A long steel rod (2 cm in diameter, 2-m-long) has been heat-treated and quenched to a temperature of 100°C in an oil bath. To cool the rod further, it is necessary to remove it from the bath and expose it to room air. Will the faster cool-down result from cooling the cylinder in the vertical or horizontal position? How long will the two methods require to allow the rod to cool to 40°C in 20°C air?

Question

GIVEN
• A long steel rod in air
• Diameter (D) = 2 cm = 0.02 m
• Length (L) = 2 m
• Initial temperature (Ts,i) = 100°C
• Air temperature (T?) = 20°C
FIND
(a) Is it faster to cool the rod vertically or horizontally?
(b) Time for rod to cool to 40°C in each position
ASSUMPTIONS
• Steel is 1% carbon
PROPERTIES AND CONSTANTS
for 1% carbon steel
Thermal conductivity (ks) = 43 W/(m K)
Specific heat (c) = 473 J/(kg K)
Density (?) = 7801 kg/m3
for dry air at the initial mean temperature of 60°C
Thermal expansion coefficient (?) = 0.003 1/K
Thermal conductivity (k) = 0.0279 W/(m K)
Kinematic viscosity (?) = 19.4 × 10–6 m2/s
Prandtl number (Pr) = 0.71

Answer 1

As the temperature of the rod decreases, the heat transfer coefficient will also decrease. Therefore, a rough numerical integration will be used to estimate the cooling time.

Note that the air properties must be evaluated at each step.



Cooling time: about 76 minutes in the vertical position, about 42 minutes in the horizontal position

An alternate method of solution uses the average heat transfer coefficients and the time-temperature history given by Equation (3.3). Evaluating the heat transfer coefficients when the rod has reached 40°C





The time required for the rod to cool to the temperature Tf is calculated by rearranging Similarly for the horizontal position: t = 2854 s = 48 min. This more approximate technique yields cooling times about 10-14% greater than the numerical technique shown above