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Home Mechanical Engineering Heat Transfer Lab Determination of the Thermal Conductivity Of The Composite Wall
Heat Transfer Lab Lab Experiments

Determination of the Thermal Conductivity Of The Composite Wall

Nov 25 2023


Aim

To determine the thermal conductivity of the composite wall.

Specifications:

Thickness of wood (L1) = 10 mm

Thickness of Asbestos (L2) = 6 mm

Thickness of MS Plate (L3) = 10 mm

Diameter of the Plate (D) = 300 mm

Diagram:

diagram

Procedure:

  1. This Apparatus is designed and fabricated mainly to study the characteristics of the composite structure. Here three slabs of different materials are provided, namely asbestos, wood & mild steel.
  2. Heat input to this composite wall is given by a nichrome wire heater bounded in mica. On both sides of the heater identical structure of composites, walls are placed. Thermocouples (Iron – Constantine) are provided at proper positions in the composite wall to record the temperature.
  3. A small hand press is binding the plates together. A digital temperature indicator with room temperature compositions is provided with selector switch.
  4. Heat input to the heater is given through a variac and measured by a wattmeter. An enclosure is given around the composite walls to ensure steady atmosphere conditions with transparent windows for visualization.
  5. Thermocouples reading are taken at regular intervals till consecutive readings are same indicating that steady state has been achieved. After establishing the steady state, the readings are tabulated and the power supply to the equipment is switched off.

Observations:

S. No. Heat input (Q) Watts/V1/Volts/I1/Amperes/Q=V*I Watts T1/(°C) T2/(°C) T3/(°C) T4/(°C) T5/(°C) T6/(°C) T7/(°C) T8/(°C) Themal Conductivity W/m K KA/KB/KC
1
2
3
4
5

Formula Used

Wood Temperature (TA) = T1 + T8(°C)/2

Asbestos Temperature (TB) = T2 + T7(°C)/2

Mild steel Temperature (TC) = T3 + T6(°C)/2

Heater Temperature (TD) = T1 + T8(°C)/2

Wood Temperature (TA) = T4 + T5(°C)/2

A = 2(πD2/4) in m2

L = L1 + L2 + L3 in m

Thermal conductivity of Wood (KA) = QL1/A(TA-TB) W/ mK

Thermal conductivity of Asbestos (KB) = QL2/A(TB-TC) W/ mK

Thermal conductivity of Mild Steel (KC) = QL3/A(TC-TD) W/ mK

Result:

Thermal conductivity of wood is (KA) __ ________________ W/ mK

Thermal conductivity of Asbestos is (KB) __ ________________ W/ mK

Thermal conductivity of Mild steel Plate is (KC) __ ________________ W/ mK

Viva-Voce Questions

What is effect of temperature on thermal conductivity of gases?

Thermal conductivity of gases increases with increasing temperature

State Fourier law?

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.

What is effect of temperature on thermal conductivity of metals?

Thermal conductivity of the metals decreases with increase in temperature

What is steady-state condition?

Steady state implies that temperature at each point of system remains constant in due course of time.

What is heat conduction?

Conduction is the transfer of heat from one part of a substance to another part or to another substance in physical contact with it.

What is thermal conductivity?

Thermal conductivity is the rate of heat transfer through a unit thickness of material per unit area per unit temperature difference

Name the material having highest & least thermal conductivity?

Diamond & Freon-12.







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