Central Air Conditioning Plants

Transcription

1 Central Air Conditioning Plants In our department, these plants (Air cooled or water cooled) are commonly available above 10 TR up to 100 TR. These types of plants are more suitable for large installations such as AIR Radio Studio / TV Studio Buildings and High Power Transmitter Buildings. In water cooled plants, external cooling towers / water spray ponds with water softening plants are the common features. These are invariably provided with AHU (Air Handling Unit) and supply & return ducts for carrying air. The central air conditioning plants or the systems are used when large buildings, hotels, theaters, airports, shopping malls etc are to be air conditioned completely. The window and split air conditioners are used for single rooms or small office spaces. If the whole building is to be cooled it is not economically viable to put window or split air conditioner in each and every room. Further, these small units cannot satisfactorily cool the large halls, auditoriums, receptions areas etc. In the central air conditioning systems there is a plant room where large compressor, condenser, thermostatic expansion valve and the evaporator are kept in the large plant room. They perform all the functions as usual similar to a typical refrigeration system. However, all these parts are larger in size and have higher capacities. The compressor is of open reciprocating type with multiple cylinders and is cooled by the water just like the automobile engine. The compressor and the condenser are of shell and tube type. While in the small air conditioning system capillary is used as the expansion valve, in the central air conditioning systems thermostatic expansion valve is used. The chilled is passed via the ducts to all the rooms, halls and other spaces that are to be air conditioned. Thus in all the rooms there is only the duct passing the chilled air and there are no individual cooling coils, and other parts of the refrigeration system in the rooms. What is we get in each room is the completely silent and highly effective air conditions system in the room. Further, the amount of chilled air that is needed in the room can be controlled by the openings depending on the total heat load inside the room. The central air conditioning systems are highly sophisticated applications of the air conditioning systems and many a times they tend to be complicated. It is due to this reason that there are very few companies in the world that specialize in these systems. In the modern era of computerization a number of additional electronic utilities have been added to the central conditioning systems. There are two types of central air conditioning plants or systems: 1) Direct expansion or DX central air conditioning plant: In this system the huge compressor, and the condenser are housed in the plant room, while the expansion valve and the evaporator or the cooling coil and the air handling unit are housed in separate room. The cooling coil is fixed in the air handling unit, which also has large blower housed in it. The blower sucks the hot return air from the room via ducts and blows it over the cooling coil. The cooled air is then supplied through various ducts and into the spaces which are to be cooled. This type of system is useful for small buildings. 2) Chilled water central air conditioning plant:

2 This type of system is more useful for large buildings comprising of a number of floors. It has the plant room where all the important units like the compressor, condenser, throttling valve and the evaporator are housed. The evaporator is a shell and tube. On the tube side the Freon fluid passes at extremely low temperature, while on the shell side the brine solution is passed. After passing through the evaporator, the brine solution gets chilled and is pumped to the various air handling units installed at different floors of the building. The air handling units comprise the cooling coil through which the chilled brine flows, and the blower. The blower sucks hot return air from the room via ducts and blows it over the cooling coil. The cool air is then supplied to the space to be cooled through the ducts. The brine solution which has absorbed the room heat comes back to the evaporator, gets chilled and is again pumped back to the air handling unit. To operate and maintain central air conditioning systems we need to have good operators, technicians and engineers. Proper preventative and breakdown maintenance of these plants is vital. Figure 5, showing control panel, compressor, condenser and accessories

3 Figure 6, Showing view of Air Conditioning Plant Room

4 Figure 7, Showing overall layout of Air Conditioning System in a multi story building

5 Air cycle Figure 8, Showing overall layout of Air Conditioning Duct System in a multi story building Indoor air may be too cold, too hot, too dry, too wet, too drafty or too still. These conditions are changed by rotating the air and these treatments are provided in the air conditioning air cycle. Air distribution system directs the treated air from the air conditioning equipment to the space to be conditioned and then back to the equipment. The main components in the air cycle are (i) Fan (iii) Supply Outlets (v) Return duct (ii) Supply duct (iv) Return outlets (vi) Filter

6 (vii) Cooling coil (or heating coil for low temperature areas). The total resistance of these components to the flow of the air plus the friction resistance caused by the air passing through the duct run are major factors in determining the size of the fan and fan motor and the amount of air pressure that is required. For a Broadcast Studio set up this resistance is of the order of 25 mm to 50 mm of water gauge. Centrifugal fan is most commonly used in commercial and residential installations. It consists of a scroll, a shaft and a wheel. The scroll is actually a housing for the shaft and wheel and the shaft serves as an axle for the wheel. The wheel is cylindrical in shape and has many blades. Centrifugal fans are available with forward or backward curved blades. A forward curved fan can deliver a required quantity of air at low fan speed. The air velocity and speed of the fan wheel (tip speed) not only play a large part in determining the efficiency of the fan but also affect the level of noise generated by the fan. High tip speed and high velocity usually result in more noise. Remote location of the fan reduces the noise but the system become more expensive. Ducts may be circular, rectangular or square in shape. From the appearance and practical point of view, rectangular ducts are generally adopted. Ducts are fabricated from a wide variety of materials. Ducts made of sheet metal are very common. The ducts are lined with glass wool or mineral wool slabs of 25 mm thickness wrapped in copper naphthanate treated cloth. Outlets are another major part of the air distribution system. They are important from the point of view of appearance, functions and performance. The primary function of the outlets is to provide properly controlled distribution of air to the room and removing the air from the room. Ceiling diffusers, grilles and registers are used as supply outlet and grilles are used as return outlets. Operation Before starting the plant, ensure that proper functioning of safety controls including interlock circuit have been checked and correctly set, and that all motors are megger tested, direction of rotation verified, all bearings lubricated and refrigeration system fully charged. The crank case heater must be energised well in advance. Proceed step by step for operating the system as follows: Start the air handling unit, ensuring that dampers in the supply duct are fully open. Open all water valves and start the water pump. Observe pressures at condenser inlet and outlet. Open hot gas valve on the condenser and the discharge service valve on the compressor. Open discharge gauge valve to read the pressure. Follow the same procedure and read the suction pressure. Open liquid line valve. Observe standing pressure on the gauges. This should be approximately 7.03 kg/cm 2 (100 psi) for R 12 and 10.5 kg/cm 2 (150 psi) for R 22 to indicate that the system is tight with no leakage. Open suction service valve and start the compressor. Observe the refrigerant and oil pressures. Check the current drawn by the compressor motor, observe the oil level in the compressor sight glass. Oil should be clear without foam after operation has stabilised.

7 Compressor Pump Down It is essential to collect the refrigerant in the condenser with isolation to prevent its loss before opening the compressor or any other part of the system. This is called pump down and the operation involves the following procedure: Short the low pressure switch with a temporary jumper wire so that the compressor does not stop before the refrigerant from it is emptied. Slowly close the suction valve with the compressor running. When the suction pressure drops to about 0.15 kg/cm 2 (2 psi), stop the compressor. Never pump the compressor below 0.15 kg/cm 2 to prevent infiltration of moisture and dirt into the crank case. After a few minutes, the dissolved refrigerant will leave the crank case raising the suction pressure. This additional refrigerant can be pumped to the condenser by operating the compressor again for a short while. Repeat the above procedure till the suction pressure does not rise above 0.15 kg/cm 2 after closing the service valves. Removing Refrigerant from the System It may be necessary to remove the refrigerant from the system into a cylinder if there is an excess charge or there is a leak in the condenser. Take the following steps for this operation: (a) Connect a suitable line between the angle valve provided for charging and an empty refrigerant cylinder. (b) Purge the air from the connection line. (c) Keep the cylinder cold by immersing it in ice cold water to ensure a faster refrigerant flow from the system. (d) Start the compressor and open the liquid line charging valve, allowing the liquid into the empty cylinder. If excess refrigerant is to be removed, hold the charging valve open only until the discharge pressure reaches the normal reading. After this operation, remove the charging line and close the charging valve. Do not overcharge the cylinder as excessive pressure is dangerous. Purging Non Condensible Gases Presence of non condensibles gases such as air causes high discharge pressure, resulting in reduction of capacity and high power consumption. In case such symptoms are present, the following check should be done: Shut down the system overnight, long enough for the temperature of all components to level off. Read the standing pressure and compare it with the refrigerant saturation pressure corresponding to the temperature of the system. If the standing pressure exceeds the saturation pressure by 0.75 kg/cm 2 (10 psi) or more, the non condensibles are excessive and must be removed.

8 For example, if R 22 is used and the system temperature is 85 o F (29.4 o C) and standing pressure is 12.8 kg/cm 2 (175 psi), then there is excess of non condensibles. Saturation pressure for R 22 corresponding to a temperature 85 o F is 11 kg/cm 2. The difference is 1.05 kg/cm 2 more than 0.75 kg/cm 2, indicating corrective purging. For purging, take the following steps: Pump down the system as described earlier. Immediately after stopping the compressor, close the compressor discharge valve. Run the water through the condensor for condensation of refrigerant vapour. Crack open the purge valve on the top of the condensor for an instant, shut it again. Allow the system to stabilize for a few minutes before reopening and closing the purge valve. Repeated purging and closing operation should clear the system of non condensible. Restore normal system operation, check the improvement in discharge pressure. Check refrigerant charge and compressor oil pressure. Refrigerant Charging A correct operating charge of refrigerant in the system is essential. Loss due to leakage in the system has to be made up. It may be necessary to replace the entire charge. An over charge results in unduly high temperatures, pressures and operating costs and may damage the system components. An undercharged system leads to insufficient cooling, high operating cost, and, in hermetic system, the compressor motor may fail. Refrigerant may be added to the system either as a vapour or liquid depending upon the location of charging point and quantity required. Generally, for adding make up refrigerant, vapour charging method is more convenient. For total system charge, liquid charging at the high side followed by vapour charging at compressor low side will be quicker. Under no circumstances should liquid refrigerant be allowed to enter the compressor to avoid damage to the compressor. The procedure for vapour charge method is described below: Open the suction and discharge shut off valves of the compressor. Install a gauge in the discharge gauge port and open the gauge line if a gauge port has not been provided. Connect a refrigerant cylinder and the connection with a compound gauge, to the charging valve provided on compressor suction line. Purge the air from the lines and tighten the connections. Admit the refrigerant by slowly opening the refrigerant cylinder. The cylinder should be kept in upright position to prevent the refrigerant from entering the compressor in liquid state. Start the compressor. As the cylinder gets emptied, its pressure will drop to the same level as the suction pressure. The remaining refrigerant can be drawn from the cylinder by closing the suction shut off valve and pulling a vacuum on the cylinder with the compressor running. Check the quantity of refrigerant charge by noting the difference in the weight of the cylinder and observing the pressure.

9 Water Treatment Algae/slime scale and corrosion on the water side of the heat transfer equipment retards heat transfer causing general loss of efficiency and breakdowns. Oxygen, Carbon Dioxide, Sulphur Dioxide absorbed from the air and dissolved in water cause corrosion, reducing the capacity of lines, increasing frictional losses and pumping cost. Hard water causes scaling problem. When heated, the minerals are left behind, which form a deposit on the heat exchanger surface. The heat transfer rating of the scale is very much lower than metal. Retarded heat transfer results in increased discharge pressure causing loss in capacity and increased power consumption. Scaling of the condensor tubes in a re circulated water system is unavoidable. De scaling has to be carried out as a preventive maintenance once every 12 months or earlier depending on the hardness of the water. De scaling can be carried out quite conveniently by circulating mild inhibited acid solution with the help of a small pump connected across the condensor inlet and the water valves are closed to confine the circulation to the condenser only. Chemical compounds are available which suspend minerals of dissolved scale. Algae attach themselves to the surfaces, and since they are living plants, they grow until they clog the passages of the system. Bacteria forms slime and close the system in much the same way as algae. Algae/Slime is controlled by use of toxic. A specialist should be consulted to determine the algae/slime. The trouble should be diagnosed as accurately as possible before any repair is attempted. Definite symptoms will accompany a faulty operation in the system. The following trouble shooting chart will help in fault location and prompt correction: ******** TONNAGE MEASUREMENT OF AC PLANTS (I) By air-flow method Tonnage of refrigeration (TR) = A x V x (H 1 - H 2 ) {FPS units} S 200

10 Where A = Cross sectional area of duct through which air is passing in sq. ft. V = Air velocity per minute, in Ft. per minute, measured by anemometer in ft./min S = Specific volume of (return) air H 1 = Enthalpy for return air, in Btu/lb H 2 = Enthalpy for supply air, in Btu/lb Note: Both H 1 and H 2 are determined from the psychometric chart with help of Dry bulb temperature (T db in deg F.) and Wet bulb temperature (T wb in deg F.) Similarly Specific volume (S) is determined from the psychometric chart Example 1: Calculate Tonnage of AC Plant having the following measurement figures: A = sq. ft. V = 293 Ft. per minute S = Specific volume of return air = 13.7 cubic Ft./ Lb. For Return duct, For Supply duct, T db = 73 0 F and T wb = 67 0 F (X) T db = 53 0 F and T wb = 49 0 F (Y) Calculations: H 1 = Enthalpy for return air, in Btu/ Lb, determined from psychometric chart in r/o (X) = 31.8 Btu/Lb. H 2 = Enthalpy for supply air, in Btu/ Lb, determined from psychometric chart in r/o (Y) = 19.8 Btu/Lb. S = Specific volume of return air = 13.7 cubic Ft./ Lb. Therefore Tonnage = Tonnage of refrigeration = A x V x (H 1 -H 2 )TR S 200 = Tonnage of refrigeration = 30.25x293 x ( ) TR = 38.8 TR (Answer) (II) By Water-flow method

11 Points to be remembered: 1Watt = 0.86 k Cal / Hr** (unit of power i.e. rate of energy) 1 Watt = Btu / Hr* or [1 Btu = Watts] 1 k Watt = 3412 Btu / Hr or [1 Btu = k Watts] # 1 Btu = k Cal 1 Ton = Btu / Hr = 200 Btu / Min = 50 k Cal / Min [200 Btu x k Cal] = 3024 k Cal / Hr [200 Btu x k Cal x 60 Min] = kw # [ = 3.561] Heat gained by water = { Q x Sp. Heat x (Th Tc) x 60} Btu/Hr ---- (A) = heat rejected by the refrigerant in the condenser Heat developed due to work done by compressor = { 3 V x I x Cos Φ} Watts = { 3 V x I x Cos Φ x 3.412}*Btu/Hr (B) Or = { 3 V x I x Cos Φ x 0.86} ** k Cal /Hr Refrigeration capacity in TR = [Heat gained by water in Btu/Hr] [Heat developed due to work done by compressor in Btu/Hr] = (A) - (B) = { Q x Sp. Heat x (Th Tc) x 60} { 3 V x I x Cos Φ x 3.412} Where Q = Quantity of water flowing through the water cooled condenser in Ltr/ Min Th = Temperature after condenser in 0 F Tc = Temperature before condenser in 0 F Sample measurements Q = Quantity of water flowing through the water cooled condenser = 620 Ltr/ Min Th = Temperature at condenser outlet = 99 0 F

12 Tc = Temperature at condenser inlet = 92 0 F V = compressor Voltage = 390 Volts I = compressor Current = 60 Amp Cos Φ = Power Factor = 0.85 Calculations: (A) = Heat rejected by the refrigerant in the condenser = Q x Sp. Heat x (Th Tc) x 60 Btu/Hr = 620 x x (99 92) x 60 = 57,3922 Btu/Hr (B) = Heat developed due to work done by compressor = { 3 V x I x Cos Φ x 3.412} Btu / Hr = { 3 x 390 x 60 x.85 x 3.412} Btu / Hr = 11, 7476 Btu / Hr Refrigeration capacity in TR = (A) - (B) = (57,3922) - (11,7476) = 38 TR, ANSWER *********** HVAC Air Conditioning Troubleshooting and Repair The following is an general A/C system troubleshooting guide. Realize that it is generic and many of the things listed here may not apply to the 944. Symptom / Possible Cause Solutions Low Compressor Discharge Pressure 1. Leak in system 2. Defective expansion valve 3. Suction valve closed 4. Freon shortage 5. Plugged receiver drier Repair 1. Repair leak in system 2. Replace valve 3. Open valve 4. Add freon 5. Replace drier

13 6. Compressor suction valve leaking 7. Bad reed valves in compressor High Compressor Discharge Pressure 1. Air in system 2. Clogged condenser 3. Discharge valve closed 4. Overcharged system 5. Insufficient condenser air 6. Loose fan belt 7. Condenser not centered on fan or too far from radiator Low Suction Pressure 1. Refrigerant shortage 2. Worn compressor piston 3. Compressor head gasket leaking 4. Kinked or flattened hose 5. Compressor suction valve leaking 6. Moisture in system 7. Trash in expansion valve or screen High Suction Pressure 1. Loose expansion valve 2. Overcharged system 3. Expansion valve stuck open 4. Compressor reed valves 5. Leaking head gasket on compressor Compressor Not Working 1. Broken belt 2. Broken clutch wire or no 12v power 3. Broken compressor piston 4. Bad thermostat 5. Bad clutch coil 6. Low Refrigerant low pressure switch has cut off clutch power Evaporator Not Cooling 1. Frozen coil, switch set too high 2. Drive belt slipping 3. Hot air leaks into car 4. Plugged receiver drier 5. Capillary tube broken 6. Shortage of refrigerant 7. High head pressure 8. Low suction pressure 9. High suction pressure Repair Repair Repair Repair Repair 6. Replace valve 7. Replace reed valves 1. Recharge system 2. Clean condenser 3. Open valve 4. Remove some refrigerant 5. Install large fan 6. Tighten fan belt 7. Center and check distance 1. Add refrigerant 2. Replace compressor 3. Replace head gasket 4. Replace hose 5. Change valve plate 6. Replace drier 7. Replace drier 1. Tighten valve 2. Remove some refrigerant 3. Replace expansion valve 4. Replace reed valves 5. Replace head gasket 1. Replace belt 2. Repair wire or check for power 3. Replace compressor 4. Replace thermostat 5. Replace clutch coil 6. Add refrigerant 1. Turn thermostat switch back 2. Tighten belt 3. Check for holes or open vents 4. Replace drier 5. Replace expansion valve 6. Add refrigerant 7. See problem #2 8. See problem #3 9. See problem #4

14 10. Defective expansion valve 11. Frozen expansion valve Frozen Evaporator Coil 1. Faulty thermostat 2. Thermostat not set properly 3. Insufficient evaporator air Repair 10. Replace expansion valve 11. Evacuate and replace drier 1. Replace thermostat 2. Set to driving condition 3. Check for excessive duct hose length, kink or bend. AC System Gauge Readings The following table is a general guideline for A/C system pressures and temperatures based on ambient outside temperature. Remember that these are a guideline and your actual temperatures and pressures will vary depending on humidity in the air and the condition of your system. A/C System Pressure Readings Ambient Temperature Low Side Pressure High Side Pressure Center Vent Temperature 60 F psi psi F 70 F psi psi F 80 F psi psi F 90 F psi psi F 100 F psi psi F 110 F psi psi F 120 F psi psi F *********