Procedure For Thıs Experıment
PROCEDURE FOR THIS EXPERIMENT
HEAT CONDUCTION UNIT
The Heat Conduction Unit consists of heated module mounted on a bench support frame. The module contains a cylindirical metal bar arrangement for a variety of linear conduction experiments. Test section is equipped with an array of temperature sensors. Cooling water, to be supplied from a standard laboratory tap, is fed to one side of the test pieces in order to maintain a steady gradient. The instrumentation permits accurate measurements of temperature and power supply. Fast response temperature probes, with a resolution of 0.1 . The power control circuit provides a continuously variable electrical output of 0-100 Watts with direct readout.
Linear Module
Fourier’s Law of Heat Conduction is most simply demonstrated with the linear conduction module. This comprises a heat section manufactured from brass fitted with an electrical heater. Three thermistor temperature sensors are installed at 10 mm intervals along the working section which has a diameter of 25 mm. A separate heat sink section, also of brass, is cooled at one end by running water whilst its working section is also fitted with thermistor temperature sensors at 10 mm intervals. The heat input section and the heat sink section may be clamped directly together to form a continous brass bar with temperature sensors at 10 mm intervals. Heat losses from the linear module are reduced to a minimum by a heat resistant casing enclosing an air space around the module. The thermistor temperature sensors are connected to miniature plugs fitted to the casing and connection from the sensors to the digital temperature readout are made via nine sensor leads fitted with approprate sockets . therefore temperature gradients can be readily plotted from rapidly acquired data . the temperature selector switch on the front panel of the electrical console permits any of the nine temperature to be displayed.
Experiment
Turn on the water supply and ensure that water is flowing from the free end of the water pipet o drain. This should be checked at intervals. Rotate the heater power control on the electrical console to the fully anti-clockwise position. Set the mains ON/OFF switch on the ON position when the digital readouts will be illuminated. The temperature will be indicated on the temperature readout. Set the heater power on the wattmeter. Select temperature 1 on selector switch. The temperature will now increase as the end of the linear module is heated. Select temperature 2 on the selector switch and note that the indicated temperature decreases towards the water-cooled end. Repeat until all nine temperatures have been selected. Temperatures should be checked at regular intervals until reaching steady state. Commissioning is now complete.
OBTAINING THERMAL CONDUCTIVITY ( k )
Our aim in this experiment is to find the thermal conductivity (k) of brass by heat conduction unit. To operate the heat conduction unit first time, power switch is turned . Then water starts flowing from the free end of the water pipe to drain.Water is used to maintain steady conditions . Wattmeter can be set to arbitrary powers that we want to use. ( In this experiment we choose three different power values: 2W,4W,6W ) Then we wait until the conditions become steady.. We understand that the conditions become steady when temperatures stops changing. Then we note the temperatures of 6 points. These temperature values can be read by using the Temperature Selection Switch. We can calculate k of brass by using these values..
Equations for calculating k
Q = kA (T / x )
k = Q/ A (T / x )
Q : Power used
A : Area of the material ( A = лD2 / 4 )
D : Diameter ( In the experiment, brass with D=25mm used )
T : Temperature difference
x : Length of the whole cyclinder (x=8cm)
Part that holds T4,T5,T6, has a cavity in the edges. As a result, air enters mechanism and helps the water to cool the brass.
a) Q = 2W
Q T1°C T2°C T3°C T4°C T5°C T6°C
2W 25,9 25,6 25,2 10,7 10,3 9,9
A = лD2 / 4
A = 4,908×10-4 m2
T = T1-T6 = 25,9-9,9 =16°C,
x = 0,08m
k = Q/ A (T / x )
k = 2 / 4,908×10-4 ( 16 / 0,08 )
k = 20,37 W/m. °C
Real k value is found by interpolating the datas in Table A-3 in the book.
The values in the table are in Kelvin so we have to find the value of T in terms of K( Kelvin).
T = 16+273 =289 K
T°C k ( W/m.K)
200 95
289 k
400 137
Real k = 114 W/m.K
We see that there is a big difference between the k value found in the experiment and real k value. This is because of the air. As mentioned above , there is a cavity in the test mechanism. Air helps the water ,to cool the brass. So heat loss occurs. then; k becomes smaller.
Temperature profile of this example is given as Chart1 .
b) Q = 4W
Q T1°C T2°C T3°C T4°C T5°C T6°C
4W 42,8 42,3 41,5 14,2 13,3 12,6
A = лD2 / 4
A = 4,908×10-4 m2
T = T1-T6 = 42,8-12,6 =30,2°C,
x = 0,08m
k = Q/ A (T / x )
k = 4 / 4,908×10-4 ( 30,2 / 0,08 )
k = 21,6 W/m. °C
Real k value is found by interpolating the datas in Table A-3 in the book.
The values in the table are in Kelvin so we have to find the value of T in terms of K( Kelvin).
T = 30,2+273 =303,2 K
T°C k ( W/m.K)
200 95
303,2 k
400 137
Real k = 117,64 W/m.K
Explanation is the same as the above one.
Temperature profile of this example is given as Chart 2.
c) Q = 6W
Q T1°C T2°C T3°C T4°C T5°C T6°C
6W 49,4 49 48,2 12,8 11,7 10,8
A = лD2 / 4
A = 4,908×10-4 m2
T = T1-T6 = 49,4-10,8 =38,6°C,
x = 0,08m
k = Q/ A (T / x )
k = 6 / 4,908×10-4 ( 38,6 / 0,08 )
k = 25,33 W/m. °C
Real k value is found by interpolating the datas in Table A-3 in the book.
The values in the table are in Kelvin so we have to find the value of T in terms of K( Kelvin).
T = 38,6+273 =311,6 K
T°C k ( W/m.K)
200 95
311,6 k
400 137
Real k = 119,32 W/m.K
Explanation is the same as the above one.
.
Temperature profile of this example is given as Chart 3.