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. controlled. Outer surface surface of the disc requires requires cooling cooling and this is done with 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 T 1-T2°C versus In r o / r i. This helps to achieve the conductivity results theoretically and from thereon comparison can be made with Fourier’s law. These results allow us to reach conclusions on how successful our experiment experiment was and account for errors.
3.0 INTRODUCTION AND THEORETICAL BACKGROUND BACKGROUND 3.1 INTRODUCTION
To start off, let’s take a look at what heat is. Heat is basically a form of energy that can be transferred from one body one body to another. It can occur when there’s a temperature difference bet ween the two bodies in question. 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 energy concerned with the science of dealing with transfers is known as heat energy. Heat transfer will always occur 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 different forms forms of heat transfer transfer are discussed.
Sheikh Shahir
Heat Conduction
KEM120702
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 between the 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 due to collision of diffusion of the fluid particles during their random motion. In solids, it is a combination of vibrations of the molecules in a lattice and the energy that is transported by free electros. A cold canned drink in a warm room, for example, eventually warms up to the room temperature as a result of heat transfer 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, its material, how thick it is and also the temperature difference across the medium. If you wrap a container of hot water with an insulating material, the rate of heat transfer is greatly diminished. The better the insulation the lower the rate of heat loss.
Fig 1-1: Heat Conduction through a large plane wall t hickness Δ 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. There’s also a temperature difference between two sides of the wall. It’s been concluded from experiments that heat transfer rate, w hich 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 conduction
∝
Area Temperatur e Difference
Equation 1
Thickness
Sheikh Shahir
Heat Conduction
.
Q kA
T 1 T 2
x
KEM120702
Where k is the thermal conductivity
K denotes a material’s excellence in the conducting warmth. Using limits the equation can be smited .
down to Q
kA
dT dx
, which is actually Fourier’s 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 experiment With the help of Fourier’s law, the Equation 2
q r where
k L r o, r i T o, T i
2 Lk
ln(r o / r i )
(T i T o )
= thermal conductivity (Wmk ) = cylinder length (m) = outer / inner radius of cylinder (m) = outer / inner temperature ( )
Figure 3 shows a hollow cylinder of which inner and outer surfaces are exposed to fluids at different temperatures. For steady-state conditions with no heat generation, the appropriate form of the heat equation is:
1 d dT kr 0 ……………………….…… (3) r dr dr
Sheikh Shahir
Heat Conduction
KEM120702
From Fourier’s law, the rate at which energy is conducted across any cylindrical surface in the solid is:
q r kA
dT dr
k (2rL)
dT dr
……………….…….. (4)
where A = 2πrL 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 q r is 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) = C1 ln r + C2 …………………………………. (5) Boundary conditions: T(r i) = Ti
and
T(r o) = To
Applying these conditions to the general solution, we then obtain: Ti = C1 ln r i + C2
and
To = C1 ln r o + C2
Solving for C1 and C2 and substituting into the general solution, we then obtain
T(r )
Ti To ln( r i / r o )
r To ……………………. (6) r o
ln
This temperature is used with Fourier’s law, Equation (4), we obtain the following expression for the heat transfer rate:
q r
2 Lk
ln( r o / r i )
(T i
T o ) ..……………………. (7)
Th e r el ation shi p between (T I – T) and ln r
From Equation (6),
T(r )
Ti To ln( r i / r o )
r To r o
ln
Ti To
Ti T(r ) Ti
ln(r i / r o )
Ti T(r )
T i T (r )
Ti To ln(r o / r i )
T i T o ln(r o / r i )
r To r o
ln
1
ln(r 0 / r i )
ln r (Ti To ) 1
ln( r 0 / r i ) ln r o ln( r 0 / r i )
ln r (T i T o )
ln r o
Sheikh Shahir
Ti T(r )
Heat Conduction
Ti To ln(r o / r i )
KEM120702
ln(r 0 / r i )
ln r (To Ti )
ln r i
From above equations, Gradient of Graph Ti – T(r) versus ln r, m =
Ti
To
ln(r o / r i )
Substitute m into Equation (7) gives Thermal conductivity, k
q r 2Lm
…………………………. (8)
4.0 RESULTS
Power input, q of 10.7 Watt Table 1 Position
Thermistor position from centre radius, r (m)
Temperature, T (°C)
Temperature difference, Ti-T
ln
( )
(°C) st
0
34.8
0.0
-
nd
0.01
33.1
1.7
0.916
rd
0.02
31.4
1.7
1.609
4
th
0.03
30.5
0.9
2.015
5
th
0.04
30.2
0.3
2.303
6
th
0.05
29.7
0.5
2.526
1 2
3
Power input, q of 30.7 Watt Table 2
Sheikh Shahir
Position
Heat Conduction
KEM120702
Thermistor position from centre radius, r (m)
Temperature, T (°C)
Temperature difference, Ti-T (°C)
st
0
48.1
0.0
-
nd
0.01
41.4
6.7
0.916
rd
0.02
36.1
5.3
1.609
4
th
0.03
33.4
2.7
2.015
5
th
0.04
32.0
1.4
2.303
6
th
0.05
30.8
1.2
2.526
1 2
3
ln
( )
Power input, q of 50.5 Watt Table 3 Position
Thermistor position from centre radius, r (m)
Temperature, T (°C)
Temperature difference, Ti-T (°C)
st
0
60.4
0.0
-
nd
0.01
49.5
10.9
0.916
rd
0.02
41.7
7.8
1.609
4
th
0.03
36.8
4.9
2.015
5
th
0.04
34.3
2.5
2.303
6
th
0.05
32.2
2.1
2.526
1 2
3
ln
( )
Sheikh Shahir
Heat Conduction
KEM120702
70 60.4 60 ) C ° ( ( T , t n e i d a r G e r u t a r e p m e T
48.1
50
49.5 41.4
40
34.8
41.7 36.1
33.1
31.4
36.8 33.4 30.5
34.3 32 30.2
32.2 30.8 29.7
Q = 10.7 W
30
Q =30.7 W Q = 50.5 W
20
10
0 0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
Thermistor postion from centre, r(m)
Graph 1: Graph of temperature, T (°C) against thermistor position from centre, r (m)
Calculations Given that: Cylinder stride, L = 0.003 m, Inner radius of cylinder, r i = 0.004 m, Outer radius, r o = 0.055 m We assume that the temperature inside the cylinder, T i is equal to temperature of thermistor at position 1, 1st and temperature outside the cylinder, T o is equal to the temperature of thermistor at position 6, 6 th. There are two methods to calculate the thermal conductivity of material, k First method In the first method, we will use Fourier’s Law equation as follows:
) (
For power input, q = 10.7 Watt From Table 1 Ti = 34.8°C; 307.8 K To = 29.7°C; 302.7 K
( ) Thermal conductivity, k = 291.7 W/m.K
Sheikh Shahir
Heat Conduction
KEM120702
For power input, q = 30 .7Watt From Table 2 Ti = 48.1°C; 321.1 K To = 30.8°C; 303.8 K
( ) Thermal conductivity, k =246.75 W/m.K For power input, q = 50.7 Watt From Table 3 Ti = 60.4°C; 333.4 K To = 32.2°C; 305.2 K
( ) Thermal conductivity, k = 249.99 W/m.K
12
10 ) C ( ) T I T ( , E C N E R E F F I D E R U T A R E P M E T
8
6
4
2
0 0
0.5
1
1.5
2
2.5
LN (RO / RI) Q = 10.7 W
Q=30.7 W
Q=50.5 W
Graph 2: A Graph of temperature difference, Ti – T (°C) vs ln
Second method In the second method, we will use this equation:
( )
3
Sheikh Shahir
Heat Conduction
KEM120702
where m is the gradient of graph T - T versus ln r ( ) i
o
For power input, q = 10.7 Watt From graph 2, the graph equation: y =- 2.315x The gradient of the graph, m = 2.31 Thus, thermal conductivity,
Thermal conductivity, k = 245.73 W/m.K For power input, q = 30.7 Watt From graph 2, the graph equation: y = -5.49x The gradient of the graph, m = 5.49 Thus, thermal conductivity,
Thermal conductivity, k = 296.66 W/m.K For power input, q = 50 Watt From graph 2, the graph equation: y = -8.53x The gradient of the graph, m = 8.53 Thus, thermal conductivity,
Thermal conductivity, k = 299 W/m.K
5.0 DISCUSSION
Pretty simply it can be noticed that there’s 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 external part 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 it can be seen, two methods can be used to determine the thermal conductivity of a material. Either by experimentally, or theoretically using the Fourier’s series law equation.
Sheikh Shahir
Heat Conduction
KEM120702
The 2nd graph which is the Graph of (T i – T) against ln r is also seen to be of negative gradient. The thermal conductivity, k, should supposedly remain unchanged for this examination of ours. It’s quite clear that the gradient increases with increase in the power supplied. Thus it can be concluded that with increase in power supply, there’s also an increase in the gradient. By using the calculated results, the data are tabulated as follows: Thermal conductivity from graph, k (W/m.K)
Power Input (W)
Derived value of thermal conductivity, k (W/m.K)
10.7
291.7
245.73
15.7
30.7
246.75
296.66
16.8
50.5
249.99
299
16.3
Percentage of Deviation (%)
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 formulas from 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 is 15.7%. So we can conclude that the two processes are equally correct to determine the value of thermal conductivity. 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 temperature profile. So perhaps it’s not entirely incorrect to assume that the graphical method is slightly superior to the 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 that practically. 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, the cylinder 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. It’s also important to keep in mind that 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 wasn’t 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 state temperature.
Sheikh Shahir
Heat Conduction
KEM120702
c) We can try to achieve the closest thing possible to a steady state environment by conducting the experiment in a vacuum chamber so that there’s 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 doesn’t exceed 20% we can assume the experiment to be successful and conclude 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,
Sheikh Shahir
Singapore, 2006. 3.
Lecture notes.
Heat Conduction
KEM120702
