INSTRUMENTATION AND CONTROL

Transcription

1

2 INSTRUMENTATION AND CONTROL This tutorial provides minimal engineering science necessary to complete the rest of the tutorials. Greater depth of the individual topics can be found on the web site. It is useful to anyone studying measurement systems and instrumentation but it is provided mainly in support of the EC module D227 Control System Engineering. This tutorial is very basic. On completion of this tutorial, you should be able to explain and use the follow ing basic relationships used in mechanical and electrical engineering science. Basic units. Displacement, velocity and acceleration. Newton s Second Law of Motion. Springs. Pressure and flow. Resistance, current, capacitance and inductance. Stress and strain. Resistivity and temperature coefficient of resistance. 1

3 1. UNITS There are only 5 basic units used in engineering. These are mass length time current temperature kilogramme (kg) metre (m) seconds (s) Ampere (A) Kelvin (K) Also used in phys ics is luminous intensity units. Candela (Cd). All other quantities used are multiples of these 2. FUNDAMENTAL MECHANICAL LAWS 2.1. VELOCITY Linear Angular Velocity = rate of change of distance (m/s) v = distance moved/time taken for constant velocity or x/t. v = dx/dt for instantaneous velocity. Angular velocity = rate of change of angle (rad/s) ω = angle turned/time taken for constant velocity or θ/t. ω = dθ/dt for varying conditions. 2.2 ACCELERATION Linear Acceleration = rate of change of velocity (m/s 2 ). a = dv/dt = d 2 x/dt 2 Angular Angular acceleration = rate of change of velocity (rad/s 2 ) α = dω/dt = d 2 θ/dt NEWTON'S 2nd LAW OF MOTION This concerns the force required to overcome the inertia of a body. Inertia (or mass) is that property of a body which opposes changes to its motion. Linear Angular Force = Mass x acceleration Torque = Moment of inertia x angular acceleration F = M dv/dt T = I dω/dt F = M d 2 x/dt 2 T = I d 2 θ/dt SPRING Linear Force = constant x change in length (N) F = kx Angular Torque = Constant x angle (N m) T = k θ 2

4 2.5. PRESSURE Pressure = Force per unit area (N/m 2 ) p = F/A 1 Pascal = 1 N/m 2 1 bar = 10 5 Pa 2.6. FLOW RATE IN PIPES Flow rate = Cross sectional area x mean velocity Q = A dx/dt 3. FUNDAMENTAL ELECTRICAL LAWS 3.1 CURRENT i = dq/dt (Coulomb/s or Amperes) 3.2 RESISTANCE V=IR (Ohm's Law) 3.3 INDUCTANCE This law is equivalent to the 2nd. law of motion. Inductance (L Henries) is that property of a component that resists changes to the flow of current. The emf (E) produced is 3.4. CAPACITANCE E = L di/dt E = L d 2 q/dt 2 This law is equivalent to the spring law. Capacitance (C Farads) is the property of a component which enables it to store electric charge. Q = C V (Coulombs) SELF ASSESSMENT EXERCISE No.1 1. A spring is compressed 30 mm by a force of 50 N. Calculate the spring constant. (1.67 N/mm) 2. A mass of 30 kg is accelerated at a rate of 4 m/s 2. Calculate the force used to do it. (120 N) 3. A flywheel is accelerated at a rate of 4 rad/s 2 by a torque of 10 Nm. Calculate the moment of inertia. (2.5 kg m 2 ) 4. A force of 5000 N acts on an area of 400 mm 2. Calculate the pressure in Pascals. (12.5 MPa) 5. A pipe 100 mm diameter carries oil at a mean velocity of 2 m/s. Calculate the flow ra te in m 3 /s. ( m 3 /s) 6. The current in an inductive coil changes at 30 A/s and the back emf is 3 V. Calculate the inductance in H. (0.1 H) 7. A capacitor has a charge of 2 Coulombs at 24 V stored on it. Calculate the capacitance in µf. (83.3 µf) 8. A flywheel turns 120 radians in 3 seconds at a constant rate. Calculate the angular velocity. (40 rad/s) 3

5 4. STRESS AND STRAIN 4.1. DIRECT STRESS s When a force F acts directly on an area A as shown in figure 1, the resu lting direct stress is the force per unit area and is given as σ = F/A. where F is the force normal to the area in Newtons A is the area in m 2 and σ (sigma) is the direct stress in N/m 2 or Pascals. Since 1 Pa is a small unit kpa, MPa and GPa are commonly used also. If the force pulls on the area so that the material is stretched then it is a tensile force and stress and this is positive. If the force pushes on the surface so that the material is co mpressed, then the force and stress is compressive and negative DIRECT STRAIN e Consider a piece of material of length L as shown in figure 1. The direct stress produces a change in length L. The direct strain produced is ε (epsilon) defined as ε = L/L The units of change in length and original length must be the same and the strain has no units. Figure 1 Strains are normally very small so often to indicate a strain of 10 write it as µε. For example we would write a strain of 7 x 10-6 as 7µε. -6 we use the name micro strain and Tensile strain is positive and compressive strain is negative MODULUS OF ELASTICITY E Many materials are elastic up to a point. This means that if they are deformed in any way, they will spring back to their original shape and size when the force is released. It has been established that so long as the material remains elastic, the stress and strain are related by the simple formula E= σ /ε and E is called the MODULUS OF ELASTICITY. The units are the same as those of stress. 4

6 ELASTIC LIMIT A material is only elastic up to a certain point. If the elastic limit is exceeded, the material becomes permanently stretched. The stress -strain graph for some metals are sh own below. The modulus of elasticity does not apply above the elastic limit. Strain gauges should not be stretched beyond the elastic limit of the strain gauge material which is approximately 3000 µε. Figure 2 A typical value for the elastic limit of strain gauges is 3000 µε. WORKED EXAMPLE No.1 A metal bar which is part of a frame is 50 mm diameter and 300 mm long. It has a tensile force acting on it of 40 kn which tends to stretch it. The modulus of elasticity is 205 GPa. Calculate the stress and strain in the bar and the amount it stretches. SOLUTION F = 40 x 10 3 N. A = πd 2 /4 = π x 50 2 /4 = 1963 mm 2 σ = F/A = (40 x10 3 )/(1963 x 10-6 ) = x 10 6 N/m 2 =20.37 MPa E = σ /ε = 205 x 10 9 N/m 2 ε = σ/e = (20.37 x 10 6 )/(205 x 10 9 ) = 99.4 x 10-6 or 99.4 µε ε = L/L L= ε x L = 99.4 x 10-6 x 300 mm = mm 4.4. POISSONS' RATIO Consider a piece of ma terial in 2 dimensions as shown in figure 2. The stress in the y direction is σ y and there is no stress in the x direction. When it is stretched in the y direction, it causes the material to get thinner in all the other directions at right angles to it. This means that a negative strain is produced in the x direction. For elastic materials it is found that the applied strain ( ε y ) is always directly proportional to the induced strain ( ε x ) such that (e x )/(e y ) = - n where ν (Nu)is an elastic constant called Poissons' ratio. The strain produced in the x direction is e x = - ne y 5 Figure 3

7 SELF ASSESSMENT EXERCISE No.2 1. A metal wire is 1 mm diameter and 1 m long. It is stretched by a force of 2 N. Calculate the change in diameter. E = 90 GPa and ν = 0.3 (Answer -8.5 x 10-6 mm). Explain what happens to the resistance of the wire. 2. Explain the significance of Poisson's ratio to the behaviour of strain gauges. 5. RESISTIVITY The resistance of a conductor increases with length L and decreases with cross sectional area A so we may say R is directly proportional to L and inversely proportional to A. R = Constant x L/A The constant is the resistivity of the material ρ hence R = ρl/a Ohms SELF ASSESSMENT EXERCISE No.3 1. Calculate the resistance o f a copper wire 5 m long and 0.3 mm diameter. The resistivity is 1.7 x 10-8 Ohm metre. (Answer Ω) 2. Calculate the resistance of a nichrome wire 2 m long and 0.2 mm di ameter given ρ = 108 x 10-8 (Answer Ω) 6. TEMPERATURE COEFFICIENT OF RESISTANCE The resistance of conductors changes with temperature. This is a problem when strain gauge devices are used. Usually the resistance increases with temperature. The amount by which the resistance changes per degree per ohm of the original resistance is called the temperature coefficient of resistance and is denoted α. The units are Ohms per Ohm per degree. Let the resistance of a conductor be Ro at 0oC. Let the resistance be R1 at θ1 oc. The change in resistance = αθ1 Ro The new resistance is R1 = Ro + αθ1 Ro Let the resistance be R2 at θ2 oc. The change in resistance = αθ2 Ro The new resistance is R2 = Ro + αθ2 Ro If the temperature changes from θ1 to θ2 the resistance changes by R = R2 - R1 = (Ro + αθ2 Ro ) - ( Ro + αθ1 Ro ) DR = Ro adq SELF ASSESSMENT EXERCISE No.4 1. A resistor has a nominal resistance of 120 Ω at 0 o C. Calculate the resistance at 20 o C. Calculate the change in resistance when the temperature drops by 5 degrees. α = 6 x 10-3 Ω/Ω o C (Answers Ω and - 3.6Ω) 6

8 INSTRUMENTATION AND CONTROL This tutorial provides an overview of instrument sensors used in process and automatic control. It is useful to anyone studying measurement systems and instrumentation but it is provided mainly in support of the EC module D227 Control System Engineering. This tutorial is mainly descriptive. Control is a broad concept and the following might apply to an automated system such as a robot or to a process control system such as a pneumatic valve controlling the flow of steam in a pipe. On completion of this tutorial, you should be able to do the following. Explain a basic measurement system. Explain the basic working principles of a variety of temperature sensors. Explain the basic working principles of a variety of pressure sensors. Explain the basic working principles of a variety of speed transducers. Explain the basic working principles of a variety of flow meters. Explain the basic working principles of a variety of force gauges. Explain the basic working principles of a variety of displacement gauges. Explain the basic working principles of a variety of level (depth) gauges. Explain in some detail the theory and use of strain gauges. In order to complete the theoretical part of this tutorial, you must be familiar with basic mechanical and electrical science.

9 1. INTRODUCTION A basic instrument system consists of three elements: i ii iii SENSOR or INPUT DEVICE SIGNAL PROCESSOR RECEIVER or OUTPUT DEVICE This tutorial is devoted to input devices but you can never separate it from the rest of the system as in many cases they are all integral (e.g. a mechanical pressure gauge incorporates all of these elements). A block diagram of a basic system is shown but they are usually more complex. Figure 1 Most modern analogue equipment works on the following standard signal ranges. Electric 4 to 20 ma Pneumatic 0.2 to 1.0 bar Older electrical equipment use 0 to 10 V. Increasingly the instruments are digital with a binary digital encoder built in to give a binary digital output. Pneumatic signals are commonly used in process industries for safety especially when there is a risk of fire or explosion. The advantage of having a standard range or using digital signals is that all equipment may be purchased ready calibrated. For analogue systems the minimum signal (Temperature, speed, force, pressure and so on ) is represented by 4 ma or 0.2 bar and the maximum signal is represented by 20 ma or 1.0 bar. This tutorial is an attempt to familiarise you with the many types of input sensors on the market today. Usually such sensors are called PRIMARY TRANSDUCERS. Things that we commonly measure are: Temperature Speed Force Stress and Strain Mass or Weight Size or Volume Pressure Flow rate Movement, Velocity and Acceleration Level or Depth Density Acidity/Alkalinity Sensors may operate simple on/off switches to detect the following: Objects(Proximity switch) Hot or cold (thermostat) Empty or full (level switch) Pressure high or low (pressure switch) The block diagram of a sensor is shown below. Figure 2

10 2 TEMPERATURE TRANSDUCERS 2.1 THERMOCOUPLES When two wires with dissimilar electrical properties are joined at both ends and one junction is made hot and the other cold, a small electric current is produced proportional to the difference in the temperature. Seebeck discovered this effect. It is true no matter how the ends are joined so the cold end may be joined at a sensitive millivolt meter. The hot junction forms the sensor end. Figure 3 The picture shows a typical industrial probe with a flexible extension and standard plug. Figure 4 Peltier showed that heat is absorbed at the hot end and rejected at the cold end. Thompson showed that part of the e.m.f. is due to the temperature gradient in the wire as well as the temperature difference between the junctions. Most thermocouple metals produce a relationship between the two temperatures and the e.m.f as follows. e = α(θ 1 - θ 2 ) + β(θ θ 2 2 α and β are constants for the type of thermocouple. The relationship is nearly linear over the operating range. The actual characteristic and suitable operating temperatures depends upon the metals used in the wires. The various types are designated in international and national standards. Typical linear operating ranges are shown for standard types. It is important that thermocouples are standard so that the same e.m.f will always represent the same temperature.

11 Type J Type K Type T Type E Type R/S Type B Type N Type L 0 to 800 o C 0 to 1200 o C -199 to 250 o C 0 to 600 o C 0 to 1600 o C 500 to 1800 o C 0 to 1200 o C 0 to 800 o C Thermocouples come in several forms. They may be wires insulated from each other with plastic or glass fibre materials. For high temperature work, the wire pairs are put inside a tube with mineral insulation. For industrial uses the sensor comes in a metal enclosure such as stainless steel. 2.2 RESISTANCE TYPE SENSORS Figure 5 These work on the principle that the electrical resistance of a conductor change with temperature. If a constant voltage is applied to the conductor then the current flowing through it will change with temperature. The resistivity of the conductor change with temperature. This usually means the resistance gets bigger as the conductor gets hotter. The following law relates the resistance and temperature. R = R o (1 + αθ) α is the temperature coefficient of resistance. R o is the resistance at 0 o C. Sometimes the equation is given as R = R o (1 αθ - βθ 2 ) A basic temperature sensor is made by winding a thin resistance wire into a small sensor head. The resistance of the wire then represents the temperature. This has an advantage over a thermocouple in that it is unaffected by the temperature of the gauge end. The main type of wire used is PLATINUM. The sensors are usually manufactured to have a resistance of 100 Ω at 0 o C and the value of α is to A typical operating range is to 400 o C. A special type of resistance sensor is called a THERMISTOR. They are made from a small piece of semiconductor material. The material is special because the resistance changes a lot for a small change in temperature and so can be made into a small sensor and it costs less than platinum wire. The temperature range is limited. They are only used for a typical range of -20 to 120 o C and are commonly used in small hand held thermometers for every day use. The relationship between resistance and temperature is of the form R = Ae B/θ

12 WORKED EXAMPLE No.1 A Platinum resistance thermometer has a resistance of 100 Ω at 0 o C and the value of α is In operation the resistance is 101 Ω. Calculate the temperature. SOLUTION Rearrange the formula to make θ the subject and evaluate. R R 100 o? = o = = C a WORKED EXAMPLE No.2 A thermocouple produces an e.m.f. in mv according to the temperature difference between the sensor tip θ 1 and the gauge head θ 2 such that e = α(θ 1 -θ 2 ) + β(θ 1 2 -θ2 2 ) α = 3.5 x 10-2 and β = 8.2 x 10-6 The gauge head is at 20 o C. The mv output is 12 mv. Calculate the temperature at the sensor. SOLUTION 10 = = 0.035? 10 = 8.2x10 6 (? 20) x ? x ? x 10? ? = 0 Solving the quadratic equation yields θ 1 = 251 o C (? 1 6 2? 1 2 ) SELF ASSESSMENT EXERCISE No.1 1. A thermocouple produces an e.m.f. in mv according to the temperature difference between the sensor tip θ 1 and the gauge head θ 2 such that e = α(θ 1 -θ 2 ) + β(θ 1 2 -θ2 2 ) Given α = 3.5 x 10-2 and β = 8.2 x 10-6 determine the mv output when the tip is at 220 o C and the gauge head at 20 o C. (Answer mv) 2. Describe the basic construction of a resistance type temperature sensor and state the reason why it is unaffected by the temperature of the gauge head. 3. State two reasons why instrument systems use standard transmission signal of either 4-20 ma or bar.

13 2.3 LIQUID EXPANSION and VAPOUR PRESSURE SENSORS These are thermometers filled with either a liquid such as mercury or an evaporating fluid such as used in refrigerators. In both cases the inside of the sensor head and the connecting tube are completely full. Any rise in temperature produces expansion or evaporation of the liquid so the sensor becomes pressurised. The pressure is related to the temperature and it may be indicated on a simple pressure gauge. Ways and means exist to convert the pressure into an electrical signal. The movement may also directly operate a thermostat. These instruments are robust and used over a wide range. They can be fitted with electric switches to set off alarms. 2.4 BIMETALLIC TYPES Figure 6 It is a well-known principle that if two metals are rigidly joined together as a two-layer strip and heated, the difference in the expansion rate causes the strip to bend. Figure 7 In the industrial type, the strip is twisted into a long thin coil inside a tube. One end is fixed at the bottom of the tube and the other turns and moves a pointer on a dial. The outward appearance is very similar to the pressure type. They can be made to operate limit switches and set off alarms or act as a thermostat. (e.g. on a boiler).

14 2.5 GLASS THERMOMETER The ordinary glass thermometer is also a complete system. Again the bulb is the sensor but the column of liquid and the scale on the glass is the processor and indicator. Mercury is used for hot temperatures and coloured alcohol for cold temperatures. The problems with glass thermometers are that they are Figure 8 Brittle Mercury solidifies at -40 o C. Alcohol boils at around 120 o C. Accurate manufacture is needed and this makes accurate ones expensive. It is easy for people to make mistakes reading them. Glass thermometers are not used much now in industry but if they are, they are usually protected by a shield from accidental breakage. In order to measure the temperature of something inside a pipe they are placed in thermometer pockets.

15 3. PRESSURE TRANSDUCERS Pressure sensors either convert the pressure into mechanical movement or into an electrical output. Complete gauges not only sense the pressure but indicate them on a dial or scale. Mechanical movement is produced with the following elements. Bourdon Tube. Spring and Piston. Bellows and capsules. Diaphragm BOURDON TUBE Figure 9 The Bourdon tube is a hollow tube with an elliptical cross section. When a pressure difference exists between the inside and outside, the tube tends to straighten out and the end moves. The movement is usually coupled to a needle on a dial to make a complete gauge. It can also be connected to a secondary device such as an air nozzle to control air pressure or to a suitable transducer to convert it into an electric signal. This type can be used for measuring pressure difference.

16 GATE Study Material Instrumentation And Control (Instrumentation Engineering) 84% OFF Publisher : Faculty Notes Author : Panel Of Experts Type the URL : Get this ebook