A simple case of unsteady state heat transfer arises from the heating or cooling of solid bodies made from good thermal conductors, for example a long cylinder, such as a meat sausage or a metal bar, being cooled in air. The rate at which heat is being transferred to the air from the surface of the cylinder is given by eqn. (5.4)
where Ta is the air temperature and Ts is the surface temperature.
Now, the heat being lost from the surface must be transferred to the surface from the interior of the cylinder by conduction. This heat transfer from the interior to the surface is difficult to determine but as an approximation, we can consider that all the heat is being transferred from the centre of the cylinder. In this instance, we evaluate the temperature drop required to produce the same rate of heat flow from the centre to the surface as passes from the surface to the air. This requires a greater temperature drop than the actual case in which much of the heat has in fact a shorter path.
Assuming that all the heat flows from the centre of the cylinder to the outside, we can write the conduction equation
where Tc is the temperature at the centre of the cylinder, k is the thermal conductivity of the material of the cylinder and L is the radius of the cylinder.
Equating these rates:
To take a practical case of a copper cylinder of 15 cm radius cooling in air kc = 380 J m-1 s-1 °C-1, hs = 30 J m-2 s-1°C-1 (from Table 5.1), L = 0.15 m,
In this case 99% of the temperature drop occurs between the air and the cylinder surface. By comparison with the temperature drop between the surface of the cylinder and the air, the temperature drop within the cylinder can be neglected. On the other hand, if the cylinder were made of a poorer conductor as in the case of the sausage, or if it were very large in diameter, or if the surface heat-transfer coefficient were very much larger, the internal temperature drops could not be neglected.
This dimensionless ratio is called the Biot number (Bi) and it is important when considering unsteady state heat flow. When (Bi) is small, and for practical purposes this may be taken as any value less than about 0.2, the interior of the solid and its surface may be considered to be all at one uniform temperature. In the case in which (Bi) is less than 0.2, a simple analysis can be used, therefore, to predict the rate of cooling of a solid body.
Therefore for a cylinder of a good conductor, being cooled in air,
dQ = hsA(Ts
-- Ta) dt
where c is the specific heat of the material of the cylinder, r is the density of this material and V is the volume of the cylinder.
Since the heat passing through the surface must equal the heat lost from the cylinder, these two expressions for dQ can be equated:
Integrating between Ts = T1 and Ts = T2 , the initial and final temperatures of the cylinder during the cooling period, t, we have:
Charts have been prepared which give the temperature relationships for solids of simple shapes under more general conditions of unsteady-state conduction. These charts have been calculated from solutions of the conduction equation and they are plotted in terms of dimensionless groups so that their application is more general. The form of the solution is:
where ƒ and F indicate functions of the terms following, Ti is the initial temperature of the solid, T0 is the temperature of the cooling or heating medium, T is the temperature of the solid at time t, (kt/crL2) is called the Fourier number (Fo) (this includes the factor k/cr the thermal conductance divided by the volumetric heat capacity, which is called the thermal diffusivity) and (hsL/k) is the Biot number.
A mathematical outcome that is very useful in these calculations connects results for two- and three-dimensional situations with results from one-dimensional situations. This states that the two- and three-dimensional values called F(x,y) and F(x,y,z) can be obtained from the individual one-dimensional results if these are F(x), F(y) and F(z), by simple multiplication:
Using the above result, the solution for the cooling or heating of a brick is obtained from the product of three slab solutions. The solution for a cylinder of finite length, such as a can, is obtained from the product of the solution for an infinite cylinder, accounting for the sides of the can, and the solution for a slab, accounting for the ends of the can.
giving rates of unsteady-state heat transfer to the centre of a slab,
a cylinder, or a sphere, are given in Fig. 5.4. On one
axis is plotted the fractional unaccomplished temperature change,
This problem can be solved by combining the unsteady-state solutions for a cylinder with those for a slab, working from Fig. 5.3.
(a) For the cylinder, of radius r = 5 cm (instead of L in this case)
(Often in these systems the length dimension used as parameter in the charts is the half-thickness, or the radius, but this has to be checked on the graphs used.)
1/(Bi) = 8 x 10-3
and so from Fig.
5.3 for the cylinder:
(b) For the slab
the half-thickness 30/2 cm = 0.15 m
and so from Fig.
5.3 for the slab:
So overall (T
- T0)/( Ti - T0
) = F(x) F(y)
Therefore T2 = 100°C
Heat-Transfer Theory > RADIATION-HEAT TRANSFER
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