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The overall heat transfer coefficient, U for an unsteady heat processing in a universal food processor was determined. Additionally, the effect of various circulating heating medium temperatures on the overall heat transfer coefficient was investigated. In this experiment, there are 3 forms of heat transfers between the water in the vessel and the heating water within the heating jacket: (1) heat convection between the water an inside surface of the vessel, (2) heat conduction through the jacket wall of the vessel and (3) heat convection between the jacket wall surface and the heating water. My report is free for all because I need good karma to be reborn as a thin woman in my next life and it's because I love you that's why. :)
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Sonia Gawade Experiment 2: Unsteady Heat Transfer Objectives: The overall heat transfer coefficient, U for an unsteady heat processing in a universal food processer was determined. Additionally, the effect of various circulating heating medium temperatures on the overall heat transfer coefficient was investigated. Notations: A Lateral surface area of a truncated cone c y-intercept of a straight line C p Specific heat capacity of water h Convective heat transfer coefficient m Gradient of a straight line equation t Time t 1 Time at an earlier interval t 2 Time at the later interval T Temperature at a specified time interval T 0 Initial temperature of the cold water T 1 Temperature at the earlier time interval T 2 Temperature at the later time interval T Temperature of the heating medium 1 | Page
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Page 1: Unsteady Heat Transfer

Sonia Gawade

Experiment 2: Unsteady Heat Transfer

Objectives:

The overall heat transfer coefficient, U for an unsteady heat processing in a universal food

processer was determined. Additionally, the effect of various circulating heating medium

temperatures on the overall heat transfer coefficient was investigated.

Notations:

A Lateral surface area of a truncated cone

c y-intercept of a straight line

Cp Specific heat capacity of water

h Convective heat transfer coefficient

m Gradient of a straight line equation

t Time

t1 Time at an earlier interval

t2 Time at the later interval

T Temperature at a specified time interval

T0 Initial temperature of the cold water

T1 Temperature at the earlier time interval

T2 Temperature at the later time interval

T∞ Temperature of the heating medium

TR Temperature ratio

U Overall heat transfer coefficient

V Volume of water in vessel

x Horizontal coordinate of a straight line graph

y Vertical coordinate of a straight line graph

ρ Density of water in vessel

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∆T Change in temperature

∆t Change in time

 π Mathematical constant that is the ratio of a circle's circumference to its diameter

Introduction:

Transient or unsteady heat transfer is a function of both time and location. On the other hand,

steady heat transfer varies solely as a function of location (Singh and Heldman, 2001).

Knowledge on unsteady heat transfer is important especially for food processes such as

sterilisation or pasteurisation as unsteady heat transfer is the dominant form of heat transfer in

these processes (Singh and Heldman, 2001). In this experiment, there are 3 forms of heat

transfers between the water in the vessel and the heating water within the heating jacket: (1)

heat convection between the water an inside surface of the vessel, (2) heat conduction

through the jacket wall of the vessel and (3) heat convection between the jacket wall surface

and the heating water. The heat transfer coefficient for these 3 heat transfers is termed as the

overall heat transfer coefficient or U. The thermal conductivity coefficient, k; was ignored as

the vessel wall was very thin.

Water has low thermal conductivity. At 2°C, thermal conductivity of water is 0.5606a W/

(m.K) and at 97°C, thermal conductivity of water is 0.6723a W/ (m.K). Hence temperatures

between 2°C and 97°C have thermal conductivities ranging between 0.5606 W/ (m.K) to

0.6723 W/ (m.K). The temperature of the heating medium used was at 35°C, 50°C and 65°C

for this experiment. Thus it is imperative to note that internal resistance to heat transfer is

negligible as water was being used as the heating medium.

Materials and Method:

a Standard Reference Data for the Thermal Conductivity of Water. (1995). American Institute of Physics and

American Chemical Society. IUPAC.

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The circulator bath was switched on. Temperature of the circulating medium was set to 35°C

using the temperature control dial. Heating control lamp flashed when temperature of 35°C

was reached. Seven litres of water was collected and its initial temperature (To), was

measured using a digital food thermometer. The vessel was filled with 7 L of water and

processor was switched on. Agitation was initiated by adjusting the motor control knob.

Initial temperature shown on the display panel of the food processor was recorded. Initial

temperatures of the cool water were recorded using different thermometers for comparison

purposes.

Temperature of the water (T), in the vessel was recorded at 30 seconds intervals for 5

minutes, then at 1 minute intervals for 10 minutes and finally at 3 minute intervals for 15

minutes. Temperature at frequent time intervals would be recorded initially as the initial

temperature gradients would be large. Hence having the temperature recorded at frequent

time intervals, would allow for better tracking of temperature changes.

After recording was done, the upper and lower circumferences of water in the vessel were

measured using a string. Height of the water was measured using a 30 cm ruler. Area and

volume were calculated using the dimensions obtained. The vessel was then emptied.

The steps above were repeated for circulating medium temperatures of 50°C and 65°C

excluding the measurement of dimensions of water in the vessel.

Results:

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[Table 1]: Dimensions of the water in the vessel.

Processing Vessel Processing Conditions

Upper diameter = Upper circumference / π

Where π was taken as 3.14159265359.

Upper circumference was 0.983 m.

Therefore, upper diameter was:

0.983 m / π = 0.313 m

Initial product temperature (T0):

26 26 28

Heating medium temperature (T∞):

35 50 65

Bottom diameter = Bottom circumference / π

Where π was taken as 3.14159265359.

Bottom circumference was 0.67 m.

Therefore, bottom diameter was:

0.67 m / π = 0.21 m

Height: 0.118 mVolume: 0.00662 m3

Area: 0.1058 m2

Area was calculated assuming the vessel took

on truncated cone geometry. Detailed

calculations available in the Appendix,

Section 1.

[Table 2]: Temperature profile when T∞ was 35°C.

Time (s) T (°C) T∞ - T (°C)

ΔT(°C)

Δt(s)

ΔT/Δt (°C/s)

TR ln TR

30 26 9 1 180

0.00556 1 0

210 27 8 1 330

0.00303 0.88889 -0.1178

540 28 7 1 120

0.00833 0.77778 -0.2513

660 29 6 1 600

0.00167 0.66667 -0.4055

1260 30 5 1 180

0.00556 0.55556 -0.5878

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Sample calculation for ΔT/Δt obtained in the first row of Table 2.

T∞ = 35°C

T0 = 26°C

∆T = T2-T1, where T2 is temperature at a later time interval and T1 is temperature at an earlier

time interval. Therefore, T2 is T from second row of Table 2 and T1 is T from first row of

Table 2.

Therefore ∆T = 27°C - 26°C = 1°C

∆t = t2-t1, where t2 is time at a later interval and t1 is time at an earlier interval. Therefore, t2 is

time from second row of Table 2 and t1 is time from first row of Table 2.

∆ T∆t =

T2−T 1

t2−t1 =

27−26210−30 = 0.005555... °C/s

~ 0.00556 °C/s

Sample calculation for TR and In (TR) obtained in the third row of Table 2.

ln (TR) = ln T ∞−TT∞−T 0

ln (TR) = ln 35−2835−26 = ln 0.77778 = -0.2513

Hence ∆ T∆t , TR and ln(TR) for all data points were calculated as shown above.

[Table 3]: Temperature profile when T∞ was 50°C.

Time (s) T (°C) T∞ - T (°C)

ΔT(°C)

Δt(s)

ΔT/Δt (°C/s)

TR ln TR

30 27 23 1 60 0.01667 0.95833 -0.0426

90 28 22 1 30 0.03333 0.91667 -0.087

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120 29 21 1 60 0.01667 0.875 -0.1335

180 30 20 1 90 0.01111 0.83333 -0.1823

270 31 19 1 90 0.01111 0.79167 -0.2336

360 32 18 1 60 0.01667 0.75 -0.2877

420 33 17 1 60 0.01667 0.70833 -0.3448

480 34 16 1 180

0.00556 0.66667 -0.4055

660 35 15 1 60 0.01667 0.625 -0.47

720 36 14 1 180

0.00556 0.58333 -0.539

900 37 13 2 180

0.01111 0.54167 -0.6131

1080 39 11 1 180

0.00556 0.45833 -0.7802

1260 40 10 1 180

- 0.41667 -0.8755

1440 40 10 0 180

0.00556 0.41667 -0.8755

1620 41 9 1 180

0.01111 0.375 -0.9808

1800 43 7 2 180

- 0.29167 -1.2321

[Table 4]: Temperature profile when T∞ was 65°C.

Time (s) T (°C) T∞ - T (°C)

ΔT(°C)

Δt(s)

ΔT/Δt (°C/s)

TR ln TR

30 30 35 1 60 0.01667 0.94595 -0.0556

90 31 34 1 30 0.03333 0.91892 -0.0846

120 32 33 2 30 0.06667 0.89189 -0.1144

150 34 31 1 60 0.01667 0.83784 -0.1769

210 35 30 1 30 0.03333 0.81081 -0.2097

240 36 29 1 60 0.01667 0.78378 -0.2436

300 37 28 1 60 0.01667 0.75676 -0.2787

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360 38 27 1 60 0.01667 0.72973 -0.3151

420 39 26 1 60 0.01667 0.7027 -0.3528

480 40 25 2 60 0.03333 0.67568 -0.392

540 42 23 1 60 0.01667 0.62162 -0.4754

600 43 22 1 60 0.01667 0.59459 -0.5199

660 44 21 1 60 0.01667 0.56757 -0.5664

720 45 20 1 120

0.00833 0.54054 -0.6152

840 46 19 1 60 0.01667 0.51351 -0.6665

900 47 18 1 180

0.01111 0.48649 -0.7205

1080 49 16 2 180

0.01111 0.43243 -0.8383

1260 51 14 2 18

0

0.01111 0.37838 -0.9719

1440 53 12 1 18

0

0.00556 0.32432 -1.126

1620 54 11 1 18

0

0.00556 0.2973 -1.213

1800 55 10 - - - 0.27027 -1.3083

Raw data for T∞ at 35°C, 50°C and 65°C are available in the Appendix, Section 2, Tables 1, 2

and 3 respectively.

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0 5 10 15 20 25 30 35 400

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

f(x) = 0.00118388414997025 x − 0.00875105435429478R² = 0.392220902933888

f(x) = 0.00104757065217392 x − 0.00396329347826091R² = 0.402751816178359

f(x) = 0.000136 x + 0.00387800000000001R² = 0.00696873563564837

35°CLinear (35°C)50°CLinear (50°C)65°CLinear (65°C)

T∞ - T (°C)

ΔT/Δ

t

[Figure 1]: Relationship between ΔT/Δt and (T∞-T) of cool water heated up by the circulating

heating water with temperatures of 35°C, 50°C and 65°C respectively.

With reference to Figure 1, the data points obtained were rather scattered. The gradient

obtained for each trend line was different.

∆ T∆t =

UAρV C p

(T∞ - T) (Equation

1)

The convective heat transfer coefficient, h was replaced with U. As seen above in equation 1,

it is in the form of y = mx. Where y is ∆ T∆t , x is (T∞ - T) and the gradient or m is

UAρV C p

.

Hence, the gradient obtained can be used to calculate the overall heat transfer coefficient, U.

A sample calculation for calculating U for T∞ at 35°C is available as follows:

m= UAρ V C p

(Equation 2)

V (actual volume of cold water) = 0.00662 m3;

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ƿ (density of water at T0 of 26°C) = 996.787 kg/m3;

Cp (specific heat capacity of water) = 4179.3 kJ/(kg ºC)

A (lateral surface area of a truncated cone) = 0.1058 m2

m (gradient for T∞ at 35°C) = 0.0007

Using Equation 2,

0.0007 ¿U (0.1058)

(996.787 ) (0.00662 )(4179.3)

U = 182.4636 W/(m2.K)

U ~ 182.46 W/(m2.K)

[Table 5]: Experimental U values calculated when gradient equals UA

ρV C p when T∞ was 35°C,

50°C and 65°C.

T∞ (°C): Gradient

[UA

ρV C p] :

T0 (°C): Density at T0

(kg/m3):

Cp [kJ/(kg ºC)]: U [W/(m2 .K)]:

35 0.0007 26 996.787b 4179.3c 182.46

50 0.00082 26 996.787b 4179.3c 213.74

65 0.00084 28 996.237b 4178.8c 218.81

b Density of water at 1 atmosphere (from: Handbook of Chemistry and Physics, CRC press, 64th Ed.)

c Osborne, Stimson, and Ginnings, B. of S. Jour. Res., 23, 238 (1939) in Handbook of Chemistry and

Physics, 53rd ed., Cleveland, Ohio, D128 (1972-1973).

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0 200 400 600 800 1000 1200 1400 1600 1800 2000

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0f(x) = − 0.000782213074585905 xR² = 0.995326349132963f(x) = − 0.000675622463981182 xR² = 0.991412631003088f(x) = − 0.000403266490247966 xR² = 0.96019332250948 35°C

Linear (35°C)

50°C

Linear (50°C)

65°C

Linear (65°C)

t (s)

In (T

R)

[Figure 2]: Relationship between ln (TR) versus time of cool water heated up by the

circulating heating water with temperatures of 45°C, 60°C and 75°C respectively.

ln (TR) = - UA

VρC pt (Equation 3)

Where ln (TR) = ln T ∞−TT∞−T 0

As seen above in equation 3, it is in the form of y = mx. Where y is ln (TR), x is time (t) and

the gradient or m is - UA

VρC p. Hence, the gradient obtained can be used to calculate the overall

heat transfer coefficient, U.

-m = - UA

VρC p (Equation 4)

A sample calculation for calculating U for T∞ at 35°C is available as follows:

V (actual volume of cold water) = 0.00662 m3;

ƿ (density of water at T0 of 26°C) = 996.787 kg/m3;

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Cp (specific heat capacity of water) = 4179.3 kJ/ (kg ºC)

A (lateral surface area of a truncated cone) = 0.1058 m2

m (gradient for T∞ at 35°C) = -0.0004

Using Equation 4,

-0.0004 = - U (0.1058)

(0.00662 ) ( 996.87 )(4179.3)

U = 104.2649 W/ (m2.K)

U ~ 104.26 W/ (m2.K)

[Table 6]: Experimental U calculated when gradient equals - UA

VρC p values when T∞ was

35°C, 50°C and 65°C.

T∞ (°C): Gradient

[ UA

V ρ C p ] :

T0 (°C): Density at T0

(kg/m3):

Cp [kJ/(kg ºC)]: U [W/(m2 .K)]:

35 0.0004 26 996.787d 4179.3e 104.26

50 0.0007 26 996.787d 4179.3e 182.46

65 0.0008 28 996.237d 4178.8e 208.39

Percentage difference between U calculated from gradient obtained from a plot of ΔT/Δt

versus (T∞-T) and ln (TR) versus time was calculated.

A sample calculation for the percentage difference when T∞ was 35°C is presented below.

U calculated when using equation 2 was 182.46 W/ (m2.K).

d Density of water at 1 atmosphere (from: Handbook of Chemistry and Physics, CRC press, 64th Ed.)

e Osborne, Stimson, and Ginnings, B. of S. Jour. Res., 23, 238 (1939) in Handbook of Chemistry and

Physics, 53rd ed., Cleveland, Ohio, D128 (1972-1973).

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U calculated when using equation 4 was 104.26 W/ (m2.K).

Formula for percentage difference:

Percentage difference was larger number−smaller number

larger number x 100 %

Percentage difference was 182.46−104.26

182.46 x 100 % = 42.8587 %

~ 42.86 %

[Table 7]: Percentage differences when U was calculated using equation 2 and 4 when T∞ was

35°C, 50°C and 65°C.

T∞ (°C): U [W/(m2 .K)] calculated using

Equation 4:

U [W/(m2 .K)] calculated using

Equation 2:

Percentage difference (%):

35 104.26 182.46 42.86

50 182.46 213.74 14.63

65 208.39 218.81 4.76

Discussion:

In this experiment, an important assumption made was that heat transfer from the heating

medium to the water was solely due to convection. Heat transfer via conduction and radiation

were ignored. The vessel used was made of stainless steel and vessel walls were very thin and

thus the heat resistance was very low. This is because for heat transfer to have negligible

resistance, conductive resistance must be low or negligible. It was also assumed that heat loss

to the surroundings wa negligible.

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The display panel of the universal food processor was only able to indicate the temperature of

water being heated up to two significant figures and zero decimal places. Hence an increment

of less than 1°C would not be displayed. This instrumental error was taken into account when

calculating the experimental U values. By logic, heat would be continuously transferred from

the heating medium to the water, resulting in a temperature increase for the water. Thus the

temperature at any appropriate time interval would have shown an increment. However this

was not the case, at several time intervals, ΔT/Δt had a magnitude of 0. Hence only some data

points could be use when plotting the graphs. This would have led to inaccuracies when

calculating the experimental U values from the gradients. To counter this, a temperature

probe that is able to display a higher amount of significant figures could be used.

The overall heat transfer coefficient U, was determined from the gradients of both graphs as

shown in Figure 1 and 2. The graph in Figure 2 had better R2 values as compared to the R2

values in Figure 1. With reference to Figure 2, data points obtained were more consistent and

R2 values obtained were close to 1, indicating that the linear model chosen was appropriate.

With reference to Table 5 and 6, both the gradients and U values increased as T∞ was

increased from 35°C to 50°C and then to 65°C. This was expected as U is directly

proportional to the gradient, m as shown in Equation 2 and 4. As U is directly proportional to

the gradient m, a linear straight line graph should have been produced by the data points

obtained for Figure 1 and 2. Convection takes place by means of macroscopic fluid motion.

Convection can be caused by a forced convection or by temperature dependent density

variations in the fluid. The latter is termed as natural convection. Temperature dependent

density variations in the water as it was being heated may have caused temperatures that were

not representative of the actual temperature of the water at a particular time interval to be

recorded instead. Thus leading to data points being scattered. However, the probability that

this was the cause for data points being scattered are minimal due to the presence of the

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revolving baffle. The heat transferred from the circulating hot medium to the water was

quickly distributed by the revolving baffle thus temperature dependent density variations

within the water were minimal.

Before the experiment commenced, the temperature of the water was measured using a

temperature probe. The initial water temperature measured using the digital food

thermometer for when T∞ was 35°C, 50°C and 65°C were 26.5°C, 26.8°C and 26.8°C

respectively. These values differed by about 1°C when compared with T0 values as shown in

Table 1. The T0 values were measured using the temperature probe located within the

processor. The reason for this discrepancy could be attributed to how the temperature probe

has been incorporated into the processor. The temperature probe is not insulated and passes

through the heating jacket of the processor; thus temperature measured by the probe is

influenced by the temperature of the heating jacket. Thus when T∞ was 65°C, the initial

temperature of the water detected by the temperature probe was almost 2°C higher than what

was measured by the digital food thermometer. This was because heat was transferred to the

temperature probe from the circulating heating medium. To minimise errors of this sort, the

processor has to be redesigned to insulate the temperature probe.

Another source of error included how the upper and bottom diameter of water in the vessel

was measured. A string was used to outline the circumference of the vessel then extended

onto a ruler to determine its value. The ruler was only accurate to 1 decimal place. Parallax

error may have occurred when reading the measurement off the ruler. A large digital vernier

calliper could have been used instead to measure the circumference and at the same time

minimise parallax error.

With reference to Table 7, percentage differences for U values decreased at T∞ increased.

This meant that either Equation 1 or 3 could be used when T∞ is higher than 65°C as

percentage difference between calculations using either Equation 1 or 3 would be less than

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5 %, leading to a better estimation of U. However, more studies have to be conducted with

the changes implemented and at T∞ higher than 65°C to make a more accurate assessment.

Conclusion:

The circulating heating medium temperature and the overall heat transfer coefficient for an

unsteady state heat processing; seemed to have a positive linear relationship as U increased

when T∞ was increased. The overall heat transfer coefficient was estimated more accurately

for the plot of In (TR) versus t as R2 values indicated that the linear model chosen was

appropriate.

References:

Singh, P. and Heldman, D. (2001). Introduction to Food Engineering. 3rd ed. London:

Academic Press. P280-300.

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APPENDIX

Section 1

Calculating area of heat transfer in the universal food processor:

[Figure 1]: A truncated cone and the various symbols denoting the various measurements.

Volume of water in the universal food processor was 0.007 m3.

Upper circumference of water in the vessel was 0.983 m.

Formula for circumference of a circle is 2πr, where r is radius of the circle.

Radius of the upper vessel (r1) is 0.983 m / 2π = 0.156 m

Bottom circumference of water in the vessel was 0.67 m.

Radius of the lower vessel (r2) is 0.67 m / 2π = 0.11 m

Surface area of water in vessel was calculated using the formula for lateral surface of a

truncated cone, assuming the volume of the water took on truncated cone geometry.

Formula for the lateral surface of a truncated (A) cone is given as:

A = [π x s2 x (r1 + r2)] m2 ---- (1)

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s2 can be calculated as:

s2 = √h2+(r1−r2)2 m

s2 = √0.1182+(0.156−0.11)2 m

s2 = 0.1266 m

s2 ~ 0.13 m

Substituting s2 = 0.1266 m into Equation (1), lateral surface area of water in the vessel was:

A = [π x 0.1266 x (0.156 + 0.11)] m2

= 0.1058 m2

~ 0.11 m2

Actual volume of water was calculated using formula for calculating the volume of a truncated cone.

Formula for volume of water (V) in a truncated cone is given as:

V = {π x h x [r1 + r2 + (r1 x r2)] x 1/3} m3

V = {π x 0.118 x [0.156 + 0.11 + (0.156 x 0.11)] x 1/3} m3

V = 0.00662 m3

V ~ 0.007 m3

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APPENDIX

Section 2

[Table 1]: Raw data obtained when T∞ was 35°C.

Time (s) T (°C) T∞ - T (°C)

ΔT/Δt (°C/s)

TR ln TR

0 26 9 - 1 030 26 9 0.00556 1 0

60 26 9 - 1 090 26 9 - 1 0120 26 9 - 1 0150 26 9 - 1 0180 26 9 - 1 0210 27 8 0.00303 0.88889 -0.1178240 27 8 - 0.88889 -0.1178270 27 8 - 0.88889 -0.1178300 27 8 - 0.88889 -0.1178360 27 8 - 0.88889 -0.1178420 27 8 - 0.88889 -0.1178480 27 8 - 0.88889 -0.1178540 28 7 0.00833 0.77778 -0.2513600 28 7 - 0.77778 -0.2513660 29 6 0.00167 0.66667 -0.4055720 29 6 - 0.66667 -0.4055780 29 6 - 0.66667 -0.4055840 29 6 - 0.66667 -0.4055900 29 6 - 0.66667 -0.40551080 29 6 - 0.66667 -0.40551260 30 5 0.00556 0.55556 -0.58781440 30 5 - 0.55556 -0.58781620 30 5 - 0.55556 -0.58781800 30 5 - 0.55556 -0.5878

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[Table 2]: Raw data obtained when T∞ was 50°C.

Time (s) T (°C) T∞ - T (°C)

ΔT/Δt (°C/s)

TR ln TR

0 26 24 - 1 030 27 23 0.01667 0.95833 -0.0426

60 28 22 - 0.91667 -0.08790 28 22 0.03333 0.91667 -0.087120 29 21 0.01667 0.875 -0.1335150 29 21 - 0.875 -0.1335180 30 20 0.01111 0.83333 -0.1823210 30 20 - 0.83333 -0.1823240 30 20 - 0.83333 -0.1823270 31 19 0.01111 0.79167 -0.2336300 31 19 - 0.79167 -0.2336360 32 18 0.01667 0.75 -0.2877420 33 17 0.01667 0.70833 -0.3448480 34 16 0.00556 0.66667 -0.4055540 34 16 - 0.66667 -0.4055600 34 16 - 0.66667 -0.4055660 35 15 0.01667 0.625 -0.47720 36 14 0.00556 0.58333 -0.539780 36 14 - 0.58333 -0.539840 36 14 - 0.58333 -0.539900 37 13 0.01111 0.54167 -0.61311080 39 11 0.00556 0.45833 -0.78021260 40 10 - 0.41667 -0.87551440 40 10 0.00556 0.41667 -0.87551620 41 9 0.01111 0.375 -0.98081800 43 7 - 0.29167 -1.2321

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[Table 3]: Raw data obtained when T∞ was 65°C.

Time (s) T (°C) T∞ - T (°C)

ΔT/Δt (°C/s)

TR ln TR

0 28 37 - 1 030 30 35 0.01667 0.94595 -0.0556

60 30 35 - 0.94595 -0.055690 31 34 0.03333 0.91892 -0.0846120 32 33 0.06667 0.89189 -0.1144150 34 31 0.01667 0.83784 -0.1769180 34 31 - 0.83784 -0.1769210 35 30 0.03333 0.81081 -0.2097240 36 29 0.01667 0.78378 -0.2436270 36 29 - 0.78378 -0.2436300 37 28 0.01667 0.75676 -0.2787360 38 27 0.01667 0.72973 -0.3151420 39 26 0.01667 0.7027 -0.3528480 40 25 0.03333 0.67568 -0.392540 42 23 0.01667 0.62162 -0.4754600 43 22 0.01667 0.59459 -0.5199660 44 21 0.01667 0.56757 -0.5664720 45 20 0.00833 0.54054 -0.6152780 46 19 - 0.51351 -0.6665840 46 19 0.01667 0.51351 -0.6665900 47 18 0.01111 0.48649 -0.72051080 49 16 0.01111 0.43243 -0.83831260 51 14 0.01111 0.37838 -0.97191440 53 12 0.00556 0.32432 -1.1261620 54 11 0.00556 0.2973 -1.2131800 55 10 - 0.27027 -1.3083

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Page 21: Unsteady Heat Transfer

Sonia Gawade

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