Unit 24: Applications of Pneumatics and Hydraulics Unit code: J/601/1496 QCF level: 4 Credit value: 15 OUTCOME 2 TUTORIAL 11 AIR COMPRESSORS AND DISTRIBUTION SYSTEM The material needed for outcome 2 is very extensive so there are 11 tutorials in this outcome. You will also be completing the requirements for outcome 1 which is integrated into it. The series of tutorials provides an extensive overview of fluid power for students at all levels seeking a good knowledge of fluid power equipment. 2 Understand the construction, function and operation of pneumatic and hydraulic components, equipment and plant Pneumatic equipment: types, construction, function and operation e.g. air compressors, coolers, dryers, receivers, distribution equipment, fluid plumbing and fittings, drain traps, FRL air service units, valves, actuators, seals Hydraulic equipment: types, construction, function and operation e.g. fluids, pumps, motors, actuators, reservoirs, accumulators, fluid plumbing and fittings, valves, filters, seals, gauges Performance characteristics: air compressors e.g. volumetric efficiency, compression ratio, isothermal efficiency; hydraulic pumps e.g. operating efficiency, losses, flow rate, operating pressure, shaft speed, torque and power On completion of this tutorial you should be able to do the following. Explain the basic properties of air. Explain how moisture is formed in compressed air. Define and calculate Free Air Delivery. Explain the basic principles of a reciprocating air compressor. Explain and calculate various efficiencies of air compressors. Explain the basics of air distribution systems. Explain the principles used to condition compressed air for use with pneumatic systems. D.J.DUNN 1
1. INTRODUCTION One advantage of air over oil as a power medium is that it can be piped from one source to all the user points. This is because air has a low viscosity and flows much more easily than oil. In addition to this, air, once used, may be vented back into the atmosphere, whereas oil must be returned to the tank with a return pipe. The disadvantage of air is that it is an expansive substance and dangerous when used at high pressures. For this reason, applications are confined to things requiring low pressures (10 bar or lower). The common source of the air is the compressor. There are many types of compressors with different working principles and working conditions. The function of all of them is to draw in air from the atmosphere and produce air at pressures substantially higher. Usually a storage vessel or receiver is used with the compressor. Atmospheric air contains WATER VAPOUR mixed with the other gases. When the air is cooled to the dew point, the vapour condenses into water and we see rain or fog. The ratio of the mass of water vapour in the air to the mass of the air, is called the ABSOLUTE HUMIDITY. The quantity of water that can be absorbed into the air at a given pressure depends upon the temperature. The hotter the air, the more water it can evaporate. When the air contains the maximum possible amount of vapour it is at its dew point and rain or fog will appear. The air is then said to have 100% humidity. When the air contains no water vapour at all (dry air), it has 0% humidity. This refers to RELATIVE HUMIDITY. For example if the air has 40% humidity it means that it contains 40% of the maximum that it could contain. There are various ways to determine the humidity of air and instruments for doing this are called HYGROMETERS. The importance of humidity to air compressors is as follows. When air is sucked into the compressor, it brings with it water vapour. When the air is compressed the pressure and the temperature of the air goes up and the result is that the compressed air will have a relative humidity of about 100% and it will be warm. When the air leaves the compressor it will cool down and the water vapour will condense. Water will then clog the compressor, the receiver and the pipes. Water causes damage to the air tools, corrodes the pipes and equipment and ruins the paint sprays. For this reason the water must be removed and the best way is to use a well designed compressor installation and distribution network. The diagram overleaf shows the layout of a two cylinder reciprocating compressor, typically for supplying a workshop. It is important to note that air supplies need to be filtered and dried to various degrees depending on the use. This is covered in the second half of the tutorial. D.J.DUNN 2
TYPICAL AIR COMPRESSOR LAYOUT FOR A WORKSHOP Figure 1 1. Induction box and silencer on outside of building with course screen. 2. Induction filter. 3. Low pressure stage. 4. Intercooler. 5. High pressure stage. 6. Silencer. 7. Drain trap. 8. After cooler 9. Pressure gauge (legal requirements for safety). 10. Air receiver. 11. Safety pressure relief valve (legal requirements for safety). 12. Stop valve (legal requirements for safety). In addition to this, the pressure vessel must be inspected regularly for corrosion and hydrostatically tested at the maximum pressure. The safe working pressure (SWP) must be shown on the storage vessel. 2. BASIC PRINCIPLES OF RECIPROCATING COMPRESSORS A simple single reciprocating compressor has a piston that is made to move up and down inside a cylinder by a crank and connecting rod. A suitable motor rotates the crank. The piston sucks air into the cylinder pulling open the inlet valve and then compresses the air until it exceeds the outlet pressure and then the outlet valve opens to allow the air out. Figure 2 D.J.DUNN 3
2.1. THE CYCLE IDEAL CYCLE Ideally the piston sucks in a full cylinder of air and then expels it by moving right up to the cylinder head. If the outlet is connected to a storage vessel, the pressure will gradually rise in the vessel as more and more air is pumped into it. The cycle may be explained with the aid of a pressure volume - diagram as follows. Figure 3 1. The piston draws in a full cylinder of air from 4 to 1. 2. The piston moves towards the cylinder head. The air will not leave the cylinder until it is compressed to a pressure equal to the outlet pressure. Compression takes place from 1 to 2. 3. When the pressure inside the cylinder is large enough to overcome the pressure outside, the outlet valve opens and air is expelled from 2 to 3 at the high pressure. The piston expels all the air by moving right up to the cylinder head. 4. The piston moves away from the cylinder head and because there is nothing inside, the inlet valve opens and air is drawn in at low pressure from 4 to 1. REALISTIC CYCLE In reality it is not possible to construct an air compressor in which the piston goes right up to the cylinder head. The piston stops short at point 3 and a small clearance volume V 3 is trapped inside at high pressure. As the piston moves away from the head, this volume must expand until the pressure is low enough to let fresh air in at point 4. The air drawn in is now V 1 V 4 and not V 1. Figure 4 The maximum volume of air that could be drawn in is V 1 V 3 that is the SWEPT VOLUME. The effect of the clearance volume is to reduce the volume of air drawn in. This gives rise to the idea of VOLUMETRIC EFFICIENCY. This is defined as follows. Induced Volume V 1 V4 vol Swept Volume V1 V3 In reality any leaks or slippage of compressed air that does not arrive at the outlet should be subtracted from the induced volume making the efficiency even lower. D.J.DUNN 4
2.2. THE TEMPERATURE OF THE AIR and ISOTHERMAL EFFICIENCY When air is compressed rapidly it gets hotter so we expect the compressed air to be hotter at the outlet. The final temperature depends on the degree of cooling that occurs during the compression stroke. If there is no cooling the process is called ADIABATIC and the compressed air is as hot as it can possibly be. If the compressor cylinder is cooled (covered later) the final temperature is reduced and in the extreme case if it is cooled so that it does not get any hotter at all, the process is called ISOTHERMAL. Figure 5 Because the air is cooler, the pressure at any point on the graph is lower. This means that the force required to compress it is reduced and so the power input to the compressor is reduced. This is used to determine the effectiveness of the cooling. It is the ratio of the indicated power with perfect (isothermal) cooling to the actual power with less cooling. (iso) = Isothermal Efficiency = Isothermal Power/Actual Power 2.3. WORK OF COMPRESSION and INDICATED POWER Work done is defined as force x distance moved. The area under a force distance graph represents the work done. The area under a p V graph is essentially the same kind of graph and the area under the graph is the work done to compress the air. It follows that the isothermal process gives the minimum work done so cooling an air compressor reduces the power input. The indicated work per cycle is the shaded area enclosed by the p - V diagram. Formulae can be derived for the work done in terms of pressures and volumes and this is covered in the tutorials on Thermodynamics for those who are interested. The ratio of the isothermal work to the actual work is called the ISOTHERMAL EFFICIENCY and the better the cooling of the compressor, the higher this will be. It follows that isothermal compression requires the least work input. This is a good reason to cool the compressor cylinder. Figure 6 If suitable equipment is used the p V diagram may be created for a given air compressor and the shaded area may be determined. This is known as an indicator diagram from old mechanical equipment originally used to draw them but now it is all done electronically. W is called the indicated work. The indicated power is found by multiplying W by the strokes per second (The crank speed). I.P. = W x N where N is the shaft speed in Rev/s D.J.DUNN 5
2.4 MECHANICAL EFFICIENCY If the actual power input to the shaft of the compressor is determined, it is called the Shaft Power S.P.). Because there are mechanical losses in the air compressor to overcome friction, operate the valves, pump the lubricating oil and so on, the power delivered into compressing the air is less. The mechanical efficiency of the air compressor is defined as: (mech) = Indicated Power/Shaft Power = I.P./S.P. WORKED EXAMPLE No. 1 The indicated work found for a single cylinder reciprocating air compressor is found to be 120 Joules per cycle when running at 360 rev/min. The shaft power is determined as 910 Watts. What is the mechanical efficiency? SOLUTION I.P. = 120 (Joules/cycle) x 360/60 (rev/s)= 720 Watts (mech) = I.P./S.P. = 720/910 = 0.79 or 79% 3. FREE AIR DELIVERY When air flows in a pipe, the mass depends upon the pressure and temperature. It would be meaningless to talk about the volume of the air unless the pressure and temperature are considered. For this reason the volume of air is usually stated as FREE AIR DELIVERY or FAD. FAD refers to the volume the air would have if let out of the pipe and returned to atmospheric pressure and temperature. The FAD is also the volume of air drawn into a compressor from the atmosphere. After compression and cooling the air is returned to the original temperature but it is at a higher pressure. Suppose atmospheric conditions are pata and Va (the FAD) and the compressed conditions are p, V and T. Applying the gas law we have pv pava pvta Va F.A.D. T Ta Tpa Usually it is assumed that the compressed air has been cooled back to the the same temperature as the atmosphere. In this case the temperature cancels in the expression and pv Va rpv F.A.D. pa FAD = Volume of compressed air x pressure ratio If the pressure is given as gauge the absolute pressure is found by adding on atmospheric pressure. Since this is approximately 1 bar pressure ratio may be defined as follows. Pressure ratio = r p = (gauge pressure + 1)/1 Note that since the air sucked in at atmospheric pressure is the same air that is delivered at the outlet pressure, the induced volume of the compressor is related to the F.A.D. F.A.D. = Induced Volume x crank speed D.J.DUNN 6
WORKED EXAMPLE No.2 An air compressor is rated at 12 dm3/s FAD. Calculate the volume of compressed air delivered at 10 bar gauge. SOLUTION pressure ratio = 10 + 1 = 11 volume of compressed air = 12/11 = 1.091 dm3/s WORKED EXAMPLE No.3 An air compressor has a swept volume of 1 litre and runs at 300 rev/min. It delivers 247 litres/min of F.A.D. Calculate the volumetric efficiency. If the delivery pressure 5 bar gauge, what is the volume of compressed air delivered? SOLUTION The swept volume is 1 litre (or 1 dm 3 ) The induced volume is the volume of atmospheric air drawn in and hence it must be the F.A.D. per cycle. Induced volume = 247/300 = 0.823 litres vol Induced Volume Swept Volume 0.823 82.3% 1 The compression ratio = (5+1)/1 = 6 Volume of compressed air (at atmospheric temperature) = 247/6 = 41.17 litres/min D.J.DUNN 7
SELF ASSESSMENT EXERCISE No.1 1. Air flows in a pipe at 12 bar gauge and with a rate of 1.3 m 3 /min. Determine the FAD equivalent to this. 2. A workshop consumes 2 m 3 /min of air at 14 bar gauge. Calculate the FAD rating of the compressor required. 3. A reciprocating air compressor has a free air delivery of 170 dm 3 /min when running at 260 rev/min. The swept volume is 0.75dm 3. The inlet is at atmospheric pressure of 1 bar. The delivery pressure is 7 bar. The input power to the compressor is 410 watts. An indicator diagram shows that the indicated work is 81 Joules per cycle. Calculate the volumetric efficiency. (87.2%) the volume of compressed air at atmospheric pressure. (24.286 dm 3 /min) the Mechanical Efficiency based on the Indicated Power. (86%) D.J.DUNN 8
4. FLOW IN PIPES First we need to understand the difference between a quantity of fluid and a flow rate of fluid. A quantity is measured in kg or cubic metres or some multiple of these. A flow rate is a quantity flowing past a given point in 1 second, minute or hour. Flow rate is found by timing the passing of a quantity. SELF ASSESSMENT EXERCISE No. 2 1. 320 litres of water flows out of a pipe into a tank in 2 minutes 20 seconds. Calculate the flow rate in litres per second. 2. 5000 kg of oil flows into a tank in 1 hour 30 minutes. Calculate the flow rate in kg per hour. 3. Air flows in a pipe at a rate of 3m 3 /minute. Calculate the quantity used in 1 hour. VELOCITY The mean velocity of a fluid flowing in a pipe is found by dividing the flow rate by the cross sectional area. velocity = Q/A where Q is the quantity in m 3 /s and A is the cross sectional area in m 2. SELF ASSESSMENT EXERCISE No. 3 1. 0.5 m 3 /s of air flows in a rectangular duct 0.5 m x 0.4 m. Determine the mean velocity. 2. 0.5 m 3 /s of air flows in a pipe 50 mm bore diameter. Calculate the mean velocity. 3. Air flows in a pipe 100 bore diameter with a mean velocity of 8 m/s. Calculate the flow rate. 4. An air compressor draws in 0.3 m 3 /minute of free air and delivers it at 8 bar gauge. Calculate the flow rate of the compressed air. D.J.DUNN 9
4. STORAGE and DISTRIBUTION The air compressor supplies the air to the receiver where it is stored. The use of a storage vessel enables cooling and water removal. It also enables the compressor to work at a steadier rate filling up the receiver when air is not being drawn off and maintaining the supply during periods of high demand. When the receiver is full, the compressor turns itself off or idles to save energy. The best way of ensuring all the users in a workshop have a constant supply of air is by using a ring main connected to the receiver. This enables the air to arrive at the point of use from two directions and so everyone has the same pressure. If a single long main was used, the user at the end of the chain would not have much pressure when the other users are taking it. Figure 7 The diagram shows a simple ring main. The points marked x are at the lowest points in the pipe runs and at the corners where water will collect. Drain traps are fitted at these points to trap the air and release the water. The take off points should be from the top of the main to reduce the risk of water and sludge being picked up from the main. D.J.DUNN 10
DRAIN TRAPS Figure 8 Drain traps are units designed to trap the air and let out the water. The diagram shows a very simple drain trap. Figure 9 D.J.DUNN 11
SELF ASSESSMENT EXERCISE No. 4 1. Air flows in a pipe at 12 bar gauge. What is the compression ratio? 2. An air receiver contains 1.3 m3 of air at 14 bar gauge. What is the free air stored? 3. An air compressor draws in 0.3 m3/s of air from the atmosphere and delivers it at 8 bar gauge. Estimate the flow rate of the compressed air. The mean velocity must not exceed 8 m/s. Determine a suitable pipe bore. 4. Air flows in a pipe 50 mm bore diameter with a pressure of 5 bar gauge and a velocity of 5 m/s. Estimate the FAD. AIR FILTERS, REGULATORS, AND LUBRICATORS. FILTER/DRYERS When air is drawn from the ring main for use in a workshop, it must be conditioned into a state fit for use. This usually involves removing the water, filtering it, regulating the pressure and adding lubricating oil. Figure 10 Water must be removed to stop corrosion and clogging of the equipment. High speed hand tools may be damaged by water droplets. Dirt must be filtered to stop wear to the equipment. The sort of contaminants removed by the filter are rust, sediment, grit, shards of thread, shards of sealing tape, scale, welding beads and so on. Most of this result from installation or repair to the pipe work. Filters and dryers are usually combined in one unit as shown. The air enters the bowl through a spiral passage which makes it spin and throws water droplets against the bowl. These run down and are collected at the bottom. The water must be drained either manually or by a simple float valve. The air passes through the filter element on the way out and this traps the dirt. D.J.DUNN 12
REGULATORS Ring mains operate typically at 14 bar but tools and pneumatic equipment work at 3 or 4 bar. The pressure must be reduced and kept constant whether a small amount of air is used or a lot. The regulator sets the pressure and keeps it constant. When the air pressure rises on the outlet, the diaphragm is pushed down onto the spring and closes the valve. When the pressure falls on the outlet, the spring pushes the valve open to let more air through. Often the regulator and filter unit are designed as one unit. LUBRICATORS Figure 11 Not all applications need lubrication. Paint spraying and shot blasting are two examples. Applications such as air drills do need it. Pneumatic cylinders and other equipment may not need lubricating if low friction materials are used in their design. The basic principle is that a small restriction in the air passage causes a slight drop in pressure which forces oil up into the dome from where it drips down into the air flow and becomes atomised in the air stream to produce air with a fine mist of oil at exit. The drip rate is controlled by a small restrictor screw in the drip tube. SYMBOLS Figure 12 Figure 13 D.J.DUNN 13
SELF ASSESSMENT EXERCISE No. 5 The diagram shows the man parts of a typical regulator. Name the components numbered and state its function. Item 1 is Its function is Item 2 is Its function is Item 3 is Its function is Item 4 is D.J.DUNN 14
The diagram shows a typical separator. Item 1 is Its function is Item 2 is Its function is Item 3 is Its function is Draw combined symbol for a filter/dryer/regulator and lubricator here. Draw the simplified combined symbol here. D.J.DUNN 15