Tutorial 5 Principles of Transducers and Temperature Measurement

Transcription

1 Tutoial 5 Pinciples of Tansuces an Tempeatue Measuement Chapte 5 Pinciples of Tansuces Example 5.1 (Johnson 26) A potentiometic isplacement senso is use to measue wokpiece motion fom to 1 cm. The esistance changes linealy ove this ange fom to 1 kω. Develop signal conitioning to povie a linea, - to 1-V output. 15V 25Ω 51Ω Senso 5.1V Op-Amp Figue 5.1 The key thing is to not lose the lineaity of the esistance vesus isplacement. We cannot put the vaying esistance in a ivie to pouce a vaying voltage because the voltage vaies nonlinealy with esistance. Remembe though that the output voltage of an inveting amplifie vaies linealy with the feeback esistance. Theefoe, let s put the senso in the feeback of a simple inveting amplifie. Then we woul have something like R 2 Vout = V in R1 We can now get i of the pesky negative by using Vin as a constant negative voltage, say 5.1 volts fom a zene ioe. Then we pick R 1 to give the esie output, 1 volts at 1 kω (1 cm), 1 1 = ( 5.1) so R 1 = 51 Ω R 1 Example 5.2 The following figue shows a capacitive isplacement senso esigne to monito small changes in wok-piece position. The two metal cylines ae sepaate by a plastic sheath/beaing of thickness 1 mm an ielectic constant at 1 khz of 2.5. If the aius of 2.5 cm, fin the sensitivity in pf/m as the uppe cyline slies in an out of the lowe cyline. What is the ange of capacity if h vaies fom 1. to 2. cm? ( ε = pf/m) The capacity is given by the following equation A C =εε 1

2 Displacement h Figue 5.2 The effective aea is the aea of the shae cylinical aea, which has a aius,, an height, h. Thus, A = 2πh, so the capacity can be expesse as h C= 2πKε The sensitivity with espect to the height, h, is efine by how C changes with h, that is, it is given by the eivative C = 2πKε h Substituting fo the given values, we get 2 C m = 2π ( 2.5)( 8.85 pf/m) = 3475 pf / m 3 h 1 m Since the function is linea with espect to h, we fin the capacity ange as C min = (3475 pf/m)(1 2 m) = pf to C max = (3475 pf/m)(2 1 2 m) = 69.5 pf. Example 5.3 An LVDT has a maximum coe motion of ± 1.5 cm with a lineaity of ±.3% ove that ange. The tansfe function is 23.8 mv/mm. If use to tack wok-piece motion fom 1.2 to cm, what is the expecte output voltage? What is the uncetainty in position etemination ue to nonlineaity? Using the known tansfe function, the output voltages can easily be foun, V( 1.2cm) = (23.8 mv/mm)( 12mm) = mv an V(1.4cm) = (23.8 mv/mm)(14mm) = 333 mv The lineaity eviation shows up in eviations of the tansfe function. Thus, the tansfe fuction has an uncetainty of (±.3)(23.8mV/mm) = ±.714mV/mm This means that a measue voltage, V m (in mv), coul be intepete as a isplacement that anges fom Vm/23.73 to Vm/23.8 mm, which is appoximately ±.3%, as expecte. Thus, if the senso output was 33 mv, which is nominally 1.4 cm, the actual coe position coul ange fom to cm. Example 5.4 The level of ethyl alcohol is to be measue fom to 5 m using a capacitive system such as shown in the following figue. 2

3 C h Figue 5.3 The following specifications efine the system: fo ethyl alcohol: ε = 26 (fo ai, ε = 1) cyline sepaation: =.5 cm plate aea: A = 2πRL whee R = 5.75 cm = aveage aius, L = istance along cyline axis Fin the ange of capacity vaiation as the alcohol level vaies fom to 5 m. The capacity is given by A C =εε Theefoe all we nee to o is to fin the capacity fo the entie cyline with no alcohol an then multiply that by 26. A = 2πRL = 2π(.575m)(5m) = 1.86 m 2 Thus, fo ai, C = (1)(8.85 pf/m)(1.86m 2 /.5m) = 3196 pf.32 µf With the ethyl alcohol, the capacity becomes C = 26(.32 µf) =.832 µf The ange is.32 to.832 µf. Execise 5.1 The output voltage of a potentiomete-type esistance tansuce is to be measue by a ecoe having an input esistance of 2 kω. If the eo of measuement is not to excee 2% at 5% f.s., etemine esistance value of the potentiomete. [1.633kΩ] Execise 5.2 The following is a typical specification fo potentiomete-type esistance tansuce. Examine the specification an explain the meaning an significance of each item. Type wie-woun esistance isplacement potentiomete Teminal esistance 1kΩ Range -25mm Resolution.4% Powe ating.25w Maximum wipe cuent 15mA Themal ift.5% pe o C Life expectancy 1 8 cycles Execise 5.3 A linea vaiable iffeential tansfome is excite with a 1 Hz 6V peak-to-peak wavefom. The input coe motion is sinusoial at 1 Hz an has a isplacement amplitue of ± 3 3

4 mm. If the l.v..t sensitivity is 2 V/mm, aw the wavefoms of the excitation voltage, input isplacement an output voltage. Execise 5.4 The specifications fo the l.v..t in Execise 5.3 ae as follows: Lineaity:.4% Resolution: infinite Resiual voltage:.5% Dift bette than.1% pe o C Output impeance: 2.5kΩ Response time: 1ms Explain the meaning an significance of the specifications. See attache sheet fo an example fo LVDT specifications. Execise 5.5 (a) Descibe the pinciple of opeation an constuction etails of the piezoelectic (quatz) tansuce. (b) A quatz pessue tansuce has a sensitivity of 8 pc/ba. If, when the input pessue is 3 bas, an output voltage of 1 V is pouce, etemine the capacitance of the evice. [24 pf] Chapte 6 Tempeatue measuement Execise 6.1 Explain the pinciple of opeation of the themocouples. Execise 6.2 Explain the pinciple of opeation of the esistance tempeatue etectos. Execise 6.3 Explain the pinciple of opeation of the themistos. Execise 6.4 Explain the pinciple of opeation of the aiation pyometes. Execise 6.5 A type-k themocouple is expose to a tempeatue of 12 o C. If the inicato is use as the col junction an its tempeatue is 5 o C. Use the following figue to calculate the e.m.f inicate. [47mV] 8 7 Type E 6 5 Type K Type T Type S Figue 6.1 Themocouple chaacteistics (efeence o C) 4

5 Execise 6.6 Using the above figue, etemine (a) the sensitivity of the type-t themocouple in the ange of o C to 3 o C, (b) the sensitivities of the type-e an type-s themocouples in the ange of 4 o C to 1 o C. [.5mV/ o C] [.8mC/ o C] [.1mV/ o C] Execise 6.7 If the ynamic elationship between the measue tempeatue θ 2 an the efeence tempeatue θ 1 fo a themomete is given by θ 2 = k ( θ 1 θ 2) whee k =.2s -1 t etemine the time constant an static sensitivity fo the themomete. [5s] Execise 6.8 A mecuy themomete use to measue the tempeatue of a liqui has ynamics epesente by the following equation: Tm 1 1 CT + Tm = Ti t R T R T whee T m ( o C) is the tempeatue of mecuy, an T i ( o C) a change in the tempeatue of the liqui, R T an C T ae constant. Fin the time constant, sensitivity an the ynamic esponse of the themomete T m. using these numeical values: T i = 1 o C, R T = 131 o C/W, an C T =.56 J/ o C. Execise 6.9 An RTD has α =.5 / o C, R = 5 Ω an a issipation constant of P D = 3 mw/ o C at 2 o C. The RTD is use in a bige cicuit as shown in the following figue, with R 1 = R 2 = 5 Ω, an R 3 is a vaiable esisto use to null the bige. If the supply is 1 V an the RTD is place in an ice bath at o C: a) Fin the value of R 3 to null the bige [R 3 = Ω]. Powe supply 1V R 1 V R 2 R 3 RTD Figue 6.2 RTD with a bige cicuit fo Execise 6.9 b) Fin the output voltage measue by a voltmete (R v = ) with the above value of R 3 if the tempeatue is 1 o C. Consie the effect of the self-heating an calculate the eo of the RTD at 1 o C. Hints: An RTD is a esistance, theefoe thee is an I 2 R powe issipate by the evice itself that causes a light heating effect, a self-heating. This may also cause an eoneous eaing o even upset the envionment in elicate measuement conition. Thus, the cuent though the RTD must be kept sufficiently low an constant to avoi self-heating. Typically, a issipation constant is povie in RTD specifications. This numbe elates the powe equie to aise the RTD 5

6 tempeatue by one egee of tempeatue. Thus, a 25-mW/ o C issipation constant shows that if I 2 R powe losses in the RTD equal 25 mw, the RTD will be heate by 1 o C. The issipation constant is usually specifie une two conitions: fee ai an a well-stie oil bath. This is because of the iffeence in capacity of the meium to cay heat away fom the evice. The self-heating tempeatue ise can be foun fom the powe issipate by the RTD, an the issipation constant fom: whee P T = P D T = tempeatue ise because of self-heating in o C P = powe issipate in the RTD fom the cicuit in W = issipation constant of the RTD in W/ o C P D Execise 6.1 (Themisto) A themisto is to monito oom tempeatue. It has a esistance of 3.5 kω at 2 o C with a slope of 1%/ o C. The issipation constant is P D = 5 mw/ o C. It is suppose to use the themisto in the ivie of the following figue to povie a voltage of 5. V at 2 o C. Evaluate the effects of selfheating [3. kω; V D = 4.6 V]. 1 V 3.5 kω Themisto R TH V D = 5 V Figue 6.3 Themisto in a ivie cicuit fo Execise 6.1 Execise 6.11 (Themocouples) Fin the emf fo a mateial with α = 5 µv/oc if the junction tempeatues ae 2 o C an 1 o C. Hint: The themoelectic effect is expesse by the following equation: T2 T1 ( ) ε= Q Q T A B whee ε = emf (also calle Seebeck emf) pouce in volts T 1, T 2 = junction tempeatues in K Q A, Q B = themal tanspot constants of the two metals This equation, which escibe the Seebect effect, shows that the emf pouce is popotional to the iffeence in tempeatue an, futhe, to the iffeence in the metallic themal tanspot 6

7 constants. Thus, if the metals ae the same, the emf is zeo, an if the tempeatue ae the same, the emf is zeo. In pactice, it is foun that the two constants, QA an QB, ae nealy inepenent of tempeatue an that an appoximate linea elationship exists as ε=α( T2 T1) whee α = constant in V/K T 1, T 2 = junction tempeatues in K Execise 6.12 A voltage of mv is measue with a type K themocouple at a o C efeence. Fin the tempeatue of the measuement junction. [572.1 o C] Hints: use the attache sheet with the following intepolation equations: Tempeatue: TH T L TM = TL + ( VM VL) VH VL whee V M = measue voltage that lies between a highe voltage, V H, an a lowe voltage, V L, which ae in the tables. The tempeatues coesponing to these voltages ae T H an T L, espectively. Voltage: V = V V + V T T ( ) H L M L M L TH TL Execise 6.13 Fin the voltage of a type J themocouple with a o C efeence if the junction tempeatue is 172 o C. [ 7.18 o C] Execise 6.14 An RTD has α (2 o C) =.4 / o C. If R = 16 Ω at 2 o C, fin the esistance at 25 o C. 7