) Coolpack and solkane

b) Copeland Selection Software 6.6

c) Solvay Flour

d) Autocad 2010

3.1 Select and Design The Compressor

To select the compressor, the software use is Copeland Selection Software

6.6. The data use for design and select the compressor is :

Refrigerant : dew point R407C

Requires capacity : 1.00 Kw

Evaporating temperature : 5oC

Condensing temperature : 45 oC

Superheat : 10 oK

Subcool : 0 oK

1) Fill the data in the software

Figure 3.1 : Fill the data in the software

31

2) Select type of compressor

Figure 3.2 : Select type of compressor

3) Export the data to Microsoft Excel document

Figure 3.3 : Export the data to Microsoft Excel

4) Create a graph and find the operation capacity of selected compressor

5) Create a graph and find the power input of selected compressor

6) Find the required data for compressor

The selected compressor is model : ZR22K3E-TFD from brand of Copeland

32

3.2 System Cycle and Air Properties

3.2.1 Air Properties

1) Find the common properties of air.

2) Create the graph for density (kg/m³)

3) From the calculation, calculate the value for density depend on temperature

ambient 32 C.⁰

4) Create the graph for specific heat (kJ/kg.K)

5) From the calculation, calculate the value for specific heat depend on temperature

ambient 32 C.⁰

3.2.2 System Cycle

1) Create a four (4) main component cycle for the system.

2) Create a mollier chart for the system cycle.

3) Then create a table and find the value for :

Tsat (⁰C)

Pressure (bar)

Enthalpy (kj/kg)

Entropy (kj/kg.K)

Specific volume (m³/kg)

3.3 Surface Coefficient (KS) and Capacity of Condenser and Evaporator

1) Create the condenser cycle.

Trefrigerant in = 87.0189 C⁰

Trefrigerant out = 45 C ⁰

Tair in = 32 C ⁰

2) Fill the data for enthalpy for condenser then calculate the Tair out.

3) Calculate the LMTD for Qc superheated and Qc condensation.

4) Calculate the surface coefficient (KS) and the total capacity required for the

condenser.

5) Create the evaporator cycle.

33

Trefrigerant in = 0.86 C⁰

Trefrigerant out = 15 C⁰

Tair in = 25 C ⁰

6) Fill the data for enthalpy for evaporator then calculate the Tair out.

7) Calculate the LMTD for Qe superheated and Qe evaporation.

8) Calculate the surface coefficient (KS) and the total capacity required for the

evaporator.

3.4 Design the Condenser and Evaporator

The step and formula used to design and estimate the sizing of both

component is using the same method except the data is different.

3.4.1 Data

a) Condenser

Oudoor Air = 32 ⁰C to

4

6 ⁰C

Air flow rate = 2000 CFM

Refrigerant in = 87.01889 ⁰C

Refrigerant out = 45 ⁰C

Air face velocity not exceed = 1000 FPM

b) Evaporator

Indoor Air =

11.00

5 ⁰C to 25 ⁰C


Refrigerant in =

0.861

5 ⁰C


Air face velocity not

exceed=

550 FPM

34

3.4.2 The Calculation

1) Compute the overall heat-transfer coefficient U, refrigerant side.

2) Then compute the overall heat-transfer coefficient U, air side.

3) Then compute the fin efficiency and the surface effevtiveness.

4) Determine the ratio of the refrigerant-side to air-side heat transfer areas.

5) Calculate the overall coefficient, U.

6) Computed the fluid capacity rate.

7) Calculate total volume of the heat exchanger.

8) Calculate the face area.

9) Then determine the depth of coil.

10) Determine the number of rows of tubes.

11) Compute the ratio, air-side.

12) Then find the mean density.

13) Calculate the lost head on the air side of the exchanger.

14) Determine the flow cross-sectional area for the refrigerant.

15) Then determine number of tubes per row.

3.5 Design Capillary Tube

To design the capillary tube, the calculation is used the Solvay Flour.

3.5.1 Data

Saturated Liquid Temp. 45 ºC

Evaporating Temp. 5 ºC

1.2 mm

ID 0.0012 m

Radius 0.0006 m

= 3.1415927

Area

(A)

0.00000113 m²

35

Mass Flow rate 0.0321 kg/s

Table 3.1 : Data for capillary tube

3.5.2 The calculation

1) Create the table for the data range 45 ºC to 5 ºC.

2) Find all the pressure for each temperature.

3) Then calculate the fluid velocity (νf) and vapour velocity (νg).

4) Then calculate the fluid enthalpy (hf) and vapour enthalpy (hg).

5) Then calculate the fluid viscosity (µf) and vapour viscosity (µg).

6) After that calculate the mass flow rate divide by area (W/A).

7) Then find the system velocity.

8) Compute the value for Reynold’s Number and f .

9) By using the formula, then calculate the increment length to find the cumulative

length.

36

CHAPTER 4

SYSTEM DESIGN

4.1 The Compressor Design

4.1.1 Data

Software = Copeland Selection Software Version 6.6 / 39463 (01/08)

Refrigerant = R407C

Power Supply = 380/420V - 3~ - 50Hz

Evaporating Temperature = 5 °C

Condensing Temperature = 45 °C

Suction Superheat = 10 K

Liquid subcooling = 0 K

4.1.2 Determine the Capacity

-20 -15 -10 -5 0 5 7 10 12.5 1530 1.88 2.39 3.02 3.77 4.65 5.66 6.11 6.81 7.43 8.0835 1.74 2.24 2.85 3.58 4.45 5.43 5.86 6.55 7.15 7.7940 1.6 2.08 2.67 3.38 4.21 5.16 5.58 6.25 6.83 7.4545 1.91 2.47 3.15 3.95 4.87 5.27 5.91 6.47 7.07

Capacity kW 50 2.28 2.91 3.67 4.54 4.92 5.53 6.07 6.6555 2.67 3.37 4.19 4.55 5.13 5.64 6.1860 3.06 3.82 4.16 4.7 5.18 5.6965 2.74 3.44 3.75 4.24 4.69 5.1667 3.28 3.58 4.06 4.49 4.94

Table 4.1 : Data of the capacity

Graph 4.1 : Temperature vs capacity

37

30 35 40 45 50 55 60 65 67

ax³

0.000003

0.000003

0.000004

0.000000

0.000003

-0.000005

-0.000007

0.000015

-0.000026

bx² 0.0026 0.0025 0.0025 0.0024 0.0024 0.0024 0.0025 0.0018 0.0027

c x 0.1892 0.1845 0.1783 0.1719 0.1623 0.1521 0.1402 0.1304 0.1195

d 4.6533 4.4455 4.2103 3.9498 3.6668 3.3694 3.0599 2.7401 2.6175

Table 4.2 : Data for temperature vs capacity

Graph 4.2 : Evaporating temperature vs capacity

a b c d x = Tc 45 °C-9E-10 1.69E-08 -1.8E-07 6.12E-06 5.02E-06

1.229E-07 2.19E-06 2.51E-06 0.00124 0.00243 2 35.3713E-06 8.25E-05 0.000233 0.019905 0.171209

7.98702E-05 0.001604 0.203466 5.006261 3.949038Te 5 °C

QѲcomp 4.86647 Kw

4.1.3 Determine the Power Input

-20 -15 -10 -5 0 5 7 10 12.5 1530 0.95 0.94 0.94 0.94 0.93 0.92 0.92 0.91 0.9 0.8835 1.07 1.07 1.07 1.06 1.06 1.05 1.04 1.03 1.01 140 1.21 1.21 1.21 1.21 1.2 1.19 1.18 1.16 1.15 1.1345 1.37 1.37 1.37 1.36 1.35 1.34 1.32 1.3 1.28

Power Input kW 50 1.55 1.55 1.54 1.53 1.52 1.5 1.48 1.4555 1.76 1.75 1.73 1.72 1.7 1.67 1.6560 1.98 1.96 1.95 1.92 1.9 1.8765 2.24 2.22 2.2 2.18 2.15 2.1267 2.33 2.31 2.29 2.26 2.23

Table 4.3 : Data for the power input

38

Graph 4.3 : Temperature vs capacity

30 35 40 45 50 55 60 65 67

a x³ -0.000004-

0.000003 -0.000003-

0.000005 -0.000009-

0.000003 0.000007 0.000001 -0.000043

b x² 0.0001 0.0001 0.0001 0.0002 0.0002 0.0002 0.0005 0.0004 0.0010

c x 0.0012 0.0017 0.0020 0.0020 0.0016 0.0028 0.0016 0.0027 0.0157

d 0.9325 1.0591 1.2015 1.3626 1.5427 1.7507 1.9800 2.2402 2.3886Table 4.4 : Data for evaporating temperature vs power input

Graph 4.4 : Evaporating temperature vs capacity

a b c d x = Tc 45 °C-5.1E-09 4.85E-08 1.5E-06 7.11E-06 -3.394E-077.1E-07 6.2311E-06 0.000204 0.00054 0.00015514 2 3

3.16E-050.00026567

4 0.008975 0.038882 0.001851240.000449 0.00360179 0.126088 0.057806 1.36174004

Te 5 °C

Wabs.comp 1.374832276 Kw

39

4.1.4 Compressor cycle

P

h

5⁰C

45 ⁰C

2is

1

4

45 ⁰C

15⁰C

SubcoolSuperheat

Chart 4.1 : Compressor cycle

5.34 m³/hDisplacement @ 50 hz

h

2is

1

15⁰CSuperheat

ref R407CTe ( ⁰C ) 5ηv 0.981207187LP ( pa ) 547098.2899HP ( pa ) 1967321.153h1 ( kj/kg ) 411.3270897h4 ( kj/kg ) 267.844709V1 ( m³/kg ) 0.04291252Vswept ( m³/h ) 5.34Vsuct ( m³/h ) 5.239646381Vsuct ( m³/s ) 0.001455457m ( kg/s ) 0.033916846Qѳcomp ( kW ) 4.866469806Wabs.comp ( kW ) 1.374832276h2is ( kj/kg ) 442.2709295ηeff 0.763378535T2is ( ⁰C ) 62.12789229S1 ( kj/kg.K ) 1.768467784

Table 4.5 : Data of the compressor cycle

40

4.1.5 Compressor Select

Type : Copeland Scroll – Compressor

Model : ZR22K3E-TFD

1) Perfomance at specified operating point ZR22K3E-TFD Data at 50 Hz

Table 4.6 :Perfomance at specified operating point ZR22K3E-TFD Data at 50 Hz

2) Compressor mechanical and physical data

Table 4.7 : Compressor mechanical and physical data

3) Compressor electrical data (380/420V - 3~ - 50Hz)

Table 4.8 : Compressor electrical data (380/420V - 3~ - 50Hz)

41

4.2 System Cycle and Air Properties

4.2.1 Determine the Density and Specific Heat of Air

Common properties for air are indicated the table below :

Temperature

Density

Specific heat capacity

Thermal conductivity

Kinematic viscosity

Expansion coefficient

Prandtl's number

- t - - ρ - - cp - - l - - ν - - b - - Pr -

(oC) (kg/m3)(kJ/kg.K) (W/m.K) x 10-6 x 10-3

(m2/s) (1/K) -150 2.793 1.026 0.0116 3.08 8.21 0.76-100 1.98 1.009 0.016 5.95 5.82 0.74-50 1.534 1.005 0.0204 9.55 4.51 0.7250 1.293 1.005 0.0243 13.3 3.67 0.71520 1.205 1.005 0.0257 15.11 3.43 0.71340 1.127 1.005 0.0271 16.97 3.2 0.71160 1.067 1.009 0.0285 18.9 3 0.70980 1 1.009 0.0299 20.94 2.83 0.708100 0.946 1.009 0.0314 23.06 2.68 0.703120 0.898 1.013 0.0328 25.23 2.55 0.7140 0.854 1.013 0.0343 27.55 2.43 0.695160 0.815 1.017 0.0358 29.85 2.32 0.69180 0.779 1.022 0.0372 32.29 2.21 0.69200 0.746 1.026 0.0386 34.63 2.11 0.685250 0.675 1.034 0.0421 41.17 1.91 0.68300 0.616 1.047 0.0454 47.85 1.75 0.68350 0.566 1.055 0.0485 55.05 1.61 0.68400 0.524 1.068 0.0515 62.53 1.49 0.68

Table 4.9 : Common properties for air

x = Tambient 32 ( ⁰C )

42

Graph 4.5 : Temperature vs density

Density = 1.140904 kg/m3

Graph 4.6 : Temperature vs specific heat

Specific Heat

= 1.004714 kJ/kg.K

43

4.2.2 System Cycle

The air-conditioning system component :

CONDENSOR

EVAPORATOR

TXV COMPRESSOR

3 2

14

5 6

1

23

4,5 5b

2is

5a

P

h

Condenser

TXV

Evaporator

Compressor

Superheat

Superheat

3a3b

Figure 4.1 : Air conditioning system component

The system cycle :

1

23

4,5 5b

2is

5a

P

h

Condenser

TXV

Evaporator

Compressor

Superheat

Superheat

3a3b

Chart 4.2 : System cycle

44

Tc 45 ⁰C

Te 5 ⁰C

Ref. R407C

Superhea

t 10 K

Subcool 0 K

Table 4.10 : Data for the system cycle

Tsat ( ⁰C) P (bar) T (⁰C) H (kj/kg) S (kj/kg.K) V (m³/kg)

1 5 5.39944031 15 420.7456446 1.801735842 0.045338828

2is 45 19.4159502 71.29131 453.5751144 1.801735842

2 45 19.4159502 87.01889 472.0087329

3 45 19.4159502 45 267.844709

4 5 5.39944031 0.861535 267.844709

5 5 5.39944031 0.861535 267.844709

5a -1.144306 5.39944031 -1.14431 198.301363

5b 5.0000031 5.39944031 5.000003 411.3270926

Table 4.11 : Table for the system cycle

3a = 456.7923 h

3b

= 0.000963 V

ŋis = 0.8-(0.004*(Ţ-5)*2)-0.5/(Ţ-0.3) = 0.640411471

Ţ = pH/pL = 3.595919032

Interpolate

5a 4 5b

198.3014

267.8447

1 411.327093

-1.14431 x 5.00000305

45

x = (271.1996-198.3014)÷(411.3271-198.3014)

= (x+1.14431)÷(5.000003+1.14431)

= 0.32645515

= 2.00584139

x = 0.86153518

4.3 Condenser and Evaporator Capacity

4.3.1 Data

Qѳcomp ( kW ) 4.866469806

Q=m*ΔH

Te 5 ⁰C

Tc 45 ⁰C

Ref. R407C

LP 5.399440315 bar

HP 19.41595019 bar

H1 411.3270897 Kj/kg

H4 267.844709 Kj/kg

m 0.033916846 kg/s

V1 (m3/kg) 0.042913

Vswept ( m³/h ) 5.34

Vsuct ( m³/h ) 5.239646

Vsuct ( m³/s ) 0.001455

Table 4.12 : Data for calculate condenser and evaporator capacity

46

4.3.2 The Condenser

Condenser

Superheat

SuperheatEvaporator

45 ⁰C 45 ⁰C

87.0189 ⁰C

32 ⁰C

0.86⁰C15⁰C

25 ⁰CT3

T4

T1T2

0.86⁰C

P

P

h

h

Chart 4.3 : Condenser system cycle

Tambient 32 oC

Qcondensation = 6.07 Qc superheated= 0.49

tliquid = t4 45 tvapour = t3a 45 tenter cond = t2 87.02

hliquid,45ºC= h4 267.84 hvapour,45ºC= h3a 456.79 henter cond = h2 472.01

t2 = 46 t1 = 39 tair in = 32.0

DT2 = 1.00 DT3 = 6.00 DT4 = 55.02

LMTDcond = 2.79 LMTDsup = 22.12

KScond = 2.174 KSsuperheated= 0.022

Table 4.13 : Table for the condenser capacity

Condenser

Vair = 3000.00 m3/h

Density air = 1.141 kg/m3

ṁair = 0.9508 kg/s

Cp air = 1.0044 kJ/kg K

Qtotal cond = 6.55 kW

KS = 2.196 kW/K

47

4.3.3 The Evaporator

SuperheatEvaporator

0.86⁰C15⁰C

25 ⁰CT3

T4

0.86⁰C

P

h

Chart 4.4 : Evaporator system cycle

Troom 25 oC

Qevaporation = 4.61 Qc superheated = 0.30

tenter evap = t5 0.86 tvapour5ºC = t5b 5.00 texit evap = t1 15.00

henter evap = h5 267.84 hvapour,5ºC=h5b 411.33 hexit evap = h1 420.75

tair in = 25.0 t3 = 18.0 t4 = 11.0

DT1 = 24.14 DT2 = 13.00 DT3 = 4.00

LMTDevap = 18.00 LMTDsup = 7.63

Ksevap = 0.256 KS superheated = 0.040

Table 4.14 : Data for the evaporator capacity

Evaporator

Vair = 3000.00 m3/h

Density air = 1.141 kg/m3

ṁair = 0.9508 kg/s

Cp air = 1.0047 kJ/kg K

Qtotal cond = 4.91 kW

KS = 0.296 kW/K

48

4.4 Sizing the Condenser

4.4.1 Data

Oudoor Air = 32 ⁰C to

4

6 ⁰C


Refrigerant in = 87.01889 ⁰C


Air face velocity not exceed = 1000 FPM

4.4.2 Calculation and Design

1. Compute the overall heat-transfer coefficient U, refrigerant side.

V = 0.000091446 m3/s

V = 1.164916855 m/s

m = 0.094931265 kg/s

ρ = 1037.586844 kg/m3

V1 = 0.00096328616 m3/kg

Dinside = 0.01 m

Rinside = 0.005 m

= 0.000117884 kg/m.s

A = 0.0000785398 m2

Cp = 1.705911601 J/kg.k

K = 0.0754295 w/m.k

ReD= ρ V Dμ

=102533

Pr= μC pK

=9.59784

49

hi=0.023KD

(Re D)0.8(Pr )0.3=3488.286Btu

ft2. hr .℃=19807.4

W

m2 . K

2. Compute the overall heat-transfer coefficient U, air side.

L/D = 60

A face velocity = 900 ft/m

Gfr=Pfr V fr=3900.8lbm

hr . ft2=5.29026

kg

m2 . s

σ = 0.555

GC=Gfr

σ=7028.46

lbmhr . ft2 =9.5320001

kgm2 . s

Doutside = 0.01715 m

μ = 0.044 lb/ft.s = 0.06548 kg/m.s

ReD=GC D

μ=8987.87

D = 0.01715 m = 0.05627 ft

Dh = 0.000333248 m

Xb = 0.0275082 m

Xa = 0.03175 m

π = 3.141592654

AA t

=4 Xb Xa

π Dh D=8.99904

JP=(Re D)−0.4 ( AA t

)−0.15

=0.018854

50

Chart 4.5 : Heat-transfer correlation for smooth plate-fin-tube coils. [12]

j-factor = 0.006

ho=( j−factor ) (GC ) (0.24 ) (0.71 )−2/3=12.71698Btu

hr . ft2 . F=72.210303

W

m2 K

3. To compute the fin efficiency and the surface effevtiveness.

Dim1=Xa

2

L=0.0159 m

Dim2=(( Xa

2)

2)+( Xb)2

1 /2

¿ ¿2

M=0.01588 m

φ= Mr

=1.851913

β= LM

=0.999676

ℜr

=(1.27 ) (φ )¿

51

ln (ℜr

)=0.676666643

∅=(ℜr −1)(1+0.35 ln (ℜr ))=1.1964

y = 0.0005 ft

k = 100 Btu.ft/ft2.hr.F

m=¿

r = 0.008575 m = 0.02813 ft

tanh ( m x r x ∅ ) = 0.640565651

Fin efficiency =

η=tanh (m×r ×∅ )

m× r×∅=0.843813

A f

A=0.919

Surface effectiveness =

ηso=1−A f

A(1−η )=0.856464

4. Determine the ratio of the refrigerant-side to air-side heat transfer areas.

⍺ = 100 ft1

Xa = 0.104166667 ft

Xb = 0.09025 ft

Dinside = 0.032808399

A i

Ao

=π D i

Xa Xb α=0.109637

52

5. Calculate the overall coefficient, U.

1U0

= 1ho ηso

+ 1

h i(A i

Ao

)=10.59004

Btu

hr . ft2. F=60.13301

W

m2. K

6. Computed the fluid capacity rate.

p = 0.07223697

Q = 120000 ft3/s

Q=p Q=8668.44lbhr

=1.092223kgs

Cair=0.24 m=2080.425Btu

hr . F

air = air

ref = refrigerant

Tao = 46 ⁰C = 114.8 ⁰F

Tai = 32 ⁰C = 89.6 ⁰F

Tri = 87.0188855 ⁰C = 188.634 ⁰F

Tao = 45 = 113 ⁰F

C ref=Cair(T ao−T ai

Tri−T ro)=693.1632

Btuhr . F

Cmin

Cmax

=0.3332

Tco = 46 ⁰C

Tci = 32 ⁰C

Thi = 87.0188855 ⁰C

ε=T co−T ci

T hi−T ci

=0.254458

53

Chart 4.6 : Effectiveness for counter-flow exchanger. [12]

NTU = 0.35

Ao=NTU (Cmin)

Uo

=22.90899 ft2

7. Calculate total volume of the heat exchanger.

V=Ao

α=0.22909 ft3

8. Calculate the face area.

A fr=QV fr

=2.22222 ft2=0.20645 m2

9. Determine the depth of coil.

L= VA fr

=0.10309 ft=0.062844 m

10. Determine the number of rows of tubes.

N r=LXb

=2.284553

So the number of rows will then be 4 because N must be integer and a

multiple of two for flow arrangement.

54

11. Compute the ratio, air-side.

AAC

= αVσ A fr

=18.5749

12. Find the mean density.

P = 2116.8

Tci = 549.27 Rankine

Tco = 574.47 Rankine

R = 53.35

Pm=P

2 R ( 1Tci

+1

Tco)=0.070653

lb

ft3 =1.131713kg

m3

13. Calculate the lost head on the air side of the exchanger.

D¿

D=

( AA t

)

1+(Xa−D)

s

=5.305619

D* = 0.090991373 m

¿ = 0.102703632

¿ = 1.517693322

(Xa−D

4 (s− y))−0.4

= 0.466203452

(Xa

D¿−1)−0.5

= 1.08381619

FP=( Re D )−0.25( DD¿ )

0.25( Xa−D4 (s− y))

−0.4

( Xa

D¿−1)−0.5

=0.0787591

55

Chart 4.7 : Correlation of friction data for smooth plate-fin-coils. [12]

f = 0.013

Gc = 9.532000148 kg/m2.s = 7028.46 lbm/hr.ft2

gc = 1 kg.m/N.s2 = 32.17 lbm.ft/lbf.s2

ƿ1 = 0.065243716 lb/ft3 = 1.04507 kg/m3

ƿ2 = 0.07606144 lb/ft3 = 1.21835 kg/m3

σ = 0.555

Tho = 572.67 Rankine

Thi = 648.3039939 Rankine

¿¿ = 0.90802

(p1

p2

−1) = 0.1658

(1+σ 2) = 1.30803

( fAAC

p1

pm

) = 0.26149

56

∆ Po=(GC)2

2 gC p12 [ (1+σ 2 )( p1

p2

−1)+( fA

AC

p1

pm

)]

¿0.4344lb

ft2

or

¿20.798 Pa

Also,

∆ Po=0.007 ft w .g

14. The flow cross-sectional area for the refrigerant.

Ch = 2080.424748 Btu/hr.F

V = 1.164916855 m/s = 3.82191 ft/s

ƿ = 1037.586844 kg/m3 = 69.0482 lb/ft3

Cp = 1.705911601 kJ/kg.K = 0.40754 Btu/lb.F

A=Ch

V pCp

=0.00537 ft2=0.00049920 m2

15. Determine number of tubes per row.

Di = 0.01 m

N= 4 A

π Di2=6.35601

Since N must be integer, 8 tubes are required.

To adapt to the flow arrangement, a coil that is 16 tubes high must be used.

H=16 Xa=0.508 m

W =A fr

H=0.4064 m

The shape of the coil, height of 0.508 m and width of 0.4064 m, may be

unacceptable.

57

Lw=2 (no . of rows ) (W )=2.43838058 m=7.99993628 ft

Chart 4.8 : Friction factors for pipe flow. [11]

moody friction = 0.024

D = 0.01 m = 0.03281 ft

V = 1.164916855 m.s = 3.82191 ft/s

g = 9.81 m/s2 = 32.1639 ft/s2

Lf w=moody friction( Lw

D )( V 2

2 g )+(no . of returnbends )( V 2

2 g ) ¿4.50783 ft w . g

Use 2-row coil, and the coil will then have 8 tubes per row.

H=8 Xa=0.381 m

W =A f r

H=0.541862m

Lw=2 (no . of rows ) (W )=4.3348988 m=14.2221 ft

58


D )( V 2

2g )+(no .of returnbends )( V 2

2g ) ¿4.633092 ft w . g

4.4.3 Drawing of Condenser

1) Isometric View

Figure 4.2 : Isometric view of condenser

2) Plant View

Figure 4.3 :Plant view of condenser

59

3) Side View

Figure 4.4 :Side view of condenser

4) Front View

Figure 4.5 : Front view of condenser

60

4.5 Sizing the Evaporator

4.5.1 Data

Indoor Air =

11.00

5 ⁰C to 25 ⁰C


Refrigerant in =

0.861

5 ⁰C


Air face velocity not

exceed=

550 FPM

4.5.2 Calculation and Design

1. Compute the overall heat-transfer coefficient U, refrigerant side.

V = 0.001455457 m3/s

V = 18.53145821 m/s

m = 0.033916845 kg/s

ρ = 23.3032221 kg/m3

V1 = 0.04291252068 m3/kg

Dinside = 0.01 m

Rinside = 0.005 m

= 0.0000113048 kg/m.s

A = 0.0000785398 m2

Cp = 0.966744498 J/kg.k

K = 0.013006608 w/m.k

ReD= ρ V Dμ

=382001

Pr= μC pK

=3.0249

61

hi=0.023KD

(Re D)0.8(Pr )0.3=1218.307Btu

ft2. hr .℃=6917.86

W

m2 . K

2. Compute the overall heat-transfer coefficient U, air side.

L/D = 60

A face velocity = 500 ft/m

Gfr=Pfr V fr=2217.99lbm

hr . ft2=3.008036

kg

m2 . s

σ = 0.555

GC=Gfr

σ=3996.38

lbmhr . ft2 =5.419885

kgm2 . s

Doutside = 0.01715 m

μ = 0.044 lb/ft.s = 0.06548 kg/m.s

ReD=GC D

μ=5110.49

D = 0.01715 m = 0.05627 ft

Dh = 0.000333248 m

Xb = 0.0275082 m

Xa = 0.03175 m

π = 3.141592654

AA t

=4 Xb Xa

π Dh D=8.99904

JP=(Re D)−0.4 ( AA t

)−0.15

=0.02363

62

Chart 4.9 : Heat-transfer correlation for smooth plate-fin-tube coils. [12]

j-factor = 0.0075

ho=( j−factor ) (GC ) (0.24 ) (0.71 )−2/3=9.038575Btu

hr . ft2 . F=51.32338

W

m2 K

3. Compute the fin efficiency and the surface effevtiveness.

Dim1=Xa

2

L=0.0159 m

Dim2=(( Xa

2)

2)+( Xb)2

1 /2

¿ ¿2

M=0.01588 m

φ= Mr

=1.851913

β= LM

=0.999676

63

ℜr

=(1.27 ) (φ )¿

ln (ℜr

)=0.676666643

∅=(ℜr −1)(1+0.35 ln (ℜr ))=1.1964

y = 0.0005 ft

k = 100 Btu.ft/ft2.hr.F

m=¿

r = 0.008575 m = 0.02813 ft

tanh ( m x r x ∅ ) = 0.5648951

Fin efficiency =

η=tanh (m×r ×∅ )

m× r×∅=0.88

A f

A=0.919

Surface effectiveness =

ηso=1−A f

A(1−η )=0.892162

4. Determine the ratio of the refrigerant-side to air-side heat transfer areas.

⍺ = 170 ft1

Xa = 0.104166667 ft

Xb = 0.09025 ft

Dinside = 0.032808399

64

A i

Ao

=π D i

Xa Xb α=0.064493

\

5. Calculate the overall coefficient, U.

1U 0

= 1ho ηso

+ 1

h i(A i

Ao

)=7.313307

Btu

hr . ft 2 . F=41.526857

W

m2 . K

6. Computed the fluid capacity rate.

p = 0.073933

Q = 35100 ft3/s

Q=p Q=2595lbhr

=0.32697593kgs

Cair=0.24 m=622.8112Btu

hr . F

air = air

ref = refrigerant

Tao = 25 ⁰C = 77 ⁰F

Tai = 11.004772 ⁰C = 51.809 ⁰F

Tri = 0.8615352 ⁰C = 33.551 ⁰F

Tao = 5.0000031 = 41 ⁰F

C ref=Cair(T ao−T ai

Tri−T ro)=2106.186

Btuhr . F

Cmin

Cmax

=0.2957

Tco = 5.0000031 ⁰C

Tci = 0.8615352 ⁰C

Thi = 25 ⁰C

65

ε=T co−T ci

T hi−T ci

=0.171447

Chart 4.10 : Effectiveness for counter-flow exchanger. [12]

NTU = 0.2

Ao=NTU (Cmin)

Uo

=17.03227 ft2

7. Calculate total volume of the heat exchanger.

V=Ao

α=1.002 ft3

8. Calculate the face area.

A fr=QV fr

=1.17 ft2=0.1086957 m2

9. Determine the depth of coil.

L= VA fr

=0.085632 ft=0.03915111m

66

10. Determine the number of rows of tubes.

N r=LXb

=1.413151

So the number of rows will then be 2 because N must be integer and a

multiple of two for flow arrangement.

11. Compute the ratio, air-side.

AAC

= αVσ A fr

=18.5749

12. Find the mean density.

P = 2116.8

Tci = 536.67 Rankine

Tco = 511.47859 Rankine

R = 53.35

Pm=P

2 R ( 1Tci

+1

Tco)=0.07058

lb

ft3 =1.21342kg

m3

13. Calculate the lost head on the air side of the exchanger.

D¿

D=

( AA t

)

1+(Xa−D)

s

=5.305619

D* = 0.090991373 m

¿ = 0.1182726

¿ = 1.517693322

(Xa−D

4 (s− y))−0.4

= 0.466203452

(Xa

D¿−1)−0.5

= 1.08381619

FP=( Re D )−0.25( DD¿ )

0.25( Xa−D4 (s− y))

−0.4

( Xa

D¿−1)−0.5

=0.0907

67

Chart 4.11 : Correlation of friction data for smooth plate-fin-coils. [12]

f = 0.013

Gc = 5.4198853 kg/m2.s = 3996.376 lbm/hr.ft2

gc = 1 kg.m/N.s2 = 32.17 lbm.ft/lbf.s2

ƿ1 = 0.0798475 lb/ft3 = 1.278997 kg/m3

ƿ2 = 0.0716598 lb/ft3 = 1.147847 kg/m3

σ = 0.555

Tho = 500.67001 Rankine

Thi = 498.22076 Rankine

¿¿ = 0.23988

(p1

p2

−1) = 0.11426

(1+σ 2) = 1.30803

( fAAC

p1

pm

) = 0.3235

68

∆ Po=(GC)2

2 gC p12 [ (1+σ 2 )( p1

p2

−1)+( fA

AC

p1

pm

)]

¿0.1135lb

ft2

or

¿5.432 Pa

Also,

∆ Po=0.0018 ft w . g

14. Determine the flow cross-sectional area for the refrigerant.

Ch = 2106.186494 Btu/hr.F

V = 18.531458210 m/s = 60.79875 ft/s

ƿ = 23.3032221 kg/m3 = 2.326135 lb/ft3

Cp = 0.96674498 kJ/kg.K = 0.692866Btu/lb.F

A=Ch

V pCp

=0.00597057 ft2=0.0005547 m2

15. Calculate number of tubes per rows.

Di = 0.01 m

N= 4 A

π Di2=7.06245164

Since N must be integer, 8 tubes are required.

To adapt to the flow arrangement, a coil that is 16 tubes high must be used.

H=16 Xa=0.508 m

W =A fr

H=0.2139679 m

69

The shape of the coil, height of 0.508 m and width of 0.2139679 m, may be

unacceptable.

Lw=2 (no . of rows ) (W )=0.428 m=1.404 ft

Chart 4.12 : Friction factors for pipe flow. [11]

moody friction = 0.023

D = 0.01 m = 0.03281 ft

V = 18.53145821 m.s = 5.066562 ft/s

g = 9.81 m/s2 = 32.1639 ft/s2


D )( V 2

2 g )+(no . of returnbends )( V 2

2 g ) ¿5.97947 ft w . g

Use 2-row coil, and the coil will then have 8 tubes per row.

H=8 Xa=0.254 m

W =A f r

H=0.4279 m

70

Lw=2 (no . of rows ) (W )=1.7117432 m=5.165955 ft


D )( V 2

2g )+(no .of returnbends )( V 2

2g ) ¿5.56157 ft w . g

4.5.3 Drawing of Evaporator

1) Isometric View

Figure 4.6 : Isometric view of evaporator

2) Plant View

71

Figure 4.7 : Plant view of evaporator

3) Side View

Figure 4.8 : Side view of evaporator

4) Front View

Figure 4.9 : Front view of evaporator

72

4.6 Sizing the Capillary Tube

4.6.1 Data

Saturated Liquid Temp. 45 ºC

Evaporating Temp. 5 ºC

1.2 mm

ID 0.0012 m

Radius 0.0006 m

= 3.1415927

Area

(A) 0.00000113 m²

Mass Flow rate 0.0321 kg/s

73

75

4.6.2 Calculation

Position

Temp.

Pressure ν1 = νf νg ν2 h1 = hf hg h2 µ1 = µf µg µ2 w/A VRe f

ºC kPa m³/kg m³/kg m³/kg kJ/kg kJ/kg kJ/kg Pa.s Pa.s Pa.s kg/s.m² m/s

1 45 1751.0063670.0009633 0.0132 267.84471

424.91157

0.0001179 28384.2104 27.342117 288936 0.0142

2 44 1706.9516050.0009584 0.0136

0.001083266.18508

424.73394

267.750.0001194 0.00001417

0.000118 28384.2104 27.2026 285277 0.0143

3 43 1663.7538680.0009536 0.014

0.001080264.5367

424.54448

266.090.0001209 0.00001407

0.000120 28384.2104 27.065983 281671 0.0143

4 42 1621.4003250.0009488 0.0144

0.001077262.8991

424.34371

264.440.0001225 0.00001398

0.000121 28384.2104 26.932151 278116 0.0144

5 41 1579.8783970.0009442 0.0148

0.001074261.27187

424.13201

262.800.000124 0.00001388

0.000123 28384.2104 26.800998 274611 0.0144

6 40 1539.175740.0009397

0.015300 0.001072259.65456

423.90978

261.170.0001256 0.00001379

0.000125 28384.2104 26.672422 271157 0.0145

7 39 1499.280230.0009352

0.015700 0.001069258.04679

423.67738

259.550.0001272 0.00001369

0.000126 28384.2104 26.546328 267750 0.0145

8 38 1460.1799570.0009309

0.016200 0.001067256.44816

423.43516

257.940.0001288 0.00001360

0.000128 28384.2104 26.422628 264392 0.0146

9 37 1421.8632060.0009266

0.016700 0.001065254.8583

423.18341

256.340.0001305 0.00001352

0.000129 28384.2104 26.301236 261080 0.0146

10 36 1384.3184560.0009224

0.017200 0.001064253.27685

422.92249

254.750.0001321 0.00001343

0.000131 28384.2104 26.182072 257815 0.0146

11 35 1347.5343610.0009183

0.017700 0.001062251.70345

422.65266

253.170.0001338 0.00001335

0.000133 28384.2104 26.065062 254594 0.0147

12 34 1311.4997510.0009142

0.018200 0.001060250.13779

422.37418

251.590.0001355 0.00001326

0.000134 28384.2104 25.950132 251417 0.0147

13 33 1276.2036180.0009103

0.018800 0.001059248.57955

422.08729

250.020.0001372 0.00001318

0.000136 28384.2104 25.837216 248284 0.0148

14 32 1241.6351080.0009064

0.019300 0.001057247.02841

421.79224

248.460.0001389 0.00001310

0.000138 28384.2104 25.726248 245193 0.0148

15 31 1207.7835210.0009025

0.019900 0.001056245.48408

421.48923

246.910.0001407 0.00001303

0.000140 28384.2104 25.617167 242145 0.0149

16 30 1174.6382970.0008987

0.020500 0.001055243.94629

421.17849

245.360.0001424 0.00001295

0.000141 28384.2104 25.509916 239137 0.0149

17 29 1142.189015 0.0008950.021100 0.001055

242.41478420.86016

243.820.0001442 0.00001287

0.000143 28384.2104 25.404437 236169 0.0150

18 28 1110.4253870.0008914

0.021800 0.001054240.88928

420.53451

242.290.000146 0.00001280

0.000145 28384.2104 25.300678 233241 0.0150

76

19 27 1079.337250.0008878

0.022400 0.001053239.36956

420.20167

240.760.0001479 0.00001273

0.000147 28384.2104 25.198589 230352 0.0151

20 26 1048.9145680.0008842

0.023100 0.001053237.85538 419.8618

239.240.0001497 0.00001265

0.000149 28384.2104 25.098121 227501 0.0151

PositionTemp. Pressure ν1 = νf νg ν2 h1 = hf hg h2 µ1 = µf µg µ2 w/A V

Re fºC kPa m³/kg m³/kg m³/kg kJ/kg kJ/kg kJ/kg Pa.s Pa.s Pa.s kg/s.m² m/s

21 25 1019.1474180.0008807

0.023800 0.000881236.34653 419.51507

236.350.0001516 #VALUE!

#VALUE!28384.2104 24.999229 224687 0.0152

22 24 990.02599690.0008773

0.024600 0.000877234.8428 419.16161

234.840.0001535 0.00001251

0.00015328384.2104 24.901867 221910 0.0152

23 23 961.54060830.0008739

0.025300 0.000874233.34399 418.80158

233.340.0001554 0.00001245

0.00015528384.2104 24.805993 219169 0.0153

24 22 933.68166470.0008706

0.026100 0.000871231.84992 418.43511

231.850.0001574 0.00001238

0.00015728384.2104 24.711568 216463 0.0153

25 21 906.43968250.0008673

0.026900 0.000867230.36041 418.06232

230.360.0001593 0.00001231

0.00015928384.2104 24.618553 213792 0.0153

26 20 879.80527870.0008641

0.027700 0.000864228.8753 417.68331

228.880.0001613 0.00001225

0.00016128384.2104 24.52691 211156 0.0154

27 19 853.76916850.0008609

0.028600 0.000861227.39444 417.29826

227.390.0001633 0.00001218

0.00016328384.2104 24.436603 208553 0.0154

28 18 828.32216260.0008578

0.029500 0.000858225.91767 416.90725

225.920.0001654 0.00001212

0.00016528384.2104 24.347599 205983 0.0155

29 17 803.45516430.0008547

0.030400 0.000855224.44486 416.51039

224.440.0001674 0.00001206

0.00016728384.2104 24.259864 203445 0.0155

30 16 779.15916760.0008516

0.031400 0.000852222.97588 416.10779

222.980.0001695 0.00001199

0.00017028384.2104 24.173366 200940 0.0156

31 15 755.4252550.0008486

0.032400 0.000849221.51062 415.69954

221.510.0001716 0.00001193

0.00017228384.2104 24.088076 198465 0.0156

32 14 732.24459510.0008457

0.033400 0.000846220.04897 415.28575

220.050.0001738 0.00001187

0.00017428384.2104 24.003965 196022 0.0157

33 13 709.60844090.0008428

0.034500 0.000843218.59082 414.86651

218.590.0001759 0.00001181

0.00017628384.2104 23.921003 193608 0.0157

34 12 687.5081280.0008399

0.035600 0.000840217.13608 414.44191

217.140.0001781 0.00001175

0.00017828384.2104 23.839163 191225 0.0158

35 11 665.9350728 0.0008370.036800 0.000837

215.68467 414.01205

215.680.0001803 0.00001170

0.00018028384.2104 23.75842 188871 0.0158

36 10 644.88077110.0008342

0.038000 0.000834214.23651 413.57701

214.240.0001826 0.00001164

0.00018328384.2104 23.678748 186545 0.0159

77

37 9 624.33679660.0008315

0.039200 0.000831212.79153 413.13683

212.790.0001849 0.00001158

0.00018528384.2104 23.600123 184248 0.0159

38 8 604.29479920.0008287

0.040500 0.000829211.34965 412.6917

211.350.0001872 0.00001152

0.00018728384.2104 23.522521 181978 0.0160

39 7 584.746504 0.0008260.041800 0.000826

209.91084 412.24164

209.910.0001895 0.00001147

0.00019028384.2104 23.445919 179736 0.0160

40 6 565.68371030.0008234

0.043200 0.000823208.47502 411.78675

208.480.0001919 0.00001141

0.00019228384.2104 23.370296 177521 0.0161

41 5 547.09828990.0008207

0.044700 0.000821207.04217 411.32709

207.040.0001943 0.00001136

0.00019428384.2104 23.29563 175332 0.0161

Position

Temp.a b c x fm Vm a′ b′ c′

Increment CumulativeºC Length m Length m

1 45

2 4464376.85 168309.76 -1663.43 0.010 0.0143 27.272

48015

0.1409.807

0.01180.0118

3 4368565.85 170030.67 -1652.09 0.010 0.0143 27.134

47076

0.1389.655

0.01160.0234

4 4272885.80 171727.32 -1641.21 0.010 0.0143 26.999

46152

0.1369.504

0.01140.0348

5 4177336.67 173400.58 -1630.76 0.009 0.0144 26.867

45245

0.1359.355

0.01120.0460

6 4083071.32 175127.05 -1620.74 0.009 0.0144 26.737

44352

0.1339.213

0.01110.0570

7 3987816.44 176755.78 -1611.13 0.009 0.0145 26.609

43475

0.1319.065

0.01090.0679

8 3893918.47 178438.60 -1601.91 0.009 0.0145 26.484

42611

0.1308.925

0.01070.0786

78

9 37100224.40 180100.57 -1593.06 0.009 0.0146 26.362

41762

0.1288.785

0.01050.0892

10 36106734.17 181742.45 -1584.58 0.009 0.0146 26.242

40927

0.1268.647

0.01040.0995

11 35113447.74 183364.92 -1576.45 0.009 0.0147 26.124

40105

0.1258.508

0.01020.1098

12 34120365.03 184968.62 -1568.65 0.008 0.0147 26.008

39297

0.1238.371

0.01000.1198

13 33128923.28 186627.52 -1561.17 0.008 0.0148 25.894

38501

0.1228.240

0.00990.1297

14 32136288.47 188195.22 -1554.00 0.008 0.0148 25.782

37718

0.1208.104

0.00970.1394

15 31145383.75 189818.66 -1547.12 0.008 0.0149 25.672

36948

0.1187.975

0.00960.1490

16 30154771.78 191425.05 -1540.53 0.008 0.0149 25.564

36189

0.1177.846

0.00940.1584

17 29164452.50 193014.89 -1534.20 0.008 0.0149 25.457

35443

0.1157.718

0.00930.1677

18 28176106.35 194660.55 -1528.13 0.008 0.0150 25.353

34709

0.1147.595

0.00910.1768

19 27186420.90 196218.56 -1522.30 0.008 0.0150 25.250

33986

0.1127.468

0.00900.1857

20 26198813.76 197832.72 -1516.70 0.008 0.0151 25.148

33274

0.1117.346

0.00880.1946

79

Position

Temp.a b c x fm Vm a′ b′ c′

Increment CumulativeºC Length m Length m

80

21 25211604.38 199431.66 -1511.33 0.008 0.0151 25.049

32574

0.1097.225

0.00870.2032

22 24226699.92 201086.48 -1506.16 0.007 0.0152 24.951

31885

0.1087.109

0.00850.2118

23 23240342.52 202655.95 -1501.19 0.007 0.0152 24.854

31207

0.1066.989

0.00840.2201

24 22256411.29 204281.55 -1496.41 0.007 0.0153 24.759

30539

0.1056.873

0.00820.2284

25 21272998.95 205892.97 -1491.80 0.007 0.0153 24.665

29882

0.1046.758

0.00810.2365

26 20290105.42 207490.53 -1487.36 0.007 0.0154 24.573

29236

0.1026.644

0.00800.2445

27 19309961.44 209144.03 -1483.08 0.007 0.0154 24.482

28599

0.1016.534

0.00780.2523

28 18330473.61 210783.86 -1478.94 0.007 0.0155 24.392

27973

0.0996.424

0.00770.2600

29 17351641.86 212410.35 -1474.94 0.007 0.0155 24.304

27357

0.0986.314

0.00760.2676

30 16375923.26 214092.41 -1471.07 0.007 0.0156 24.217

26751

0.0976.209

0.00750.2750

31 15401014.16 215761.25 -1467.32 0.007 0.0156 24.131

26155

0.0956.103

0.00730.2824

32 14426914.48 217417.13 -1463.68 0.007 0.0157 24.046

25568

0.0945.998

0.00720.2896

33 13456331.77 219128.24 -1460.14 0.007 0.0157 23.962

24991

0.0935.896

0.00710.2966

34 12486727.99 220826.47 -1456.69 0.007 0.0158 23.880

24423

0.0915.795

0.00700.3036

35 11 520996.44 222579.51 -1453.33 0.006 0.0158 23.799 2386 0.090 5.69 0.0068 0.3104

81

5 6

36 10556429.42 224319.71 -1450.05 0.006 0.0159 23.719

23316

0.0895.598

0.00670.3172

37 9593026.85 226047.27 -1446.84 0.006 0.0159 23.639

22776

0.0875.500

0.00660.3238

38 8633980.79 227829.29 -1443.70 0.006 0.0160 23.561

22245

0.0865.405

0.00650.3302

39 7676300.87 229598.74 -1440.62 0.006 0.0160 23.484

21723

0.0855.309

0.00640.3366

40 6723397.11 231422.16 -1437.58 0.006 0.0161 23.408

21209

0.0845.217

0.00630.3429

41 5775608.44 233299.13 -1434.59 0.006 0.0161 23.333

20705

0.0835.126

0.00620.3490

From the table, the final length of capillary tube is 0.3490 meter.

CHAPTER 5

RESULT AND ANALYSIS

5.1 Effect of Evaporating Temperature On Cooling Capacity

5.1.2 Evaporating Temperature Vs Cooling Capacity

The main objective for this project is to to analyse the effect of evaporating

temperature on cooling capacity. That is mean by using varies evaporating

temperature (constant condensing temperature Tc=45⁰C) that is 2⁰C to 13⁰C then to

look out the result of the cooling capacity for each temperature tested.

Temperature (⁰C) 2 3 4 5 6 7 8 9 10 11 12 13

Cooling Capacity (kW) 4.3 4.48 4.67 4.87 5.06 5.27 5.48 5.69 5.91 6.13 6.36 6.59Table 5.1 : Data for evaporating temperature vs cooling capacity

0 2 4 6 8 10 12 140

1

2

3

4

5

6

7 Te vs Cooling Capacity

Evaporating Temperature (⁰C)

Cooling Capacity (kW)

Graph 5.1 : Evaporating temperature Vs cooling capacity

The graph show that when the evaporating temperature is increased, the

cooling capacity will also increased.

82

Then from the data above, there are the factor that need to be calculated to

determine the optimum evaporating temperature for Malaysia conditions. That is :

a) Compression ratio

b) Coefficient of performance (C.O.P)

c) Power consumption

5.1.2 Compression Ratio and Coefficient of Perfomance Vs Cooling

Capacity

Temperature (⁰C) 2 3 4 5 6 7 8 9 10 11 12 13Cooling Capacity (kW) 4.3

4.48

4.67

4.87

5.06

5.27

5.48

5.69

5.91

6.13

6.36

6.59

Compression Ratio3.98

3.85

3.72

3.59

3.48

3.36

3.26

3.15

3.05

2.95

2.86

2.77

C.O.P2.81

2.87

2.92

2.98

3.04 3.1

3.17

3.23

3.29

3.36

3.43 3.5

Table 5.2 : Data for evaporating temperature Vs cooling capacity Vs compression ratio Vs

C.O.P

4.3 4.48 4.67 4.87 5.06 5.27 5.48 5.69 5.91 6.13 6.36 6.590

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Compression Ratio & C.O.P vs Cooling Capacity

Compression RatioC.O.P


Chart 5.1 : Cooling capacity Vs compression ratio & C.O.P

The chart contain two (2) type of data versus cooling capacity which is the

compression ratio and C.O.P. From the chart, its show that when the cooling

capacity is increased, the compression ratio is decreased and meanwhile, the

C.O.P value will be increased.

83

5.1.3 Power Consumption Vs Cooling Capacity

For the power consumption, the calculation is make for eight (8) hours per

day and calculate for a month (20 days of working period). From the latest electricity

tariff by Tenaga Nasional Berhad (TNB), the rate of charge per kw/h is RM 0.218.


4.3 4.484.67

4.87 5.06 5.27 5.48 5.69 5.91 6.13 6.36 6.59

Rate (RM)150.02

156.43

163169.74

176.67

183.76

191.03

198.5

206.11

213.92

221.91

230.08

Table 5.3 : Data for cooling capacity vs rate

2 3 4 5 6 7 8140150160170180190200210220230240

Power Consumption vs Cooling Capacity


Rate (RM)

Graph 5.2 : Cooling capacity Vs rate

The graph show that the power consumption will be increased when the

cooling cacacity increased.

From the calculation and comparison have been done, the optimum evaporating

temperature for Malaysia conditions is about 8 C to 9 C. Its because from the criteria⁰ ⁰

for compression ratio and C.O.P, its about the ideal value compare the other tested

temperature.

After design all the required component for the system, then by using the formula

and calculation that have been used in Microsoft Excel, the next step is :

Research the effect of condensing temperature on condenser sizing.

Research the effect of evaporating temperature on evaporator sizing.

84

Research the effect of condensing temperature and evaporating temperature

on capillary tube length.

5.2 Effect of Condensing Temperature On Condenser Sizing

5.2.1 Temperature Vs Depth

Temperature (⁰C) 35 37 39 41 43 45 47 49

Depth (mm)81.2 72.1

68.5 66.7

62.9 62.8

62.8 62.7

Table 5.4 : Data for condensing temperature vs depth

34 36 38 40 42 44 46 48 5060

65

70

75

80

85 Tc vs Depth

Condensing Temperature (⁰C)

Wid

th (m

m)

Graph 5.3 : Condensing temperature Vs depth

The graph show that the depth design for the coil is become more smaller

when the condensing temperature is increased.

5.2.2 Temperature Vs Height

Temperature (⁰C) 35 37 39 41 43 45 47 49

Height (mm) 286 286286 286 381

381 381 381

Table 5.5 : Data for condensing temperature vs height

34 36 38 40 42 44 46 48 50280

300

320

340

360

380

400 Tc vs Height

Graph 5.4 : Condensing temperature Vs height

85

The graph show that the height design of the coil is same for the range

temperature 35 C to 41 C and 43 C to 49 C.⁰ ⁰ ⁰ ⁰

5.2.3 Temperature Vs Width

Temperature (⁰C) 35 37 39 41 43 45 47 49

Width (mm) 722 722722 722 542

542 542 542

Table 5.5 : Data for condensing temperature vs width

34 36 38 40 42 44 46 48 50530

560

590

620

650

680

710Tc vs Width


Width (mm)

Graph 5.5 : Condensing temperature Vs width

The graph show that the width design of the coil is same for the range


5.2.4 Temperature Vs Rows of Tube

Temperature (⁰C) 35 37 39 41 43 45 47 49Rows of tube 4 4 4 4 4 4 4 4

Table 5.6 : Data for condensing temperature vs rows of tube

34 36 38 40 42 44 46 48 502

4

6 Tc vs Rows of tube


Rows of tube

Graph 5.6 : Condensing temperature Vs rows of tube

86

The graph show that the number rows of tube is same although the

condensing temperature is increased.

5.2.5 Temperature Vs Tube per Rows

Temperature (⁰C) 35 37 39 41 43 45 47 49Tube per row 6 6 6 6 8 8 8 8Table 5.7 : Data for condensing temperature vs tube per rows

34 36 38 40 42 44 46 48 504

5

6

7

8

9

10 Tc vs Tube per row


Tube per row

Graph 5.7 : Condensing temperature Vs tube per rows

The graph show that the width number of tube per row of the coil is same for

the range temperature 35 C to 41 C and 43 C to 49 C.⁰ ⁰ ⁰ ⁰

5.3 Effect of Evaporating Temperature On Evaporator Sizing

5.3.1 Temperature Vs Depth

Temperature (⁰C) 2 3 4 5 6 7 8 9 10 11 12 13

Depth (mm) 35.6 37.4 39.239

40 41 41 45 46 48.1 50 52

Table 5.8 : Data for evaporating temperature vs depth

1 2 3 4 5 6 7 8 9 10 11 12 13 1434

38

42

46

50Te vs Depth


Wid

th (m

m)

87

Graph 5.8 : Evaporating temperature Vs depth

The graph show that the depth design for the coil is become more bigger

when the evaporating temperature is increased.

5.3.2 Temperature Vs Height

Temperature (⁰C) 2 3 4 5 6 7 8 9 10 11 12 13

Height (mm)318

254

254

254

254

254

254

191

191

191

191

191

Table 5.9 : Data for evaporating temperature vs heigth

1 2 3 4 5 6 7 8 9 10 11 12 13 14180200220240260280300320

Te vs Height

Graph 5.9 : Evaporating temperature Vs heigth

The graph show that the height design of the coil is same for the range


5.3.3 Temperature Vs Width

Temperature (⁰C) 2 3 4 5 6 7 8 9 10 11 12 13

Width (mm)342

428

428

428

428

428

428

571

571

571

571

571

Table 5.10 : Data for evaporating temperature vs width

88

1 2 3 4 5 6 7 8 9 10 11 12 13 14330380430480530580

Te vs Width


Width (mm)

Graph 5.10 : Evaporating temperature Vs width

The graph show that the width design of the coil is same for the range


5.3.4 Temperature Vs Rows of Tube

Temperature (⁰C) 2 3 4 5 6 7 8 9 10 11 12 13Rows of tube 2 2 2 2 2 2 2 2 2 2 2 2

Table 5.11 : Data for evaporating temperature vs rows of tube

1 2 3 4 5 6 7 8 9 10 11 12 13 140

1

2

3

4Te vs Rows of tube


Rows of tube

Graph 5.11 : Evaporating temperature Vs rows of tube

The graph show that the number rows of tube is same although the

evaporating temperature is increased.

5.3.5 Temperature Vs Tube per Rows

Temperature (⁰C) 2 3 4 5 6 7 8 9 10 11 12 13Tube per row 10 8 8 8 8 8 8 6 6 6 6 6

89

Table 5.12 : Data for evaporating temperature vs tube per rows

1 2 3 4 5 6 7 8 9 10 11 12 13 14456789

10

Te vs Tube per row


Tube per row

Graph 5.12 : Evaporating temperature Vs tube per rows

The graph show that the width number of tube per row of the coil is same for

the range temperature 3 C to 8 C and 9 C to 13 C.⁰ ⁰ ⁰ ⁰

From the calculation and comparison have been for that two (2) component, the

design of the coil is related and depend on the setpoint temperature have been set for

the design. When the temperature is different, the depth, number rows of tube and

number of tube per row will be changed. The height and width of the coil design is

change according to the number rows of tube and number of tube per row.

5.4 Effect of Condensing and Evaporating Temperature On Capillary Tube

Length

5.4.1 Condensing Temperature Vs Length

Temperature (⁰C) 35 37 39 41 43 45 47 49

Length (mm) 347 347347 347 348

349 351 353

Table 5.13 : Data for condensing temperature vs length

90

34 36 38 40 42 44 46 48 50340

344

348

352

356

360

Tc vs Length


Length (mm)

Graph 5.13 : Condensing temperature Vs length

The graph show that the length of the capillary tube is same for the

temperature range between 35 C to 41 C and will be increased after that⁰ ⁰

temperature is increased.

5.4.2 Evaporating Temperature Vs Length

Temperature (⁰C)

2 3 4 5 6 7 8 9 10 11 12 13

Depth (mm)419

394

370

349

329

311

294

279

264

251

238

227

Table 5.14 : Data for evaporating temperature vs Length

91

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15200

240

280

320

360

400

440

Te vs Length


Length (mm)

Graph 5.14 : Evaporating temperature Vs Length

The graph show that the length of the capillary tube is decreased when the

evaporating temperature is increased.

So, the length of capillary tube is change varies to the temperature. Basically, the

function of the capillary tube is to regulate the flow of the refrigerant and then to

drop the temperature and pressure that flow to the evaporator.

92

CHAPTER 6

CONCLUSION AND RECOMMENDATIONS

6.1 Conclusion

The design of air conditioning system is involve a many aspect that need to

focused in term to make that system is running properly and efficient. The most

important thing in designing and building the air conditioning system is the four (4)

main component in the system cycle. That is compressor, condenser, evaporator and

capillary tube. If the design of this component is wrong, the system will become a

problem and also risk to the system damage.

The main objective of this project is to analyse the effect of evaporating

temperature on cooling capacity and then to determine the evaporating temperature

for Malaysia conditions. This mean that after designing and calculating by using

Microsoft Excel and Solvay Flour, and the varies evaporating temperature that have

been selected, than compare for each evaporating temperature with cooling capacity.

For the result show that evaporating temperature is increased, the cooling capacity

will also be increased. Its because to managed the system achieve the design setpoint

of evaporating temperature, the cooling capacity must be achieved the criteria to

running correctly.

Then, from the analyse effect of evaporating temperature on cooling capacity,

there we can make a calculation and make a comparison about the compression ratio,

coefficient of performance (C.O.P) and power consumption. Then from that data, we

can conclude which is the optimum evaporating temperature for Malaysia conditions

by look at the margin between the compression ratio ang coefficient of performance

(C.O.P). From this project, the optimum evaporating temperature is between 8 C and⁰

9 C. Its because the margin for the compression ratio and coefficient of performance⁰

(C.O.P) at this temperature is the closest and lower.

93

The other research involve in this project is to analyse the effect of the

evaporating temperature and condensing temperature on the coil design. From the

research, what can be concluded is the design of the coil for condenser and

evaporator is depend on the evaporating and condensing temperature. The

temperature is effect directly on the sizing depth of the coil, number rows of tube and

number of tube per row. Then from the design of number rows of tube and number of

tube per row, it will determine the result of the design sizing for the height and width

of the coil.

For the capillary tube, there also have been make a research to see the effect

of the evaporating and condensing temperature on the length of capillary tube. So

from the result, it can be conclude that when the condensing temperature increased,

the length of capillary tube is increased. Its become more longer because to achieve

the indoor setpoint, its require more length for capillary tube to drop the temperature

and pressure. For the evaporating temperature, the length is decreased because when

the evaporating temperature is increased, the capillary tube is required less operation

and not need to drop the pressure and temperature to the lower value.

What can be concluded in design and select the compressor is, the main

factor involved in this section is the type of refrigerant use, the evaporating and

condensing temperature and the value of the subcool and superheat. For this design,

the refrigerant use is R407c.

That’s all what can be concluded from this project and the last word that can

be said is that to design the air conditioning system, there are a lot of factor must be

review in designing the component.

6.2 Recommendations

After finished doing this final year project, there is some recommendations

that can be make and must be review for the future benefit. For the first thing I want

to recommend is about the knowledge to finish the project. For your information,

there are a lot of difficulties I have challenged in the period to finish this project. The

lack of knowledge is one of the factor. During the learning process in the class, I’m

recommend to the department of Air Conditioning and Refrigeration to make a

discussion about add the topic into the syllabus for student.

94

The other recommendations is about the development of the software. The

industry and responsible department should build and create a one software about to

design a complete air conditioning system. That involve design and select the

compressor, to design evaporator and condenser and design the capillary tube or

expansion valve.

For this project, what I hope in the future is that my project can help and be

the reference to the other student who want to design the air conditioning system.

Although my project is not perfect, but hopefully this can guide student in designing

their own air conditioning system.

95

REFERENCES

1. Continuing Education and Development, Inc., (2009), “Description of Cooling

Systems,” Selection Tips for Air Conditioning Systems, page 2 – 4.

2. Dr. Sam C M Hui, (2007), “Basic Concepts,” Brief Notes on Air Conditioning

System Design, page 1.

3. Michelle Addington, (2010), “Split-System Air Conditioners,” Yale University

Design Standards, page 25 - 26.

4. Carlyle Compressor Company, (2000), “Application Guide,” Millenium Scroll

Compressor Catalogue, page 1 - 21.

5. Emerson Climate Technologies, (2003), “Scroll Compressor for Air-

Conditioning,” Copeland Scroll, page 1 - 22. .

6. Gartner and Harrison, (2010), “Lowering Condensing Pressure,” Gartner –

Refrigeration & MFG. Inc, page 1 - 4.

7. SPX Brand, (2009), “Evaporators Type Selection,” Evaporator Handbook,

page 4 - 33.

8. IIT Kharagpur, (2008), “Lesson 19 – Perfomance of Reciproating Compressor,”

Refrigeration and Air Conditioning, Ver.6, page 2 - 24.

9. Department of Standards Malaysia, (2007), “ACMV system equipment,

electrically operated, cooling mode & system cooling equipment/component,

heat-operated,” Malaysia Standard (MS1525), page 35 - 38.

10. “Temperature, Density, Specific Heat, Thermal Conductivity, Expansion

Coefficient, Kinematic Viscosity and Prandtl's Number for Temperatures

Ranging 150 - 400 C,” in ⁰ Air Properties, from

http://www.engineeringtoolbox.com.

11. Faye C. McQuiston, Jerald D.Parker and Jeffrey D.Spitler, (2005) “Chapter 10 –

Flows, Pumps, and Piping Design,” Heating, Ventilating and Air Conditioning

Analysis and Design, Vol.6, page 299 – 307.

12. Faye C.McQuiston, Jerald D.Parker and Jeffrey D.Spitler, (2005), “Chapter 14 –

Extended Surface Heat Exchangers,” Heating, Ventilating and Air Conditioning

Analysis and Design, Vol.6, page 482 – 513.

13. W.F.Stoecker and J.W.Jones, (1982), “Chapter 13 – Expansion Devices,”

Refrigeration and Air Conditioning, Vol.2, page 264 – 269.

96

design and simulation of air cooledconditioner

Documents

fixed angular

tco tci thi

mass flow

volume flow

air face velocity

wet bulb temperature

optimum evaporating

clearance