Thermal Conductivity of Composite Slab
Aim: To determine thermal conductivity of composite slab
Theory: Many engineering applications require use of composite structures to reduce the heat transfer rate like walls of cold storage, fridge, furnace wall etc, in which different materials are arranged in series or parallel. A hot fluid flow inside the tube covered with a layer of thermal insulation is an example of composite system because in this case the thermal conductivity of tube metal and insulation are different. Heat transfer rate can be estimated through the composite system by using the thermal resistance concept.
Thermal Conductivity of Composite Slab
Figure shows experimental set up. The composite slab consists of mild steel, bakelite and wood of same diameter and or different thicknesses. All the three plates are clamped in pairs in consecutive sequence. The series starts with mild steel plate, bakelite, press wood with aluminum on one side of heater in the centre and the same on the other side of the heater. The mica plate heater has been provided to supply heat input across these composite walls.
Thermocouples have been placed at proper locations in the composite structure to record temperatures. Multi channel temperature indicator has been used to indicate these temperatures. The difference in the thermal conductivity of different materials is observed with this experimental set up.
Specifications:
1. Dia. of slab = 300 mm each.
2. Thickness of slab (MS) = 12.5 mm
(Bakelite) = 12.5mm
(Pressed Wood) = 25mm
Safety Precautions:
1) Keep the dimmer stat at zero position before switching on the power supply.
2) Increase the voltage gradually.
3) Operate selector switch of temperature indicator gently.
4) Do not exceed 100 volts for safety reasons.
Procedure:
1) Switch on the supply.
2) Give steady energy input to the heater with the help of dimmerstat from 80 to 100V.
3) Record the input to the heater with selector switch, voltmeter & ammeter.
4) Note down the temperature every 15 minutes till a steady condition is reached.
Observation Table:
1) Input Voltage V = ………volt
2) Input Current I = ……….amp.
THERMOCOUPLES
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TIME(min)
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15
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30
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45
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60
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75
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90
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105
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120
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T1
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T2
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T3
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T4
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T5
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T6
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T7
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T8
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Calculations:
Heat input Q = (V x I )/ 2
= (∆T) / R ;
Q =
(k1, k2 & k3 can be found out)
Where Ta = (T1 + T2) / 2 ; Tb = (T3 + T4) / 2
Tc = (T5 + T6) / 2 ; Td = (T7 + T8) / 2
A1, A2, A3 are equal = A =
Where, kcompo = Thermal conductivity of composite slab
R = Total thermal resistance
= (L1+L2+L3/A.kcompo)
R=R1+R2+R3
Result Table:
Material
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Thermal conductivity (W/ m K)
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Mild steel
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Bakelite
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Press Wood
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Composite slab
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Plot temp distribution across thicknesses of different slabs.
Conclusion:
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