Experiment 10:

Response of the 2nd order over damped system

Thermowell/Pseudo first order

Aim: To determine the time constant of the given thermowell from its response to the step change in the

surrounding temperature.

Apparatus:

A thermowell made-up from about 50c.c. Volume test tube filled with water, with small bulb

mercury in glass thermometer, fitted at its center

Oil bath with a heater

Stop watch

Tissue paper to wipeout oil from the thermowell

Procedure: 1. Heat the oil in the oil bath to the temperature such that the temperature of thermowell dipped in the oil

bath rises to about 90-95 oc

2. Take out the thermowell from oil bath and wipeout oil from the surface of test tube with tissue paper

3. As soon as the temperature reaches 90=95oc start the stop watch

4. Note down the time taken for the temperature reading of the thermowell to fall by every 2oc interval

Theory: The majority of thermal systems contain two or more capacities are interacting, since the heat flux

depends on temperature difference and not just on the temperature in one part of the system. The

simplest type of interaction is that typified by thermometer bulb placed inside a thermowell. This is

called “dead end system” since the equation for the bub has no output term. In the following derivation

the only resistances are assumed to be those in the fluid outside the thermowell and the fluid in between

the thermowell and the thermometer bulb. The thermal capacities of envelop of thermometer as well as

that of thermowell are neglected. The heat balance equation for thermometer can be written as

Thermal energy in = Accumulation of thermal energy (1)

H2*A2*(T1 - T2) = M2*Cp2*𝑑𝑡2

𝑑𝑡 ………….(2)

[h1*A1*(Ti - T1)] – [h2*A2*(T1 - T2)] = M1*Cp1* 𝑑𝑡1

𝑑𝑡 ………..(3)

Taking Laplace transform of equations (2) and (3) and rearranging the equations to get a relation

Between input temperature Ti and the temperature indicated by the thermometer T2 gives

𝑇2(𝑆)

𝑇1(𝑆) =

1

(𝑇1∗𝑇2)∗𝑆2+(𝑇1+𝑇2+𝑅1∗𝐶2)∗𝑆+1 …….(4)

In using equation (4), remember that τ2 is the time constant of the thermometer bulb in the well, which

is not the same as time constant of the bare bulb in the surrounding air. It may be noted that time

constant for larger bulb in air was only about 150 sec. Since the thermometer bulb in the present case is

smaller and it is immersed in water both capacitance and resistance terms will become small and hence τ

2 will be of the order of about 5-19 sec. The time constant τ1 Is that of thermowell, which is very large

in comparison to τ2. The term R1*C2 Has the dimensions of time and shows the interaction between the

two first order systems. Since C2 is very small R1*C2 Have a very small value. Equation (4) can be

factored to give two effective time constants. Therefore equation (4) can be written as

𝑇2(𝑆)

𝑇1(𝑆) =

1

(𝑇𝐴∗𝑆+1)∗(𝑇𝐵∗𝑆+1) ………(5)

The quadratic formula can be used to get TA And TB . Because of interaction, the ration of effective

time constants, TA/TB Is always greater than T2/T1 . This makes transient response more like that of a

single first-order element and is known as pseudo first order system.

Observations

Time from

beginning of

step change (s)

Temp indicated

by thermometer

(oC) Observed

Time

constant

(s)

Temp of

thermometer

predicted (oC)

0 90 86.87

39 88 3019.23 84.93

68 86 1919.29 83.52

119 84 1758.65 81.13

142 82 1462.18 80.09

164 80 1273.52 79.11

203 78 1222.62 77.42

233 76 1144.83 76.15

270 74 1105.55 74.64

311 72 1081.81 73.02

351 70 1055.29 71.50

392 68 1031.36 69.98

437 66 1015.22 68.39

483 64 998.87 66.81

538 62 995.38 65.02

591 60 984.28 63.36

650 58 978.05 61.61

707 56 964.86 60.00

783 54 970.04 57.97

857 52 966.20 56.11

940 50 965.17 54.17

1033 48 966.06 52.15

1112 46 947.64 50.56

1222 44 946.44 48.52

1362 42 955.60 46.19

1488 40 942.62 44.31

1638 38 930.64 42.33

1818 36 917.13 40.27

2148 34 945.46 37.23

Graph

Calculations: (for reading no.4)

Beginning temperature (Y0) = 90°C

Room Temperature = 28°C

Y (t) = Y0 - ΔY (1 -𝑒−𝑡/𝜏 )

𝜏 = −𝑡

ln (1−(𝑌𝑖𝑛𝑖−𝑌(𝑇)

𝛥𝑌)

Say for reading number 4:

84 = 90 - 62 (1 - 𝑒−𝑡/𝜏)

62 (1 -𝑒−𝑡/𝜏 ) = 90 – 84

𝑒−𝑡/𝜏= 1- ((90−84)

62)

Here t = 119 seconds

So, we get 𝜏 = 1169.16 s

Similarly after calculating all time constant values and averaging them, we get = 932.388 s

Now, by putting Value in Y (t) = Y0 - 𝛥Y (1 - 𝑒−𝑡/𝜏)

Where, Y0 = 90 and Y = 68

By putting value of time in the above equation, we can predict the temperature values.

So, for reading 4

Ypred. = 90 – 62 (1-𝑒−119/932.38 )

Ypred. = 82.57°C

y = 2E-12x4 - 1E-08x3 + 4E-05x2 - 0.072x + 90.824

0

10

20

30

40

50

60

70

80

90

100

0 500 1000 1500 2000 2500

Tem

pe

ratu

re (

°C)

Time (s)

Graph of Temperature vs. time

Y(t) obs

Y(t) pred

Poly. (Y(t) obs)

Results:

Equation of response can be written as y = 2E-12x4 - 1E-08x3 + 4E-05x2 - 0.0771x + 95.293 with one

point being (0, 90). 𝒅𝒀

𝒅𝒕At x = 0 gives slope of tangent at x = 0 and with y = mx + c

Where m = 𝒅𝒀

𝒅𝒕

And one point is (0, 90) so c = 90

M = -0.072

For x intercept, y = 0

So, x-intercept of tangent is 1250

Time constant of thermometer from tangent to the response curve at t (= 0) is = 1250 s

Time constant of thermometer from 63.2% response (0.632*2148) = 1357.53 s

Time constant of thermometer from analytical method 1169.16 s