Download - Heat Conduction 2014-15

8/10/2019 Heat Conduction 2014-15

1/12

Sheikh Shahir Heat Conduction KEM120702

1.0 OBJECTIVE

The objective of this experiment was to scrutinize the temperature gradient and determine material

thermal conductivity when the radial conductivity is constant.

2.0 ABSTRACT

We carried out the experiment in a mannerism so as to ensure that the objectives of our experiment was

met. For our equipment a cylindrical brass disc is used. It has an inner radius of 4 mm and an outer radius

of 55mm. Its length is 3mm. We attach thermistor sensors interspaced at every 10mm along the radius.

Heat intake is taken care of with the aid of an electric consul. We apply heat to the disc electrically and

power flow to the disc can be controlled. Outer surface of the disc requires cooling and this is done with

a steady influx of water. Experiment is started off by controlling the heat intake using an electric consul.

Heat applied to the cylinder can be varied with a varied power output. Having done that, the temperature

at each position on the cylinder was recorded. This was repeated using 30W and 50W power supply. The

sensor is to be deployed onward from the centre of the cylinder radius at a spacing of 10mm. The readings

are taken at six variable marks. The readings of the power input and the temperature is tabulated. After

the data is in tabulated form, we are required to plot two graphs. One that correlates the relation between

Temperature (T) and the distance (r ) .This allows us to find out the theoretical value of conductivity. .

Another plot is made using T1-T2C versus In ro/ ri. This helps to achieve the conductivity results

theoretically and from thereon comparison can be made with Fouriers law.These results allow us to reach

conclusions on how successful our experiment was and account for errors.

3.0 INTRODUCTION AND THEORETICAL BACKGROUND

3.1 INTRODUCTION

To start off, lets take a look at what heat is. Heat is basically a form of energy that can be transferred

from onebody to another. It can occur when theres a temperature difference between the two bodies inquestion. Thermodynamic analysis can be said to be concerned with the amount of heat transfer a body

undergoes whence it undergoes transformation from one equilibrium state to another. The energyconcerned with the science of dealing with transfers is known as heat energy. Heat transfer will alwaysoccur from a body of higher temperature to a body of lower temperature. The heat transfer will stop when

the two bodies reach the same temperature.

Heat can be transferred in three different ways. Those being conduction, radiation and convection. Also,for all three modes, heat transfer is always from a higher temperature body to a lower temperature one.The SI unit of heat is the Joule. We can measure heat by Calirometry or determine it indirectly based on

calculations on other methods. In Physics, the concepts of latent heat and sensible heat are used.

Now, the different forms of heat transfer are discussed.


2/12


Conduction can be considered to be the transfer of energy from particles that are at a higher energy level

to those adjacent particles that are at a lower energy level. This occurs as a result of interaction betweenthe said particles. Conduction can take place in solids, liquids, or gases.In gases and liquids, conduction has a slightly different meaning. Conduction in gas and liquids occur dueto collision of diffusion of the fluid particles during their random motion. In solids, it is a combination of

vibrationsof the molecules in a lattice and the energy that is transported by free electros. A cold canneddrink in a warm room, for example, eventually warms up to the room temperature as a result of heattransfer from the room to the drink through the aluminum can by conduction.

Rate of heat conduction through a medium will depend on the geometrical shape of the medium, itsmaterial, how thick it is and also the temperature difference across the medium. If you wrap a container ofhot water with an insulating material, the rate of heat transfer is greatly diminished. The better theinsulation the lower the rate of heat loss.

Fig 1-1: Heat Conduction through a large plane wall thickness x and area A.

Heat conduction can be understood using the figure above. The above figure shows a plane wall. Its

thickness and area are labelled in the diagram. Theres also a temperature difference between two sides of

the wall. Its been concluded from experiments that heat transfer rate, which is denoted by Q, through the

wall becomes double of its usual value when the area A which is perpendicular in the direction of motion

of heat or the temperature gradient across the wall is doubled. Heat conduction rate is halved when the

thickness between the wall layers is doubled.

Rate of Heat conductionThickness

DifferenceeTemperaturArea

Equation 1


3/12


x

TTkAQ

21.

Where k is the thermal conductivity

K denotes a materials excellence in the conducting warmth. Using limits the equation can be smited

down todx

dTkAQ

.

, which is actually Fouriers law of heat conduction. The gradient of the

temperature is denoted by dx/dT.

3.2 THEORETICAL BACKGROUND

To start off, we assume the shape of the brass disc to be cylindrical.

Fig 1-3: Apparatus of the experimentWith the help of Fouriers law, the

Equation 2

)()/ln(

2oi

io

r TTrr

Lkq

where k = thermal conductivity (Wmk)

L = cylinder length (m)

ro, ri = outer / inner radius of cylinder (m)To, Ti = outer / inner temperature ( )

Figure 3 shows a hollow cylinder of which inner and outer surfaces are exposed to fluids at differenttemperatures. For steady-state conditions with no heat generation, the appropriate form of the heat

equation is:

01

dr

dTkr

dr

d

r. (3)


4/12


From Fouriers law, the rate at which energy is conducted across any cylindrical surface in the solid is:

dr

dT)rL2(k

dr

dTkAq r ... (4)

whereA = 2rL is the heat transfer area or area normal to the direction of heat transfer. From Equation(3), the quantity kr (dT/dr) is independent of r. Therefore, for equation (4), conduction heat transfer rate qris a constant in the radial direction.Assuming the value of k to be constant, Equation (3) is integrated twice to obtain the general solution:

T(r) = C1ln r + C2. (5)

Boundary conditions:T(ri) = Ti and T(ro) = To

Applying these conditions to the general solution, we then obtain:

Ti= C1ln ri+ C2 and To= C1ln ro+ C2

Solving for C1and C2and substituting into the general solution, we then obtain

o

ooi

oi Tr

rln)r/rln(

TT)r(T

. (6)

This temperature is used with Fouriers law, Equation (4), we obtain the following expression for the heattransfer rate:

)()/ln(

2oi

io

r TTrr

Lkq

... (7)

The relationship between (TIT) and ln r

From Equation (6),

o

ooi

oi Tr

rln)r/rln(

TT)r(T

o

ooi

oi

ii T

r

rln

)r/rln(

TTT)r(TT

o

i0

oi

io

oi

i rln)r/rln(

11)TT(rln

)r/rln(

TT)r(TT

)/ln(

ln)/ln()(ln

)/ln()(

0

0

i

oioi

io

oii

rr

rrrTTr

rr

TTrTT


5/12


)r/rln(

rln)TT(rln

)r/rln(

TT)r(TT

i0

i

io

io

oi

i

From above equations,

Gradient of Graph TiT(r) versus ln r, m =)r/rln(

TT

io

oi

Substitute m into Equation (7) gives

Thermal conductivity,Lm2

qk r

. (8)

4.0 RESULTS

Power input, qof 10.7 WattTable 1

Position Thermistor

position from

centre radius, r

(m)

Temperature, T

(C)

Temperature

difference, Ti-T

(C)

ln ( )

1st 0 34.8 0.0 -

2nd

0.01 33.1 1.7 0.916

3rd

0.02 31.4 1.7 1.609

4th

0.03 30.5 0.9 2.015

5th

0.04 30.2 0.3 2.303

6th

0.05 29.7 0.5 2.526



6/12


Position Thermistor

position from

centre radius, r

(m)

Temperature, T

(C)

Temperature

difference, Ti-T

(C)

ln ( )

1st 0 48.1 0.0 -

2nd

0.01 41.4 6.7 0.916

3rd

0.02 36.1 5.3 1.609

4th

0.03 33.4 2.7 2.015

5th

0.04 32.0 1.4 2.303

6th

0.05 30.8 1.2 2.526


Position Thermistor

position from

centre radius, r

(m)

Temperature, T

(C)

Temperature

difference, Ti-T

(C)

ln ( )

1st 0 60.4 0.0 -

2nd

0.01 49.5 10.9 0.916

3rd 0.02 41.7 7.8 1.609

4th

0.03 36.8 4.9 2.015

5th

0.04 34.3 2.5 2.303

6th

0.05 32.2 2.1 2.526


7/12


Graph 1: Graph of temperature, T (C) against thermistor position from centre, r (m)

CalculationsGiven that:Cylinder stride, L = 0.003 m,Inner radius of cylinder, ri= 0.004 m,

Outer radius, ro= 0.055 mWe assume that the temperature inside the cylinder, Tiis equal to temperature of thermistor at position 1,

1st

and temperature outside the cylinder, Tois equal to the temperature of thermistor at position 6, 6th

.There are two methods to calculate the thermal conductivity of material, k

First method

In the first method, we will use Fouriers Lawequation as follows:

( )

For power input, q= 10.7 Watt

From Table 1Ti= 34.8C; 307.8 KTo= 29.7C; 302.7 K

()

Thermal conductivity, k= 291.7 W/m.K

34.833.1

31.4 30.5 30.2 29.7

48.1

41.436.1

33.432 30.8

60.4

49.5

41.736.8

34.332.2

0

10

20

30

40

50

60

70

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

TemperatureGradient,T((C)

Thermistor postion from centre, r(m)

Q = 10.7 W

Q =30.7 W

Q = 50.5 W


8/12


For power input, q= 30 .7Watt

From Table 2Ti = 48.1C; 321.1 KTo = 30.8C; 303.8 K

()

Thermal conductivity, k=246.75 W/m.K

For power input, q= 50.7 Watt

From Table 3Ti = 60.4C; 333.4 K

To = 32.2C; 305.2 K

()


Graph 2: A Graph of temperature difference, TiT (C) vs ln ( )

Second method

In the second method, we will use this equation:

0

2

4

6

8

10

12

0 0.5 1 1.5 2 2.5 3TEMPERATUREDIFFERENCE,

(TI-T)(C)

LN (RO / RI)

Q = 10.7 W Q=30.7 W Q=50.5 W


9/12


where mis the gradient of graph T i- Toversus ln r

( )

For power input, q= 10.7 WattFrom graph 2, the graph equation: y =- 2.315x

The gradient of the graph, m = 2.31Thus, thermal conductivity,


For power input, q= 30.7 WattFrom graph 2, the graph equation: y = -5.49x

The gradient of the graph, m = 5.49

Thus, thermal conductivity,


For power input, q= 50 WattFrom graph 2, the graph equation: y = -8.53xThe gradient of the graph, m = 8.53Thus, thermal conductivity,

Thermal conductivity, k= 299 W/m.K

5.0 DISCUSSION

Pretty simply it can be noticed that theres a penchant for negative gradient in case of all the graphs in

question. With increase in radius, the difference in temperature is seen to decrease. This has a very simple

explanation. When we are heating the innards of the cylinder with the aid of a power source, the externalpart of the cylinder is apparently at a subversive temperature due to a constant flow of water. However,

this results in a variation of temperature between the external and inner part resulting in conduction.

. Using table 4, we can calculate the theoretical values and compare it with the experimental values. As itcan be seen, two methods can be used to determine the thermal conductivity of a material. Either byexperimentally, or theoretically using the Fouriers series law equation.


10/12


The 2nd graph which is the Graph of (T iT) against ln r is also seen to be of negative gradient. Thethermal conductivity, k, should supposedly remain unchanged for this examination of ours. Its quite clear

that the gradient increases with increase in the power supplied. Thus it can be concluded that withincrease in power supply, theres also an increase in the gradient.By using the calculated results, the data are tabulated as follows:

Power Input (W)

Derived value of

thermal conductivity, k

(W/m.K)

Thermal conductivity

from graph, k (W/m.K)Percentage of Deviation

(%)

10.7 291.7 245.73 15.7

30.7 246.75 296.66 16.8

50.5 249.99 299 16.3

Table 4

The table above compares the different values of thermal conductivity obtained at different power inputs.The graphical value is obtained from the experiment whereas the derived value is obtained using formulasfrom theory. The percentage deviation in the two values is less than 20% so we can successfully conclude

that our experiments were more or less successful. The highest deviation is 16.8% whereas the lowest is15.7%. So we can conclude that the two processes are equally correct to determine the value of thermalconductivity. Also speaking from another viewpoint, the theoretical value only considers the inner and

outer temperature for the metal disc whereas the graphical value makes use of the entire temperatureprofile. So perhaps its not entirely incorrect to assume that the graphical method is slightly superior tothe theoretical one.Since the deviation is a little over 15%, we could have done certain things to improve the experiment.Some possible errors are listed below.

We make the assumption that the system to be in a steady state but that is impossible to achieve thatpractically. Steady-state condition cannot be achieved as many disturbances outside the system

The theoretical formula we are using only applies for one dimensional bodies. In reality, thecylinder in question is not one dimensional. Therefore inaccuracy creeps into our results.

Also we make the assumption that the power input is constant. This is not possible in reality. There

is energy loss to the surroundings through sound and vibration. Its also important to keep in mindthat the power supply used to heat up the room has a certain efficiency. Therefore the power

supplied is always lower than the power indicator.

Effect of convection was ignored in this experiment. This is inaccurate as convection takes place

when the cylinder is exposed to the surrounding air. The first sensor in the experiment was worn out due to years of use and it wasnt possible to get a

very accurate temperature reading from the first sensor.

We can take a few precautionary steps to cut down on our errors:a) We have to ensure that the probe touches the tiny pin properly so that we can accurately measure

the temperature of that position.b)

We should wait for some time after the cylinder is heated to get the closest possible steady statetemperature.


11/12


c) We can try to achieve the closest thing possible to a steady state environment by conducting theexperiment in a vacuum chamber so that theres no outside interference.

6.0 CONCLUSION

It can be concluded that the cylinder with constant radial conduction heat transfer displays a

Behavior where the temperature decreases as radial distance increases.

The calculated and graphical thermal conductivity of the brass plate used in the experiment is as follows:

Derived value

of thermal

conductivity,

k (W/m.K)

Thermal conductivity from graph,

k (W/m.K)

291.7 245.73

246.75 296.66

249.99 299

Since the discrepancy in the data doesnt exceed 20% we can assume the experiment to be successful andconclude that both of these two methods can be used to determine the thermal conductivity of a material.

7.0 REFERENCES

1. Laboratory worksheet

2. Yunus A. Cengel, Heat and Mass Transfer: A Practical Approach, 3rd Edition, McGraw-Hill,


12/12


Singapore, 2006.

3. Lecture notes.

Download - Heat Conduction 2014-15

Top Related