# Temperature Measurement Lab Report.pdf

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MJ 2440 – MEASURING TECHNIQUES Lab No 01- temperature measurement

Prepared by: Raveendra Sampath Kumara Sarukkali Kankanamge 800226-A298 rsksk[email protected] International College for Business and Technology Sri Lanka

26-05-2011 1

Q 01) Finding the ambient temperature

Measured voltage (given)

= 0.7896 mV

= 0.7896 mV

= V1 – V2

Temperature at J2

= 0 °C

Thus, V2

= 0 mV

Hence, V1

= 0.7896mV

From thermocouple reference table Temperature ( oC ) 19 20

2

Voltage (mV) 0.749 0.79

o

Voltage (mV) vs Temperature ( C) 0.795 0.79 0.785

y = 0.041x - 0.03 0.78

Voltage

0.775 0.77 0.765 0.76 0.755 0.75 0.745 18.8

19

19.2

19.4

19.6

19.8

Temperature

Voltage

= 0.041×Temperature - 0.03

0.7896

= 0.041×Temperature - 0.03

Ambient Temperature

= 19.99 °C

Q 02) Finding the temperature of hot water From the previous measurement, the following is written down. Ambient air temperature

= 19.99 °C

Hence temperature of J2

= 19.99 °C

V2

= 0.7896 mV

= 2.568

= V1 – V2

2.568

= V1 – 0.7896

V1

= 3.358

Hence, temperature of J1 = 80 °C 3

20

20.2

Q 03) If the polarity of the thermocouple placed in ice water had been reversed, what would be the Voltage Temperature

: :

By changing the polarity of TC on channel on logger Voltage Temperature

: :

Q 04)

The equivalent circuit is as shown in the following figure.

4

a) If the junction is at ambient air

= V1 – V2

Both the junctions are in ambient condition. i.e 19.99 °C Thus, V1 = 0.7896

V2 = 0.7896

= 0.7896 - 0.7896

=0

b) If the junction is at ice In this case junction J1 is in the ice and junction J2 is in the ambient condition, which is 19.99 °C. Thus, V1 = 0

V2 = 0.7896

= V1 – V2

= 0 - 0.7896

= - 0.7896

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c) If the junction is at hot water In this case junction J1 is in hot water and it is 80 °C, and junction J2 is in ambient condition and it is 19.99 °C. Thus, V1 = 3.358

V2 = 0.7896

= V1 – V2

= 3.358 – 0.7896

= 2.5684

Q 05) Connect the single thermocouple with the extremely long junction to a channel of the instrument, and configure the logger reading voltage on the display. By inserting the junction only partly into the hot or cold water, or by heating the wire in between your fingers, try to figure out which part of the junction has the temperature displayed by the instrument. Is it the tip, the end towards the insulation or perhaps an average of the whole length? It is whole length

Q 06) Insert the thermocouple with a short junction to the bottom of the thin copper tube (which could be a thermometer well of a test rig). Insert the copper tube two, five, ten and 20 cm into the ice-water and read the temperature for each insertion length, waiting till the temperature has stabilized completely before each reading. Redo the test using the brass tube. Finally, redo the test with only the (coiled) thermocouple wire without a tube. Compare the results. Which of the two tubes is the better thermometer well? Why? What insertion length is necessary to make a correct reading in the three cases? In this case junction J1 is in ice water (0 °C), and junction J2 is in ambient condition (19.99 °C) Thus, V1 = 0

V2 = 0.7896

= V1 – V2

= 0 – 0.7896

= – 0.7896

Correct reading is (– 0.7896) mV

6

Thermocouple Copper tube Brass tube Bare wire

2cm 5cm 10cm -0.7516mV -0.77022mV -0.78921mV -0.7586mV -0.78185mV -0.78921mV -0.78573mV -0.78921mV -0.78921mV

Errors of above values are shown in the following table. Thermocouple 2cm 5cm Copper tube 0.038mV 0.01938mV Brass tube 0.031mV 0.00775mV Bare wire 0.00387mV 0.00039mV

10cm 0.00039mV 0.00039mV 0.00039mV

Error (mV) vs Inserted Length (cm) 0.05 Copper Tube Brass tube Bare wire

0.04

Error (mV)

0.03

0.02

0.01

0 0

0.5

1

1.5

2

2.5

3

3.5

-0.01 Inserted Length (cm)

Error decreases with the insertion length increases. Minimum error happens when bare wire is used. But Brass is better for thermometer well.

7

Q 08) Connect a Pt100 sensor to the logger. Make a drawing below, and explain how the measurement is done. Read the manual of the logger to see how Pt100 measurements are conducted. Illustrate the use both 2 wire as well as 4 wire measurements. What is the difference between these measurements?

Two Wire Configuration

The above figure illustrates two wires configuration. It is only used when high accuracy is not required as the resistance of the connecting wires is always included with that of the sensor leading to errors in the signal. The values of the lead resistance can only be determined in a separate measurement without the resistance thermometer sensor and therefore a continuous correction during the temperature measurement is not possible. Four Wire Configuration

The four wire resistance thermometer configuration even further increases the accuracy and reliability of the resistance being measured. In the diagram above a standard two terminal RTD is used with another pair of wires to form an additional loop that cancels out the lead resistance.

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Q 09) Configure the logger to read the temperature of the Pt100 sensor while having the sensor in the air. What is the temperature for the 2 wire setup? Measured resistance is 107.8 ohm PT100 has a resistance of 100 ohms at 0 °C and 138.4 ohms at 100 °C.

resistance =

107.8 =

(138.4 − 100) Temperature + 100 100

(138.4 − 100) Temperature + 100 100

Temperature = 20.31 °C Q 10) Redo point 9 above but insert resistance of the each of the wire, simulating (very) long connection cables. What is the new reading? Conclusion? What would be an alternative connection which may be better? Measured resistance

resistance =

= (107.8+1+1) ohm = 109.8 ohm

(138.4 − 100) Temperature + 100 100 9

Temperature

= 25.52 °C

Conclusion: When long connection cables are used, two wires configuration should not be used, since resistance of the cables cause an error to the final reading. Alternative connection Four wires configuration can be used, if long cables are used. Q 11) Read the temperature of ice-water with the illustrated setup using thermocouple and a Pt100-element. What is reading of the thermocouple? ………………………mV ………………………°C -What is reading of the Pt100-element? ………………………Ω ………………………°C

Resistance of Pt100 element

= 107.8 ohm

Temperature corresponds to 107.8 ohm

= 25.52 °C. (Calculated in question 9)

Thus, temperature of J2 and J3

= 25.52 °C 10

= V3 + V1 – V2

= -0.7896 mV

V3

=0

since the junction is made up of similar metal.

11

Calculated values of V2 Temperature ( oC ) 25 26

Voltage (mV) 0.992 1.033

o

Voltage (mV) vs Temperature ( C) 1.04 1.035 1.03 1.025

y = 0.041x - 0.033

Voltage

1.02 1.015 1.01 1.005 1 0.995 0.99 0.985 24.8

25

25.2

25.4

25.6

25.8

26

Temperature

From the above graph Voltage

= 0.041×temperature + 0.033

V2

= 0.041×25.52 + 0.033

V2

= 1.07932

= V3 + V1 – V2

-0.7896

= 0 + V1 – 1.07932

V1

= 0.28972

Temperature ( oC ) 7 8

12

Voltage (mV) 0.273 0.312

26.2

Voltage (mV) vs Temperature (oC) 0.315

0.31

0.305

0.3

Voltage

y = 0.039x 0.295

0.29

0.285

0.28

0.275

0.27 6.8

7

7.2

7.4

7.6

7.8

8

8.2

Temperature

Voltage

= 0.039×temperature

0.28972

= 0.039×temperature

Temperature of junction 1

= 7.429 °C

When thermocouple is used in practical application, it is not simple to maintain reference temperature. RTD is integrated with the thermocouple to measure the temperature of the secondary junction of the thermocouple. Thus, reference temperature is measured by the RTD. Since, reference temperature is known; it is possible to calculate the actual temperature.

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