engg measurements

Mechanical Measurements Prof. S.P.Venkatesan

Indian Institute of Technology Madras

Mechanical Measurements

Module 2:

1. Thermometry

Formally we start the study of “Mechanical Measurements” now!

Module 2 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 2.1

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 100000 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



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.



Figure 1 Principle of thermometry explained

Assume that one of the coordinates of the system (Y) is fixed at a value

equal to Y0. 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 X1, X2 and so on. These will then correspond to the respective

temperatures T1, T2 and so on. We refer to X as the thermometric property

and the system as a thermometer.

Y = Y0

X2 X3 X1 X4 X

Isotherms Y

T2 T3 T1 T4



Table 1 Thermometers and thermometric properties

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 2 Schematic of a constant volume gas thermometer

Thermometer Thermometric property

Symbol

Gas at constant volume Pressure P Electric resistance under constant tension

Electrical resistance R

Thermocouple Thermal electromotive force

E

Saturated vapor of a pure substance

Pressure P

Blackbody radiation Spectral emissive power

,bE λ

Acoustic thermometer Speed of sound a



We look at the constant volume gas thermometer in some detail now.

Schematic of such a thermometer is shown in Fig. 2. 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.



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 tpp 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

by sttp

tp

p 1.366049 as p 0p

→ → . The gas thermometer temperature scale is

defined based on this unique ratio and by assigning a numerical value of

0tpT 273.16 K or 0.01 C= . The defining relation is

st sttp

tp tp

T p as p 0T p

= → (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 610.65 Pa. The ice point is at 273.15 K or 0°C

and was used in early times as the primary fixed point in thermometry. In

00.20.40.60.8

11.21.41.61.8

2

0 100 200 300 400 500

ptp

pst/

ptp

Gas A Gas B Gas C



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

tptp tp

T p as p 0T p

= → (2)

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.



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 ptp at the triple point of water has the indicated

value.

ptp 100 200 300 400 p 233.4 471.6 714.7 962.9

If the triple point of water is taken as 273.16 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 tp

pp as tpp 0→ . This value is obtained by extrapolation. The

ratios are calculated and are given by the following table.

p 100 200 300 400

tp

pp

2.334 2.358 2.382 2.407

Difference 0.024 0.024 0.025

o Since the common differences are constant we may extrapolate

linearly to get

31.2024.0334.2pp

limtp0ptp

=−=→



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 T90 is defined in terms of the vapor-pressure

temperature relations 3He and 4He.

• Between 3.0 K and the triple point of neon (24.5561 K) T90 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 (13.8033 K) and the

freezing point of silver (961.78 ºC) T90 is defined by means of platinum



resistance thermometers calibrated at specified sets of defining fixed

points and using specified interpolation procedures

• Above the freezing point of silver (961.78ºC) T90 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 2.

Table 2 Secondary fixed points used in ITS90

Equilibrium state

T90 K T90oC Equilibrium state T90 K T90

oC

Triple point of H2

13.8033

-259.3467 Triple point of Hg 234.3156

-38.8344

Boiling point of H2 at 250 mm Hg

17 -256.15 Triple point of H2O

273.16 0.01

Boiling point of H2 at 1 atmosphere

20.3 -252.85 Melting point of Ga

302.9146

29.7646

Triple Point of Ne

24.5561

-248.5939 Freezing point of In

429.7483

156.5985

Triple point of O2

54.3584

-218.7916 Freezing point of Sn

505.078

961.928

Triple Point of Ar

83.8058

-189.3442 Freezing point of Al

933.473

660.323

Freezing point of Ag

1234.93

961.78

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.



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.



Sub Module 2.2

Thermoelectric thermometry

Thermoelectric thermometry is based on thermoelectric effects or

thermoelectricity discovered in the 19th century. They are:

• Seebeck effect discovered by Thomas Johann Seebeck in 1821

• Peltier effect discovered by Jean Charles Peltier in 1824

• 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:

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

Current I

T1 T2

A

B



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 ABPP Q I= = ±π where ABπ is the Peltier voltage (this

expression defines the Peltier emf), PQ 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

( )1 21 2 T T

T,Tπ = π − π (3)

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 2.

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.

Current I

T1 T2

A

B



Note that if the direction of the current is as shown the electrons move in the

opposite direction. If 2 1T T> , 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 is2

1

T

T AT

Q I dTσ= ∫ where TQ the

Thomson heat is and Aσ is the Thomson coefficient for the material. A

similar expression may be written for the Thomson heat in conductor B.

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

( ) ( )

( ) ( ) ( )

2 1

2 1 1 2

2

2 1 1

T TS P T A B A B A BT T T T

TA B A B A BT T T

V V V σ dT σ dT

σ -σ dT

= + = π − π + +

= π − π +

∫ ∫

∫ (4)

We define the Seebeck coefficient ABα through the relation sAB

dVdT

= α . In

differential form, equation 4 may then be rewritten as (assume 2 1T T dT− = )

A

B

T1 T2

VS



( )

( )

S ABAB A B

S AB A B

dV ddT dT

dV d dT

π= α = + σ − σ

= π + σ − σ

(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 P11

ABP

1 1

Qs I

T Tπ

= = . Similarly at

junction 2 the entropy change is P 2

1

BA ABP

2 2 2

Qs I I

T T Tπ π

= = = − . Again if we assume

that the temperature difference is 2 1T T dT− = the net change in entropy is

Pds IdTπ⎛ ⎞= ⎜ ⎟

⎝ ⎠. The net change in entropy due to Thomson heat in the two

conductors may be written as A B

T

ds I dTT

σ − σ= . Combining these two we get

A BP T

AB A BAB 2

ds=ds ds I d dT 0T T

d dTI dT 0T TT

σ − σ⎡ π ⎤⎛ ⎞+ = + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦π σ − σ⎡ ⎤= − π + =⎢ ⎥⎣ ⎦

(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.



From equation 5 ( )A B s ABσ σ dT dV d− = − π . Introducing this in equation 6 we

get

sAB ABAB 2

dVd ddT 0T T TT

π π− π + − =

or

sAB AB S

dVT T TdT

π = = α = α (7)

We have renamed the Seebeck coefficient as Sα according to normal practice.

Differentiating equation 7 we get AB s sd Td dTπ = α + α . Introduce this in

equation 5 to get ( )s s s A B sdV Td dT dT dT= α + α + σ − σ = α or

( ) sA B

dTdTα

σ − σ = − (8)

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 2 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.



For example, with Chromel (material A) and Alumel (material B) as the

two wire materials, the expression is a quartic of

form 2 3 4S 1 1 1 1V 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

2 3 41 S 2 S 3 S 4 St 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.



Example 2

For the Chromel-Alumel pair the Seebeck voltage varies with temperature

according to the fourth degree polynomial

3 2S

6 3 10 4

V 39.44386 t 5.8953822 10 t

4.2015132 10 t 1.3917059 10 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

. s

sdV Vα 39.4 Cdt

µ= ≈ °

o Using the Kelvin relations, we also have the following:

1. ( ) ( )2

s SA B Chromel Alumel 2

dα d Vσ σ σ σ T -T dT dT

− = − = − =

2.

A B Chromel Alumel

S

T 39.44386T

− −π = π

= α ≈



o Note that T in the Kelvin relation is in Kelvin and t in the polynomial is

in 0°C. Also note that d ddT dt

≡ . Hence the second derivative of the

Seebeck voltage is given by

2 22S S

2 3 42 2d V d V 2a 6a t 12a tdT dt

= = + +

o Near the ice point we may take t = 0 and write

( )Chromel Alumelσ σ ( 2)(0.0058953822)(273.15) 3.2206 µV.

− ≈ −

= −

and

Chromel Alumel ST (273.15)(39.44386) 10774.1 V 0.0108 V−π = α = = µ ≈



Example 3

The thermocouple response shown below (Copper Constantan

thermocouple with the cold junction at the ice point) follows the law VS

= a t + b t2. Obtain the parameters a and b by least squares. Here t is

in °C and VS is in mV.

Temperature, °C 37.8 93.3 148.9 204.4 260

VS, mV 1.518 3.967 6.647 9.523 12.572

o Since the fit follows the form specified above, it is equivalent to a linear

relation between E= VS/t and t. Since VS/t is a small number we shall

work with 100 VS/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=100VS/t x2 y2 xy yfit

37.8 4.015873 1428.84 16.12724 151.8 4.036073

93.3 4.251876 8704.89 18.07845 396.7 4.24048

148.9 4.46407 22171.21 19.92792 664.7 4.445255

204.4 4.659002 41779.36 21.7063 952.3 4.649661

260 4.835385 67600 23.38094 1257.2 4.854436

Sum: 744.4 22.22621 141684.3 99.22085 3422.7 6.638481

Mean = 148.88 4.445241 28336.86 19.84417 684.54 4.445181

(Note: I have used EXCEL to solve the problem)

o The statistical parameters are calculated and presented in the form of a

table:



Variance of x= 6171.606

Variance of y= 0.084001

Covariance = 22.73252

Slope of fit line = 0.003683

Intercept of fit line = 3.896856

o Using the fit parameters the calculated values of yfit are shown in the

last column of the first table. Reverting back to t and VS, we get the

following relation:

( ) 25S t10683.3t03897.0100/3.896856t0.003683tV −×+≈+=

o We compare the data with the fit in the table below:

t VS VS(fit)

37.8 1.518 1.526

93.3 3.967 3.956

148.9 6.647 6.619

204.4 9.523 9.504

260 12.572 12.622

o Standard error of the fit with respect to data may be calculated as

mV035.03

003774.03

))fit(VV( 2SS

Err ==−

=σ



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!

o The inverse relation may similarly be obtained by fitting a linear relation

between S

tV

and SV . This is left as an exercise to the student. The

result obtained is 2S St 25.173V 0.3769V= − with a standard error

of t 2.209 Cσ = ± ° .

0

2

4

6

8

10

12

14

0 50 100 150 200 250 300Temperature oC

Seeb

eck

volta

ge m

V

VS VS(fit)



Example 2 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 2. As an

example, the interpolating polynomial recommended for the K type

thermocouple with the reference junction at the ice point is given as:

3 2 6 3s

10 4

V 39.44386 t 5.8953822 10 t 4.2015132 10 t

1.3917059 10 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:

OSS

AB S

SA B

dV 39.444 V / CdtT 273.15 39.444

10774.1 VdT 273.15 2 0.005895dt

3.22064 V

α = = µ

π = α = ×

= µα

σ − σ = − = × ×

= − µ

The variation of the Seebeck coefficient over the range of this thermocouple is

given in Fig. 7 below



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 VS VS(fit) 37.8 1.52 1.499 93.3 3.819 3.728 148.9 6.092 5.990 204.4 8.314 8.273 260 10.56 10.581

371.1 15.178 15.237537.8 22.251 22.276815.6 33.913 33.874

1093.3 44.856 44.8791371.1 54.845 54.827

The maximum deviation is some 0.102 mV. The standard error is

approximately ±0.053 mV! This translates to roughly an error of ±1.2°C.

0

5

10

15

20

25

30

35

40

45

0 200 400 600 800 1000 1200 1400

Temperature oC

Seeb

eck

coef

ficie

nt µ

V/o C



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 2% 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 -3.22 V at 0°C to a maximum value of +31.16 V at

1350°C.



Sub Module 2.3

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

Element Resistance in Ohms at 100°C

Nickel -60 to 180 100 152 Copper -30 to 220 100 139 Platinum -200 to 850 100 136

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:

( )( )

2 3 O Ot 0 1 2 3

2 O Ot 0 1 2

R R 1 K t K t K t 100 t , 200 C t 0 C

R R 1 K t K t ,0 C t 250 C

= + + + − − < <

= + + < < (9)



Where

3 O 7 O 2 12 O 41 2 2K 3.90802 10 / C; K 5.802 10 / C ;K 1.2735 10 / C− − −= × = − × = − ×

The ratio 100 0

0

R R100R

− is denoted by and is given as 0.00385/°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.2 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.

0

0.5

1

1.5

2

2.5

3

3.5

-200 -100 0 100 200 300 400 500 600

Temperature, oC

Res

ista

nce

ratio

, R(t)

/R0

Figure 1 Characteristics of a Platinum resistance thermometer



Figure 2 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]

t 0Pt

100 0

R Rt 100R R

−= ×

− (10)

The quantities appearing in the above are:

RPt = Platinum resistance temperature, Rt = Resistance of sensor at

temperature t, R0 = Resistance of the sensor at the ice point and R100 =

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

Ceramic Powder

Protective SheathPlatinum Element

1 2 3



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

( )2100 0 1 2R R 1 100K 100 K= + + .

Hence

2100 0 1 2R R 100K 100 K− = + .

We then have

2t 0 1 2

2100 0 1 2

2 2 2 21 2 1 2 2 2

1 2 1 22 2

22 2 2 2 2

1 2 1 1

R R K t K t100 100R R 100K 100 K

K t K t K t 100K t 100K t K tK 100K K 100K

100K t K t 100K t K t K t tt t t 100 1K 100K K K 100 100

− +× = ×

− +

+ + − += =

+ +

⎛ ⎞− − ⎡ ⎤= − ≈ − = + × −⎜ ⎟ ⎢ ⎥+ ⎣ ⎦⎝ ⎠

With the K’s given earlier, we

have7

2 423

1

K 5.802 10100 10 1.485K 3.90802 10

−

−− ×

× = × = − = −δ×

. Thus we have

t 0

100 0

R R t t100 t 1R R 100 100

− ⎛ ⎞⎛ ⎞× = − δ −⎜ ⎟⎜ ⎟− ⎝ ⎠⎝ ⎠

This may be rephrased as (using the definition given in Equation 10)

t 0 Pt PtPt

100 0Correction c

R R t tt tt 100 1 t 1R R 100 100 100 100

− ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞≈ × + δ − = + δ −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ (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.



Figure 10 Callendar correction as a function of tPt

-101234567

0 50 100 150 200 250 300

Platinum resistance temperature, tPt

Cor

rect

ion,

c



Example 4

The resistance of a Platinum resistance sensor of 0R 100= Ω was

measured to be119.4Ω . This sensor has α value of 0.00385. What is

the corresponding temperature without and with correction?

o We have R0 = 100Ω, α = 0.00385 and δ = 1.485.

o Hence ( ) ( )100 0R R 1 100 100 1 100 0.00385 13.5= + α = × + × = Ω

o The measured sensor resistance is given as tR 119.4= Ω .

o By definition the Platinum resistance temperature is

Ot 0Pt

100 0

R R 119.4 100t 100 100 50.39 CR R 138.5 100

− −= × = × =

− −

o This is also the uncorrected value of the temperature.

o The Callendar correction is calculated as

OPt Ptt t 50.39 50.39c 1 1.485 1 0.37 C100 100 100 100

⎛ ⎞ ⎛ ⎞= δ − = × − × = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

o The corrected temperature is thus given by

OPtt t c 50.39 0.37 50.02 C= + = − =



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 2) 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.



Bridge circuit for resistance thermometry:

Three wire arrangement for lead wire compensation

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 R1 and R3 are chosen to be equal and the same as R0 of the

RTD. Two lead wires (labeled 2 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 (milli-

ammeter will indicate zero) when R2 = R0 when the RTD is maintained at the

ice point. During use, when the RTD is at temperature t, the resistance R2 is

adjusted to restore balance. If the lead wires have resistances equal to Rs2

and Rs3, we have

RTD

1

3

Battery

2

Milli- Ammeter

Compensating Leads

R3

R1

R2



( )t s3 2 s2 t 2 s2 s3R R R R or R R R R+ = + = + − (12)

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 12 is the bridge

arrangements that are used for this purpose.

Figure 12 Bridge circuit with lead wire compensation

(four wire arrangement)

(a) (b)

R1

R2 R2 R3

3 42 1

Milli- Ammeter

Battery

Compensating Leads

RTD

1 243

Milli- Ammeter

Battery

Compensating Leads

RTD

R1

R3



The choice of the resistances is made as given for the three wire

arrangement. If the lead wires have resistances equal to Rs1 - Rs4, we have

the following.

Condition for bridge balance in arrangement shown in Figure 12(a):

t s4 2(a) s2

For balancarrangement a

R R R R+ = + (12)

Condition for bridge balance in arrangement shown in Figure 12(b):

t s2 2(b) s4

For balancearrangement b

R R R R+ = + (13)

We see that by addition of Equations 5 and 6, we get

2(a) 2(b)t

R RR

2+

= (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.



Example 5

An RTD has C/004.020 °=α . If 20R 106= Ω (resistance at 20°C),

determine the resistance at 25°C. The above RTD is used in a bridge

circuit with 1 2 3R R R 100= = = Ω and 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 20°C.

o Note the definition of α viz. tt

tt

s1 dRR dt R

α = = at any temperature.

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.

R1 Voltmeter

Battery

RTD Rt

R3 R2

10 V

VA

VB



o With the given data of 20 0.004 / Cα = ° the slope may be determined as

20 20 20s R 0.004 106 0.424 / C= α = × = Ω °

o Assuming the response of the sensor to be linear over small changes

in temperature, the resistance of the sensor at 25°C may be

determined as

( ) Ω=×+=×−+= 12.108424.05106s2025RR 202025

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 21°C or 19°C.

o Case (a): t = 21°C. The potentials are given by the following:

A 21 2

10 10V 10 R 10 100 5 VR R 100 100

= − = − × =+ +

o If there is a change of 1.0oC in temperature the resistance changes by

0.424 Ω as given by the slope. The resistance of the RTD will be

106.424 Ω in this case.

o The potential VB is then given by

B 2121 3

10 10V 10 R 10 106.424 4.844 VR R 106.424 100

= − = − × =+ +

o The voltmeter should read

A BV V 5 4.844 0.156 V or 156 mV− = − =

o Case (b): t = 19°C. The potentials are given by the following:

V5100100100

1010RRR

1010V 221

A =×+

−=+

−=

o If there is a change of -1.0oC in temperature the resistance changes by



-0.424 Ω as given by the slope. The resistance of the RTD will be

105.576 Ω in this case.

o The potential VB is then given by

B 2119 3

10 10V 10 R 10 105.576 4.864 VR R 105.576 100

= − = − × =+ +

o The voltmeter should read

A BV V 5 4.864 0.136 V or 136 mV− = − =

o The smaller of these or 0.136 V or 136 mV is the resolution of the

voltmeter required for 1oC resolution. Practically speaking we may

choose a voltmeter with 100 mV resolution for this purpose.



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 PD W/K. The temperature excess of the RTD is given

by CPRItD

t2

°⎥⎦

⎤⎢⎣

⎡=∆ . Example 6 demonstrates this.



Example 6

An RTD has 20 0.005 / Cα = ° , 20R 500= Ω and a dissipation constant of

DP 30 mW / C= ° at 20oC. The RTD is used in a bridge circuit with

1 3R R 500= = Ω and 2R 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 C0° , find

the value of R3 to null the bridge. Take the effect of self heating into

account.

o Note: Figure in Example 2 is appropriate for this case also.

o Given data: 320 20 D s0.005 / C, R 500 , P 30 10 W / C, V 10 V−α = ° = Ω = × ° =

o Since 1 3R R 500= = Ω , at null RTD 2R R= . Thus the current through the

RTD is ( )

s

2

VR 500+

and hence the dissipation in the RTD is

( )

2s

22

V RR 500

⎡ ⎤⎢ ⎥+⎣ ⎦

. The self heating leads to a temperature change

of( )

2s 2

2 D

V RR 500 P

⎡ ⎤⎢ ⎥+⎣ ⎦

C° . The temperature of the RTD is thus

( )

2s 2

2 D

V RR 500 P

⎡ ⎤⎢ ⎥+⎣ ⎦

C° instead of 0oC as it should have been.

o The resistance of the RTD is thus given by (assuming linear variation

of resistance with temperature)

( )

2s 2

2 20 202 D

V RR R [1 (20 ]R 500 P

⎡ ⎤= − α − ⎢ ⎥+⎣ ⎦

o This has to be solved for 2R to get the variable resistance which will

null the bridge.



o The solution may be obtained by iteration. The iteration starts with the

trial value

02 20 20R R (1 20 ) 500 (1 0.005 20) 450= − α = × − × = Ω

o Substitute this in the right hand side of the previous expression to get

( )

20

1 s 22 20 20 0

D2

2

3

V RR R [1 (20 ]PR 500

10 450500 1 0.005 20 454.16450 500 30 10−

⎡ ⎤⎢ ⎥= − α −⎢ ⎥+⎣ ⎦

⎡ ⎤⎧ ⎫⎪ ⎪⎡ ⎤⎢ ⎥= × − × − × = Ω⎨ ⎬⎢ ⎥+⎣ ⎦ ×⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦

o It so happens that we may stop after just one iteration! Thus the

required resistance to null the bridge is 454.16 Ω.



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 130oC. Assume 0T 115 C= ° .

t°C 100 105 110 115 120 125 130 R Ω 573.40 578.77 584.13 589.48 594.84 600.18 605.52

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

0594.84 584.13 0.00182

120 110−

α = =−

o The linear fit is:

( )ft 115 0R R 1 t 115= + α −⎡ ⎤⎣ ⎦

o We make a table to compare the linear fit with the data.

t Rft R(data) Difference 100 573.39 573.40 0.01 105 578.75 578.77 0.02 110 584.12 584.13 0.01 115 589.48 589.48 0.00 120 594.84 594.84 0.00 125 600.21 600.18 -0.03 130 605.57 605.52 -0.05

Linear fit appears to be very good



Sub Module 2.4

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

0

1 1T T

T 0R R e⎛ ⎞

β −⎜ ⎟⎝ ⎠= (15)

Here is a constant and all the temperatures are in Kelvin, To is the ice point

temperature and R0 the corresponding resistance of the thermistor.

Figure 13 Typical thermistor types

Disc type Thermistor

Rod type Thermistor

Bead type Thermistor

Glass Envelope

Bead



A typical thermistor has the following specifications:

025

70

RR 2000 , 18.64R

= Ω = (16)

This thermistor has a value of 3917 K and a resistance of 6656 at

0oC. We may compare the corresponding numbers for a standard Platinum

resistance element.

0 70 0 0R 100 , R R (1 0.00385 70) 1.2695R= Ω = + × =

Hence, we have

0

70

R 1 0.788R 1.2695

= = (17)

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 14.

Figure 14 Variation of thermistor resistance with temperature

0

1000

2000

3000

4000

5000

6000

7000

0 20 40 60 80 100Temperature, oC

Res

ista

nce,

Ω



Typical thermistor circuit for temperature measurement

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 (2). A parallel

resistance of 500 is chosen for the simulation. The equivalent resistance of

the thermistor in parallel with Rp 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.

Battery Thermistor

Voltmeter

Rs

Rp



0

50

100

150

200

250

300

350

400

450

500

0 10 20 30 40 50 60 70 80 90 100

Temperature, oC

Equi

vale

nt R

esis

tanc

e, Ω

Data Linear (Data)

Figure 16 Equivalent resistance variations for the circuit shown in

Figure 3

If the series resistance is chosen as sR 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.

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 10 20 30 40 50 60 70 80 90 100

Temperature, oC

Out

put a

cros

s th

erm

isto

r, V

Figure 17 Variation of voltage across thermistor as a function of its

temperature



Example 8

A thermistor has a resistance temperature relationship given by

00

1 1R R expT T

⎡ ⎤⎛ ⎞= β −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦. For a certain thermistor R0 =80000 Ω where

T0 = 273.16 K. The resistance of the thermistor has been measured

accurately at three other temperatures as given below:

t,°C 50 100 150

T, K 323.16 373.16 423.16

R, Ω 10980 2575 858

Using the above data estimate β. Use this β to estimate the resistance

of the thermistor at 10 and 110 oC. Compare the data with the values

calculated using the β determined above.

o Given Data: T0 = 273.16 K, R0 = 80000 Ω

o Taking logarithms and rearranging, we have

( )[ ]

0

0

ln R R1 T 1/ T

β =−

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.



t, °C 50 100 150

R (measured), Ω 10980 2575 858

R (calculated), Ω 11070 2602 861

o The match is very good!

Sub Module 2.5

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



2

1b 5 C

T

C 1E (T)

e 1λ

λ

=λ

−

(18)

C1=First radiation constant = 3.742 x 108 W µm4 / m2 and C2 = Second

radiation constant = 14390 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<<C2. 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

2C1 T

b 5CE (T) e

−λ

λ =λ

(19)

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.

0.00E+00

2.00E-01

4.00E-01

6.00E-01

8.00E-01

1.00E+00

1.20E+00

1.40E+00

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000Temperature, K

% D

iffer

ence

Wavelength = 0.65 M icron



The spectral Black body emissive power has strong temperature

dependence. In fact the emissive power peaks at a wavelength temperature

product of

maxT 2898 m Kλ = µ − (20)

This is referred to as Wein’s displacement law. Figure 19 shows this

graphically.

1.E-04

1.E-03

1.E-02

1.E-01

1.E+00

1.E+01

1.E+02

1.E+03

1.E+04

1.E+05

1.E+06

1.E+07

1.E+08

0.1 1 10 100

Wavelength , mm

Mon

ochr

omat

ic e

mis

sive

pow

er E

b, W

/m2

m

T = 300 K

T = 5800 K

T = 1500 K

T = 2898 m K

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



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 λε as the ratio of the emissive power of the actual

surface to that from a black surface at the same temperature and wavelength.

a

b

E (T)E (T)

λλ

λε = (21)

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

TB. Thus:

a b BE (T) E (T )λ λ= (22)

If the emissivity of the surface (referred to as the target) is λε , we use

Equation 22 to write b b BE (T) E (T )λ λ λε = . Using Wein’s approximation this may

be re written as22B

CCT1 1T

5 5C Ce e

−− λλλε =

λ λ. We may cancel the common factor on

the two sides, take natural logarithms, and rearrange to get

( )B 2

1 1 lnT T C λ

λ− = ε (23)

Equation 23 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 23 indicates that BT T≤ since

1λε ≤ for any surface. In actual practice the intervening optics may introduce



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 1λτ ≤ to get

( )B 2

1 1 lnT T C λ λ

λ− = ε τ (24)

We refer to Equation 28 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 20. Since no surface has zero emissivity

we allow it to vary from 0.02 to 1 in this figure. As the spectral emissivity

decreases the actual temperature increases as shown.

1300

1400

1500

1600

1700

1800

0 0.2 0.4 0.6 0.8 1

Spectral emissivity, ελ

Act

ual t

empe

ratu

r T, K

Figure 20 Temperature a surface whose brightness temperature equals

gold point temperature of 1337.6 K. λ = 0.66 µm.



Figure 20 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 20 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 21.

Figure 21 Schematic of a black body furnace

Electrically Heated Refractory Sphere 0.3 m diameter

Temperature Controller

Insulation

Temperature Sensor

Opening – 50 mm diameter, 150 mm long



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 22 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 21 is

2opening

sphere

A 1 0.05 0.007A 4 0.3

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

(25)

The effective emissivity for this area ratio is about 0.995!

0.9

0.92

0.94

0.96

0.98

1

0 0.2 0.4 0.6 0.8 1

Ratio of Opening Area to Cavity Surface Area

Eff

ect

ive

E

mis

siv

ity

Figure 22 Variation of effective emissivity of a cavity

radiator with area ratio



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 23.

Figure 23 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



through a red filter and is viewed by an observer as indicated in Figure 23.

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 24.

Figure 24 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 25). The

average eye peaks around 0.6 µm and responds very poorly beyond 0.67 µm.

The red filter transmits radiation beyond about 0.62 µm. On the whole the red

High Correct Low



filter – eye combination uses the radiation inside the red triangle shown in

Figure 25. This region corresponds to a mean 0f 0.66 µm with a spectral

width of approximately 0.03 µm.

00.10.20.30.40.50.60.70.80.9

1

0.45 0.5 0.55 0.6 0.65 0.7 0.75Wavelength, Micron

Tra

nsm

ittan

ce

Red FilterEye Response

Figure 25 Effective wavelength for the pyrometer



Example 9

A pyrometer gives the brightness temperature of an object to be

800oC. The optical transmittance for the radiation collected by the

pyrometer is known to be 0.965 and the target emissivity is 0.260.

Estimate the temperature of the object. Take λ = 0.655 µm as the

effective wavelength for the pyrometer.

Given data:

2 b0.655 m, C 14390 m K, 0.965, T 800 C 1073 Kλ = µ = µ − τ = = ° =

Target emissivity is

0.260ε =

The emissivity transmittivity product is

0.260 0.965 0.251ετ = × =

Using the pyrometer equation, we get:

( ) ( )

a

b 2

1 1T 1151 K1 1 0.655ln ln 0.251T C 1073 14390

= = =λ

+ ετ + ×



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 21 2C 3.743 10 W m , C 14387 m K−= × = µ −

o The data specifies the actual (Ta) as well as the brightness (Tb)

temperatures and it is desired to determine the emissivity of the metal

block at the stated wavelength.

a b0.65 m, T 1015 C 1288K,T 900 C 1173 Kλ = µ = ° = = ° =

o We make use of the pyrometer equation to estimate ε.

a b

2

1 1 1 1T T 1288 1173exp exp 0.1850.65

C 14387

⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎢ ⎥ε = = =⎢ ⎥λ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

o This is the nominal value of the emissivity.

o The error in ε is due to error in Ta. This is calculated using the

standard method discussed in the class. We need to calculate the

derivative of ε with respect to Ta.



a b2 22

a a a b a2

2

1 1T TC Cd d 1 1 1exp ( )

dT dT T T TC

14387 1 0.185 0.00247 / K0.65 1288

⎡ ⎤−⎢ ⎥⎡ ⎤⎧ ⎫ε ⎢ ⎥= − × = −⎢ ⎨ ⎬⎥ λλ λ⎢ ⎥⎩ ⎭⎣ ⎦⎢ ⎥⎣ ⎦

= − × × = −

o Since K10Ta ±=∆ , the error in emissivity is given by

aa

d T 0.00247 10 0.025dT

ε∆ε = ∆ = − × ± ≈ ∓



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 2.

Table 5 Approximate emissivity values

Temperature °C

Surface 600 1200 1600 1800

Iron, Un-oxidized 0.2 0.37

Iron, Oxidized 0.85 0.89

Molten Cast Iron 0.29

Molten Steel 0.28 0.28

Nickel, Oxidized 0.75 0.75

Fireclay 0.52 0.45

Silica Bricks 0.54 0.46

Alumina Bricks 0.23 0.19

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 26 shows a plot of temperature error as a function

of emissivity error, both in percent. For small errors the relationship is almost

linear with aT 0.65∆ = ∆ε . The case considered is a vanishing filament

pyrometer operating at 0.65 µm.



0

2

4

6

8

10

0 2 4 6 8 10Error in Emissivity, %

Err

or

in

Tem

pera

ture

, %

Figure 26 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 1 2andλ λ close to each other. Let

the corresponding emissivity values of the target be 21 andεε . The color

temperature Tc of the target is defined by the equality of ratios given by

1 1

2 2

a a b c

a a b c

E (T ) E (T )E (T ) E (T )

λ λ

λ λ= (26)

Thus it is the temperature of a black body for which the ratio of spectral

emissive powers at two chosen wavelengths λ1 and λ2 is the same as the

corresponding ratio for the actual body.



We may use the Wein’s approximation and write the above as

2 21

1 a 1 c 1 2 2

2 a 1 2 b 1 22 22

2 a 2 c

C Cexp expT T C C1 1 1 1or exp exp

T TC Cexp expT T

⎡ ⎤ ⎡ ⎤ε − −⎢ ⎥ ⎢ ⎥ ⎡ ⎤ ⎡ ⎤λ λ ⎧ ⎫ ⎧ ⎫ε⎣ ⎦ ⎣ ⎦= − − = − −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ε λ λ λ λ⎡ ⎤ ⎡ ⎤ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦ε − −⎢ ⎥ ⎢ ⎥λ λ⎣ ⎦ ⎣ ⎦

Taking natural logarithms, we have

1 2 2

2 a 1 2 c 1 2

C C1 1 1 1lnT T

⎡ ⎤ ⎡ ⎤ε− − = − −⎢ ⎥ ⎢ ⎥ε λ λ λ λ⎣ ⎦ ⎣ ⎦

This may be rearranged as

1

2

a c 2

1 2

ln1 1 1

1 1T T C

εε

− =−

λ λ

(27)

This equation that links the color and actual temperatures of the target

is the counterpart of Equation 23 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 27 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 26. Thus we have

1

2

25

a 1 a1 25

a 2 21

2 a

CexpE TE Cexp

T

λ

λ

⎡ ⎤⎢ ⎥λε λ ⎣ ⎦=

ε ⎡ ⎤λ⎢ ⎥λ⎣ ⎦

Again we may take natural logarithms to get



1

2

22 1

a 5a 1 2

5a 2 1

1 1CT

Eln

Eλ

λ

⎛ ⎞−⎜ ⎟λ λ⎝ ⎠=

⎡ ⎤λ ε⎢ ⎥

λ ε⎢ ⎥⎣ ⎦

(28)

Similarly we may show that

1

2

22 1

c 5a 1

5a 2

1 1CT

Eln

Eλ

λ

⎛ ⎞−⎜ ⎟λ λ⎝ ⎠=

⎡ ⎤λ⎢ ⎥

λ⎢ ⎥⎣ ⎦

(29)

Equation 28 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 27.

Figure 27 Schematic of a two color pyrometer

SM

Amplifier

Amplifier

SC Display

HT O

M2

M1 RF

Sh BF

Ph2

Ph1

W

HT – hot target; O – objective lens; W – wedge; M1 and M2 – mirrors; Sh – shutter; SM – synchronous motor; RF – red filter;

BF – blue filter; Ph1, Ph2 – photocells; SC – scaler



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 28. The

scaler introduces the other ratios that appear in the same equation. The

display then indicates the actual temperature.



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

if21 20.66 m, 0.5 m, 0.50λλ = µ λ = µ ε = ?

Given Data: 1B 1T 1000 K, 0.66 m, 0.8λ= λ = µ ε =

The ideal pyrometer equation may be recast to read as

( )1

a1

B 2

1T 1 lnT C λ

=λ

+ ε

Thus the true target temperature is

( )a

1T 1010.3 K1 0.66 ln 0.81000 14390

= =+

For determining the colour temperature the required data is:

1

2

1

2

0.66 m, 0.8

0.5 m, 0.5λ

λ

λ = µ ε =

λ = µ ε =

We determine the color temperature by using the relation

2

1

c

a2

2 1

1T

ln1T 1 1C

1 969.7K0.76ln

1 0.861 11010.3 14390

0.5 0.66

λ

λ

=⎡ ⎤⎛ ⎞ε⎢ ⎥⎜ ⎟⎜ ⎟ε⎢ ⎥⎝ ⎠−⎢ ⎥⎛ ⎞⎢ ⎥−⎜ ⎟⎢ ⎥λ λ⎝ ⎠⎣ ⎦

= =⎡ ⎤⎛ ⎞

⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥−⎛ ⎞⎢ ⎥× −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦



Sub Module 2.6

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 28 we show the schematic of the thermal model

appropriate for this study.



Figure 28 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

T0. Let us assume that the probe is characterized by the following physical

parameters:

Density of the probe material = ρ kg/m3, Volume of the probe = V m3,

Surface area of the probe that is exposed to the flowing fluid = S m2, 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/m2°C. By

conservation of energy, we have

Rate of change of rate of heat transfer betweenint ernal energy of probe the probe and the fluid

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ (30)

Probe at uniform temperature T(t), Characteristic length Lch.

Fluid stream at T∞(t)°C and h W/m2 °C



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

introduced above, the left hand side of Equation 30 is given by dTVCdt

−ρ . The

right hand side of Equation 30 is given by hS(T T )∞− . With these, after some

rearrangement, Equation 30 takes the form

dT hS hST Tdt VC VC ∞+ =

ρ ρ (31)

Note that this equation holds even when the probe temperature is

lower than the fluid temperature. The quantity VChS

ρ has the unit of time and

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 Lch in Figure 28. Noting that the product of

density and volume is the mass M of the probe, the first order time constant

may also be written as MChS

τ = . The time constant may be interpreted in a

different way also, using electrical analogy. The quantity MC represents the

thermal capacity and the quantity 1hS

represents the thermal resistance.

Based on this interpretation an electric analog may be made as shown in

Figure 29.



Figure 29 Electrical analog of a first order thermal system

In the electric circuit shown in Figure 29 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

TdT Tdt

∞+ =τ τ

(32)

Note that Equation 32 may be simplified using the integrating factor τt

e to write

it as

t td Te T edt

τ τ∞

⎛ ⎞⎜ ⎟ =⎜ ⎟⎝ ⎠

(33)



This may be integrated to get t tt

0

Te T e dt Aτ τ∞= +∫ where A is a constant of

integration. Using the initial condition 0T(t 0) T= = we get, after minor

simplification

t t tt

00

T T e e T e dt− −

τ τ τ∞= + ∫ (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

0

T T eT T

−∞ τ

∞

−= φ =

− (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.



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5

t/τ

φ

0.367

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 12 shows how this is done.



Example 12

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 Tamb =

25°C. The temperature difference between the probe and the medium

in which it is immersed is recorded as given below:

t (s) 0.35 0.6 0.937 1.438 2.175 3.25 T-Tamb 60 50 40 30 20 10

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-Tamb) = -0.6065t + 4.2837R2 = 0.9989

00.5

11.5

22.5

33.5

44.5

0 0.5 1 1.5 2 2.5 3 3.5

t,s

ln(T

-Tam

b)

ln(T - Tamb) Linear (ln(T - Tamb))

o The slope of the line is -0.6065 and hence the time constant is

1 1 1.6487 1.65 sslope 0.6065

τ = − = − = ≈−



o The correlation coefficient of the linear fit is -0.9994. 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 Lch 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



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

( ) 0T t T R t∞ = + (36)

The general solution to the problem is given by (using Equation 34)

t tt t t0

0 0

T RT e A e dt te dtτ τ τ= + +τ τ∫ ∫ (37)

where A is a constant of integration. The first integral on the right hand side is

easily obtained as ( )te 1 .ττ − Second integral on the right hand side is

obtained by integration by parts, as follows.

( )t ttt t t t 2 t

00 0

t e dt t e e dt t e e 1τ τ τ τ τ= τ − τ = τ − τ −∫ ∫ (38)

If the initial temperature of the first order system is iT , then iTA = , since both

the integrals vanish for t = 0 (the lower and upper limit will be the same). On

rearrangement, the solution is

( ) ( ) ( )t

i 0 0T t T T R e T R t R− τ= − + τ + + − 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

oT Rt T(t) R+ − = τ . (40)

The steady state response has a lag equal to R with respect to the input.



2025303540455055

0 20 40 60 80 100

Time t, s

Te

mp

era

ture

T,

oC

Response

Input

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= 20°C, T0 = 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 = 0.15×10 = 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



component and combine such responses to get the actual response. Hence

we look at a periodic sinusoidal input given by

( )aT T cos t∞ = ω (41)

In the above expression Ta 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

( )( )( )

t1

at ao 2 2 2 2

Steady state responseTransient response

T cos t tanT eT T e1 1

− −τ

− τω − ωτ

= − ++ ω τ + ω τ

(42)

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 32.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.01 0.1 1 10 100Circular frequency time constant product, ωτ

Amplitude reduction factor Phase lag/ (90 degree)

Figure 32 Response of a first order system to periodic input



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.

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15 20

Time, s

Tem

pera

ture

ratio

T/T

0

Output response

Input

Time lag

Figure 33 Response of a first order system to periodic input

The case shown in Figure 33 corresponds

to a

o

T 0.25; 1rad s and 1 sT

= ω = τ = . 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.



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 45oC. 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

( )( )

tT te

T 0−

τ= 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 0.55 min 0.55 60 s 33 sτ = = × =

o We need the temperature excess at s50t = from the start. Hence

( ) ( )t 50

33T 50 T 0 e 45e 9.89 C− −

τ= = = °

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.

0.1 min 0.1 60 0.6 s∆τ = ± = ± × = ±

o The influence coefficient Iτ is given by

5033

2t 50

T 50I 45 e 0.454 C / s33

−τ

=

∂ ⎛ ⎞= = × = °⎜ ⎟∂τ ⎝ ⎠

Hence the uncertainty in the estimated temperature excess is:

( )T 50 I 0.454 0.6 0.272 Cτ∆ = ± ∆τ = ± × = °



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

appropriate correlation for the Nusselt number is given by 4/1Ra43.02Nu +=

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

mD 12 mm 0.012 m, T (50 30) / 2 40 C= = = + = °

The air properties required are read off a table of properties:

6 216.96 10 m / s, Pr 0.71, k 0.027 W / m C−ν = × = = °

The isobaric compressibility of air is calculated based on ideal gas

assumption. Thus 3 1

amb

1 1 3.3 10 KT 30 273

− −β = = = ×+

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



40°C. Hence the mean shell temperature is Shell50 40T 45 C

2+

= = ° . The

temperature difference is Shell ambT T T 45 30 15 C∆ = − = − = ° . The value of the

acceleration due to gravity is taken as 2g 9.8 m / s= . The Rayleigh number is

then calculated as

( )3 3 3

2 26

g TD 9.8 3.3 10 15 0.012Ra Pr 0.71 206916.96 10

−

−

β∆ × × × ×= = × =

ν ×

o The Nusselt number is then calculated as

1/ 4 1/ 4Nu 2 0.43Ra 2 0.43 2069 4.9= + = + × =

The heat transfer coefficient is then calculated as

2Nu k 4.9 0.027h 11.07 W / m KD 0.012

×= = =

o Time constant calculation:

Copper shell properties are

38954 kg / m , C 383.1 J / kg Cρ = = °

Copper shell thickness is 0.001 mδ =



Mass of the copper shell is calculated as

2 2 3M D 8954 0.012 0.001 4.051 10 kg−= ρπ δ = × π× × = ×

Surface area of shell exposed to the fluid is

2 2 4 2S D 0.012 4.524 10 m−= π = π× = ×

The time constant is then estimated as

3

4MC 4.051 10 383.1 310 shS 11.07 4.524 10

−

−× ×

τ = = =× ×

o Cooling follows an exponential process. Hence we have, the time t40 at

which the shell temperature is 40°C,

4040 30t 310ln 214.9 s50 30

−⎛ ⎞= − =⎜ ⎟−⎝ ⎠



Sub Module 2.7

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.



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.

Lead Wire Conduction

Ts

Heat flux paths

Tt < Ts



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 H3 to an environment.

The probe would involve as additional heat loss due to conduction. If we

supply heat H2 by heating the probe such that there is no temperature

gradient along the thermocouple probe then H1-H2 =H3 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.



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



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.

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.

I

Thermocouple junction

IIL

Solid at Ts

h, Tamb Lead Wire

Heat conducting cement



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.

Measuring junction

Flow

Lead wireDuct wall

Thermometer well



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

Gas

Conduction

Radiation

Gas convection + radiation Sensor

Lead/Support

Wall Visible surfaces

Gas



• In case of high speed flow, viscous dissipation effects may be

important

Conduction error in thermocouple temperature measurement:

Lead wire model

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

L1

Insulation

Wire 1 rw1,k1

Wire 2 rw2,k2

L2 r2 r1

1 21 22,

4wL Lr r r +

= =



the two wires as indicated. We replace the two wires and the insulation by

a single wire of radius r1 and a coaxial insulation layer of outer radius r2.

The single wire model:

1) The area thermal conductivity product must be the same for the two

wires and the single wire. Thus

( ) ( ) 1 w1 2 w2two wires one wirekA kA k A k A= = + . If the two wires have the

same diameter (this is usually the case) we may replace this

by ( ) ( ) ( )1 21 w1 2 w2 w1one wire

k kkA k A k A 2A

2+

= + = . 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 1 wr 2r= as

indicated in Figure 40.

2) The insulation layer is to be replaced by a coaxial cylinder of inner

radius r1 and the outer radius r2. The outer radius is taken

as 1 22

L Lr4+

= . Note that if 1 2L L 2r= = (true for a circle of radius r),

this formula gives 2r 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.



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 42 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

Heat flow direction in insulation

Heat flow direction in wire



temperature of the junction. In order to reduce the effect of this

thermocouple lead wire conduction, the junction is attached to a heat

collecting pad of copper as indicated in the figure.

Now consider the typical application presented in Figure 42. Figure 43

explains the nomenclature employed for the analysis of this case. The heat

conducting pad receives heat from the front face of area S and loses heat

only through the thermocouple due to lead wire conduction. The appropriate

thermal parameters are as shown in Figure 43.

Figure 43 Nomenclature for lead wire conduction analysis

Heat loss through the lead wire is modeled by using fin type analysis,

familiar to us from the study of heat transfer. Since the wire is usually very

long, it may be assumed to infinitely long. The heat loss from the wire to the

ambient is modeled as that due to an overall heat transfer coefficient given by

Overall2 2

i 1

1hr r1 ln

h k r

=⎛ ⎞

+ ⎜ ⎟⎝ ⎠

(43)

hf, Tf

h, Tamb

Area, S



The perimeter of the wire is 1P 2 r= π and the area thermal conductivity product

for the wire is ( ) ( ) ( )1 2 2 21 1 1 2

k kkA 2 r r k k

2+

= π = π + . The appropriate fin

parameter m is then given by

( ) ( )Overall Overall 1 Overall

21 2 11 2 1

h P h 2 r 2hmkA k k rk k r

π= = =

++ π (44)

Assuming the lead wire to be infinitely long, the heat loss through the lead

wire is given by

( )Lead wire t ambQ kAm T T= − (45)

Under steady conditions this must equal the heat gained from the fluid by the

pad given by

( )Gain by pad f f tQ h S T T= − (46)

Equating (3) and (4) we solve for the sensor indicated temperature as

f f ambt

f

h ST kAmTTh S kAm

+=

+ (47)

Equation 47 shows that the sensor temperature is a weighted mean of the

fluid temperature and the ambient temperature. It is clear that the smaller the

weight on the ambient side better it is from the point of view of temperature

measurement. This is a general feature, as we shall see later, in all cases

involving temperature measurement. The thermometric error is then given by

Error t fT T T= − (48)

Example 15 below demonstrates the use of the above analysis in a typical

situation.



Example 15

A copper constantan thermocouple of wire diameter each of 0.25 mm

is used for measuring the temperature of a surface which is

convectively heated by a fluid with a heat transfer coefficient of 67

W/m2°C. The area of the surface exposed to the fluid is 10 cm

2. The

thermocouple has an insulation of thickness 1 mm all round and the

overall size is 5 mm x 2.5 mm. The thermal conductivity of the

insulation is 1 W/moC. The thermocouple is exposed to an ambient at

a temperature of 30°C subject to a heat transfer coefficient of 5

W/m2°C. If the fluid temperature is 200°C what is the temperature

indicated by the thermocouple? Take thermal conductivity of copper as

386 W/moC and the thermal conductivity of constantan as 22.7 W/m°C.

Wire side calculation:

Diameter of each thermocouple wire d 0.00025 m=

Thermal conductivities of the thermocouple wires

1 2 tan tan386 / , 22.7 /= = ° = = °Copper Consk k W m C k k W m C

The area of cross section of each wire is

2 28 2d 0.00025A 4.909 10 m

4 4−= π = π = ×

Effective thermal conductivity area product for the thermocouple pair is

( ) ( )1 2 8 52 (386 22.7) 4.909 10 2.006 10 /2

− −+= = + × × = × °

k kkA A W m C

Overall heat transfer coefficient is now calculated:



The overall heat transfer coefficient is calculated by combining the

insulation and film resistances. We have 2h 5 W / m C= ° . The radius of

the single wire equivalent is 10.00025 0.000177

2 2dr m= = = . The outer

radius of effective insulation layer is 20.005 0.0025 0.001875

4r m+

= = .

Thermal conductivity of insulation material is ik 1 W / m C= ° . The

overall heat transfer coefficient is

2Overakl

2 2

i 1

1 1h 4.892 W / m C1 0.001875 0.001875r r1 lnln 5 1 0.000177h k r

= = = °⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

The overall heat transfer coefficient perimeter product is thus given by

( )Overall 1 Overallh P 2 r h 2 0.000177 4.892 0.005433 W / m C= π = × π× × = °

The fin parameter is calculated as

1Overall5

h P 4.892m 16.457 mkA 2.006 10

−−= = =

×

The surface temperature may now be calculated by equating the heat

transfer from the fluid to surface to that lost through the thermocouple

insulation. The appropriate data is:

2 2 3 2f f ambh 100 W / m C, S 10 cm 10 m , T 200 C and T 30 C−= ° = = = ° = °

From the material presented earlier, assuming the thermocouple wires

to be very long, the surface temperature is given by

f f ambt

f5

5

h ST kAm TTh S kAm

100 0.001 200 2.006 10 16.457 30 199.2 C100 0.001 2.006 10 16.457

−

−

+=

+

× × + × × ×= = °

× + × ×

The thermometer error is t fT T 199.2 200 0.8 C− = − = − °



Temperature error due to radiation:

Errors in temperature measurement may occur due to surface

radiation, especially at elevated temperatures. We consider the same

example that was considered above. Assume that the copper disk has a

surface emissivity of . Let it also view a cold background at Tbkg. The heat

loss is now due to lead wire conduction along with radiation to the ambient.

Heat loss due to radiation is given by

( )4 4Radiation t bkgQ S T T= εσ −

Note that the temperatures are to be expressed in Kelvin in Equation 49 and

is the Stefan Boltzmann constant. The temperature of the sensor is

determined by equating heat gain by convection to heat loss by conduction

and radiation. Thus

Lead wire Radiation Gain by padQ Q Q+ = (50)

Using Equations 45, 46 and 49 we then have

( ) ( ) ( )4 4t amb t bkg f f tkAm T T S T T h S T T− + εσ − = − (51)

The above non-linear algebraic equation needs to be solved to arrive at the

value of the measured temperature.



Example 16

Reconsider Example 1 with the following additional data:

The copper pad has a surface emissivity of 0.05 and views a cooler

background at a temperature of 450 K. What is the thermometric error

in this case?

In addition to the heat loss by lead wire conduction we have to include

that due to radiation. This is given by

8 4 4 12 4 4Radiation t tQ 0.05 5.67 10 (T 450 ) 2.84 10 (T 450 )− −= × × − = × −

Using the material already available in Example 1 the equation that

governs the sensor temperature is given by

( )12 4 4t t t0.00033(T 30) 2.84 10 (T 450 ) 0.067 473 T−− + × − = −

This equation may be solved by Newton Raphson method. Alternately the

solution may be obtained by making a plot of the difference between the

left hand side and right hand side of this equation and locate the point

where it crosses the temperature axis. Such a plot is shown below.



-0.15

-0.1

-0.05

0

0.05

0.1

466 468 470 472 474

Tt

LHS

-RH

S

It is clear that the sensor temperature is now 470.2 K or 197.2°C. The

temperature error has changed to -2.8°C! Error due to radiation is, in fact,

more than that due to lead wire conduction.



Measurement of temperature within a solid:

Now we shall look at the situation depicted in Figure 37. Temperature

error is essentially due to conduction along the lead wires. However, the

portion embedded within the solid (II) has a different environment as

compared to the part that is outside (I). Both of these may be treated by the

single wire model introduced earlier. Assume that the solid is a temperature

higher than the ambient. The thermocouple junction will then be at an

intermediate temperature between that of the solid and the ambient. Heat

transfer to the embedded thermocouple is basically by conduction while the

heat transfer away from the part outside the solid is by conduction and

convection. The embedded part is of finite length L while the portion outside

may be treated as having an infinite length.

Represent the temperature of the single wire equivalent as Ti in a plane

coinciding with the surface of the solid. Let Tt be the temperature of the

junction while Ts is the temperature of the solid. Let the ambient temperature

be Tamb. The fin parameter for the embedded part may be calculated based

on the overall heat transfer coefficient given by

Overall,II3 3 2 2

c 2 i 1

1hr r r rln lnk r k r

=⎛ ⎞ ⎛ ⎞

+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(52)

In the above, kc is the thermal conductivity of the heat conducting cement, r3 is

the radius of the hole and the other symbols have the earlier meanings. Note

that expression 10 is based on two conductive resistances in series. The

corresponding fin parameter is Overall,IIII

h Pm

kA= . The overall heat transfer

coefficient for the exposed part of the thermocouple is given by the expression



given earlier, viz. Overakl,I2 2

i 1

1hr r1 ln

h k r

=⎛ ⎞

+ ⎜ ⎟⎝ ⎠

. The corresponding fin parameter

value is Overall,II

h Pm

kA= .

Figure 44 Nomenclature for thermal analysis

Referring now to Figure 44 we see that the heat transfer across the surface

through the thermocouple should be the same i.e. II IQ Q= . Using familiar fin

analysis, we have

( ) ( )IIII Overall,II s i

II

tanh m LQ h P T T

m= − (53)

For the exposed part, we have

( )I I i ambQ kAm T T= − (54)

Equating the above two expressions we solve for the unknown temperature Ti.

Thus

( )

1 s 2 ambi

1 2

II1 Overall,II 2 I

II

w T w TT wherew w

tanh m Lw h P and w kAm

m

+=

+

= = (55)

I

Ts

Tamb

QII

Ti

Tt

II

QI Surface



Having found the unknown temperature Ti, we make use of fin analysis for the

embedded part to get the temperature Tt. Using familiar fin analysis, we have

i st s

II

(T T )T Tcosh(m L)

−= + (56)

Note that the fin analysis assumes negligible heat transfer near the bottom of

the hole!

Following points may be made in summary:

1) The longer the depth of embedding smaller the thermometric error

2) Higher the thermal conductivity of the epoxy filling the gap between the

thermocouple and the hole the smaller the thermometric error

3) The smaller the diameter of the thermocouple wires smaller is the

thermometric error

4) Smaller the thermal conductivity of the thermocouple wires smaller the

thermometric error

5) If it is possible the insulation over the thermocouple wires should be as

thin as possible in the embedded portion and as thick as possible in the

portion that is outside the hole



Example 17

Thermocouple described in Example 1 is used to measure the

temperature of a solid by embedding it in a 6 mm diameter hole that is

15 mm deep. The space between the thermocouple and the hole is

filled with a heat conducting epoxy that has a thermal conductivity of 10

W/m°C. The lead wires coming out of the hole are exposed to an

ambient at 30°C with a heat transfer coefficient of 5 W/m2°C. If the

temperature of the solid is 80°C, estimate the temperature indicated by

the thermocouple.

From the results in Example 1, the following are available:

5kA 2.006 10 W m / C−= × ° , 1Im 16.457 m−=

The weight w2 is then given by

52 Iw kAm 2.006 10 16.457 0.00033 W m / C−= = × × = °

For the embedded part, the following calculations are made:

30.006r 0.003 m

2= = , ck 10 W / m C= ° ,

2Overall,II

1h 218.88 W / m C0.003 0.003 0.001875 0.001875ln ln

10 0.001875 1 0.00177

= = °⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The fin parameter is then calculated as

1II 5

218.8 2 0.00177m 110.08 m2.006 10

−−

× × π×= =

×

With L = 0.015 m, we have IIm L 110.08 0.015 1.651= × =

The weight w1 is then given by

( )1

tanh 1.651w 218.8 2 0.00177 0.002052

110.08= × × π× × =



The unknown temperature Ti is now calculated as

0.002052 80 0.00033 30 73.070.002052 0.00033iT C× + ×

= = °+

The sensor temperature is then calculated as

(73.07 80)80 77.4cosh(1.651)tT C−

= + = °

The thermometric error is thus equal to -2.6°C.



The thermometer well problem

This is a fairly common situation as has been mentioned earlier. The

well acts as a protection for the temperature sensor but leads to error due to

axial conduction along the well. It is easily recognized that the well may be

treated as a fin and the analysis made earlier will be adequate to estimate the

thermometric error.

Figure 45 Nomenclature for the thermometer well problem

Assumptions:

1) Since the thermometer well has a much larger cross section area than the

thermocouple wires conduction along the wire is ignored.

2) The thermometer well is heated by the gas while it cools by radiation to the

walls of the duct (based on g t wT T T> > ).

Velocity U Temperature Tg

Temperature Tw

Measuring junction

Flow

Lead wireDuct wall

Thermometer well

Temperature Tt

ID = di, OD = do, kw L



3) Well is treated as a cylinder in cross flow for determining the convection

heat transfer coefficient between the gas and the well surface.

The heat transfer coefficient is calculated based on the Zhukaskas correlation

given by

m nNu C Re Pr= (57)

In this relation Re, the Reynolds number is based on the outside

diameter of the well and all the properties are evaluated at a suitable mean

temperature. The constants C, m and n are given in Table 6.

Radiation heat transfer may be based on a linearised model if the gas

and wall temperatures are close to each other. In that case the well

temperature variation along its length is also not too big. Thus we

approximate the radiant flux ( )4 4R wq T T= εσ − by the relation

( ) ( )3R w w R wq 4 T T T h T T≈ εσ − = − where 3

R wh 4 T= εσ is referred to as the

radiation heat transfer coefficient.

Table 6 Constants in the Zhukaskas correlation

Re C m 1-40 0.75 0.4 40-103 0.51 0.5 103-2×105 0.26 0.6 2×105-106 0.076 0.7 m Pr < 10 0.36 Pr > 10 0.37



Analysis:

Figure 46 Thermometer well analysis schematic

Refer to Figure 46 and the inset that shows an expanded view of an elemental

length of the well. Various fluxes crossing the boundaries of the element are:

1) ( )Con gQ hP x T T= ∆ − 2) ( )R R wQ h P x T T= ∆ − 3) , =Cond in wx

dTQ k Adx

and

4) Cond,in wx x

dTQ k Adx −∆

=

In the above the perimeter P is given by oP dπ= and area of cross section A is

given by ( )2 2

o id dA

4

−= π . Energy balance requires

that Cond,in con Cond,out RQ Q Q Q+ = + . Substituting the expressions for the fluxes

and using Taylor expansion of the derivative, we have

( )2

w g R w w w 2x x x

dT dT d Tk A hP x T T h P x(T T ) k A k A xdx dx dx

+ ∆ − = ∆ − + − ∆

This equation may be rearranged as

Tt

Tw

U,Tg

QCon

QCond,out

QCond,in

QR

x

∆x



2R R

g w2w w w

(h h )P h Pd T hPT T T 0k A k A k Adx+

− + + = (58)

Let g R wref

R

hT h TT

h h+

=+

be a reference temperature. Then Equation 58 is

rewritten as

22eff2

d m 0dx

θ− θ = (59)

where refT Tθ = − and ( )Reff

w

h h Pm

k A+

= is the effective fin parameter.

Equation 59 is the familiar fin equation whose solution is well known.

Assuming insulated boundary condition at the sensor location, the indicated

sensor temperature is given by

( ) ( )( )

w reft t ref

eff

T TT T

cosh m L−

θ = − = (60)

The thermometric error is thus given by

( ) ( ) ( )( )

w reft g ref g

eff

T TT T T T

cosh m L−

− = − + (61)

Following points may be made in summary:

1) The longer the depth of immersion L smaller the thermometric error

2) Lower the thermal conductivity of the well material the smaller the

thermometric error

3) Smaller the emissivity of the well and hence the hR smaller the

thermometric error

4) Larger the fluid velocity and hence the h smaller the thermometric error



Example 18

Air at a temperature of 373 K is flowing in a tube of diameter 10 cm at

an average velocity of 0.5 m/s. The tube walls are at a temperature of

353 K. A thermometer well of outer diameter 4 mm and wall thickness

1 mm made of iron is immersed to a depth of 5 cm, perpendicular to

the axis. The iron tube is dirty because of usage and has a surface

emissivity of 0.85. What will be the temperature indicated by a

thermocouple that is attached to the bottom of the thermometer well?

What is the consequence of ignoring radiation?

Step wise calculations are shown below:

Step 1.Well outside convective heat transfer coefficient:

Given data:

0 f wd 0.004 m, U 0.5 m / s, T 373 K, T 353 K= = = =

The fluid properties are taken at the mean temperature given

by m373 353T 363 K

2+

= =

From table of properties for air the desired properties are:

6 223.02 10 m / s, k 0.0313 W / m K, Pr 0.7−ν = × = =

The Reynolds number based on outside diameter of thermometer well

is

06

Ud 0.5 0.004Re 86.923.02 10−

×= = =

ν ×

Zhukaskas correlation is used now. For the above Reynolds number

the appropriate constants in the Zhukaskas correlation are

C 0.51, m 0.5 and n 0.37= = = .



The convection Nusselt number is

m n 0.5 0.37Nu C Re Pr 0.51 86.9 0.7 4.17= = × × =

Hence the convective heat transfer coefficient is

2

0

Nu k 4.17 0.0313h 32.6 W / m Kd 0.004

×= = =

Step 2 Radiation heat transfer coefficient:

Linear radiation is used since the fluid and wall temperatures are close

to each other.

The pertinent data is: 8 2 45.67 10 W / m K , 0.85−σ = × ε =

The radiation heat transfer coefficient is

3 8 3 2r wh 4 T 4 0.85 5.67 10 353 8.5 W / m K−= εσ = × × × × =

Step 3 Calculation of reference temperature

f r wref

r

hT h T 32.6 373 8.5 353T 368.9 Kh h 32.6 8.5

+ × + ×= = =

+ +

Step 4 Well treated as a fin:

Well material has a thermal conductivity of wk 45 W / m K=

Internal diameter of well is equal to outside diameter minus twice the

wall thickness and is given by m002.0001.02004.0t2dd 0i =×−=−=

The fin parameter

( )( )

( )10 r

f 2 2 2 20 i

w

0.004 32.6 8.5d (h h )m 34.87 md d 0.004 0.002

k 454 4

−× +π += = =

− −π ×

Since the well length is L = 0.05 m the non-dimensional fin parameter

is



f fm L 34.87 0.05 1.74µ = = × =

Step 5 Non-dimensional well bottom temperature

It is given by tf

1 1 0.339cosh( ) cosh(1.74)

θ = = =µ

Hence the temperature indicated by the sensor attached to the well is

( ) ( )t ref t w refT T T T 368.9 0.339 353 368.9 363.5 K= + θ − = + × − =

The thermometric error is some 9.5°C.

If radiation is ignored the above calculations should be done by taking

hr= 0 and Tref = Tw. This is left as an exercise to the student.

engg measurements

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