A long steel slab, having 8-inch thickness, is initially at
200 degrees Fahrenheit. Its surfaces
are suddenly changed to 900 degrees F.
Determine the temperature at the center of the slab after 10
minutes. Thermal diffusivity of steel
is given as 0.41 square feet per hour.
Solution:
A wall of finite thickness and at a uniform original
temperature is subject to surroundings with constant temperature Ts. It is assumed that there is no contact
resistance between the medium and the surface it contacts, so that the face
temperature of the slab is also Ts.
This differs form ordinary quenching in which there is a very definite
contact resistance. The conduction
equation is represented by partial t over partial q is equal
to a partial
square t over partial x square. The
boundary conditions for finite slab heated on both faces are that, when x
is equal to x and q is equal
to zero, t is equal to t0 and, when x is equal to zero and q is equal
to zero, t is equal to Ts, where t0 is the initial uniform
temperature of the solid.
Y is equal to f2 of c, where c is given
by 4 a q over L
squared. Further, Y is the
dimensionless temperature and is given by the ratio of Ts minus T
and Ts minus T0.
For this case, chi is calculated to be 0.615. Use the Figure 18.9 to read the value of the
function, f2, as 0.279.
Plugging in the values of Ts and T0, we get T as
704.6 degrees Fahrenheit.