To find the thermal conductivity of the specimen by two slab guarded hot plate method.
S. No. | V1/Volts | I1/Amps | T1/(°C) | T2/(°C) | T3/(°C) | T4/(°C) | T5/(°C) | T6/(°C) | T7/(°C) | T8/(°C) |
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5 |
Applying the Fourier's Law of conduction for a Composite wall structure.
Thermal conductivity (K) = QL W/mK/2A (Th-Tc)
Th= Specimen bottom temperature in (°C)
Tc= Specimen top temperature in(°C)
Q1= heat input to central heater in watts
L2= Thickness of specimen in m
A1= Metering area of specimen in m 2
Central heater input (q) = V * I Watts
Metering area (A) = π/4(10+x) 2in mmWhere, X = gap between heater plate = 0.003 m.
Specimen hot side temperature (Th) = [T1+T2/2](°C)
Specimen cold side temperature (Tc) = [T5+T6/2](°C)
Thermal conductivity of specimen (K) = q*L /2A (Th-Tc) in W/ mK
The thermal conductivity of insulating slab is ............................ W/mK
Thermal conductivity of gases increases with increasing temperature
Fourier law: the rate of heat conduction through a material depends on geometry of medium, its thickness & material of the medium as well as temperature across the medium.
Thermal conductivity of the metals decreases with increase in temperature.
a.Moisture b. Density of material c. Pressure d. Temperature e. Structure of material
Thermal conductivity is the rate of heat transfer through a unit thickness of material per unit area per unit temperature difference
Diamond & Freon-12.
Heat conduction is a mechanism of heat transfer from a region of high temperature to a region of low temperature within a medium [solid, liquid or gases] or different medium in direct physical contact.
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