ME2355 HEAT TRANSFER THROUGH COMPOSITE WALLS - Computer Programming

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Sunday, March 13, 2011

ME2355 HEAT TRANSFER THROUGH COMPOSITE WALLS


HEAT TRANSFER THROUGH COMPOSITE WALLS
Aim:
            To determine the rate of heat transfer through different layers of composite wall
Description of Apparatus:
When heat conduction takes place through two or more solid materials of different thermal conductivities, the temperature drop across each material depends on the resistance offered to heat conduction and the thermal conductivity of each material.
The experimental set-up consists of test specimen made of different materials aligned together on both sides of the heater unit.  The first test disc is next to a controlled heater.  The temperatures at the interface between the heater and the disc is measured by a thermocouple, similarly temperatures at the interface between discs are measured.  Similar arrangement is made to measure temperatures on the other side of the heater.  The whole set-up is kept in a convection free environment.  The temperature is measured using thermocouples (Iron-Cons) with multi point digital temperature indicator.  A channel frame with a screw rod arrangement is provided for proper alignment of the plates.
The apparatus uses a known insulating material, of large area of heat transfer to enable unidirectional heat flow.  The apparatus is used mainly to study the resistance offered by different slab materials and to establish the heat flow is similar to that of current flow in an electrical circuit.
The steady state heat flow Q = Δt/R
Where Δt = is the overall temperature drop and
R is the overall resistance to heat conduction.
Since the resistance are in series
R = R1 + R2 
Where R1, R2 are resistance of each of the discs.

Tabulation:

Sl.No.

Voltmeter reading

Ammeter reading

T1

T2

T3

T4

T5

T6

T7

T8

 







 

 

 

 

 

 

 

 

 

 



Specification:

  1. Thermal conductivity
Of sheet asbestos                                = 0.116 W/MK
Thickness                                            = 6mm
       2.  Thermal conductivity of wood           = 0.052W/MK
            Thickness                                            = 10mm
       3.  Dia. Of plates                                      = 300mm
4.      The temperatures are measured from bottom to top plate T1,T2,………….T8.         

PROCEDURE

1.      Turn the screw rod handle clockwise to tighten the plates.
2.      Switch on the unit and turn the regulator clockwise to provide any desired heat input.
3.      Note the ammeter and voltmeter readings.
4.      Wait till steady state temperature is reached.
5.      (The steady state condition is defined as the temperature gradient across the plates does not change with time.)
6.      When steady state is reached note temperatures and find the temperature gradient across each slab.
7.      Since heat flow is from the bottom to top of the heater the heat input is taken as Q/2 and the average temperature gradient between top and bottom slabs from the heater to be taken for calculations.  Different readings are tabulated as follows.

 

CALCULATION:

Now the resistance ( R ) offered by individual plates for heat flow.
R1 = L1/AK1                R2 = L2 / AK2             R3 = L3/AK3
Where A = Area of the plate
            K = Thermal Conductivity
            L = Thickness of the plate.
Knowing the thermal conductivities
Q = (T4 – T1)/R =(T2 –T1)/R1=(T3 – T2)/R2=(T4 – T3)/R3 

 

COMPOSITE WALLS

V      A      T1    T2       T3    T4       T5    T6       T7       T8          Time for 1 Rev.
182   0.5    76     75        72     71        66     67        50        51           E.M
                  heater          ms 71.5     ashess 66.5      wood 50.5
Area of the plate п / 4 (0.3)2 = 0.07m2
Resistance of Asbestos (R1) = L1 /A1K1 = 0.005/0.07 X 69 X 10-3 =1.03
Resistance of Wood (R2) = L2/A2K2 = 0.008/0.07 X 52 X 10-3 = 2.19
Heat flow Q1 = Temp. across Asbestos / R1 = 5/1.03 =4.85 Watts
                Q2 = Temp. across Wood / R2 = 16/2.19= 7.3 Watts
As per electrical anology Q1 = Q2 = Q3
Total Resistance R3 = 1.03 + 2.19 = 3.22
                 Q3 =(Temp. across Asbestos + Wood) / R3 = 21/3.22 = 6.521
As we have find the inside heat transfer co-efficient for heat flow from heater to MS plate, we consider only the second and third layer.

 

 

 

 

 

result:

The rate of heat transfer through different materials are found to be
  1. MS section                  = ------------- W
  2. Wood section              = ------------- W
  3. Asbestos section         = --------------W

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