Dasar Teori Pendingin

Embed Size (px)

Citation preview

  • 8/10/2019 Dasar Teori Pendingin

    1/64

    Dasar Teori Pendingin

    Beban pendingin:

    1. Sensible beban panas yang dipengaruhi oleh perbedaan suhu yang lewat konstruksi

    bangunan, electronic appliances, lampu, faktor manusia, dll.2. Latent beban yang dipengaruhi karena kelembapan udara

    A.Prinsip kerja mesin pendingin udara:

    The machine has 3 main parts: compressor, condenser and evaporator. Compressor and

    condenser are outdoor units, while evaporator (sometimes as part of a furnace for

    heating) is indoor unit.

    The working fluid arrives at the compressor as a cool, low-pressure gas. The compressor

    squeezes the fluid. This packs the molecule of the fluid closer together. The closer the

    molecules are together, the higher its energy and its temperature.

    The working fluid leaves the compressor as a hot, high pressure gas and flows into the

    condenser. In the condensing coil, the refrigerant releases its heat. A fan draws draws

    outdoor air in trough louvers surrounding the outdoor cabinet and blows air across the

    hot coil. As the air blows across the coil, it transfers heat to the outside air and

    cools the refrigerant inside the coil. If you looked at the air conditioner part outside

    a house, look for the part that has metal fins all around. The fins act just like a

    radiator in a car and it helps the heat go away, or dissipate, more quickly.

  • 8/10/2019 Dasar Teori Pendingin

    2/64

    When the working fluid leaves the condenser, its temperature is much cooler and it has

    changed from a gas to a liquid under high pressure. The refrigerant then flows indoors

    through tubing and passes through a small opening in the expansion valve. In this valve,

    the refrigerant expands and the liquid's pressure drops and becomes a low-temperature,low-pressure liquid that flows into the evaporator coil where it evaporates into a gas

    by absorbing and extracting heat from the return air blown over the indoor coil by a

    fan. The evaporator also has metal fins to help in exchange the thermal energy with the

    surrounding air. At the same time, moisture in the air is removed as it condenses on the

    indoor coil. Humidity removed from the air, as it makes contact with the indoor coil, is

    collected in a pan at the bottom of the coil and flows to a drain.

    By the time the working fluid leaves the evaporator, it is a cool, low pressure gas. It

    then returns to the compressor to begin its trip all over again.

    Connected to the evaporator is a fan that circulates the air inside the house to blow

    across the evaporator fins. Hot air is lighter than cold air, so the hot air in the room

    rises to the top of a room. There is a vent there where air is sucked into the air

    conditioner and goes down ducts. The hot air is used to cool the gas in the evaporator.

    As the heat is removed from the air, the air is cooled. It is then blown into the house

    through other ducts usually at the floor level.

    This continues over and over and over until the room reaches the temperature you want

    the room cooled to. The thermostat senses that the temperature has reached the right

    setting and turns off the air conditioner. As the room warms up, the thermostat turns

    the air conditioner back on until the room reaches the temperature.

    The liquid line is a small pipe with pencil size. The suction line is a larger pipe with

    broom-handle size and it should be insulated by being covered with black foam-rubber

    insulation type.

    B.Jenis-jenis pendingin ruangan

    1.AC Split

    - Komponen AC dibagi 2 : indoor (filter udara, evaporator, evaporator blower,

    expansion valve dan control unit) dan outdoor (compressor, condenser, condenser

    blower dan refrigerant filter).

    - Indoor dan outdoor dihubungkan oleh 2 saluran refrigerant 1 untuk menghubungkan

    evaporator dan compressor, 1 untuk menghubungkan refrigerant filter dan expansion

    valve.

    - Ada juga kabel power untuk memasok listrik ke compressor dan condenser blower.

    - Cocok untuk ruangan yang butuh ketenangan seperti bedroom, library or study room.

  • 8/10/2019 Dasar Teori Pendingin

    3/64

    - Advantages:

    o Bisa dipasang di ruangang yang tidah berhubungan dengan udara luar.

    o Suara di dalam ruangan tidak berisik.

    - Disadvantages:

    o Assembly and disassembly membutuhkan tenaga terlatih.o Maintenance membutuhkan peralatan khusus dan tenaga terlatih.

    o Harga lebih mahal.

    2.AC Window

    - Semua komponen AC (filter udara, evaporator, evaporator blower, condenser,

    condenser blower, compressor, expansion valve, refrigerant filter dan control unit)

    terpasang pada satu base plate dan dimasukkan ke dalam kotak plat sehingga menjadi

    satu kesatuan unit yang compact.

    - Biasa dipilih karena space limitation seperti di rumah susun.

    - Advantages:

    o Assembly dan disassembly lebih mudah.

    o Maintenance lebih mudah.

    o Lebih murah harganya.

    - Disadvantages:

    o Karena semua komponen terpasang pada base plate yang posisinya dekat dengan

    ruangan yang didinginkan, akan timbul suara berisik terutama dari compressor.

    o Tidak semua ruangan dapat menggunakan AC window karena harus dipasang dengan

    condenser menghadap ke open space supaya udara panas dapat terbuang.

    3.AC Central

  • 8/10/2019 Dasar Teori Pendingin

    4/64

    - Udara dari ruangan didinginkan pada cooling plant di luar ruangan tersebut,

    kemudian udara dingin dialirkan kembali ke dalam ruangan tersebut,

    - Cocok untuk gedung bertingkat berlantai banyak sepeti hotel atau mal.

    - Advantages:

    o Suara dalam ruangan tidak berisik sama sekali.o Estetika ruangan bagus karena tidak ada unit indoor.

    - Disadvantages:

    o Planning, installation, operation dan maintenance membutuhkan skilled workers.

    o Apabila ada kerusakan pada saat beroperasi, dampaknya akan terasa ke seluruh

    ruangan.

    o Pengaturan temperatur udara hanya dapat dilakukan pada centrak cooling plant.

    o High investment cost, as well as operation and maintenance costs.

    4.AC Standing Cocok untuk situational and mobile occasions seperti pengajian atau seminar outdoor

    karena unitnya yang movable.

    C.Komponen utama sistem pendingin

    1.Compressor

    Berfungsi untuk mengalirkan refrigerant ke seluruh sistem pendingin. Sistem kerjanya

    dengan mengubah tekanan sehingga perbedaan tekanan memungkinkan refrigerant mengalir

    dari area bertekanan tinggi ke area bertekanan rendah.

    Low T, low P refrigerant dari evaporator di-compress sehingga T dan P naik dan ditekan

    keluar dari compressor menuju ke condenser.Tinggi rendahnya suhu diatur dengan

    thermostat.

    Jenis-jenis compressor:

    Reciprocating compressor (compressor torak)

    Pada saat langkah hisap piston, gas refrigerant yang bertekanan rendah ditarik

    masuk melalui katup hisap yang terletak pada piston atau di kepala compressor. Pada

    saat langkah buang, piston menekan refrigerant dan mendorongnya keluar melalui

    katup buang yang terletak pada kepala silinder.

  • 8/10/2019 Dasar Teori Pendingin

    5/64

    Rotary compressor

    Rotor terdiri dari 2 baling-baling. Langkah hisap terjadi saat katup pemasukan

    mulai terbuka dan berakhir saat katup pemasukan tertutup. Pada saat katup pemasukantertutup, dimulai langkah tekan sampai katup pengeluaran membuka, sedangkan pada

    katup pemasukan secara bersamaan sudah terjadi langkah hisap dan seterusnya.

    Blade compressor (kompresor sudu)

    Biasa digunakan untuk lemari es,

    freezer dan pengkondisian rumah tangga.

    Juga untuk kompresor pembantu pada

    bagian tekanan rendah sistem kompresi

    bertingkat besar.

    Tips:

    Compressor is the most costly part of the air conditioning system to replace. Its

    projected life is about 8 to 10 years. After the system has been turned on, the

    compressor should start up smoothly. A straining, grunting, groaning or squealing noise

    indicates problem condition that should be checked and investigated. It should operate

    continuously without any noise (except a low hum) or squeaks. The compressor should also

    not operate in short cycles (on and off repeatedly).

  • 8/10/2019 Dasar Teori Pendingin

    6/64

    2.Condenser

    Berfungsi untuk membuang kalor yang diserap dari evaporator dan panas yang diperoleh

    dari compressor dan mengubah wujud gas menjadi cair.

    Jenis-jenis condenser:

    Air-cooled condenser

    Kalor dipindahkan dari refrigerant ke udara dengan menggunakan natural or forced

    circulation. Condenser dibuat dari pipa baja, tembaga dengan diberi sirip untuk

    memperbaiki transfer kalor pada sisi udara. Refrigerant mengalir di dalamnya dan

    udara megalir di luarnya. Digunakan untuk kapasitas kecil seperti refrigerator dan

    small water cooler.

    Water-cooled condenser

    Ada 3 jenis:

    Shell and tube

    Sekumpulan pipa (tubes) dipasangkan di dalam shell (pipa galvanis) yang

    berbentuk silinder di mana kedua jenis fluida saling bertukar kalor yang

    mengalir secara terpisah (air dan freon). Air harus mengalir dengan

    kecepatan tinggi untuk mendapatkan high heat transfer.

    Shell and coil

    Terdiri dari sebuah cangkang yang dilas elektrik dan berisi coil air,

    kadang-kadang juga denga pipa bersirip.

    Advantage:low manufacturing cost.

    Disadvantage: difficult to service in the field. If a leak develops in the

    coil, the shell head must be removed and the entire coil pulled from the

    coil to find and repair the leak. Nuisance to clean

  • 8/10/2019 Dasar Teori Pendingin

    7/64

    Double tube

    Biasa digunakan dalam hubungan dengan cooling tower dan spray pond. Water

    passes through the inner tube along with the exterior air condenses the

    refrigerant in the outer tube creates double cooling that improves

    efficiency of the condenser. Has counterflow design.

    Advantage: easy to clean.

    Evaporative-cooled condenser

    Refrigerant melepaskan kalor ke air, lalu air melepaskan kalor ke udara dalam

    bentuk uap air. Udara meniggalkan uap air dengan kelembapan tinggi seperti dalam

    coling tower. Sistem ini menggabungkan fungsi condenser dan cooling tower. Banyak

    digunakan di pabrik amoniak.

  • 8/10/2019 Dasar Teori Pendingin

    8/64

    Recommendation:

    Water-cooled condensers are generally not used for residential structures because they

    are quite wasteful and costly to operate. If the central air-conditioning system

    operates for twelve hours a day, this type of system can waste several thousand gallonsof water a day. If you find this type of condenser in the air-conditioning system, you

    should consider its replacement.

    In large air-conditioning systems such as those found in apartment buildings, water-

    cooled condensers are not wasteful or costly to operate because the cooling water is

    recirculated. After the water absorbs heat in the condenser, it is pumped up to a

    cooling tower, often located on the roof, where it loses its heat and is then

    recirculated to the condenser.

    Diagnosis:

    When the air-conditioning system is turned on, the fan associated with the condenser

    should begin to operate at the same time as the compressor. Check if the fan is turning

    or not by placing your hand over the unit to feel air rushing over your hand. After the

    system has been operational for about fifteen minutes, the air being discharged through

    the condensing coil should be warm since it removes heat that has been generated during

    the compression of the refrigerant. Air that is not warm is usually an indication that

    the compressor is not operating properly.

    After the system has been operational for about fifteen minutes, look at the low-

    pressure refrigerant line (the broom-handle size pipe) that is covered with insulation.

    If a section of the pipe is exposed, grab it with your hand. If the compressor is

    working properly and there is an adequate refrigerant charge, the pipe will be quite

    cool to the touch. On many occasions, the pipe and end fittings will be "sweating" as a

    result of condensation. This is a normal operating condition. However, a low-pressure

    line covered with frost usually indicates a deficiency in refrigerant. Even though

    cooling can be obtained with an air-conditioning system that is deficient in

    refrigerant, the efficiency of the system is greatly reduced.

    Occasionally you will find a sight glass on the small-diameter refrigerant line. It is a

    small device installed directly into the high-pressure liquid refrigerant line that

    allows the homeowner to see whether there is a problem with the flow of the liquid

    refrigerant. The refrigerant is colourless, and if the system is operating properly,

    nothing unusual will be noted when looking at the sight glass. However, when the system

    is low in refrigerant, bubbles will show up in the liquid as it passes under the glass.

  • 8/10/2019 Dasar Teori Pendingin

    9/64

    The location of the compressor-condenser is important for efficient operation. The

    compressor should be located where it will receive a minimum of direct sunlight, since

    the cooler the air flowing across the condenser, the more efficient the cycle. The

    compressor should be positioned so that the condenser air intake is at least 12 inches

    away from any obstruction or dense shrubbery. If not, there will not be an adequateairflow for condenser cooling.

    Airflow through unit must not be obstructed

    The compressor-condenser can be a noisy piece of equipment and should be vibration-

    mounted on a concrete slab or precast concrete blocks that will not settle. The unit

    should be level. Excessive uneven settlement can cause fractures in refrigerant-line

    fittings and thereby allow the refrigerant to escape.

    Look for an electrical disconnect switch on the exterior wall near the compressor. The

    purpose of this switch is to allow the maintenance man to disconnect the unit so that if

    someone in the house unknowingly turns the thermostat down, the unit will not be

    activated while he is making repairs.

    The overall compressor-condenser unit should be checked to see whether it is in need of

    a cleaning. These units require periodic cleaning because leaves, seed pods, twigs, and

    dust tend to clog the condenser, thus restricting the airflow. The condenser has fins

    like an automobile radiator's, which can clog easily.

    3.Katup ekspansi (expansion valve)

    Dipergunakan untuk menurunkan tekanan dan mengekspansi secara adiabatik (no heat

    transfer - occurs because of ) cairan dengan P dan T tinggi sampai mencapai

    tingkat P dan T rendah.Dalam hal ini, refrigerant ekspansi dari tekanan kondensasi ke

    tekanan evaporasi. Refrigerant cair diinjeksikan melalui orifice dan segera berubah

    menjadi kabut dengan P dan T rendah.

  • 8/10/2019 Dasar Teori Pendingin

    10/64

    Expansion valve juga berfungsi sebagai alat kontrol refrigerasi yang berfungsi:

    Mengatur jumlah refrigerant yang mengalir dari pipa cair menuju evaporator sesuai

    dengan laju penguapan pada evaporator.

    Mempertahankan perbedaan tekanan antara condenser dan evaporator agar penguapan

    pada evaporator berlangsung pada tekanan kerjanya.

    Pipa kapiler

    - Merupakan salah satu alat ekspansi dengan 2 kegunaan menurunkan tekanan refrigerant

    cair dan mengatur aliran refrigerant ke evaporator.

    - Refrigerant cair masuk ke kapiler dan mengalir sehingga P berkurang akibat gesekan dan

    percepatan refrigerant.

    - Pipa kapiler melayani semua sistem refrigerasi berukuran kecil (hingga kapasitas

    refrigerasi 10 kW).

    -

    Mempunyai ukuran panjang 1-6 meter dengan diameter 0.5-2 mm.- Ukuran pipa tergantung dari: kapasitas pendinginan, kondisi operasi dan jumlah

    refrigerant dari mesin refrigerasi yang bersangkutan.

    - Pada waktu compressor berhenti bekerja, pipa kapiler menghubungkan bagian tekanan

    tinggi dengan tekanan rendah sehingga menyamakan tekanannya dan memudahkan start

    berikutnya.

    4.Evaporator (penguap)

    Berfungsi sebagai penukar kalor dan menguapkan refrigerant dalam sistem sebelum dihisap

    oleh compressor. Panas udara sekeliling dihisap oleh evaporator sehingga T udara

    sekeliling turun. Suhu udara rendah ini dipindahkan ke tempat lain dengan dihembus oleh

    kipas sehingga terjadi aliran udara.

    Jenis-jenis evaporator sesuai dengan refrigerant yang ada di dalamnya:

    Jenis ekspansi kering

    Refrigerant yang diekspansikan melalui katup ekspansi pada waktu masuk ke

    evaporator sudah dalam keadaan campuran cair dan uap, sehingga keluar dari

    evaporator dalam keadaan uap air.

  • 8/10/2019 Dasar Teori Pendingin

    11/64

    Jenis setengah basah

    Kondisi refrigerant di antara evaporator jenis kering dan basah. Dalam evaporator

    ini selalu terdapat refrigerant cair dalam pipa penguapnya.

    Jenis basahSebagian besar dari evaporator terisi oleh cairan refrigerant.

    Perpindahan kalor di dalam evaporator

    Heat transfer within evaporator occurs is the sum of forced convection happening inside

    and outside of tubes and conduction across the tubes. Combined heat transfer coefficient

    can be determined by first calculating the heat transfer coefficient of the refrigerant

    side and the air side. The total heat transfer coefficient is then calculated based on

    the outer and inner surface areas of the pipes.

    Diagnosis:

    The evaporator will usually be located in the attic or the basement. When the unit is

    located in the attic, you will often find the refrigerant lines from the compressor

    running up along the outside of the structure and entering the building at the attic

    level. When the evaporator coil is located in the basement, either as a separate unit or

    in the furnace plenum, the refrigerant lines from the compressor are short and run

    directly into the structure. If possible, the evaporator coil should be observed after

    the unit has been operational for about thirty minutes. The coil might not always be

    accessible because it may be covered with a sheet-metal casing that cannot be easily

    disassembled.

    If the evaporator is accessible, look at the coil and the associated refrigerant tubing.

    The evaporator is used to cool and dehumidify the circulating air. If you notice a

    frosting condition (a build-up of ice) on portions of the coil and refrigerant tubing

    rather than dripping water, the system is not operating properly. The frosting is

    usually the result of an insufficient airflow through the evaporator coil or an

    inadequate amount of refrigerant in the system.

    When the house is heated by forced warm air, the most common location for the evaporator

    is in the furnace plenum hence called furnace-mounted evaporator. A furnace-mounted unit

    takes advantage of the ducts that have been installed for heating the house and also

    uses the heating-system blower to circulate the cool air. You can tell if the evaporator

    coil is located inside the furnace plenum by whether there are refrigerant lines

    entering the sheet-metal casing of the plenum. The refrigerant lines inside the casing

    are connected directly to the evaporator coil. The most common type of coil that is

    found in a furnace plenum is a two-section design, an A-coil (because of its shape).

    Sometimes an inclined or horizontal coil is used.

  • 8/10/2019 Dasar Teori Pendingin

    12/64

    Evaporator (cooling) coil mounted in furnace plenum.

    Below the evaporator coil is a pan that collects the water condensing out of the

    circulating air. The water is then removed by means of a plastic drain line that will be

    visible when looking at the furnace plenum. Depending on the location of the furnace,

    the condensate drain line will run to a nearby sink where the condensate drips down the

    drain or will run through the foundation wall, where the condensate drips on the

    outside. When the condensate drain line is extended through the foundation wall, there

    should be a splash plate below the end of the pipe so that the dripping water can be

    directed away from the foundation.

    Sometimes the condensate drain line runs from the furnace down to a small hole in the

    floor slab. The condensate trickling out of the drain line accumulates below the slab.

    This method of removing the condensate is not very desirable in those areas where the

    water will not readily drain because of high water table or high clay content in the

    soil. Even though the amount of water discharging from the condensate drain is small,

    the introduction of additional water could aggravate a condition that makes the lower

    level of the structure vulnerable to water seepage.

    The condensate drain line should terminate at the plenum with a U-shaped trap. Since the

    condensate drain line is an open pipe leading directly into the cooling coil, the trap

    (which is partially filled with water) prevents any of the cool air from escaping

    through the pipe. Look for a trap.

    Occasionally the condensate drain discharges into a small rectangular box located near

    the furnace. This box is the reservoir for a lift pump. The purpose of the pump is to

    lift the condensate to a level where it can then flow to any desired location. Without a

    pump, it is often necessary to position the drain line so that it blocks a portion of

    the room or interferes with foot traffic. Check the pump's operation. These pumps have a

  • 8/10/2019 Dasar Teori Pendingin

    13/64

    float control that is activated when the water reaches a pre-set level. If there is only

    a small amount of water in the reservoir, the pump can be checked by pouring water from

    a glass into the reservoir.

    Now look at the overall furnace plenum around the evaporator coil. Rust and mineraldeposits indicate a past or present problem in condensate removal. Water overflowing the

    condensate drain pan can damage the heat exchanger below. If you see this condition, you

    should have the furnace heat exchanger checked by a heating contractor for signs of

    deterioration.

    When the evaporator coil is located inside the furnace plenum, the blower for the

    heating system is also used as the blower for circulating the cool air. Because cool air

    is heavier than warm air, when the blower is used for air-conditioning, it should

    operate at a higher speed. Most often, however, the furnace is equipped with a one-speedmotor. Consequently, the air-conditioning system is often not as effective as it might

    be. A pair of double-or triple-sheaved hubs can be installed to allow multispeed

    operation. When the blower is turned off by the master switch, check the tension in the

    fan belt. There should be no excessive slack. Press the belt midway between the pulleys.

    If the belt gives more than 34 inch to an inch, it is too loose, and adjustment is

    needed. When the blower is operating, listen for any unusual noises or vibrations. You

    might also ask the owner when the unit was last servicedthere is no substitute for

    periodic maintenance.

    5.Blower Coil

    When the evaporator coil is housed in a separate casing that contains a blower for

    circulating the cool air, the coil is commonly referred to as a blower coil. Most often,

    the blower coil is located in the attic. However, it can be located in a closet or in

    the basement. The blower coil should be vibration-mounted to prevent the noise of the

    blower unit from being transmitted into the living area. Vibration mounting can be

    achieved by placing the unit on rubber, cork, or styrofoam pads. The vibrations might

    also be isolated in the attic by suspending the unit from the roof rafters.

    The base of the blower coil is basically a condensate collection pan. The accumulated

    condensate is removed by means of a drain line that will extend through the exterior

    wall, terminating on the outside, or extend through the lower portion of the roof,

    terminating in the gutter. Sometimes the condensate drain line terminates in the

    plumbing vent stack located in the attic. In many communities, this type of termination

    is not permitted because it is not in compliance with the plumbing code.

  • 8/10/2019 Dasar Teori Pendingin

    14/64

    Attic-mounted blower coil. Unit is resting on a styrofoam pad to minimize vibrations.

    Below the unit is an auxiliary condensate drain pan and associated drain line. Note

    that the main condensate drain line does not have a U-shaped trap.

    The purpose of the vent stack is to channel sewer gases in the plumbing system to the

    outside. If the condensate drain line is connected to the vent stack and there is no

    trap on the drain line, the sewer gases may back up into the condensate drain line,

    enter the blower coil, and be circulated throughout the house. The condensate drain line

    should have a U-shaped trap near its connection to the blower-coil housing. On many

    installations, this trap is omitted. Look for it. If it is missing, one should be

    installed.

    Air-conditioning condensate drain line terminating in plumbing vent stack. In many

    communities, this type of termination does not comply with the plumbing code.

    When the blower coil is located in the attic, certain steps must be taken to prevent

    cosmetic damage to the ceiling below in the event of a blockage in the main condensate

  • 8/10/2019 Dasar Teori Pendingin

    15/64

    drain line. Some blower-coil housings have a fitting for an auxiliary drain line that is

    located just above the main condensate drain fitting. If the main drain becomes clogged,

    the level of the condensate will rise and be drawn off by the auxiliary drain.

    For those blower coils that do not have a fitting in the housing for an auxiliary drainline, there should be an auxiliary drain pan below the unit. The auxiliary pan will

    collect any condensate that overflows from the main pan when there is a blockage in the

    main drain line. Look for an auxiliary drain pan. In some parts of the country auxiliary

    drain pans are installed when the blower coil is located over any furred space, even

    when the blower housing has an auxiliary drain fitting. Unfortunately, many air-

    conditioning contractors do not install the auxiliary drain or drain pan. Because rising

    costs make it difficult to remain competitive, they cut costs wherever they can.

    The auxiliary drain pan must have a separate drain line that discharges to the outside.It should not be connected to the main drain line. If it is, it reflects poor-quality

    workmanship; if the main drain line becomes clogged near the discharge end, the

    auxiliary drain line will also not function.

    Auxiliary condensate drain line connected to main condensate line. This negates the use

    of an auxiliary drain and reflects poor-quality workmanship. If the main drain becomes

    clogged near the discharge end, the auxiliary drain will not function. Note that the U-

    shaped trap is missing.

    Recommendation:

    If the evaporator coil is accessible, it should be inspected for frost build-up. From an

    efficiency point of view, the attic is the least desirable area for locating the blower

    coil because of the high temperatures, easily reaching 140 F to 150 F that normally

    occur during the summer. Even though the blower coil is insulated, there will be a heat

    gain because of this high temperature. The overall attic temperature, however, can be

    lowered by increasing the number or size of the attic vent openings. A ridge vent isquite effective, as is a thermostatically controlled power ventilator.

  • 8/10/2019 Dasar Teori Pendingin

    16/64

    General recommendation:

    a. After the air-conditioning system has been operational for about fifteen minutes,

    the temperature of the air discharging from the supply registers should be about 15

    degrees lower than the temperature in the room. If the air does not have a slightchill, it might be because there is a heat gain along the duct leading to that

    register as a result of inadequate insulation, or the system may be undersized or

    low in refrigerant.

    b. While checking the temperature of the air leaving the supply registers, also check

    the airflow. If the air discharging from the registers has a low flow and appears to

    be sluggish, it might indicate that there is an obstruction within the system caused

    by dirty filters or icing on the evaporator coils. Sometimes the condition is caused

    by an undersized fan or the need for balancing the airflow between the registers.

    c.The location of the supply registers is important for effective air-conditioning.Being heavier than warm air, the cool air will tend to accumulate near the lower

    portion of the room and the warm air near the top. As a result, there is usually a

    temperature difference between the ceiling and floor. This stratification of heat

    layers can be minimized by adequate circulation in the room. Adequate circulation

    can be achieved by locating the supply registers on the opposite side of the room

    from the return grille. When the return grille is near the supply register, the air

    discharging from the supply is drawn in by the return grille and does not have a

    chance to circulate adequately around the room.

    d. In many houses, the rooms do not have individual return grilles. Instead there might

    be a large central return located in the hall. In these cases, the supply registers

    should be located on a wall that will allow the supply air to circulate completely

    prior to being drawn off and returned to the central grille. Also, the doors to the

    individual rooms must be undercut so that when they are closed, the supply air will

    be able to flow to the return grille.

    e. Ideally, air-conditioning supply registers should be located in or near the ceiling.

    To minimize air-conditioning installation costs, rather than install new ducts many

    new homes use the ducts and registers provided with a forced-air heating system.

    These registers are usually located near or at the floor level and are quite

    effective for heating purposes. However, when they are used for air-conditioning,

    they are less effective and tend to increase the stratification effect. Some houses

    with forced-warm-air heating have what are called high-low registers. The duct

    supplying the heat register is extended vertically to a point near the ceiling level

    where it terminates at another register. When the system is used for heating, the

    damper controlling the upper register is manually closed, and the lower register is

    opened. When the system is used for cooling, the damper controlling the lower

    register is closed, and the upper register is opened. This type of arrangement is

    very desirable and is often found in high-quality construction.

  • 8/10/2019 Dasar Teori Pendingin

    17/64

    6.Ducts

    There are two basic types of ducts used in residential structuressheet metal and glass

    fibre. While checking the distribution portion of the air-conditioning system, look for

    exposed ducts. The glass-fibre type of duct is by its very nature insulated. However,the metal duct may or may not be insulated. The fact that there is exposed metal on the

    outside does not mean that the duct is not insulated. The insulation might be located

    inside the duct. Whether the metal duct is insulated can be determined by feeling the

    duct when the system is operating (if no insulation, the duct will be quite cool) or by

    striking the duct with your fingernail. If there is no insulation, you will hear a

    ringing sound, and if there is insulation, you will hear a dull thud.

    All ducts that lead through unfinished areas such as crawl spaces and attics must be

    insulated so that the cool air flowing through the ducts will not absorb heat from itssurroundings. Pay particular attention to the joints for indications of air leakage. Any

    open joints should be sealed with inexpensive duct tape. Also, whether the evaporator

    coil is located inside the furnace plenum or in the attic, check the joints around the

    housing for air leakage. Very often, there are open joints that must be sealed.

    D.Termodinamika Sistem Refrigerasi

    1.Siklus refrigerasi Carnot

    Merupakan kebalikan dari mesin carnot. Mesin carnot menerima energi kalor dari

    temperature tinggi, yang kemudian energy tersebut dirubah menjadi suatu kerja dan sisa

    energi tersebut dibuang ke sumber panas temperature rendah. Siklus refrigerasi carnot

    menerima energi pada temperature rendah dan mengeluarkan energi pada temperature tinggi.

    Oleh sebab itu, pada siklus pendingin diperlukan penambahan kerja dari luar.

  • 8/10/2019 Dasar Teori Pendingin

    18/64

    Proses-proses yang membentuk Carnot refrigeration cycle:

    Adiabatic compression (1-2)

    Isothermal heat release (2-3)

    Adiabatic expansion (3-4)

    Isothermal heat gain (4-1)

    Tujuan utama: penyerapan kalor dari sumber bersuhu rendah pada proses 4-1 (isothermal

    heat gain). Seluruh proses lainnya dibuat sedemikian rupa sehingga energi bersuhu rendah

    dapat dikeluarkan ke lingkungan yang bersuhu tinggi.

  • 8/10/2019 Dasar Teori Pendingin

    19/64

    2.Siklus kompresi uap standar (teoritis)

    Asumsi yang digunakan:

    a.Proses kompresiKompresi terjadi dari titik 1 ke 2. Pada siklus sederhana, refirigerant diasumsikan

    tidak mengalami perubahan kondisi selama mengalir di jalur hisap. Proses kompresi

    diasumsikan isentropic sehingga pada diagram berada pada satu garis entropi konstan,

    dan titik 2 berada pada kondisi super panas. Proses ini memerlukan kerja dari luar dan

    entalpi uap naik dari h1 ke h2. Besarnya kenaikan ini sama dengan besarnya kerja

    kompresi yang dilakukan uap refrigerant.

    b.Proses kondensasi

    Proses 2-3 ini terjadi di kondensor di mana uap panas refrigerant dari kondensordidinginkan oleh air sampai pada temperature kondensasi, kemudian uap tersebut

    dikondensasikan. Pada titik 2, refrigerant pada kondisi uap jenuh pada tekanan dan

    temperature kondensasi. Proses ini terjadi pada isobar (constant P) dan jumlah panas

    yang dipindahkan selama proses ini adalah beda entalpi antara titik 2 dan 3.

    c.Proses ekspansi

    Proses berlangsung dari titik 3 ke 4. Pada proses ini terjadi penurunan tekanan

    refrigerant dari tekanan kondensasi (di titik 3) menjadi tekanan evaporasi (di titik

    4). Pada waktu cairan di ekspansi melalui katup ekspansi atau pipa kapiler keevaporator, temperature refrigerantjuga turun dari temperature kondensat ke temperatur

  • 8/10/2019 Dasar Teori Pendingin

    20/64

    evaporasi. Ini merupakan proses adiabatic dimana entalpi fluida konstan sepanjang

    proses. Refrigerant pada titik 4 berada pada kondisi campuran-uap.

    d.Proses evaporasi

    Proses 4-1 berlangsung pada evaporator dan dalam tekanan konstan. Pada titik 1 seluruhrefrigerant berada dalam kondisi uap jenuh. Selama proses 4-1, entalpi refrigerant

    naik akibat penyerapan kalori dari ruang refrigerasi. Besarnya kalor yang diserap

    adalah beda entalpi antara titik 1 dan 4 yang biasa disebut dengan efek pendinginan.

    3.Siklus kompresi uap aktual

    Perbedaan antara siklus standard dan actual terjadi karena asumsi yang diterapkan dalam

    siklus standard.

    Pada siklus actual terjadi pemanasan lanjut (superheating) uap refrigerant yang

    meninggalkan evaporator sebelum masuk ke kondensor. Hal ini mungkin terjadi akibat 2

    hal: tipe peralatan ekspansi yang digunakan atau karena penyerapan panas di jalur masuk

    (suction line)antara evaporator dan kompresor,

    Demikian juga pada refrigerant cair mengalami pendinginan lanjut atau bawah dingin (sub-

    cooling) sebelum masuk katup ekspansi atau pipa kapiler.

    Dua keadaan di atas adalah normal bahkan memiliki fungsi yang berguna yaitu memastikan

    agar seluruh refrigerant yang memasuki kompresor atau alat ekspansi dalam keadaan 100%

    uap atau 100% cair.

    Perbedaan yang paling penting antara siklus standard dan aktual adalah pada siklus

    standard, dianggap tidak ada penurunan tekanan pada kondensor dan evaporator sedangkan

    pada siklus aktual terjadi penurunan tekanan karena adanya gesekan antara dinding pipa

    dengan refrigerant. Akibatnya, compressor di antara titik 1 dan 2 memerlukan lebih

    banyak kerja dibandingkan dengan siklus standard. Jadi pada siklus aktual, proses

    kompresi bukan isentropic (tidak ada heat transfer antara refrigerant dan dinding pipa)

    dan juga bukan polytropic.

    E.Klasifikasi Sistem Refrigerasi

    Berdasarkan prinsip kerjanya, sistem refrigerasi terbagi menjadi 3 jenis:

    1.Sistem refrigerasi kompresi uap

    Mengambil keuntungan dari kenyataan bahwa fluida yang bertekanan tinggi pada suhu

    tertentu cenderung menjadi lebih dingin jika dibiarkan mengembang.Jika perubahan tekanan

    cukup tinggi, maka gas yang ditekan akan menjadi lebih panas daripada sumber dingin di

  • 8/10/2019 Dasar Teori Pendingin

    21/64

    luar (contoh: udara luar) dan gas yang mengembang akan menjadi lebih dingin daripada

    suhu dingin yang dikehendaki. Dalam kasus ini, fluida digunakan untuk mendinginkan

    lingkungan bersuhu rendah dan membuang panas ke lingkungan bersuhu tinggi.

    1-2 Cairan refrigerant dalam evaporator menyerap panas dari sekitarnya, biasanya

    udara, air atau cairan proses lain. Selama proses ini cairan merubah bentuknya dari cair

    menjadi gas, dan pada keluaran evaporator gas ini diberi pemanasan berlebih/ superheated

    gas.

    2-3 Uap yang diberi panas berlebih masuk menuju kompresor dimana tekanannya dinaikkan.

    Suhu juga akan meningkat, sebab bagian energi yang menuju proses kompresi dipindahkan ke

    refrigerant.

  • 8/10/2019 Dasar Teori Pendingin

    22/64

    3-4 Superheated gas bertekanan tinggi lewat dari kompresor menuju kondenser. Bagian

    awal proses refrigerasi (3-3a) menurunkan panas superheated gas sebelum gas ini

    dikembalikan menjadi bentuk cairan (3a-3b). Refrigerasi untuk proses ini biasanya

    dicapai dengan menggunakan udara atau air. Penurunan suhu lebih lanjut terjadi pada

    pekerjaan pipa dan penerima cairan (3b - 4), sehingga cairan refrigeran didinginkan ketingkat lebih rendah ketika cairan ini menuju alat ekspansi.

    4-1 Cairan yang sudah didinginkan dan bertekanan tinggi melintas melalui peralatan

    ekspansi, yang mana akan mengurangi tekanan dan mengendalikan aliran menuju Kondenser

    harus mampu membuang panas gabungan yang masuk evaporator dan kondenser.

    Dengan kata lain: (1 - 2) + (2 - 3) harus sama dengan (3 - 4). Melalui alat ekspansi

    tidak terdapat panas yang hilang maupun yang diperoleh.

    2.Sistem refrigerasi absorbsi

    Dalam siklus refrigerasi absorbsi, dipergunakan penyerap untuk menyerap refrigerant yang

    diuapkan di dalam evaporator sehingga menjadi suatu larutan absorbsi. Kemudian, larutan

    absorbsi tersebut dimasukan ke dalam sebuah generator untuk memisahkan refrigerant dari

    larutan absorbsi tersebut dengan cara memanasi, yang sekaligus akan menaikan tekanannya

    sampai mencapai tingkat keadaan mudah diembunkan.

    Dalam sistem ini terdapat 2 siklus siklus refrigerant (A-B-C-D) dan siklus pelarut

    (B-C-E-F).

    Pada titik A, refrigerant dalam keadaan tekanan dan temperatur rendah dan berfase cair.

    Kemudian, di evaporator refrigeran menyerap kalor dari objek yang didinginkan sehingga

    berubah menjadi gas (titik B). Gas refrigerant ini mengalir ke absorber sehingga

    diabsorpsi oleh larutan LiBr, akbatnya larutan kaya akan refrigeran, keadaan ini disebut

    dengan larutan kuat

    (strong solution), kemudian larutan kuat ini dipompakan ke generator

    (titik C). Pada generator, kalor digunakan untuk memisahkan antara refrigerant dan

  • 8/10/2019 Dasar Teori Pendingin

    23/64

    pelarut. Karena titik didih refrigerant lebih rendah dari pada pelarut maka refrigerant

    menguap menuju kondenser. Uap refrigeran ini kemudian terkondensasi pada kondenser

    dengan membuang kalor sehingga fasanya menjadi cair (titik D). Setelah itu refrigerant

    cair ini diekspansi sehingga tekanannya menjadi rendah (titik A). Siklus ini terus

    berlangsung sehingga efek refigerasi (proses A-B) terjadi secara kontinu. Disisi lain,pada siklus pelarut (B-C-E-F), larutan yang miskin akan refrigerant (titik C)

    diekspansikan untuk dialirkan ke absorber sehingga mengabsorb refrigerant menjadi

    larutan kuat. Siklus ini pun berlangsung terus menerus.

    Untuk keperluan pengkondisian udara biasanya digunakan sistem air (H2O) sebagai

    refrigeran dan Larutan Litium Bromida (LiBr) sebagai pelarut. Sedangkan untuk keperluan

    yang membutuhkan temperatur lebih rendah, seperti pembuatan es, digunakan ammonia (NH3)

    sebagai refrigerant dan air(H2O) sebagai pelarut.

    3.Sistem refrigerasi udara

    Pada siklus ini, udara bertindak sebagai refrigerant di mana udara menyerap panas pada

    tekanan konstan P, di dalam refrigerator. Udara panas keluar refrigerator, dikompresi

    untuk dibuang panasnya ke lingkungan melalui cooler pada tekanan konstan P2 (P2 > P1).

    Udara keluar cooler dikembalikan ke keadaan awal oleh mesin ekspansi untuk dapat

    melakukan langkah awal pada siklus berikutnya.

    F.

    Prosedur Pengisian Refrigerant

    Pengisian refrigerant harus sesuai dengan takaran. Kelebihan refrigerant dalam sistem

    dapat menyebabkan temperatur evaporasi yang tinggi akibat tekanan dari refrigerant yang

    tinggi. Selain itu dapat menyebabkan kompresor rusak akibat kerja kompresor yang terlalu

    berat dan adanya kemungkinan liquid suction. Sebaliknya bila jumlah refigeran yang

  • 8/10/2019 Dasar Teori Pendingin

    24/64

    diisikan sedikit, dengan kata lain kurang dari yang ditentukan, maka sistem akan

    mengalami kekurangan pendinginan.

    Ada beberapa cara pengisian refrigerant di antaranya adalah:

    a.Mengisi sistem berdasarkan berat refrigerantb.Mengisi sistem berdasarkan banyaknya bunga es yang terjadi di evaporator

    c.Mengisi sistem berdasarkan temperatur dan tekanan

    CAUTION:

    Wear safety goggles and protective gloves.

    If air is mixed in refrigeration cycle, poor cooling may result and also if moisture

    is mixed in refrigeration cycle, clogging (freezing) or rust may result.

    Before charging the refrigerant, evacuate the system using vacuum pump to remove air

    and moisture in the system. Moisture can be evaporated and removed easily even atnormal temperature, if the system is evacuated using vacuum pump.

  • 8/10/2019 Dasar Teori Pendingin

    25/64

    1) Close all valves of manifold gauge

    2) Install the low-/high-pressure hoses

    to corresponding service ports on

    vehicle.

    CAUTION: Be sure that hoses are securely

    connected.

    3) Connect the centre hose of manifold

    gauge with vacuum pump.

    4) Activate the vacuum pump and open the

    valves on low-/high-pressure sides.

    CAUTION: Be sure to evacuate the system

    using vacuum pump.

    5) After at least 5 mins of evacuation,

    if the low-pressure gauge readingshows 100.0 kPa (750 mmHg, 29.5 inHg)

    or higher, close the valves on centre

    hose to stop the vacuum pump.

    6) Leave it at least 5 to 10 mins after

    closing the valves on low-/high-

    pressure sides, and then check the

    low-pressure gauge reading for any

    changes. When the gauge readingchanges, this is a sign of leakage.

    Check the pipe or hose connector

    points, and repair it if necessary.

    Repeat the procedure from 1) after

    repairing the faulty part.

    7) If there are no leaks, further

    evacuate the system for another 20 to

    30 mins.

  • 8/10/2019 Dasar Teori Pendingin

    26/64

    8) Close all valves and stop the vacuum

    pump.

    9) Following the can tap operation

    manual instructions, install it torefrigerant can.

    10) Disconnect the vacuum pump from

    centre hose and connect the hose to

    tap valve.

    11) When a refrigerant recovery container

    is used, measure the refrigerant

    amount in use using a weighting scale

    before connecting to centre hose.

    12) Open the valve on refrigerant (HFC-

    134a) source.

    13) Loosen the centre hose connection on

    manifold gauge (if applicable, press

    a purge valve on manifold gauge) only

    for a couple of seconds to allow the

    air in the centre hose to escape by

    the refrigerant (the process is

    called purging).

    14)Ensure that high-pressure valve ofmanifold gauge is closed, and then

    open the low-pressure side valve only

    to charge the refrigerant.

    CAUTION: Do not open high-pressure

    valve. Be sure to open the low-pressure

    valve.

    15) Close the low-pressure valve when the

    low-pressure gauge reading reaches

    200 kPa (1500 mmHg, 59.1 inHg).

    16) Using a leak tester, check the system

    for refrigerant leaks

    17) After confirming that there are no

    leaks with the leak test, charge therequired amount of refrigerant.

  • 8/10/2019 Dasar Teori Pendingin

    27/64

    18) If the refrigerant (HFC-134a) is

    empty, close the low-pressure valve

    and then close the valve on can tap

    before replacing the empty source.

    Restart charging operation afterreplacing the refrigerant source with

    a new one and purging.

    19) Close the low-pressure valve if the

    charge rate of refrigerant becomes

    worse.

    20) Confirm that both the low-/high-

    pressure valves are closed. Start theengine with A/C switch OFF.

    21) Quickly repeat A/C switch ON-OFF

    cycles a few times to prevent initial

    compressor damage.

    22) Set up the vehicle to the following

    status:

    A/C switch ON

    Engine running at 1500 rpm

    Blower speed setting to Hi Temp. setting to MAX COOL

    Air inlet setting to RECIRC

    Window open

    23) Open the low-pressure valve and

    charge the specified amount of

    refrigerant.

    24)Close all valves and disconnect thehoses from service port after

    charging the refrigerant.

    25) Install the cap to service port.

    G.Tipe Refrigerant

    R22 (CFC) tidak lagi digunakan karena termasuk Ozone Depleting Substance (ODS). R134a

    sebagai alternatif sudah cukup baik akan tetapi tidak bias menggantukan R-12 secara

    langsung tanpa melakukan modifikasi sistem refrigerasi (drop in substitute), relatif

    mahal, masih berpotensi sebagai ODS dan sangat bergantung kepada pelumas sintetik yang

    sering menyebabkan masalah dengan sifat higroskopis (kemampuan menyerap air baik melalui

    absorbsi atau adsorpsi.

    Alternatif lain adalah refrigerant hydrocarbon seperti propana (R-290), isobutana (R-

    600a), n-butana (R-600). Campuran yang sering digunakan adalah R-290/600a, R-290/600 dan

    R-290/600/600a.

    Keuntungan:

    Ramah lingkungan karena memiliki nilai Ozone Depleting Potential (ODP) 0 dan Global

    Warming Potential (GWP) yang hampir dapat diabaikan.

    Properti termofisika dan karakteristik perpindahan kalor yang baik.

    Kerapatan fase uap yang rendah.

    Kelarutan yang baik dengan pelumas mineral.

    Kerugian:

    Mudah terbakar.

  • 8/10/2019 Dasar Teori Pendingin

    28/64

    Salah satu refrigerant yang dapat digunakan adalah MUSICOOL. Hidrokarbon MUSICOOL (MC)

    mampu menggantikan refrigerant sintetik (CFC, HCFC, HFC) secara langsung tanpa

    penggantian komponen sistem refrigerasi. MC-12 menggantikan R-12, MC-22 menggantikan R-

    22 dan MC-134 menggantikan R-134a. Sifat fisika dan termodinamik hidrokarbon MUSICOOL

    memberikan kinerja sistem refrigerasi yang lebih baik, keawetan umur kompresor dan hematenergy.

    H.Commercial Cooling Load

    The room cooling load

    is the rate at which heat must be removed from the room air to

    maintain it at the design temperature and humidity.

    The amount of heat that must be removed (the cooling load) is not always equal to the

    amount of heat received at a given time (called the instantaneous heat gain). This

    difference is a result of the heat storage and time lag effects. Of the total amount of

    heat entering the building at any instant, only a portion of it heats the room air

    immediately; the other part (the radiation) heats the building mass-the roof, walls,

    floors, and furnishings. This is the heat storage effect. Only at a later time does the

    stored heat portion contribute to heating the room air.

    During the time of day at which the instantaneous heat gain is the highest (the

    afternoon), the cooling load is less than the instantaneous heat gain. This is because

    some of this heat is stored in the building mass and is not heating the room air. Later

    in the day, the stored heat plus some of the new entering heat is released to the roomair, so the cooling load becomes greater than the instantaneous heat gain.

    This effect is noticed in the huge southern European cathedrals built of massive, thick

    stone walls. Even on a sunny, very hot day the church interior remains quite cool,

    though it is not air-conditioned. The entering heat doesn't reach the interior it

    merely heats the walls (heat storage). By the time the heat reaches the interior (time

    lag), night has come. In this extreme example of time delay, the building may even have

    a reverse heat flow at night-heat flows out from the hot walls to the cool outdoors.

    The cooling load calculation procedure that will be explained here is called theCLF/CLTD method since this procedure is relatively easy to understand and use.

  • 8/10/2019 Dasar Teori Pendingin

    29/64

  • 8/10/2019 Dasar Teori Pendingin

    30/64

    The cooling load temperature difference (CLTD) is not the actual temperature difference between the outdoor and indoor air. It

    is a modified value that accounts for the heat storage/time lag effects. Tables below list CLTD values for some wall and roof

    constructions.

  • 8/10/2019 Dasar Teori Pendingin

    31/64

  • 8/10/2019 Dasar Teori Pendingin

    32/64

  • 8/10/2019 Dasar Teori Pendingin

    33/64

  • 8/10/2019 Dasar Teori Pendingin

    34/64

  • 8/10/2019 Dasar Teori Pendingin

    35/64

    The CLTD values in Tables 6.1 and 6.2 above are based on the following conditions:

    1.Indoor temperature is 78oF DB

    2.Outdoor average temperature on the design day is 85oF DB

    3.Date is July 21st

    4.Location is 40oN latitude

    If the actual condition differs from any of the above, the CLTD must be corrected as

    follows:

    () ( )

    Where

    = corrected value of CLTD (oF)

    = temperature from Table 6.1, 6.2, 6.5

    = correction for latitude and month, from Table 6.4

    = room temperature (oF)

    = average outside temperature on a design day (oF)

    The temperature can be found as follows:

    (

    )

    Where

    = outside design dry bulb temperature (oF)

    = daily temperature range (oF)

    Both values of and for various locations in Indonesia can be found in the table

    below.

  • 8/10/2019 Dasar Teori Pendingin

    36/64

    The hours listed in Tables 6.1 and 6.2 are Solar Time which is approximately equal to Standard Time.

  • 8/10/2019 Dasar Teori Pendingin

    37/64

    c) Conduction Through Interior Structure

    The heat that flows from interior unconditioned spaces to the conditioned space through

    partitions, floors, and ceilings can be found from equation:

    Where

    = heat gain (cooling load) through partition, floor or ceiling (BTU/hr)

    = overall heat transfer coefficient for partition, floor or ceiling

    (BTU/hr-ft2-oF)

    = temperature difference between unconditioned and conditioned space (oF)

    If the temperature of the unconditioned space is not known, an approximation often used

    is to assume that it is at 5oF less than the outdoor temperature. Spaces with heat

    sources, such as boiler rooms, may be at a much higher temperature.

    d) Solar Radiation Through Glass

    Radiant energy from the sun passes through transparent materials such as glass and

    becomes a heat gain to the room. Its value varies with time, orientation, shading, and

    storage effect. The solar cooling load can be found from the following equation:

    Where

    = radiation cooling load for glass (BTU/hr)

    = maximum solar heat gain factor (BTU/hr-ft2)

    = area of glass (ft2)= shading coefficient

    = cooling load factor for glass

    The maximum solar heat gain factor (SHGF) is the maximum solar heat gain through single

    clear glass at a given month, orientation, and latitude. Values are shown in Table 6.6

    for the 21stday of each month.

  • 8/10/2019 Dasar Teori Pendingin

    38/64

    The SHGF gives maximum heat gain values only for the type of glass noted and without any

    shading devices. To account for heat gains with different fenestration arrangements, the

    shading coefficient SC is introduced. Table 6.7 lists some values of SC.

    The cooling load factor CLF accounts for the storage of part of the solar heat gain.

    Values of CLF to be applied to the solar load calculation are shown in Tables 6.8, 6.9,and 6.10. Note that there are separate listings for Light (L), Medium (M), and Heavy (H)

  • 8/10/2019 Dasar Teori Pendingin

    39/64

  • 8/10/2019 Dasar Teori Pendingin

    40/64

  • 8/10/2019 Dasar Teori Pendingin

    41/64

  • 8/10/2019 Dasar Teori Pendingin

    42/64

  • 8/10/2019 Dasar Teori Pendingin

    43/64

  • 8/10/2019 Dasar Teori Pendingin

    44/64

    External Shading Effect

    The values for the SHGF shown in Table 6.6 are for directsolar radiation-when the sun

    shines on the glass. External shading from building projections (or other objects) may

    shade all or part of the glass. In these cases, only an indirectradiation reaches the

    glass from the sky and ground. The SHGF values for any shaded glass are the same as theN (north) side of the building, which also receives only indirect radiation.

    In order to find the total radiation through partly shaded glass, the shaded area

    portion must first be found. Table 6.11 can be used to find the shading from overhead

    horizontal projections. The values in the table are the vertical feet of shade for each

    foot of horizontal projection. The following example illustrates the use of Table 6.11.

  • 8/10/2019 Dasar Teori Pendingin

    45/64

    e)

    Design Conditions

    The cooling load calculations are based on inside and outside design conditions. Some

    recommended values for inside design conditions are given in tables below.

    Suhu Netral dan Batas Nyaman Responden Berdasarkan Kelompok Suku

    Kelompok Suku

    Suhu Netral ) Batas Nyaman )

    o

    C) o

    C) o

    C) o

    C) o

    C) o

    C)

    1.Aceh (n=6) 24.3 24.3 23.4 20.5-27.3 20.7-27.9 20.2-26.6

    2.Tapanuli (n=23) 25.9 26.2 24.6 22.5-29.2 22.9-29.4 20.2-28.9

    3.Minang (n=27) 26.9 27.4 25.7 23.7-30.1 24.1-30.6 21.7-29.6

    4.Sumatera yang

    lain (n=16)27.0 27.3 25.9 23.7-30.3 23.9-30.7 21.8-30.1

    5.Betawi (n=23) 27.0 27.3 25.9 23.7-30.3 23.9-30.7 21.8-30.1

    6.Sunda (n=86) 26.4 26.6 25.0 23.9-28.9 23.9-29.3 21.8-28.3

    7.Jawa (n=232) 26.4 26.7 25.2 22.8-29.9 23.2-30.2 21.0-29.4

    8.Indonesia yang

    lain (n=62)26.9 27.4 26.2 22.6-31.2 22.5-32.2 21.3-31.1

    The outdoor design conditions are based on reasonable maximums, using weather records.

    The DB (dry bulb temperature) and coincidentWB (wet bulb temperature) occurring at thesame time should be used as the corresponding design values. The separate DB and WB

  • 8/10/2019 Dasar Teori Pendingin

    46/64

  • 8/10/2019 Dasar Teori Pendingin

    47/64

    g) People

    The heat gain from people comprises 2 parts, sensible heat and the latent heat resulting

    from perspiration. Some of the sensible heat may be absorbed by the heat storage effect,

    but not the latent heat. The equations for cooling loads from from people are

    Where

    ,= sensible and latent heat gains (loads) (BTU/hr)

    ,= sensible and latent heat gains per person (BTU/hr)

    = number of people

    = cooling load factor for people

    The rate of heat gain from people depends on their physical activities. Table 6.13 listsvalues for some typical activities. The rates are suitable for a 75

    oF DB room

    temperature. Values vary slightly at other temperatures, as noted.

  • 8/10/2019 Dasar Teori Pendingin

    48/64

  • 8/10/2019 Dasar Teori Pendingin

    49/64

    The heat storage effect factor CLF applies to the sensible heat gain from people. If the

    air conditioning system is shut down at night, however, no storage should be included,

    and CLF = 1.0. Table 6.14 lists values of CLF for people.

  • 8/10/2019 Dasar Teori Pendingin

    50/64

    h) Equipment and Appliances

    The heat gain from equipment may sometimes be found directly from the manufacturer or

    the nameplate data, with allowance for intermittent use. Some equipment produces both

    sensible and latent heat. Some values of heat output for typical appliances are shown in

    Table 6.15. CLF factors (not shown) apply if the system operates 24 hours.

  • 8/10/2019 Dasar Teori Pendingin

    51/64

    The heat output from motors and the equipment driven by them results from the conversion

    of the electrical energy to heat. The proportion of heat generated that is gained by the

    air-conditioned space depends on whether the motor and driven loads are both in the

    space or only one of them is. Table 6.16 lists heat outputs for each condition.

    For any lighting and equipment that operates on a periodic intermittent basis, the heatgains should be multiplied by the proportion of operating time. However, it is often not

    possible to guarantee predicted operations, so using such factors should be approached

    with caution.

    i) Infiltration

    Infiltration of air through cracks around windows or doors results in both a sensible

    and latent heat gain to the rooms. Procedures and equations for calculating infiltration

    heat gains are explained below.

    Sensible heat gain effect of infiltration air

    Infiltration occurs when outdoor air enters through building openings, due to wind

    pressure. The openings of most concern to us are cracks around window sashes and door

    edges, and open doors. Infiltration air entering a space in hot days would increase the

    room air temperature. Therefore, heat must be lost from the room to overcome this

    effect.

    The amount of sensible heat gained from infiltrating air can be determined from the

    sensible heat equation:

    Where

    = sensible heat gain from infiltration air (cooling loads) (BTU/hr)

    = weight flow rate of outdoor air infiltration (lb/hr)

    = specific heat air (BTU/lb-oF)

    = temperature change between indoor and outdoor air (oF)

    The weight-flow rate of air () in above equation is expressed in lb/hr. However, air

    flow rates in HVAC work are usually measured in () (CFM). If the units are

    converted, using the appropriate specific heat of air, the sensible heat equation is:

  • 8/10/2019 Dasar Teori Pendingin

    52/64

    Where

    = sensible heat gain from infiltration air (cooling loads) (BTU/hr)

    = air infiltration (or ventilation) flow rate (ft3/min)

    = temperature change between indoor and outdoor air (o

    F)

    Latent heat gain effect of infiltration air

    If infiltration air is more humid than the room air, the room air humidity may increase

    to an unacceptable level for comfort. If the room air humidity is to be maintained,

    water vapour must be removed. The removal of this moisture requires heat (latent heat of

    condensation of water). This is expressed by the following equation:

    (

    )

    Where

    = latent heat required for infiltration or ventilation air (BTU/hr)

    = air infiltration (or ventilation) flow rate (ft3/min)

    ,

    = higher (outdoor) and lower (indoor) humidity ratio in grains water/lb dry

    air (gr w/lb d.a.)

    If the change in humidity is acceptable, then the latent heat gain can be neglected. The

    humidity ratios can be read from psychometric chart.

    Finding the infiltration rate

    There are 2 methods used to estimate CFM of infiltration air: the crack methodand the

    air change method.

    Crack Method The crack method assumes that a reasonably accurate estimate of the rate

    of air infiltration per foot of crack opening can be measured or established. Energy

    codes list maximum permissible infiltration rates for new construction or renovation

    upgrading. Table 3.4 lists typical allowable infiltration rates, based on a 25 MPH wind.

    The crack lengths and areas are determined from architectural plans or field

    measurements. Example below illustrates use of the crack method. The quality of

    installation and the maintenance of windows and doors greatly affect the resultant crack

    infiltration. Poorly fitted windows may have up to five times the sash leakage shown in

    Table 3.4.

  • 8/10/2019 Dasar Teori Pendingin

    53/64

    Corner Room Infiltration When the infiltration rate is calculated for a room with two

    adjacent exposed walls (a corner room) with door or window openings on both sides, we

    assume that infiltration air comes through cracks on one side only, since the wind can

    only come from one direction at any given time. The wind changes direction, of course,

    but the infiltration effects cannot be additive, since they occur at different times.

    If the wind comes obliquely (toward the corner), the projected crack lengths for each

    side are less; the overall effect is the same as if the wind came directly from one side

    only (using its actual crack lengths). If there are different types or sizes of openings

    on each side, the side that has the greater CFM should be used for the calculation.

    Door Usage For buildings that have frequent door usage (e.g., department stores), the

    infiltration that results from door opening should be included. The rate of door usage

    (number of people per minute) is first determined, with the advice of the architect or

    owner. Some average infiltration rates are shown in Table 3.5.

    For doors that are left indefinitely open, special means may be used to try to offset

    infiltration. Unit heaters, which can blow warm air directly at the opening, and air

  • 8/10/2019 Dasar Teori Pendingin

    54/64

    curtains, which direct a vertical warm air barrier across the opening, are two such

    methods. However, it is difficult to determine the effect on the building heating load

    of these methods.

    Additional air infiltration may occur through a porous wall. If the walls have

    significant porosity, sealant coatings or other coverings may be applied to them.

    In high-rise buildings, a thermal stack effect may increase infiltration through

    existing cracks. This occurs when the warmer inside air, which rises through the

    building and exits out through cracks on upper stories, is replaced by colder outside

    air entering through cracks on lower floors. Consult the ASHRAE Handbook for more

    information.

    Much publicity has been given to reducing infiltration in existing buildings by use of

    weather-stripping and the sealing of cracks around frames, sillplates, wall

    penetrations, and other openings.

    Air Change Method This procedure for finding the infiltration rate is based on the

    number of air changes per hour (ACH) in a room caused by the infiltration.

    One air change is defined as being equal to the room air volume.

    Determination of the expected number of air changes is based on experience and testing.

    Suggested values range from 0.5 ACH to 1.5 ACH for buildings ranging from "tight" to

    "loose" construction. Using the definition of an air change, equation below can be used

    to find the air infiltration rate in CFM:

    Where

    = air infiltration (or ventilation) flow rate to room (ft3/min)

    = number of air changes per hour for room

    = room volume (ft3)

    Crack Method vs Air Change Method

    Some suggestions in choosing between these 2 methods:

    1.The air change method is used primarily in residential construction heating load

    estimates, but there is no reason why the "rack method cannot be used if reliable data

    are available.

    2.The crack method is generally used in non-residential construction. Reliable data from

    window manufacturers and quality control of installation and maintenance may provide

    good estimates using this method.

    Ventilation Outside Air) Load

    Some outside air is usually brought into non-residential buildings through the

    mechanical ventilation equipment (air handling units) in order to maintain the indoor

    air quality. The outside ventilation air will be an additional part of the building

    cooling load, since the entering air is at the outdoor temperature and humidity.

  • 8/10/2019 Dasar Teori Pendingin

    55/64

  • 8/10/2019 Dasar Teori Pendingin

    56/64

    cooling equipment. Many modern buildings have fixed (sealed) windows and therefore have

    no infiltration loss, except for entrances.

    j)

    Room Cooling Load

    The room cooling load is the sum of each of the cooling load components (roof, walls,

    glass. solar, people, equipment, and infiltration) in the room.

    When calculating cooling loads, a prepared form is useful. The following abbreviations

    will be used for convenience.

    TCL, SCL, LCL = component total, sensible, latent cooling loads

    RTCL, RSCL, RLCL = room total, sensible, latent cooling loads

    BTCL, BSCL, BLCL = building total, sensible, latent cooling loads

    CTCL, SCSL, CLCL = coil total, sensible, latent cooling loads

    k)

    Room Peak Cooling Load

    Because the air conditioning system must be sized to handlepeak loads, we must know how

    to find the room peak cooling load.

    The external heat gain components vary in intensity with time of day and time of year

    because of changing solar radiation as the orientation of the sun changes and because of

    outdoor temperature changes. This results in a change in the total room cooling load.

    Sometimes it is immediately apparent by inspecting the tables at what time the peak load

    occurs, but often calculations are required at a few different times.

    Some general guidelines can be offered to simplify this task. From the CLTD (cooling

    load temperature difference), SHGF (sensible heat gain factor) and CLF (cooling load

    factor) tables we can note the following:

    1. For west-facing glass, maximum load is in mid-summer in the afternoon.

    2. For east-facing glass, maximum solar load is in early or mid-summer in the morning

    3. For south-facing glass, maximum solar load is in the fall or winter in early

    afternoon.

    4. For southwest-facing glass, maximum solar load is in the fall in the afternoon.

    5. For roofs, maximum load is in the summer in the afternoon or evening.

    6. For walls, maximum load is in the summer in the afternoon or evening.

    These generalizations can be used to localize approximate times of room peak loads. For

    instance, we might expect a south-facing room with a very large window area to have a

    peak load in early afternoon in the fall-not in the summer! If the room had a small

    glass area, however, the wall and glass heat conduction might dominate and the peak load

    time would be a summer afternoon. Once the appropriate day and time are located, a few

    calculations will determine the exact time and value of the peak load.

  • 8/10/2019 Dasar Teori Pendingin

    57/64

    l)

    Building Peak Cooling Load

    The building cooling load is the rate at which heat is removed from all air-conditionedrooms in the building at the time the building cooling load is at its peak value.

    If peak cooling loads for each room were added, the total would be greater than the peak

    cooling load required for the whole building, because these peaks do not occur at the

    same time. Therefore, the designer must also determine the time of year and time of day

    at which the building cooling load is at a peak, and then calculate it. A reasoning and

    investigation similar to that carried out in finding room peak loads is used.

  • 8/10/2019 Dasar Teori Pendingin

    58/64

    From our previous discussion and a study of the tables, the following guidelines emerge:

    1.For buildings that are approximately square-shaped in plan with similar construction

    on all four walls, the peak load is usually in late afternoon in summer. This is

    because the outside temperature is highest then, and there is no differential

    influence of solar radiation on one side of a building.

    2.For buildings with a long south or southwest exposure having large glass areas, thepeak load may occur in the fall, around mid-day, because radiation is highest then.

    This case requires careful analysis.

    3.For one-story buildings with very large roof areas, the peak load usually occurs in

    the afternoon in summer.

    Once the peak load time is determined, the total building heat gains can be calculated.

    The search for the time and value of peak room

    Diversity

    On some projects, the actual building peak load may be less than the calculated valuebecause of load diversity. In some buildings, at the time of peak load, usage practice

    may be such that all of the people are not present and some of the lights and equipment

    are not operating. In these cases, a diversity factor or usage factor is sometimes

    estimated and applied to the calculated building peak load in order to reduce it. For

    example, if it is estimated that only 90% of the lighting is actually on at peak load

    time, the calculated lighting load would be multiplied by a factor of 0.90. Choosing

    proper diversity factors requires both experience and judgment about building use

    practices.

    m)

    Cooling Coil Load

    After the building cooling load is determined, the cooling coil load is found.

    The cooling coil load is the rate at which heat must be removed by the air conditioning

    equipment cooling coil(s).

    The cooling coil load will be greater than the building load because there are heat

    gains to the air conditioning system itself. These gains may include:

    1.Ventilation (outside air)

    2.Heat gains to ducts

    3.Heat produced by the air conditioning system fans and pumps4.Air leakage from ducts

    n) Ventilation

    Some outside air is generally brought into a building for health and comfort reasons.

    The sensible and latent heat of this air is usually greater than that of the room air,

    so it becomes part of the cooling load. The excess heat is usually removed in the

    cooling equipment, however, so it is part of the cooling coil load but not the building

    load.

  • 8/10/2019 Dasar Teori Pendingin

    59/64

    The equations for determining the sensible and latent cooling loads from ventilation air

    are

    (

    )

    Where

    = sensible heat gain from infiltration air (cooling loads) (BTU/hr)

    = air infiltration (or ventilation) flow rate (ft3/min)

    = temperature change between indoor and outdoor air (oF)

    ,

    = higher (outdoor) and lower (indoor) humidity ratio in grains water/lb dry

    air (gr w/lb d.a.)

    The total heat removed from the ventilation air is .

    Recommended outdoor air ventilation rates for some applications are listed in Table 6.

    17. This table has ventilation rates similar to many state codes and standards. The

    ventilation rates in Table 6.17 are often higher than the minimum listed in earlier

    standards.

    If the peak load does not occur at the time of the day that the outdoor temperature is

    at a maximum, a correction must be made to the outdoor temperature used for calculating

    ventilation and infiltration loads. Table 6.18 lists this correction.

  • 8/10/2019 Dasar Teori Pendingin

    60/64

    o) Heat Gain To Ducts

    The conditioned air flowing through ducts will gain heat from the surroundings. If the

    duct passes through conditioned spaces, the heat gain results in a useful cooling

    effect, but for the ducts passing through unconditioned spaces it is a loss of sensible

    heat that must be added to the building sensible cooling load (BSCL). The heat gain can

    be calculated from the heat transfer equation below:

    Where

    = duct heat gain (BTU/hr)= overall coefficient of heat transfer (BTU/hr)

  • 8/10/2019 Dasar Teori Pendingin

    61/64

    = duct surface area (ft2)

    = temperature difference between air in duct and surrounding air (oF)

    It is recommended that cold air ducts passing through unconditioned areas be insulated

    to at least an overall value of R-4 (U = 0.25).

    If there is significant heat gain to return air ducts, it should also be calculated, but

    it is only added to the coil sensible cooling load (CSCL), not the building sensible

    cooling load (BSCL). Although the heat gain to supply ducts in conditioned spaces is notwasted, care should be taken that it does not affect the distribution of cooling. If

    there is a long run of duct with a number of outlets, the heat gains in the first

    sections of duct might be significant enough such that the air temperature at the last

    outlets can be too high. In this case, it might be useful to insulate the duct even

    though it is in the conditioned area.

    Some designers find it reasonably accurate to add a percentage to the supply duct heat

    gain, rather than going through elaborate calculations. For insulated supply ducts, 1-3%

    of the building sensible load (BSCL) is suggested, depending on the extent of ductwork.

    p)

    Fan and Pump Heat

    Some of the energy from the system fans and pumps is converted into heat through

    friction and other effects, and becomes part of the sensible heat gain that should be

    added to the load. For a draw-through fan arrangement (fan downstream from the cooling

    coil), the heat is added to the BSCL, whereas for a blow-through arrangement (fan

    upstream from the coil) the heat gets added to the CSCL load.

    An approximate allowance for fan heat can be made as follows:

    For 1 in. w.g. pressure add 2.5% to BSCL

    For 2 in. w.g. pressure add 5% to BSCL For 4 in. w.g. pressure add 10% to BSCL

  • 8/10/2019 Dasar Teori Pendingin

    62/64

    The heat from the chilled water pump on small systems is generally small and may be

    neglected, but for large systems it may range from 1-2% of sensible load. For central

    systems with remote chilled water cooling coils, the pump heat is a load on the

    refrigeration chiller, but not the cooling coil. This leads to a new term, the

    refrigeration load.

    The refrigeration load (RL) is the load on the refrigeration equipment.

    For a direct expansion system, the refrigeration load and cooling coil load are equal.

    For a chilled water system, the refrigeration load is the cooling coil(s) load plus the

    chilled water pump heat.

    q)

    Duct Air Leakage

    Duct systems will leak air at joints. Unfortunately, many systems have unnecessarily

    high air leakage due to sloppy installation. However, a careful job should limit ductleakage to 5% or less of the total CFM. If ducts are outside the conditioned space, the

    effect of leakage must be added to the BSCL and BLCL. If the air leaks into the

    conditioned space, then it does useful cooling, but care should be taken that it is not

    distributed to the wrong location.

    r) Supply Air Conditions

    After the sensible and latent heat gains are calculated, the required supply air

    conditions (flow rate, temperature, and humidity) necessary to satisfy room conditions

    are determined.

  • 8/10/2019 Dasar Teori Pendingin

    63/64

    SUMMARY OF COMMERCIAL COOLING LOAD CALCULATION PROCEDURES

    1. Determine indoor and outdoor design conditions based on selected locations

    2. Use architectural plans to measure dimensions of all surfaces through which there

    will be external heat gains, for each room.

    3. Calculate areas of all these surfaces.

    4. Select heat transfer coefficient V-values for each element from appropriate tables,

    or calculate from individual R-values.

    5. Determine time of day and month of peak load for each room by calculating external

    heat gains at times that they are expected to be a maximum. Search Tables 6.1

    (Cooling Load Temperature Differences (CLTD) for Calculating Cooling Load from Flat

    Roofs), 6.2 (CLTD for Calculating Cooling Load from Sunlit Walls), 6.6 (MaximumSolar Heat Gain Factor (SHGF) for Sunlit Glasses, North Latitudes), and 6.8(CLF for

    Glass without Interior Shading, in North Latitude Spaces Having Carpeted Floors) to

    find maximum values. Often calculations at a few different times will be required,

    but the suggestions in Room Peak Cooling Load Sectionshould be helpful.

    6. Calculate each room peak load, using the values for the external heat gains

    determined above and by calculating and adding the internal heat gains from people,

    lights, and equipment. The architect or building owner will furnish the data needed

    for the calculations. If there is infiltration, this must be added to the room load.

    7. Find the time of building peak load using a similar search process as in item 5 and

    the suggestions inBuilding Peak Cooling Load Section

    .

    8. Calculate the building load at peak 'time, adding all external and internal gains

    and infiltration, if any. Add supply duct heat gain (Heat Gain to Ducts Section

    ),

    duct heat leakage (Duct Air Leakage Section

    ), and draw-through supply fan heat gain

    (Fan and Pump Heat Section

    ), if significant.

    9. Find the cooling coil and refrigeration load by adding the ventilation load (Table

    6.17(Minimum Mechanical Ventilation Requirement Rates)) to the building heat gains;

    add blow-through fan, return air fan, and pump heat gains, if significant.

    10. Calculate required supply air conditions.

  • 8/10/2019 Dasar Teori Pendingin

    64/64