Question

discussion and conclusion for the thermal conductivity of solids Lab for the thermodynamics course

Answer 1

Recognise that heat transfer involves an energy transfer across boundary. A logical place to begin studying such process is conservation of energy (First Law of Thermodynamics) for a closed system.

The sign convention on work is such that negative work out is positive work in:

The work in term could describe an electric current flow across the system boundary and through a resistance inside the system. Alternatively it could describe a shaft turning across the system boundary and overcoming friction within the system. The net effect in either case would cause the internal energy of the system to rise. In heat transfer we generalize all such terms as “heat sources”.

The energy of the system will in general include internal energy, (U), potential energy, ( 0.5 mgz), or kinetic energy,(½ mv² ). In case of heat transfer problems, the latter two terms could often be neglected. In this case,

Where Tref is the reference temperature at which the energy of the system is defined as zero. When we differentiate the above expression with respect to time, the reference temperature, being constant disappears:

In the equation above we substitute the 6-heat inflows/outflows using the appropriate sign:

Substitute for each of the conduction terms using the Fourier Law:

Where q_gen is defined as the internal heat generation per unit volume. The above equation reduces to:

Dividing by the volume

Which is the general conduction equation in three dimensions. In the case where k is independent of x, y and z then

For one dimension Heat Transfer (in on solid slab or single plate):

Please refer to the image1 attached.

For this case there is no heat generation inside the slab (qg = 0).

Then the differential equation governing heat diffusion is:

With constant k, the above equation may be integrated twice to obtain the general solution:

Where C1 and C2 are constants of integration. To obtain the constants of integration, we apply the boundary conditions at x = 0 and x = L, in which case

Once the constants of integration are substituted into the general equation, the temperature distribution is obtained:

The heat flow rate across the wall is given by:

Electrical Analogy:

I=V/R