Mechanical Measurements. Module 2:

Transcription

1 Mechanical Measurements Module : 1. Thermometry Formally we start the study of Mechanical Measurements now! Module will consider the measurement of field quantities like temperature, pressure and fluid velocity. First topic to be covered is Thermometry or the Science and Art of Temperature Measurement Sub Module.1 1. Thermometry or the science and art of temperature measurement Preliminaries Temperature along with pressure is an important parameter that governs many physical phenomena. Hence the measurement of temperature is a very important activity in the laboratory as well as in industry. The lowest temperature that is encountered is very close to 0 K and the highest temperature that may be measured is about K. This represents a very large range and cannot be covered by a single measuring instrument. Hence temperature sensors are based on many different principles and the study of these is the material of this sub module. We take recourse to thermodynamics to provide a definition for temperature of a system. Thermodynamics is the studies of systems in equilibrium and temperature is an important intensive property of such systems. Temperature is defined via the so called zeroth law of

2 thermodynamics. A system is said to be in equilibrium if its properties remain invariant. Consider a certain volume of an ideal gas at a specified pressure. When the sate of this volume of gas is disturbed it will eventually equilibrate in a new state that is described by two new values of volume and pressure. Even though we may not be able to describe the system as it is undergoing a change we may certainly describe the two end states as equilibrium states. Imagine two such systems that may interact through a wall that allows changes to take place in each of them. The change will manifest as changes in pressure and/or volume. If, however, there are no observable changes in pressure and volume of each one of them when they are allowed to interact as mentioned above, the two systems are said to be in equilibrium with each other and are assigned the same temperature. The numerical value that is assigned will have to follow some rule or convention as we shall see later. The zeroth law of thermodynamics states that is a system C is in equilibrium separately with two thermodynamic systems A and B then A and B are also in equilibrium with each other. At once we may conclude that systems A and B are at the same temperature! Thermometry thus consists of using a thermometer (system C) to determine whether or not two systems (A and B in the above) are at the same temperature. Principle of a thermometer Principle of any thermometer may be explained using the facts indicated in Fig. 1. Consider a system whose state is fixed by two properties coordinates X and Y. It is observed that several pairs of values of X, Y will be in equilibrium with a second system of fixed temperature (or a fixed state). These multiplicity of sates must all be characterized by the same temperature and hence represent an isotherm.

3 Y Isotherms T 1 T T 3 T 4 Y = Y 0 X 1 X X 3 X 4 X Figure 1 Principle of thermometry explained Assume that one of the coordinates of the system (Y) is fixed at a value equal to Y 0. Then there is only one sate that will correspond to any given isotherm. If the system is allowed to equilibrate with a system characterized by different isotherms, the property X will change as indicated by the points of intersection X 1, X and so on. These will then correspond to the respective temperatures T 1, T and so on. We refer to X as the thermometric property and the system as a thermometer.

4 Table 1 Thermometers and thermometric properties Thermometer Thermometric Symbol property Gas at constant volume Pressure P Electric resistance under Electrical resistance R constant tension Thermocouple Thermal electromotive E force Saturated vapor of a pure Pressure P substance Blackbody radiation Spectral emissive power Eb, λ Acoustic thermometer Speed of sound a Table 1 shows several thermometers that are actually used in practice. The thermometric property as well as the symbol that is used to indicate it is also shown in the table. Constant volume gas Thermometer Figure Schematic of a constant volume gas thermometer

5 We look at the constant volume gas thermometer in some detail now. Schematic of such a thermometer is shown in Fig.. It consists of a certain volume of a gas contained in a rigid vessel. The vessel is connected to a U tube manometer with a side tube and height adjustable reservoir as shown. The volume is kept constant by making the meniscus in the left limb of the U tube always stay at the mark mad3e on the left limb of the U tube. The right limb has a graduated scale attached to it as shown. The gas containing vessel is immersed in a constant temperature environment. The graduated scale helps in determining the pressure of the confined gas in terms of the manometer head. The following experiment may be performed. Choose the pressure of the gas to have a definite value when the constant temperature environment corresponds to standard fixed state such as the triple point of water (or the ice point at one atmosphere pressure). Now move the thermometer into an environment at the steam point (boiling point of water at one atmosphere). The pressure of the gas has to be adjusted to a higher value than it was earlier by adjusting the height of the reservoir suitably so as to make the meniscus in the left limb of the U tube stay at the mark. The above experiment may be repeated by taking less and less gas to start with by making the pressure at the triple point of water to have a smaller and smaller value (the vessel volume is the same in all the cases). The experiment may also be repeated with different gases in the vessel. The result of such an experiment gives a plot as shown in Fig. 3.

6 pst/ptp Gas A Gas B Gas C p tp Figure 3 Gas thermometer characteristics The ratio of the pressure of the gas corresponding to the steam point to that at the triple point of water tends to a unique number as p tp 0 (the intercept on the pressure ratio axis) independent of the particular gas that has been used. This ratio has been determined very accurately and is given p by st tp ptp as p 0. The gas thermometer temperature scale is defined based on this unique ratio and by assigning a numerical value of Ttp 0 = K or 0.01 C. The defining relation is Tst pst as p tp 0 Ttp ptp = (1) This last value is referred to as the single fixed point for thermometry or the primary fixed point. At this temperature ice, liquid water and water vapor all coexist, if in addition, the pressure is maintained at 4.58 mm Mercury column or Pa. The ice point is at K or 0 C and was used in early times as the primary fixed point in thermometry. In

7 ordinary laboratory practice the ice point is easier to achieve and hence is commonly used. Equation 1 may be generalized to define the constant volume gas thermometer temperature scale as T p = as p tp 0 () T p tp tp Thus the temperature ratio and pressure ratios are the same in the case of a constant volume gas thermometer. The latter is measured while the former is inferred. The message thus is clear! A measurable property that varies systematically with temperature is used to infer the temperature! The measured property is termed the thermometric property.

8 Example1 In determining the melting point of a certain alloy with a gas thermometer, an investigator finds the following values of the pressure p when the pressure p tp at the triple point of water has the indicated value. p tp p If the triple point of water is taken as K, what is the melting point of the alloy? o In order to determine the melting point we need the limiting value of the ratio p p tp as ptp 0. This value is obtained by extrapolation. The ratios are calculated and are given by the following table. p p p tp Difference o Since the common differences are constant we may extrapolate linearly to get p lim = =.31 p tp 0 p tp

9 Practical thermometry We have mentioned earlier that the range of temperatures encountered in practice is very wide. It has not been possible to device a single thermometer capable of measuring over the entire range. Since all thermometers must be pegged with respect to the single fixed point viz. the temperature at the triple point of water it is necessary to assign temperature values to as many reproducible states as possible using the constant volume gas thermometer. Subsequently these may be used to calibrate other thermometers that may be used to cover the range of temperatures encountered in practical thermometry. These ideas are central to the introduction of International Temperature Scale 1990 (or ITS90, for short). The following is a brief description of ITS90. Specification of ranges and corresponding thermometers according to ITS 90 Between 0.65 K and 5.0 K T 90 is defined in terms of the vapor-pressure temperature relations 3He and 4He. Between 3.0 K and the triple point of neon ( K) T 90 is defined by means of a helium gas thermometer calibrated at three experimentally realizable temperatures having assigned numerical values (defining fixed points) and using specified interpolation procedures. Between the triple point of equilibrium hydrogen ( K) and the freezing point of silver ( ºC) T90 is defined by means of platinum

10 resistance thermometers calibrated at specified sets of defining fixed points and using specified interpolation procedures Above the freezing point of silver (961.78ºC) T 90 is defined in terms of a defining fixed point and the Planck radiation law. It is noted that the above uses several secondary fixed points to define the temperature scale. These are shown in Table. Table Secondary fixed points used in ITS90 Equilibrium state Triple point of H Boiling point of H at 50 mm Hg Boiling point of H at 1 atmosphere Triple Point of Ne Triple point of O Triple Point of Ar T 90 K T 90 o C Equilibrium state T 90 K T 90 o C Triple point of Hg Triple point of H O Melting point of Ga Freezing point of In Freezing point of Sn Freezing point of Al Freezing point of Ag Even though the ITS90 specifies only a small number of thermometers, in practice we make use of many types of thermometers. These are discussed in detail below.

11 How do we make a thermometer? Properties that vary systematically with temperature may be used as the basis of a thermometer. Several are listed here. Thermoelectric thermometer Based on thermoelectricity - Thermocouple thermometers using two wires of different materials Electric resistance Resistance thermometer using metallic materials like Platinum, Copper, Nickel etc. Thermistors consisting of semiconductor materials like Manganese- Nickel-cobalt oxide mixed with proper binders Thermal expansion Bimetallic thermometers Liquid in glass thermometer using mercury or other liquids Pressure thermometer Pyrometry and spectroscopic methods Radiation thermometry using a pyrometer Special methods like spectroscopic methods, laser based methods, interferometry etc.

12 Sub Module. Thermoelectric thermometry Thermoelectric thermometry is based on thermoelectric effects or thermoelectricity discovered in the 19 th century. They are: Seebeck effect discovered by Thomas Johann Seebeck in 181 Peltier effect discovered by Jean Charles Peltier in 184 Thomson effect discovered by William Thomson (later Lord Kelvin) in 1847 The effects referred to above were all observed experimentally by the respective scientists. All these effects are reversible unlike heat diffusion (conduction of heat) and Joule heating (due to electrical resistance of the material) which are irreversible. In discussing the three effects we shall ignore the above mentioned irreversible processes. It is now recognized that these three effects are related to each other through the Kelvin relations. Thermoelectric effects: A Current I T 1 T B Figure 4 Sketch to explain Peltier effect Consider two wires of dissimilar materials connected to form a circuit with two junctions as shown in Fig. 4. Let the two junctions be maintained at

13 1 T T different temperatures as shown by the application of heat at the two junctions. An electric current will flow in the circuit as indicated with heat absorption at one of the junctions and heat rejection at the other. This is referred to as the Peltier effect. The power absorbed or released at the junctions is given by P= Q =±π ABI where π AB is the Peltier voltage (this expression defines the Peltier emf), P Q P is rate at which the heat absorbed or rejected. The direction of the current will decide whether heat is absorbed or rejected at the junction. For example, if the electrons move from a region of lower energy to a region of higher energy as they cross the junction, heat will be absorbed at the junction. This again depends on the nature of the two materials that form the junction. The subscript AB draws attention to this fact! The above relation may be written for the two junctions together as ( T,T) 1 π =π π (3) A Current I T 1 T B Figure 5 Sketch to explain Thomson effect Note that the negative sign for the second term on the right hand side is a consequence of the fact that the electrons move from material A to material B at junction 1 and from material B to material A at junction. Consider now a single conductor of homogeneous material (wire A alone of Fig. 4) in which a temperature gradient exists. The current I is maintained by heat absorption or heat rejection along the length of the wire.

14 Note that if the direction of the current is as shown the electrons move in the opposite direction. If T > T1, the electrons move from a region of higher temperature to that at a lower temperature. In this case heat will be rejected from the wire. The expression for heat rejected isqt = I σ AdT where QT the T T1 Thomson heat is and σ A is the Thomson coefficient for the material. A similar expression may be written for the Thomson heat in conductor B. A T 1 T B V S Figure 6 Sketch to explain the Seebeck effect If we cut conductor B (or A) as indicated the Seebeck emf appears across the cut. This emf is due to the combined effects of the Peltier and Thomson effects. We may write the emf appearing across the cut as T T1 S = P + T = ( πab) ( π T AB) + A B T1 + T 1 T V V V σ dt σ dt T AB T AB T A B 1 T 1 ( ) ( ) ( ) = π π + σ -σ dt (4) We define the Seebeck coefficient dvs α AB through the relation dt =α AB. In differential form, equation 4 may then be rewritten as (assume T T 1 = dt)

15 T B dv dt S dπ =α = + σ σ dt ( ) ( ) AB AB A B dv = dπ + σ σ dt S AB A B (5) Kelvin relations: Since the thermoelectric effects (Peltier and Thomson effects) are reversible in nature there is no net entropy change in the arrangement shown in Fig. 4.The entropy changes are due to heat addition or rejection at the junctions due to Peltier effect and all along the two conductors due to Thomson effect. The entropy change due to Peltier effect may be obtained as follows: At junction 1, the entropy change is s P 1 Q P 1 πab = = I. Similarly at T T 1 1 Q π P BA πab junction the entropy change is sp = = I = I. Again if we assume 1 T T T that the temperature difference is T T 1 = dtthe net change in entropy is dsp π = Id. The net change in entropy due to Thomson heat in the two T σ σ A conductors may be written as ds = I dt. Combining these two we get T π σa σb ds=dsp + dst = I d + dt 0 T T = dπab dt σ I A σb = π AB + dt = 0 T T T (6) The total entropy change is equated to zero since both the thermoelectric processes are reversible. The current I can have arbitrary value and hence the bracketed term must be zero.

16 From equation 5 ( ) σ σ dt = dv dπ. Introducing this in equation 6 we A B s AB get dπab dt dvs dπab π AB + = 0 T T T T or dvs π AB = T = Tα AB = Tα S (7) dt We have renamed the Seebeck coefficient as αs according to normal practice. Differentiating equation 7 we get dπ AB = Tdα s +α s dt. Introduce this in equation 5 to get ( ) ( ) s A B dv = Tdα +α dt + σ σ dt =α dt or s s s A B s dα σ σ = T (8) dt Equations 7 and 8 constitute the Kelvin relations. How do we interpret the Kelvin relations? The Seebeck, Peltier and Thomson coefficients are normally obtained by experiments. For this purpose we use the arrangement shown in Fig. 6 with the junction labeled maintained at a suitable reference temperature, normally the ice point (0 C). The junction labeled 1 will then be called the measuring junction. Data is gathered by maintaining the measuring junction at different temperatures and noting down the Seebeck voltage. If the measuring junction is also at the ice point the Seebeck voltage is identically equal to zero. The data is usually represented by a polynomial of suitable degree.

17 For example, with Chromel (material A) and Alumel (material B) as the two wire materials, the expression is a quartic of 3 4 S form V = a t+ a t + a t + a t where The Seebeck voltage is in µv and the temperature is in C. An inverse relation is also used in practice in the form S S 3 S 4 S t = A V + A V + A V + A V. Two examples follow. We shall see later that the coefficients in the polynomial are related to the three thermoelectric effects.

18 Example For the Chromel-Alumel pair the Seebeck voltage varies with temperature according to the fourth degree polynomial S 3 V = t t t t The Seebeck voltage is in µv while the temperature is in C. Discuss the behavior of this thermocouple near the ice point. o It is clear that, near the ice point, the Seebeck coefficient (it is also called the thermoelectric power) is. dvs αs = 39.4 dt µ V C o Using the Kelvin relations, we also have the following: dαs dvs A B = Chromel Alumel = = σ σ σ σ T -T dt dt 1. ( ) ( ). π =π A B Chromel Alumel = Tα T S

19 o Note that T in the Kelvin relation is in Kelvin and t in the polynomial is in 0 C. Also note that d d. Hence the second derivative of the dt dt Seebeck voltage is given by S dvs dv dt = = a + 6a3t + 1a4t dt o Near the ice point we may take t = 0 and write ( ) σ σ ( )( )(73.15) Chromel Alumel = 3.06 µv. and π = T α = (73.15)( ) = µ V V Chromel Alumel S

20 Example 3 The thermocouple response shown below (Copper Constantan thermocouple with the cold junction at the ice point) follows the law V S = a t + b t. Obtain the parameters a and b by least squares. Here t is in C and VS is in mv. Temperature, C V S, mv o Since the fit follows the form specified above, it is equivalent to a linear relation between E= V S /t and t. Since V S /t is a small number we shall work with 100 V S /t and denote it as y. We shall denote the temperature as x. The following table helps in evolving the desired linear fit. x=t y=100v S /t x y xy y fit Sum: Mean = (Note: I have used EXCEL to solve the problem) o The statistical parameters are calculated and presented in the form of a table:

21 Variance of x= Variance of y= Covariance =.735 Slope of fit line = Intercept of fit line = o Using the fit parameters the calculated values of y fit are shown in the last column of the first table. Reverting back to t and V S, we get the following relation: V = t S 5 ( t ) / t t o We compare the data with the fit in the table below: t V S V S (fit) o Standard error of the fit with respect to data may be calculated as σ (V V 3 (fit)) S S Err = = = mv

22 o The 3 in the denominator is the degrees of freedom. A plot is made to compare the data and the fit. It is clear that the thermocouple behavior is mildly nonlinear. The standard error of fir translates to approximately ±1 C! VS VS(fit) Seebeck voltage mv Temperature o C o The inverse relation may similarly be obtained by fitting a linear relation between t V S and V S. This is left as an exercise to the student. The result obtained is S S t = 5.173V V with a standard error of σ t =±.09 C.

23 Example has shown that the thermoelectric data may be expressed in terms of global polynomial to facilitate interpolation of data. A simple quadratic fit has been used to bring home this idea. In practice the appropriate interpolating polynomial may involve higher powers such that the standard error is much smaller than what was obtained in Example. As an example, the interpolating polynomial recommended for the K type thermocouple with the reference junction at the ice point is given as: s V = t t t t This is a fourth degree polynomial and passes through the origin. The Seebeck voltage is given in µv and the temperature is in C. Using Kelvin relations, the appropriate parameters are calculated near the ice point as: AB dvs V / O C α S = = µ dt π = Tα = S = µ V dαs σa σ B = T = dt = 3.064µ V The variation of the Seebeck coefficient over the range of this thermocouple is given in Fig. 7 below

24 Seebeck coefficient µv/ o C Temperature o C Figure 7 Variation of Seebeck coefficient with temperature for K type thermocouple A small excerpt from a table of Seebeck voltages is taken and the corresponding fit values as calculated using the above fourth degree polynomial is given in Table 1. Note that the voltages in this table are in mv. Table 3 comparison of actual data with fit t V S V S (fit) The maximum deviation is some 0.10 mv. The standard error is approximately ±0.053 mv! This translates to roughly an error of ±1. C.

25 The above shows that the three effects are related to the various terms in the polynomial. The Seebeck coefficient and the Peltier coefficients are related to the first derivative of the temperature. The contributions to the first derivative from the higher degree terms are not too large and hence the Seebeck coefficient is a very mild function of temperature. In the case of K type thermocouple this variation is less than some % over the entire range of temperatures. The Thomson effect is related to the second derivative of the polynomial with respect to temperature. The value of this is again small and varies from -3. V at 0 C to a maximum value of V at 1350 C.

26 Sub Module.3 Resistance Thermometry Resistance thermometry depends on the unique relation that exists between resistance of an element and the temperature. The resistance thermometer is usually in the form of a wire and its resistance is a function of its temperature. Material of the wire is usually high purity Platinum. Other materials also may be used. The resistance variation of different materials is indicated by Table 1. Table 1 Resistance variation of different wire materials Material Temperature Range C Element Resistance in Ohms at 0 C Nickel -60 to Copper -30 to Platinum -00 to Element Resistance in Ohms at 100 C Platinum resistance thermometer is also referred to as (Platinum Resistance Thermometer) PRT (or PT) or Resistance Temperature Detector (RTD). Usually the resistance of the detector at the ice point is clubbed with it and the thermometer is referred to as, for example, PT100, if it has a resistance of 100 Ω at the ice point. The resistance of standard high purity Platinum varies systematically with temperature and it is given by the International standard calibration curve for wire wound Platinum elements: ( { } ) 3 O O t = < < R R 1 K t K t K t 100 t, 00 C t 0 C ( ) t 0 1 O O R = R 1+ K t+ K t,0 C< t < 50 C (9)

27 Where 3 O 7 O 1 O 4 1 = = = K / C; K / C ; K / C R The ratio 100 R 0 100R 0 is denoted by and is given as / C. It is seen that the resistance temperature relationship is non linear. The response of a Platinum resistance thermometer is usually plotted in the form of ratio of resistance at temperature t to that at the ice point as a function of temperature as shown in Fig. 1. The sensor in the Fig. is shown with a three wire is shown with three wire arrangement. The resistance sensor is also available with four wire arrangement. These two aspects will be discussed later Resistance ratio, R(t)/R Temperature, o C Figure 1 Characteristics of a Platinum resistance thermometer

28 Ceramic Powder 1 3 Platinum Element Protective Sheath Figure Typical PRT sensor schematic with three wire arrangement Platinum resistance thermometer and the Callendar equation As seen from equation (1) the Platinum resistance thermometer has essentially a non-linear response with respect to temperature. We define a temperature scale defined as the Platinum resistance temperature that is basically given by a linear scale defined through the relation] Rt R0 tpt = 100 R R (10) The quantities appearing in the above are: R Pt = Platinum resistance temperature, R t = Resistance of sensor at temperature t, R 0 = Resistance of the sensor at the ice point and R 100 = Resistance of the sensor at the ice point. Obviously the non-linearity will have to be taken into account to get the correct temperature from the linear value

29 obtained by Equation 10. This is done by applying a correction to the Platinum resistance temperature as suggested by Callendar. From Equation 9 we have Hence We then have ( ) R = R K K R R = 100K K. t = Kt 1 + Kt Kt Kt 100Kt + Kt R R Kt K t R R 100K 100 K = = K + 100K K + 100K Kt Kt 100Kt Kt K t t = t t = t K1+ 100K K1 K With the K s given earlier, we 7 K 4 have = 10 = = δ. Thus we have K Rt R0 t t 100 = t δ 1 R R This may be rephrased as (using the definition given in Equation 10) Rt R0 t t tpt tpt t 100 +δ 1 = tpt +δ 1 R100 R Correction c (11) This is referred to as the Callendar equation and the second term is the Callendar correction, represented as c. The Callendar correction is evidently zero at both the ice and steam points. The correction is non-zero at all other temperatures. Figure 10 shows the Callendar correction as a function of the Platinum resistance temperature over a useful range of the sensor.

30 Correction, c Platinum resistance temperature, t Pt Figure 10 Callendar correction as a function of t Pt

31 Example 4 The resistance of a Platinum resistance sensor of R 0 = 100Ω was measured to be119.4ω. This sensor has α value of What is the corresponding temperature without and with correction? o We have R 0 = 100Ω, α = and δ = o Hence ( ) ( ) R = R α = = 13.5Ω o The measured sensor resistance is given as R = 119.4Ω. o By definition the Platinum resistance temperature is Rt R O tpt = 100 = 100 = C R R o This is also the uncorrected value of the temperature. o The Callendar correction is calculated as tpt tpt O c =δ C 100 = = o The corrected temperature is thus given by t = t + c = = 50.0 C Pt O t

32 RTD measurement circuits The resistance of the resistance sensor is determined by the use of a DC bridge circuit. As mentioned earlier there are two variants, viz. the three wire and the four wire systems. These are essentially used to eliminate the effect of the lead wire resistances that may adversely affect the measurement. There are two effects due to the lead wires: 1) they add to the resistance of the Platinum element ) the resistance of the lead wires may also change with temperature. These two effects are mitigated or eliminated by either the three or four wire arrangements. The lead wires are usually of higher diameter than the diameter of the sensor wire to reduce the lead wire resistance. In both the three and four wire arrangements, the wires run close to each other and pass through regions experiencing similar temperature fields (refer Figure 9). Hence the change in the resistance due to temperature affects all the lead wires by similar amounts. The resistances of the lead wires are compensated by a procedure that is described below.

33 Bridge circuit for resistance thermometry: Three wire arrangement for lead wire compensation R 1 Battery R Milli- Ammeter R 3 Compensating Leads 3 1 RTD Figure 11 Bridge circuit with lead wire compensation (Three wire arrangement) Figure 11 shows the bridge circuit that is used with three lead wires. The resistances R 1 and R 3 are chosen to be equal and the same as R 0 of the RTD. Two lead wires (labeled and 3) are connected as indicated adding equal resistances to the two arms of the bridge. The third lead wire (labeled 1) is used to connect to the battery. Thus the bridge will indicate null (milliammeter will indicate zero) when R = R 0 when the RTD is maintained at the ice point. During use, when the RTD is at temperature t, the resistance R is adjusted to restore balance. If the lead wires have resistances equal to R s and R s3, we have

34 ( ) R + R = R + R or R = R + R R (1) t s3 s t s s3 If the two lead wires are of the same size the bracketed terms should essentially be zero and hence the lead wire resistances have been compensated. Four wire arrangement for lead wire compensation The four wire arrangement is a superior arrangement, with reference to lead wire compensation, as will be shown below. Figure 1 is the bridge arrangements that are used for this purpose. R 1 R 1 Battery R R R 3 Milli- Ammeter Battery Milli- Ammeter R 3 1 Compensating Leads Compensating Leads 4 1 (a) RTD (b) RTD Figure 1 Bridge circuit with lead wire compensation (four wire arrangement)

35 The choice of the resistances is made as given for the three wire arrangement. If the lead wires have resistances equal to R s1 - R s4, we have the following. Condition for bridge balance in arrangement shown in Figure 1(a): Rt + Rs4 = R(a) + Rs (1) For balanc arrangement a Condition for bridge balance in arrangement shown in Figure 1(b): Rt + Rs = R(b) + Rs4 (13) For balance arrangement b We see that by addition of Equations 5 and 6, we get R t R(a) + R(b) = (14) The lead wire resistances thus drop off and the correct resistance is nothing but the mean of the two measurements. Since the lead wire resistances actually drop off, the four wire scheme is superior to the three wire scheme.

36 Example 5 An RTD has α 0 = 0.004/ C. If R0 = 106 Ω (resistance at 0 C), determine the resistance at 5 C. The above RTD is used in a bridge circuit with R 1 = R = R 3 = 100 Ωand the supply voltage is 10 V. Calculate the voltage the detector must be able to resolve in order to measure a 1 C change in temperature around 0 C. o Note the definition of α viz. α 1dR st t = R dt = R at any temperature. t t Symbol s stands for the slope of the resistance versus temperature curve for the sensor. (The earlier definition assumes that α is constant and is evaluated using the resistance values at the ice and steam points.) o The circuit used for measurement is shown in the following figure. V A R 1 Voltmeter Battery R R 3 10 V V B RTD R t

37 o With the given data of α 0 = / C the slope may be determined as s0 = α 0R0 = = 0.44 Ω / C o Assuming the response of the sensor to be linear over small changes in temperature, the resistance of the sensor at 5 C may be determined as ( 5 0) s = = 108. Ω R = R + 1 o Infer all voltages with reference to the negative terminal of the battery taken as zero (ground). The voltmeter reads the potential difference between A and B. If there is change of temperature of 1 C the temperature of the RTD may either be 1 C or 19 C. o Case (a): t = 1 C. The potentials are given by the following: V = 10 R = = 5 V R + R A 1 o If there is a change of 1.0 o C in temperature the resistance changes by 0.44 Ω as given by the slope. The resistance of the RTD will be Ω in this case. o The potential V B is then given by V = 10 R = = V R + R B o The voltmeter should read V V = = V or 156 mv A B o Case (b): t = 19 C. The potentials are given by the following: = 10 R = A R + R = V 1 5 V o If there is a change of -1.0 o C in temperature the resistance changes by

38 -0.44 Ω as given by the slope. The resistance of the RTD will be Ω in this case. o The potential V B is then given by V = 10 R = = V R + R B o The voltmeter should read V V = = V or 136 mv A B o The smaller of these or V or 136 mv is the resolution of the voltmeter required for 1 o C resolution. Practically speaking we may choose a voltmeter with 100 mv resolution for this purpose.

39 Effect of self heating The bridge arrangement for measuring the sensor resistance involves the passage of a current through the sensor. Heat is generated by this current passing through the RTD. The heat has to be dissipated by an increase in the sensor temperature compared to the medium surrounding the sensor. Thus the self heating leads to a systematic error. Assume that the conductance (dissipation constant) for heat transfer from the RTD to the surrounding medium is P D W/K. The temperature excess of the RTD is given I R by t = PD t C. Example 6 demonstrates this.

40 Example 6 An RTD has α 0 = / C, R0 = 500 Ω and a dissipation constant of PD = 30 mw/ C at 0 o C. The RTD is used in a bridge circuit with R1 = R3 = 500Ω and R is a variable resistor used to null the bridge. If the supply voltage is 10 V and the RTD is placed in a bath at 0 C, find the value of R 3 to null the bridge. Take the effect of self heating into account. o Note: Figure in Example is appropriate for this case also. o Given data: D s α = 0.005/ C, R = 500 Ω, P = W / C, V = 10 V o Since R 1 = R 3 = 500Ω, at null RRTD = R. Thus the current through the RTD is s ( R + 500) V and hence the dissipation in the RTD is s ( R + 500) of V ( + ) R s R V R 500 P D s R V ( R + 500) P D. The self heating leads to a temperature change C. The temperature of the RTD is thus C instead of 0 o C as it should have been. o The resistance of the RTD is thus given by (assuming linear variation of resistance with temperature) s R V R = R 0[1 α0(0 ] ( R + 500) PD o This has to be solved for R to get the variable resistance which will null the bridge.

41 o The solution may be obtained by iteration. The iteration starts with the trial value 0 R = R 0 (1 0 α 0 ) = 500 ( ) = 450 Ω o Substitute this in the right hand side of the previous expression to get 0 1 Vs R = 0 α0 0 PD R R [1 (0 ] ( R + 500) = = Ω + o It so happens that we may stop after just one iteration! Thus the required resistance to null the bridge is Ω.

42 Example 7 Use the values of RTD resistance versus temperature shown in the table to find the equation for the linear approximation of resistance between 100 and 130 o C. Assume T 0 = 115 C. t C R Ω o For the linear fit we calculate the α value by using the mean slope near the middle of the table. We use the values shown in blue to get α 0 = = o The linear fit is: ( ) Rft = R α0 t 115 o We make a table to compare the linear fit with the data. t R ft R(data) Difference Linear fit appears to be very good

43 Sub Module.4 Thermistors Resistance thermometry may be performed using thermistors. Thermistors are many times more sensitive than RTD s and hence are useful over limited ranges of temperature. They are small pieces of ceramic material made by sintering mixtures of metallic oxides of Manganese, Nickel, Cobalt, Copper, Iron etc. They come in different shapes as shown in Figure 13. Resistance of a thermistor decreases non-linearly with temperature. Thermistors are extremely sensitive but over a narrow range of temperatures. The resistance temperature relation is known to follow the law R 1 1 β T T T 0 0 = R e (15) Here is a constant and all the temperatures are in Kelvin, T o is the ice point temperature and R 0 the corresponding resistance of the thermistor. Glass Envelope Bead Disc type Thermistor Rod type Thermistor Bead type Thermistor Figure 13 Typical thermistor types

44 A typical thermistor has the following specifications: R0 R5 = 000 Ω, = (16) R 70 This thermistor has a value of 3917 K and a resistance of 6656 at 0 o C. We may compare the corresponding numbers for a standard Platinum resistance element. R0 = 100 Ω, R70 = R 0( ) = R0 Hence, we have R0 1 = = (17) R The resistance change is thus very mild in the case of a PRT as compared to a thermistor. The variation of the resistance of the thermistor described above is shown plotted in Figure Resistance, Ω Temperature, o C Figure 14 Variation of thermistor resistance with temperature

45 Typical thermistor circuit for temperature measurement Battery R s Thermistor R p Voltmeter Figure 15 Typical thermistor circuit Since a thermistor has a highly non-linear temperature response it is necessary to use some arrangement by which the resistance variation with temperature is nearly linear. One way of achieving this is to connect a suitable parallel resistance as shown in Figure 15. As an example we consider the thermistor whose characteristics are given by (). A parallel resistance of 500 is chosen for the simulation. The equivalent resistance of the thermistor in parallel with R p varies nearly linearly as shown in Figure 16. The circuit is basically a voltage divider circuit. The potential difference across the series resistance or the thermistor provides an output that is related to the temperature of the thermistor.

46 Equivalent Resistance, Ω Data Linear (Data) Temperature, o C Figure 16 Equivalent resistance variations for the circuit shown in Figure 3 If the series resistance is chosen as Rs = 500 Ω and the battery voltage is 9 V, the voltmeter reading varies as shown in Figure 17. The output decreases monotonically with temperature and the non-linearity is mild. Output across thermistor, V Temperature, o C Figure 17 Variation of voltage across thermistor as a function of its temperature

47 Example 8 A thermistor has a resistance temperature relationship given by 1 1 R = R0 exp β. For a certain thermistor R 0 T T0 =80000 Ω where T 0 = K. The resistance of the thermistor has been measured accurately at three other temperatures as given below: t, C T, K R, Ω Using the above data estimate β. Use this β to estimate the resistance of the thermistor at 10 and 110 o C. Compare the data with the values calculated using the β determined above. o Given Data: T 0 = K, R 0 = Ω o Taking logarithms and rearranging, we have β= ln ( R R0 ) [ 1T 1/T ] o The tabulated data in the above expression will thus give three values for β. The average of these three values should give the best estimate for β. The values of β are 3506, 3503 and 3495 K. The mean value is thus equal to 3501 K (whole number retained). o Now we calculate the resistances at the three temperatures using the value of β obtained above. The data is conveniently tabulated. 0

48 t, C R (measured), Ω R (calculated), Ω o The match is very good! Sub Module.5 Pyrometery Pyrometry is the art and science of measurement of high temperatures. According to the International Practical Temperature Scale 1968 (IPTS68) Pyrometry was specified as the method of temperature measurement above the gold point. Pyrometry makes use of radiation emitted by a surface (usually in the visible part of the spectrum) to determine its temperature. The measurement is thus a non-contact method of temperature measurement. We shall introduce basic concepts from radiation theory before discussing Pyrometry in detail. Radiation fundamentals Black body radiation exists inside an evacuated enclosure whose walls are maintained at a uniform temperature. The walls of the enclosure are assumed to be impervious to heat transfer. The black body radiation is a function of the wavelength λ and the temperature T of the walls of the enclosure. The amount of radiation heat flux leaving the surface, in a narrow band dλ aroundλ, of the enclosure is called the spectral emissive power and is given by the Planck distribution function

49 E C 1 = λ e 1 1 bλ (T) 5 C λt (18) C 1 =First radiation constant = 3.74 x 10 8 W µm 4 / m and C = Second radiation constant = µm K. It is seen that the wavelength is specified in µm (= 10-6 m). It is to be noted that there is no net heat transfer from the surface of an enclosure and hence it receives the same flux as it emits. It may also be seen that the -1 in the denominator is much smaller than the exponential term as long as λt<<c. This is indeed true in Pyrometry applications where the wavelength chosen is 0.65 µm and the measured temperature may not be more than 5000 K. It is then acceptable to approximate the Planck distribution function by the Wein s approximation given by C C E (T) e λ λ = (19) λ 1 T b 5 % Difference 1.40E E E E E E-01.00E E+00 Wavelength = 0.65 Micron Figure 18 Error in using approximate Wein s approximation instead of the Planck function It is clear from Figure 18 that the error in using the Wein s approximation in lieu of the Planck function is around 1% even at a temperature as high as 5000 K Temperature, K

50 The spectral Black body emissive power has strong temperature dependence. In fact the emissive power peaks at a wavelength temperature product of λ max T = 898 µ m K (0) This is referred to as Wein s displacement law. Figure 19 shows this graphically. Monochromatic emissive power E b, W/m m 1.E+08 1.E+07 T = 898 m K 1.E+06 T = 5800 K 1.E+05 1.E+04 1.E+03 1.E+0 1.E+01 T = 1500 K 1.E+00 1.E-01 T = 300 K 1.E-0 1.E-03 1.E Wavelength, mm Figure 19 Black body characteristics and the Wein s displacement law (Red line corresponds to a wavelength of 0.65 µm normally used in Pyrometry) If you imagine keeping the wavelength fixed at say 0.65 µm, we see that the ordinate is a strong function of temperature! The surface will appear brighter to the eye higher the temperature. This is basically the idea central to

51 Pyrometry. Actual surfaces, however, are not black bodies and hence they emit less radiation than a black surface at the same temperature. We define the spectral emissivity ε λ as the ratio of the emissive power of the actual surface to that from a black surface at the same temperature and wavelength. E aλ (T) ε λ = (1) E (T) bλ Brightness temperature It is defined such that the spectral emissive power of the actual surface is the same as that of a hypothetical black body at the brightness temperature T B. Thus: E (T) = E (T ) () aλ bλ B If the emissivity of the surface (referred to as the target) is ε λ, we use Equation to write ε E b (T) = E b (T B). Using Wein s approximation this may λ λ λ C C be re written as T C ε e λ λ = λ λ e C λt B. We may cancel the common factor on the two sides, take natural logarithms, and rearrange to get 1 1 λ = ln ( ε λ ) (3) T T C B Equation 3 is referred to as the ideal pyrometer equation. This relation relates the actual temperature of the target to the brightness temperature of the target. The brightness temperature itself is measured using a vanishing filament pyrometer. Equation 3 indicates that TB ελ Tsince 1 for any surface. In actual practice the intervening optics may introduce

52 attenuation due to reflection of the radiation gathered from the target. It is also possible that one introduces attenuation intentionally as we shall see later. We account for the attenuation by multiplying the emissivity of the surface by a transmission factor τ 1to get B λ 1 1 λ = ln ( ελτ λ) (4) T T C We refer to Equation 8 as the pyrometer equation. We infer from the above that the brightness temperature of a surface depends primarily on the surface emissivity. If a surface is as bright as a black body at the gold point, the actual temperature should vary with spectral emissivity of the surface as indicated in Figure 0. Since no surface has zero emissivity we allow it to vary from 0.0 to 1 in this figure. As the spectral emissivity decreases the actual temperature increases as shown Actual temperatur T, K Spectral emissivity, ε λ Figure 0 Temperature a surface whose brightness temperature equals gold point temperature of K. λ = 0.66 µm.

53 Figure 0 also may be interpreted in a different way. Consider an actual surface whose spectral emissivity is known. If we introduce a transmission element with variable attenuation before making a comparison with a reference black body at the gold point, the abscissa in Figure 0 may be interpreted as the spectral emissivity transmission factor product. In that case the comparison is with respect to a single reference temperature, the gold point temperature. If this fixed point is determined with great precision on the ideal gas scale we have achieved measurement of an arbitrary temperature higher than the gold point temperature with reference to this single fixed point. This has the advantage that the pyrometer reference source (usually a standard tungsten filament lamp) runs at a constant temperature resulting in long life for the lamp. Black body reference Cavity radiator Pyrometry requires a reference black body source that may be maintained at a desired temperature. This is achieved by making use of a black body cavity shown schematically in Figure 1. Opening 50 mm diameter, 150 mm long Electrically Heated Refractory Sphere 0.3 m diameter Temperature Controller Insulation Temperature Sensor Figure 1 Schematic of a black body furnace

54 The black body reference consists of an electrically heated refractory sphere with a small opening as shown. The surface area of the sphere is much larger than the area of the opening through which radiation will escape to the outside. The radiation leaving through the opening is very close to being black body radiation at a temperature corresponding to the temperature of the inside of the sphere. Figure shows that the effective emissivity of the opening is close to unity. The emissivity of the surface of the sphere is already high, 0.96, for the case shown in the figure). The area ratio for the case shown in Figure 1 is Aopening = A sphere (5) The effective emissivity for this area ratio is about 0.995! Effective Emissivity Ratio of Opening Area to Cavity Surface Area Figure Variation of effective emissivity of a cavity radiator with area ratio

55 Many a time the cavity radiator has the sphere surrounded on the outside by a material undergoing phase change (solid to liquid). The melting point of the material will decide the exact temperature of the black body radiation leaving through the opening, as long as the material is a mixture of solid and liquid. Vanishing filament pyrometer This is fairly standard equipment that is used routinely in industrial practice. The principle of operation of the pyrometer is explained by referring to the schematic of the instrument shown in Figure 3. Figure 3 Schematic of a vanishing filament pyrometer The pyrometer consists of collection optics (basically a telescope) to gather radiation coming from the target whose temperature is to be estimated. The radiation then passed through an aperture (to reduce the effect of stray radiation), a neutral density or grey filter (to adjust the range of temperature) and is brought to focus in a plane that also contains a source (tungsten filament standard) whose temperature may be varied by varying the current through it. The radiation from the target and the reference then passes

56 through a red filter and is viewed by an observer as indicated in Figure 3. The observer will adjust the current through the reference lamp such that the filament brightness and the target brightness are the same. The state of affairs in the image seen by the eye of the observer is shown in Figure 4. High Correct Low Figure 4 Pyrometer adjustments If the adjustment is such that the filament is brighter than the target the setting is referred to as high. The filament appears as a bright object in a dull background. If the adjustment is such that the filament is duller than the target the setting is referred to as low. The filament appears as a dull object in a bright background. If the adjustment is correct the filament and the background are indistinguishable. Thus the adjustment is a null adjustment. The filament vanishes from the view! In this setting we have a match between the brightness of the target and the filament. The temperature of the filament is in deed the brightness temperature of the object. The wavelength corresponds to roughly a value of 0.66 µm. This is achieved by a combination of the response of the red filter and the eye of the observer (Figure 5). The average eye peaks around 0.6 µm and responds very poorly beyond 0.67 µm. The red filter transmits radiation beyond about 0.6 µm. On the whole the red

57 filter eye combination uses the radiation inside the red triangle shown in Figure 5. This region corresponds to a mean 0f 0.66 µm with a spectral width of approximately 0.03 µm. Transmittance Eye Response Red Filter Wavelength, Micron Figure 5 Effective wavelength for the pyrometer

58 Example 9 A pyrometer gives the brightness temperature of an object to be 800 o C. The optical transmittance for the radiation collected by the pyrometer is known to be and the target emissivity is Estimate the temperature of the object. Take λ = µm as the effective wavelength for the pyrometer. Given data: λ= µ m, C = µ m K, τ= 0.965, Tb = 800 C = 1073 K Target emissivity is ε = 0.60 The emissivity transmittivity product is ετ= = 0.51 Using the pyrometer equation, we get: Ta 1 1 = = = 1151 K 1 λ ln ( ετ ) + ln ( 0.51) T C b

59 Example 10 The brightness temperature of a metal block is given as 900 C. A thermocouple embedded in the block reads 1015 C. What is the emissivity of the surface? The pyrometer used in the above measurement is a vanishing filament type with an effective λ of 0.65 µm. Assuming that the thermocouple reading is susceptible to an error of ±10 C while the brightness temperature is error free determine an error bar on the emissivity determined above. o The first and second radiation constants, in SI units are: 16 1 C = W m, C = 14387µ m K o The data specifies the actual (T a ) as well as the brightness (T b ) temperatures and it is desired to determine the emissivity of the metal block at the stated wavelength. λ= 0.65 µ m, T = 1015 C = 188K,T = 900 C = 1173 K a b o We make use of the pyrometer equation to estimate ε Ta T b exp exp ε= = = λ C o This is the nominal value of the emissivity. o The error in ε is due to error in T a. This is calculated using the standard method discussed in the class. We need to calculate the derivative of ε with respect to T a.

60 1 1 dε d C 1 1 T a T b C 1 = exp = ( ) dt a dta λ Ta Tb λ λ Ta C = = / K o Since T a = ±10 K, the error in emissivity is given by dε ε = Ta = ± dt a

61 Emissivity values Temperature measurement using a pyrometer requires that the surface emissivity of the target be known. Useful emissivity data is given in Table 5. These are representative values since the nature of the surface is application specific and much variation is possible. One way out of this is to perform an experiment like the one presented in Example. Table 5 Approximate emissivity values Temperature C Surface Iron, Un-oxidized Iron, Oxidized Molten Cast Iron 0.9 Molten Steel Nickel, Oxidized Fireclay Silica Bricks Alumina Bricks In fact one may vary the surface temperature over a range of values that is expected to occur in a particular application and the measure the emissivity values over this range. An emissivity table may then be made for later use. Uncertainty in emissivity affects the measurement of temperature using a pyrometer. Figure 6 shows a plot of temperature error as a function of emissivity error, both in percent. For small errors the relationship is almost linear with Ta = 0.65 ε. The case considered is a vanishing filament pyrometer operating at 0.65 µm.

62 10 Error in Temperature, % Error in Emissivity, % Figure 6 Effect of uncertainty in emissivity on pyrometer measurement of temperature Ratio Pyrometry and the two color pyrometer Uncertainty in target emissivity in the case of vanishing filament pyrometer is a major problem. This makes one look for an alternate way of performing pyrometric measurement, based on the concept of color or ratio temperature. Consider two wavelengths λ 1 and λ close to each other. Let the corresponding emissivity values of the target be ε andε 1. The color temperature T c of the target is defined by the equality of ratios given by E aλ (T 1 a) E bλ (T 1 c) = (6) E (T ) E (T ) aλ a bλ c Thus it is the temperature of a black body for which the ratio of spectral emissive powers at two chosen wavelengths λ 1 and λ is the same as the corresponding ratio for the actual body.

63 We may use the Wein s approximation and write the above as C C ε1 exp exp λ1ta λ1tc ε 1 C 1 1 C 1 1 = or exp = exp C C ε Ta λ1 λ Tb λ1 λ ε exp exp λ T λ T a c Taking natural logarithms, we have ln This may be rearranged as ε1 C 1 1 C 1 1 = ε Ta λ1 λ Tc λ1 λ = T T C a c ε ln 1 ε 1 1 λ λ 1 (7) This equation that links the color and actual temperatures of the target is the counterpart of Equation 3 that linked the brightness and actual temperatures of the target. In case the emissivity does not vary with the wavelength we see that the color and actual temperatures are equal to each other. We also see that the actual temperature may be either less than or greater than the color temperature. This depends solely on the ratio of emissivities at the two chosen wavelengths. Equation 7 by itself is not directly useful. We measure the ratio of emissive powers of the target, at its actual temperature, at the two chosen wavelengths. Thus what is measured is the ratio occurring on the left hand side of Equation 6. Thus we have E E C ε λ = ε exp λta exp 5 aλ1 1 λ1ta 5 aλ λ1 C Again we may take natural logarithms to get

64 T 1 1 C λ λ1 a = 5 Eaλ λ 1 1ε ln E 5 aλ λ ε1 (8) Similarly we may show that T C 1 1 λ λ1 c = 5 Eaλ λ 1 1 ln E 5 aλ λ (9) Equation 8 is useful as a means of estimating the actual temperature for a target, whose emissivities at the two chosen wavelengths are known, and the measured values of emissive powers at the two wavelengths are available. The instrument that may be used for this purpose is the two color pyrometer whose schematic is shown in Figure 7. M 1 RF Amplifier Ph 1 W SM SC Display HT O M Sh BF Ph Amplifier HT hot target; O objective lens; W wedge; M 1 and M mirrors; Sh shutter; SM synchronous motor; RF red filter; BF blue filter; Ph 1, Ph photocells; SC scaler Figure 7 Schematic of a two color pyrometer

65 The radiation from the target is split up in to two beams by the use of a wedge, as indicated. Two mirrors redirect the two beams as shown. These pass through a rotating wheel with openings, as indicated. The beams then pass through two filters that allow only a narrow band around a well defined wavelength (indicated as red and blue). The rotating wheel chops the two beams and creates ac signal to be detected by the photocells. The ratio of the two signals is equal to the ratio of emissive powers in Equation 8. The scaler introduces the other ratios that appear in the same equation. The display then indicates the actual temperature.

66 Example 11 A certain target has a brightness temperature of 1000 K when viewed by a vanishing filament pyrometer. The target emissivity at 0.66 m is known to be 0.8. What is the true temperature of the target? What is the colour temperature of the same target if λ = 0.66 µ m, λ = 0.5 µ m, ε = 0.50? 1 λ Given Data: T = 1000 K, λ = 0.66 µ m, ε = 0.8 B 1 λ1 The ideal pyrometer equation may be recast to read as T a = 1 T 1 λ + 1 ln ε C B ( λ ) 1 Thus the true target temperature is T a 1 = = K ln ( 0.8 ) For determining the colour temperature the required data is: λ = 0.66 µ m, ε = λ = 0.5 µ m, ε = 0.5 We determine the color temperature by using the relation 1 Tc = ε λ ln 1 ελ 1 Ta 1 1 C λ λ1 1 = = 969.7K 0.76 ln λ λ 1

67 Sub Module.6 Measurement of transient temperature Many processes of engineering relevance involve variations with respect to time. The system properties like temperature, pressure and flow rate vary with time. These are referred to as transients and the measurement of these transients is an important issue while designing or choosing the proper measurement technique and the probe. Here we look at the measurement of temperature transients. Temperature sensor as a first order system - Electrical analogy Let us look at a typical temperature measurement situation. We visualize the temperature probe as a system that is subject to the temperature transient. The probe is exposed to the environment whose temperature changes with time and it is desired to follow the temperature change as closely as possible. In Figure 8 we show the schematic of the thermal model appropriate for this study.

68 Probe at uniform temperature T(t), Characteristic length L ch. Fluid stream at T (t) C and h W/m C Figure 8 Schematic of a temperature probe placed in a flowing medium The model assumes that the probe is at a uniform temperature within it at any time t. This means that the probe is considered to be thermally lumped. The medium that flows over the probe is at a temperature that may vary with respect to time. Initially the probe is assumed to be at temperature T 0. Let us assume that the probe is characterized by the following physical parameters: Density of the probe material = ρ kg/m 3, Volume of the probe = V m 3, Surface area of the probe that is exposed to the flowing fluid = S m, Specific heat of the probe material = C J/kg C, Heat transfer coefficient for heat transfer between the probe and the surrounding medium = h W/m C. By conservation of energy, we have Rate of change of rate of heat transfer between int ernal energy of probe = the probe and the fluid (30)

69 If we assume that the probe is at a higher temperature as compared to the fluid heat transfer will be from the probe to the fluid and the internal energy of the probe will reduce with time. Using the properties of the probe dt introduced above, the left hand side of Equation 30 is given by ρ VC. The dt right hand side of Equation 30 is given by hs(t T ) rearrangement, Equation 30 takes the form. With these, after some dt hs hs + T= T dt ρvc ρvc (31) Note that this equation holds even when the probe temperature is lower than the fluid temperature. The quantity ρvc has the unit of time and hs is referred to as the time constant τ of the first order system (first order since the governing differential equation is a first order ordinary differential equation). The first order time constant involves thermal and geometric properties. The volume to surface area ratio is a characteristic length dimension and is indicated as L ch in Figure 8. Noting that the product of density and volume is the mass M of the probe, the first order time constant MC may also be written as τ=. The time constant may be interpreted in a hs different way also, using electrical analogy. The quantity MC represents the thermal capacity and the quantity 1 represents the thermal resistance. hs Based on this interpretation an electric analog may be made as shown in Figure 9.

70 Figure 9 Electrical analog of a first order thermal system In the electric circuit shown in Figure 9 the input voltage represents the temperature of the fluid, the output voltage represents the temperature of the probe, the resistance R represents the thermal resistance and the capacitance C represents the thermal mass (mass specific heat product) of the probe. Equation 31 may be rewritten as dt dt T + = τ T τ (3) t τ Note that Equation 3 may be simplified using the integrating factor e to write it as d t Te τ = T e dt t τ (33)

71 This may be integrated to get t τ t 0 t τ Te = T e dt + A where A is a constant of integration. Using the initial condition T(t = 0) = T0 we get, after minor simplification t t t t τ τ τ 0 0 T= T e + e T e dt (34) This is the general solution to the problem. If the variation of fluid temperature with time is given, we may perform the indicated integration to obtain the response of the probe as a function of time. Response to step input If the fluid temperature is constant but different from the initial temperature of the probe, the solution is easily shown to be represented by T T T T 0 =φ= e t τ (35) The temperature difference between the probe and the fluid exponentially decreases with time. The variation is indicated in Figure 30. At the end of one time constant the temperature difference is some 37% of the initial temperature difference. After about 5 time constants the temperature difference is quite negligible.

72 φ t/τ Figure 30 Response of a first order system to a step input A step input may be experimentally realized by heating the probe to an initial temperature in excess of the fluid temperature and then exposing it quickly to the fluid environment. The probe temperature is recorded as a function of time. If it is plotted in the form ln(φ ) as a function of t, the slope of the line is negative reciprocal of the time constant. In fact, this is one method of measuring the time constant. Example 1 shows how this is done.

73 Example 1 A temperature probe was heated by immersing it in boiling water and is then quickly transferred in to a fluid medium at a temperature of T amb = 5 C. The temperature difference between the probe and the medium in which it is immersed is recorded as given below: t (s) T-T amb What is the time constant of the probe in this situation? o The data is plotted on a semi-log graph as shown here. It is seen that it is well represented by a straight line whose equation is given as an inset in the plot. EXCEL was used to obtain the best fit. Semi-log plot ln(t-t amb ) ln(t - Tamb) Linear (ln(t - Tamb)) ln(t-t amb ) = t R = t,s o The slope of the line is and hence the time constant is 1 1 τ= = = s slope

74 o The correlation coefficient of the linear fit is This shows that the data has been collected carefully. A note on time constant It is clear from our discussion above that the time constant of a system (in this case the temperature probe) is not s property of the system. It depends on parameters that relate to the system as well as the parameters that define the interaction between the system and the surrounding medium (whose temperature we are trying to measure, as it changes with time). The time constant is the ratio of thermal mass of the system to the conductance (reciprocal of the thermal resistance) between the system and the medium. It is also clear now how we can manipulate the time constant. Thermal mass reduction is one possibility. The other possibility is the reduction of the thermal resistance. This may be achieved by increasing the interface area between the system and the medium. In general this means a reduction in the characteristic dimension L ch of the system. A thermocouple attached to a thin foil will accomplish this. The characteristic dimension is equal to half the foil thickness, if heat transfer takes place from both sides of the foil. Another way of accomplishing this is to use very thin thermocouple wires so that the bead at the junction has very small volume and hence the thermal mass. Indeed these are the methods used in practice and thin film sensors are commercially available. Response to a ramp input In applications involving material characterization heating rate is controlled to follow a predetermined program heating. The measurement of the corresponding temperature is to be made so that the temperature sensor follows the temperature very closely. Consider the case of linear heating and

75 possibly linear temperature rise of a medium. Imagine an oven being turned on with a constant amount of electrical heat input. We would like to measure the temperature of the oven given by T () t = T0 + Rt (36) The general solution to the problem is given by (using Equation 34) t t t τ T0 t τ R t τ (37) T e = A + e dt + te dt τ τ 0 0 where A is a constant of integration. The first integral on the right hand side is t τ easily obtained as ( ) τ e 1. Second integral on the right hand side is obtained by integration by parts, as follows. t t t t τ t τ t τ t τ t τ ( ) te dt= tτe τ e dt= tτe τ e 1 (38) If the initial temperature of the first order system is T i, then A = Ti, since both the integrals vanish for t = 0 (the lower and upper limit will be the same). On rearrangement, the solution is t τ i 0 0 () ( ) ( ) T t = T T + R τ e + T + Rt R τ (39) We notice that as t the transient part tends to zero (transient part is the exponential decaying part) and the steady part (this part survives for t >> ) yields To + Rt T(t) = Rτ. (40) The steady state response has a lag equal to R with respect to the input.

76 Temp erature T, o C Input Response Time t, s Figure 31 Typical response of a first order system to ramp input Figure 31 shows the response of first order system to a ramp input. The case shown corresponds to Ti= 0 C, T 0 = 35 C, R = 0.15 C/s and τ = 10 s. For t (t > 5τ = 50 s) the probe follows the linear temperature rise with a lag of Rτ = = 1.5 C. In this case it is advisable to treat this as a systematic error and add it to the indicated temperature to get the correct oven temperature. Response to a periodic input There are many applications that involve periodic variations in temperature. For example, the walls of an internal combustion engine cylinder are exposed to periodic heating and hence will show periodic temperature variation. Of course, the waveform representing the periodic temperature variation may be of a complex shape (non sinusoidal). In that case the waveform may be split up in to its Fourier components. The response of the probe can also be studied as that due to a typical Fourier

77 component and combine such responses to get the actual response. Hence we look at a periodic sinusoidal input given by a ( ) T = T cos ω t (41) In the above expression T a is the amplitude of the input wave and is the circular frequency. We may use the general solution given by Equation 34 and perform the indicated integration to get the response of the probe. The steps are left as exercise to the student. Finally the response is given by 1 ( ω ( ωτ) ) t t Ta e τ Tcos a t tan τ o T= T e + ( 1+ω τ ) 1+ω τ Transient response Steady state response (4) Again for large t, the transient terms drop off and the steady sate response survives. There is a reduction in the amplitude of the response and also a time lag with respect to the input wave. Amplitude reduction and the time lag (or phase lag) depend on the product of the circular frequency and the time constant. The variations are as shown in Figure Amplitude reduction factor Phase lag/ (90 degree) Circular frequency time constant product, ωτ Figure 3 Response of a first order system to periodic input

78 In order to bring out the features of the response of the probe, we make a plot (Figure 33) that shows both the input and output responses, for a typical case Temperature ratio T/T Output response Time lag -0. Input Time, s Figure 33 Response of a first order system to periodic input The case shown in Figure 33 corresponds T to a T o = 0.5; ω= 1rad s and τ= 1 s. The output response has an initial transient that adjusts the initial mismatch between the probe temperature and the imposed temperature. By about 4 to 5 time constants (4 to 5 seconds since the time constant has been taken as 1 second) the probe response has settled down to a response that follows the input but with a time lag and an amplitude reduction as is clear from Figure 33.

79 Example 13 The time constant of a first order thermal system is given as 0.55 min. The uncertainty in the value of the time constant is given to be ± 0.01 min. The initial temperature excess of the system over and above the ambient temperature is 45 o C. It is desired to determine the system temperature excess and its uncertainty at the end of 50 s from the start. Hint: It is known that the temperature excess follows the formula () ( ) T t T 0 t τ = e where T(t) is the temperature excess at any time t, T(0) is the temperature excess at t = 0 and τ is the time constant. o We shall convert all times given to s so that things are consistent. The time constant is τ= 0.55 min = s = 33 s o We need the temperature excess at ( ) ( ) t = 50 t 50 τ 33 T 50 = T 0 e = 45e = 9.89 C s from the start. Hence o We would like to calculate the uncertainty in this value. We shall assume that this is due to the error in the time constant alone. τ = ± 0.1 min = ± = ± 0.6 s o The influence coefficient Iτ is given by t= T 50 Iτ = = 45 e = C / s τ 33 Hence the uncertainty in the estimated temperature excess is: ( ) T 50 =± I τ=± = 0.7 C τ

80 Example 14 o A certain first order system has the following specifications: Material: copper shell of wall thickness 1 mm, outer radius 6 mm Fluid: Air at 30 C Initial temperature of shell: 50 C o How long should one wait for the temperature of the shell to reach 40 C? Assume that heat transfer is by free convection. Use suitable correlation (from a heat transfer text) to solve the problem. o Heat transfer coefficient calculation: Heat transfer between the shell and the air is by natural convection. The 1/ 4 appropriate correlation for the Nusselt number is given by Nu = Ra where Ra is the Rayleigh number. The characteristic length scale is the sphere diameter. The air properties are calculated at the mean temperature at t = 0. From the given data, we have D = 1 mm = 0.01 m, T m = ( ) / = 40 C The air properties required are read off a table of properties: 6 ν= m / s, Pr = 0.71, k = 0.07 W / m C The isobaric compressibility of air is calculated based on ideal gas assumption. Thus 1 1 β= = = K T amb 3 1 The temperature difference for calculating the Rayleigh number is taken as the mean shell temperature during the cooling process minus the ambient temperature. We are interested in determining the time to cool from 50 to

81 C. Hence the mean shell temperature is T Shell = = 45 C. The temperature difference is T= TShell Tamb = 45 30= 15 C. The value of the acceleration due to gravity is taken as g = 9.8 m/s. The Rayleigh number is then calculated as gβ TD Ra = Pr = 0.71 = 069 ν 6 ( ) o The Nusselt number is then calculated as 1/4 1/4 Nu = Ra = = 4.9 The heat transfer coefficient is then calculated as o Time constant calculation: Copper shell properties are Nu k h = = = W / m K D ρ= 8954 kg / m, C = J / kg C Copper shell thickness is δ = m

82 Mass of the copper shell is calculated as 3 M =ρπd δ= 8954 π = kg Surface area of shell exposed to the fluid is 4 S =π D =π 0.01 = m The time constant is then estimated as 3 MC τ= = = 310 s hs o Cooling follows an exponential process. Hence we have, the time t 40 at which the shell temperature is 40 C, t40 = 310 ln = 14.9 s 50 30

83 Sub Module.7 Systematic errors in temperature measurement Systematic errors are situation dependent. We look at typical temperature measurement situations and discuss qualitatively the errors before we look at the estimation of these. The situations of interest are: Measurement of temperature o at a surface o inside a solid o of a flowing fluid Surface temperature measurement using a compensated probe: Consider the measurement of the temperature of a surface by attaching a thermocouple sensor normal to it, as shown in Figure 34.

84 Lead Wire Conduction T s T t < T s Heat flux paths Figure 34 Temperature measurement of a surface Lead wires conduct heat away from the surface and this is compensated by heat transfer to the surface as shown. This sets up a temperature field within the solid such that the temperature of the surface where the thermocouple is attached is depressed and hence less than the surface temperature elsewhere on the surface. This introduces an error in the surface temperature measurement. One way of reducing or altogether eliminating the conduction error is by the use of a compensated sensor as indicated in Figure 35.

85 Figure 35 Schematic of a compensated probe This figure is taken from Industrial measurements with very short immersion & surface temperature measurements by Tavener et al. The surface temperature, in the absence of the probe is at an equilibrium temperature under the influence of steady heat loss H 3 to an environment. The probe would involve as additional heat loss due to conduction. If we supply heat H by heating the probe such that there is no temperature gradient along the thermocouple probe then H 1 -H =H 3 and the probe temperature is the same as the surface temperature. Figure 3 (taken from the same reference) demonstrates that the compensated probe indicates the actual surface temperature with negligible error.

86 Figure 36 Comparison between the thermally compensated probe and two standard probes Compensated probes as described above are commercially available from ISOTECH (Isothermal Technology Limited, Pine Grove, Southport, Merseyside, England) and described as 944 True Surface temperature measurement systems. Figure 37 shows how one can arrange a thermocouple to measure the temperature inside a solid. The thermocouple junction is placed at the bottom of a blind hole drilled into the solid. The gap between the thermocouple lead wires and the hole is filled with a heat conducting cement. The lead wire is exposed to the ambient as it emerges from the hole. It is easy to visualize that the lead wire conduction must be compensated by heat conduction into

87 the junction from within the solid. Hence we expect the solid temperature to be greater than the junction temperature (this is the temperature that is indicated) which is greater than the ambient temperature. This assumes that the solid is at a temperature higher than the ambient. h, T amb I Lead Wire Solid at T s L II Thermocouple junction Heat conducting cement Figure 37 Measurement of temperature within a solid Often it is necessary to measure the temperature of a fluid flowing through a duct. In order to prevent leakage of the fluid or prevent direct contact between the fluid and the temperature sensor, a thermometer well is provided as shown in Figure 38. The sensor is attached to the bottom of the well as indicated. The measured temperature is the temperature of the bottom of the well and what is desired to be measured is the fluid temperature.

88 Duct wall Lead wire Flow Thermometer well Measuring junction Figure 38 Measurement of temperature of a moving fluid If the duct wall is different from the fluid temperature, heat transfer takes place by conduction between the fluid and the duct wall and hence the well bottom temperature will be at a value in between that of the fluid and the wall. There may also be radiation heat transfer between surfaces, further introducing errors. If the fluid flows at high speed (typical of supersonic flow of air) viscous dissipation conversion of kinetic to internal energy may also be important. With this background we generalize the thermometer error problem in the case of measurement of temperature of a gas flow as indicated in Figure 39.

89 Visible surfaces Wall Radiation Lead/Support Conduction Gas Sensor Gas convection + radiation Figure 39 Heat transfer paths for a sensor in gas flow The temperature of the sensor is determined, under the steady state, by a balance of the different heat transfer processes that take place, as indicated in Figure 39. Not all the heat transfer processes may be active in a particular case. The thermometer error is simply the difference between the gas temperature and the sensor temperature. Estimation of the error will be made later on. Summary of sources of error in temperature measurement: Sensor interferes with the process Conduction error in surface temperature measurement Sensor interferes with the process as well as other environments Radiation error While measuring temperature of moving fluids convection and conduction processes interact and lead to error

90 In case of high speed flow, viscous dissipation effects may be important Conduction error in thermocouple temperature measurement: Lead wire model Insulation L 1 r 1 L r Wire 1 r w1, k 1 Wire r w, k L + L r1 = rw, r = 4 1 Figure 40 Single wire equivalent of a thermocouple Heat transfer through the lead wires of a thermocouple leads to error in the measured temperature. Since a thermocouple consists of two wires of different materials covered with insulation, and since the error estimation should involve a simple procedure, we replace the actual thermocouple by a single wire thermal equivalent. How this is accomplished is indicated by referring to Figure 40. The cross section of an actual thermocouple is shown at the left in Figure 40. It consists of two wires of different materials with the indicated radii and thermal conductivity values. The insulation layer encloses

91 the two wires as indicated. We replace the two wires and the insulation by a single wire of radius r 1 and a coaxial insulation layer of outer radius r. The single wire model: 1) The area thermal conductivity product must be the same for the two wires and the single wire. Thus ( ka) ( ka) k1a w1 kaw two wires = = +. If the two wires have the one wire same diameter (this is usually the case) we may replace this ( ) ( ) k1+ k by( ka) = k one wire 1Aw1+ kaw = Aw1. Thus the thermal conductivity of the single wire equivalent is equal to the mean of the thermal conductivities of the two wires and the area of cross section of the single wire is twice the area of cross section of either wire. Hence the radius of the single wire equivalent is given by r 1 = r w as indicated in Figure 40. ) The insulation layer is to be replaced by a coaxial cylinder of inner radius r 1 and the outer radius r. The outer radius is taken L + L =. Note that if L 1 = L = r(true for a circle of radius r), 4 as r 1 this formula gives r = r, as it should. 3) Since the insulation layer is of low thermal conductivity while the wire materials have high thermal conductivities, it is adequate to consider heat conduction to take place along the single wire and radially across the insulation, as indicated in Figure 41.

92 Heat flow direction in insulation Heat flow direction in wire Figure 41 Heat flow directions Figure 34 shows a typical application where the temperature of a surface exposed to a moving fluid is being measured. The solid is made of a low thermal conductivity plastic. Figure 4 Surface temperature measurement The thermocouple lead wires conduct away some heat that is gathered by the thermocouple in contact with the solid. This will tend to depress the