Heat conduction in an insulated

Assume that the following equation describes the two-dimensional heat conduction in an insulated pipe... 

We consider the unsteady heat conduction in an insulated bar. The thermal energy balance can be transformed into the dimensionless form... 

Example 3.14 Entropy production and dissipation function in heat conduction Consider one-dimensional heat conduction in an isotropic solid rod. The surface of the rod is insulated and the cross-sectional area is constant (Figure 3.2). Describe the entropy production and the dissipation function for the heat conduction in an isotropic rod. The entropy change of the rod element is... 

Figure 9.8(a) shows how the conduction band C and the empty valence band V are not separated in a conductor whereas Figure 9.8(c) shows that they are well separated in an insulator. The situation in a semiconductor, shown in Figure 9.8(b), is that the band gap, between the conduction and valence bands, is sufficiently small that promotion of electrons into the conduction band is possible by heating the material. For a semiconductor the Fermi energy E, such that at T= 0 K all levels with E < are filled, lies between the bands as shown. 

Since heat transfer through these insulations can occur by several different mechanisms, the apparent thermal conductivity of an insulation that incorporates all of these heat-transfer possibilities offers the best means of comparing these difference types. Table III provides a listing of some accepted k values for popular insulations used in cryogenic storage and transfer systems. 

Consider condensate forming in an insulated container with the lower surface held at constant temperature, Tw, as shown in Ftg. PI 1.1. Since the condensate cannot drain away, the rate of condensation will decrease as the layer thickness increases. Assuming that the heat transfer is by quasi-steady conduction derive an expression for film thickness as a function of time and for the heat transfer rate as a function of time ... 

An electric heater in the form of a 50-by-l00-cm plate is laid on top of a semiinfinite insulating material having a thermal conductivity of 0.74 W/m °C. The heater plate is maintained at a constant temperature of 120°C over all its surface, and the temperature of the insulating material a large distance from the heater is 1S°C. Calculate the heat conducted into the insulating material. 

Considep two-dimensional transient heat transfer in an L-shaped solid body that is initially at a uniform temperalure of 90°C and whose cross section is given in Fig. 5-51. The thermal conductivity and diffusivity of the body are k = 15 W/m C and a - 3.2 x 10 rriVs, respectively, and heat is generated in Ihe body at a rate of e = 2 x 10 W/m. The left sutface of the body is insulated, and the bottom surface is maintained at a uniform temperalure of 90°C at all times. A1 time f = 0, the entire top surface is subjected to convection to ambient air at = 25°C with a convection coefficient of h = 80 W/m C, and the right surface is subjected to heat flux at a uniform rate of r/p -5000 W/m. The nodal network of the problem consists of 15 equally spaced nodes vrith Ax = Ay = 1.2 cm, as shown in the figure, Five of the nodes are at the bottom surface, and thus their temperatures are known. Using the explicit method, determine the temperature at the top corner (node 3) of the body after 1,3, 5, 10, and 60 min. 

In certain instances, the heat loss from an insulated pipe may exceed that from an uninsulated one. In these systems, the insulator has a relatively high thermal conductivity and its resistance to heat flow is insufficient to compensate for the additional heat loss resulting from the increased exposed area. The heat loss increases to a maximum, then it decreases with an increase in thickness. The thickness of the insulator which corresponds to the maximum heat loss is called the critical insulation thickness Xcr, ft.). It is verified by the following equation ... 

The basic method of determining the thermal conductivity is to place a sample of known dimensions in a temperature gradient and measure the rate of the resulting heat flow through it. Suitable apparatus consists of a disc of ice cream sandwiched between two plates made from a material of known thermal conductivity and placed in an insulated cylinder. One plate is heated and the other cooled to produce the temperature gradient, which is measured with thermocouples embedded in the plates and sample. 

The system would consist of a solid (solidified gas or liquid) in an insulated container, an evaporation path.to space, and a conduction path from the solid to the device to be cooled (see Fig. 1). The obtainable temperature depends upon the choice of the solid and the pressure maintained in the system. The operating time depends only on the amount of the solid and the heat input. Such a system should be very reliable, as there is only one moving part, and cooling would continue as long as any solid material remained. 

EXAMPLE 43-1. Heat Flow Through an Insulated fVall of a Cold Room A cold-storage room is constructed of an inner layer of 12.7 mm of pine, a middle layer of 101.6 mm of cork board, and an outer layer of 76.2 mm of concrete. The wall surface temperature is 255.4 K inside the cold room and 297.1 K at the outside surface of the concrete. Use conductivities from Appendix A.3 for pine, 0.151 for cork board, 0.0433 and for concrete, 0.762 W/m K. Calculate the heat loss in W for 1 and the temperature at the interface between the wood and cork board. 

In an insulator all low energy bands i.e., valence bands are completely occupied. The energy gap between this band and the next conduction band is too large and even on heating, electrons are not able to jump the forbidden zone and move into the conduction band. Thus they are not able to conduct electricity. 

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