MAT6679: Heat and Materials with Application
Magnus Anderson
1

Quench of a cylindrical bar
In this problem we model a 1D problem where heat is lost from the
inner and outer diameter of a cylinder during quenching, considering
heat loss through convection. We compare the numerical solution to
a benchmark solution.
There are two datasets provided. The second includes radiative heat
exchanges with the environment. We assume an isothermal initial
temperature of 980°C and model the cooling of the component to
room temperature (20°C). Note that when including radiation, it is
convenient to model the problem in units of Kelvin opposed to
Celsius. The thermophysical properties are given in Table 1.
The component geometry is shown in Figure 1, where the inner radius
is 0.27m. The outer radius is 0.3m. Time-temperature data is known at
the centre point within the cylinder wall and 5mm from the inner and
outer walls.
The heat transfer is approximated using a one-dimensional model in the radial direction of the
cylinder. The temperature dependent thermo-physical properties are
Table 1: Thermo-physical properties
(kg/m3) (W/m/K) (J/Kg/K)
8344.2 − 0.4 6.952 + 0.0157 351.21 + 0.216

The heat transfer coefficients and emissivity vary for the two sets of benchmark data, as described
below:
Parameter Benchmark example 1 Benchmark example 2
Inner heat transfer coefficient (W/m^2/K) 7000 500
Outer heat transfer coefficient (W/m^2/K) 2000 200
emissivity 0 0.5

This problem will be solved using the finite difference schemes for the 1st and 2nd spatial derivatives,
and the 1st temporal derivative.



≅
+1
− −1

2Δr
1
2

2
≅
+1
− 2
+ −1

(Δr)2
2


≅

+1 −

Δ
3
The finite difference scheme for heat transfer within the interior is


Figure 1: Component geometry
MAT6679: Heat and Materials with Application
Magnus Anderson
2


+1 = (1 − 20,)
+ (0, + 0, + 0,)+1
+ (0, − 0, − 0,) −1

4
where
=

,
5
0, =

()2
6
=
+1 − −1
4
7
=

2
8

The finite difference schemes without radiation for the inner and outer diameter are given below
1
+1 = (1 − 2 − 2)1
+ 22
+ 2∞ 9

+1 = (1 − 2 − 2)
+ 2−1
+ 2∞ 10
where
=
ℎ

11
Please update this to include radiation.