Question 7.8.2: Consider the quenching process treated in Example 7.8.1. If ......
Consider the quenching process treated in Example 7.8.1. If the thermal capacitance of the liquid bath is not large, the heat energy transferred from the cube will change the bath temperature, and we will need a model to describe its dynamics. Consider the representation shown in Figure 7.8.3. The temperature outside the bath is {T}_{o}, which is assumed to be known. The convective resistance between the cube and the bath is R_{1}, and the combined convective/conductive resistance of the container wall and the liquid surface is R_{2}. The capacitances of the cube and the liquid bath are C and C_{b}, respectively.
a. Derive a model of the cube temperature and the bath temperature assuming that the bath loses no heat to the surroundings (that is, R_{2}=\infty).
b. Obtain the model’s characteristic roots and the form of the response.
a. Assume that T\gt T_{b}. Then the heat flow is out of the cube and into the bath. From conservation of energy for the cube,
C{\frac{d T}{d t}}=-{\frac{1}{R_{1}}}(T-T_{b}) (1)
and for the bath,
C_{b}{\frac{d T_{b}}{d t}}={\frac{1}{R_{1}}}(T-T_{b}) (2)
Equations (1) and (2) are the desired model. Note that the heat flow rate in equation (2) must have a sign opposite to that in equation (1) because the heat flow out of the cube must be the same as the heat flow into the bath.
b. Applying the Laplace transform to equations (1) and (2) with zero initial conditions, we obtain
(R_{1}C s+1)T(s)-T_{b}(s)=0 (3)
(R_{1}C_{b}s+1)T_{b}(s)-T(s)=0 (4)
Solving equation (3) for {\mathbf{}}T_{b}(s) and substituting into equation (4) gives
[(R_{1}C_{b}s+1)(R_{1}C s+1)-1]T(s)=0from which we obtain
R_{1}^{2}C_{b}C s^{2}+R_{1}(C+C_{b})s=0So the characteristic roots are
s=0,\quad s=-{\frac{C+C_{b}}{R_{1}C C_{b}}}Because equations (3) and (4) are homogeneous, the form of the response is
\begin{array}{l l}{{T(t)=A_{1}e^{0t}+B_{1}e^{-t/\tau}=A_{1}+B_{1}e^{-t/\tau}\quad}}&{{\tau=\frac{R_{1}C C_{b}}{C+C_{b}}}}\\ {{T_{b}(t)=A_{2}e^{0t}+B_{2}e^{-t/\tau}}}=A_{2}+B_{2}e^{-t/\tau}&{{}}\end{array}where the constants A_{1},\,A_{2},\,B_{1},\,\mathrm{and}\ B_{2} depend on the initial conditions. The two temperatures become constant after approximately 4τ. Note that T(t)\to A_{1} and T_{b}(t)\to A_{2}\,\mathrm{as}\,t\to\infty. From physical insight we know that T and {{T}}_{b} will become equal as t\rightarrow\infty.\mathrm{Therefore},A_{2}=A_{1}.
The final value of the temperatures, A_{1}, can be easily found from physical reasoning using conservation of energy. The initial energy in the system consisting of the cube and the bath is the thermal energy in both; namely, C T(0)+C_{b}T_{b}(0). The final energy is expressed as C A_{1}+C_{b}A_{1}, and is the same as the initial energy. Thus,
C T(0)+C_{b}T_{b}(0)=C A_{1}+C_{b}A_{1}\qquad\mathrm{or}\qquad A_{1}={\frac{C T(0)+C_{b}T_{b}(0)}{C+C_{b}}}=A_{2}Note also that T(0)=A_{1}+B_{1},\;\mathrm{and}\;T_{b}(0)=A_{2}+B_{2},\;\mathrm{Thus},\;B_{1}=T(0)-A_{1}\;\mathrm{and}\; B_{2}=T_{b}(0)-A_{2}.