Question 7.8.1: Hardness and other properties of metal can be improved by th......
Hardness and other properties of metal can be improved by the rapid cooling that occurs during quenching, a process in which a heated object is placed into a liquid bath (see Figure 7.8.1). Consider a lead cube with a side length of d = 20 mm. The cube is immersed in an oil bath for which h = 200 W/(m²·°C). The oil temperature is T_{b}.
Thermal conductivity varies as function of temperature, but for lead the variation is relatively small (k for lead varies from 35.5 W/m·°C at 0 °C to 31.2 W/m·°C at 327 °C). The density of lead is 1.134 × 10^{4} kg/m³. Take the specific heat of lead to be 129 J/kg·°C.
(a) Show that temperature of the cube can be considered uniform; and (b) develop a model of the cube’s temperature as a function of the liquid temperature T_{b}, which is assumed to be known.
a. The ratio of volume of the cube to its surface area is
L=\frac{d^{3}}{6d^{2}}=\frac{d}{6}=\frac{0.02}{6}and using an average value of 34 W/m·°C for k, we compute the Biot number to be
N_{B}={\frac{200(0.02)}{34(6)}}=0.02which is much less than 0.1. According to the Biot criterion, we may treat the cube as a lumped-parameter system with a single uniform temperature, denoted T.
b. If we assume that T\gt T_{b}, then the heat flows from the cube to the liquid, and from conservation of energy we obtain
C{\frac{d T}{d t}}=-{\frac{1}{R}}(T-T_{b}) (1)
When deriving thermal system models, you must make an assumption about the relative values of the temperatures, and assign the heat flow direction consistent with that assumption. The specific assumption does not matter as long as you are consistent. Thus, although the bath temperature will be less than the cube temperature in the quenching application, you may still assume that T_{b}\gt T and arrive at the correct model as long as you assign the heat flow direction to be into the cube. However, when making such assumptions, your physical insight is improved if you assume the most likely situation; the nature of the quenching process means that T\gt T_{b}, so this is the logical assumption to use. The thermal capacitance of the cube is computed as
C=m c_{p}=\rho V c_{p}=1.134\times10^{4}(0.02)^{3}(129)=11.7\,\mathrm{J/}^{\circ}\mathrm{C}The thermal resistance R is due to convection, and is
R=\frac{1}{h A}=\frac{1}{200(6)(0.02)^{2}}=2.08^{\circ}C\cdot\mathrm{s}/JThus the model is
11.7\frac{d T}{d t}=-\frac{1}{2.08}\left(T-T_{b}\right)or
24.4\frac{d T}{d t}+T=T_{b}The time constant is τ = RC = 24.4 s. If the bath is large enough so that the cube’s energy does not appreciably affect the bath temperature {\mathbf{}}T_{b}, then when the cube is dropped into the bath, the temperature {\mathbf{}}T_{b} acts like a step input. The cube’s temperature will reach the temperature {\mathbf{}}T_{b} in approximately 4τ = 98 s.