Lab ReportThermal Conduction in Solids ( Aluminum and Composite)
Aim
Aim of this lab work is to understand the thermal conduction in solids
Objectives
Following are the objective of this lab work of thermal conduction in solids
1. To measure the thermal conduction of copper
2. Plotting the thermal conductivity characteristic curve for composite wall comprising of aluminum and copper
Theory
According to the law of energy or heat transfer, heat or energy always from the area of high potential to the area of low potential in free natural state or conditions. Heat from area of high potential to area of low potential in three different ways conduction, convection and radiation from these three methods of heat transfer conduction is the methods of heat transfer which transfer heat between metals.
Conduction
Conduction is mode of heat transfer from high potential source to low potential source of single body with the help of its vibrating molecules
Convection
Convection is mode of the transfer from high potential source to low potential source in fluid or from fluids.
Radiation
Radiation is mode of the transfer from high potential source to low potential source without the help of any medium.
Rate of heat flow
I the case when there is a heat transfer between region of high potential to the region of low potential trough the solid then it is important to calculate the amount of heat transfer Q in time t through a particular cross section area A of solid having x thickness.
Q= KA*t*∆T/∆x= KA/x(T_2T_1)
Here
U= k/x
For composite wall
U= K_1/x_1 + K_2/x_2
So
Q= UA(T_2+T_1)
Apparatus
Following are the apparatuses which are required for carrying out the experiment of find the thermal conductivity of solids
1. Stop watch
2. Thermal conductivity of solids equipment’s
3. Beaker
4. Specimens of aluminum and copper both have 30 mm length and 40 mm diameter
5. Heat conductive paste
Apparatus Setup
1. The heater of the apparatus must be switched on
2. All the thermocouples should be on front panel and in Dewar vessel
3. The flowing water which is used for cooling must covers the overflow present on the constant water pressure device
For Aluminum
Time

0

3

6

9

12

15

18

27

24

T2

32.5

34.6

35.4

36.9

37.9

38.6

38.7

38.4

38.4

T1 = 43.4 T2 = 38.4
t1 = 19 t2 = 15
Mass of empty beaker = 0.104 Kg
Time to get 200 ml of water = 80 sec
Beaker mass with water= 0.29 kg
Mass flow rate =(Mass of water and beaker – Mass of beaker)/time
Mass flow rate=(0.29 – 0.104)/80 =2.325*〖10〗^(3)
Calculating K
Q= m ̇*Cp*(t_1t_2 )
Cp = 4190 J/KgK
Q= 2.32*〖10〗^(3)*4190*(t_1t_2 )
Q= 38.967
Ka= Q/A*∆x/∆T=Q/A*((x_1x_2))/((T_2T_1))
Ka=38.967/0.001256*0.025/((38.443.4) )=168.5
For composite Wall
Time

0

3

6

9

12

15

18

21

24

27

T2

31.3

33

33.7

34.5

34.8

35

35.3

35.4

35.5

35.6

T1 = 50.3 T2 = 40.4T3 = 38.3 T4 = 35.7t1 = 19 t2 = 15
Q= 38.967
Kc= Q/A*∆x/∆T=Q/A*((x_1x_2))/((T_4T_3))
Kc=38.967/0.001256*0.025/((35.738.3) )=298.16
Composite Wall Theory
U=Kc/xc+Ka/xa
U=298.16/0.025+168.5/0.025 =16848.4
Q= UA(T_4T_1)
Q= 16848.4*0.001256(35.750.3)
Q= 309.11
Q= 38.967
Kc= Q/A*∆x/∆T=Q/A*((x_1x_2))/((T_4T_3))
Kc=38.967/0.001256*0.025/((35.738.3) )=298.16
Composite Wall Theory
U=Kc/xc+Ka/xa
U=298.16/0.025+168.5/0.025 =16848.4
Q= UA(T_4T_1)
Q= 16848.4*0.001256(35.750.3)
Q= 309.11
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