Question 7.7: Temperature Dynamics of a Heated Object Consider a steel sph...
Temperature Dynamics of a Heated Object
Consider a steel sphere with a radius of r=0.01 \mathrm{~m} submerged in a hot water bath with a heat transfer coefficient of h=350 \mathrm{~W} /\left(\mathrm{m}^{2 .}{ }^{\circ} \mathrm{C}\right). For steel, the density is \rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, the specific heat capacity is \mathrm{c}=440 \mathrm{~J} /\left(\mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right), and the thermal conductivity is k=43 \mathrm{~W} /\left(\mathrm{m} \cdot{ }^{\circ} \mathrm{C}\right). The temperature of the water T_{\mathrm{f}} is 100^{\circ} \mathrm{C} and the initial temperature of the sphere T_{0} is 25^{\circ} \mathrm{C}.
a. Determine whether the sphere’s temperature can be considered uniform.
b. Derive the differential equation relating the sphere’s temperature T(t) and the water’s temperature T_{\mathrm{f}}.
c. Using the differential equation obtained in Part (b), construct a Simulink block diagram to find the sphere’s temperature T(t).
a. The characteristic length of the sphere is
L_{\mathrm{c}}=\frac{V_{\text {body }}}{A_{\text {surface }}}=\frac{4 / 3 \pi r^{3}}{4 \pi r^{2}}=\frac{1}{3} r=\frac{0.01}{3} .
The Biot number of the steel sphere is
B i=\frac{h L_{c}}{k}=\frac{350(0.01)}{43(3)}=2.71 \times 10^{-2}<0.1.
Thus, the sphere can be treated as a lump-temperature system, and its temperature can be considered uniform within the body.
b. The dynamic model of the sphere’s temperature can be derived using the law of conservation of energy,
\frac{\mathrm{d} U}{\mathrm{~d} t}=q_{\mathrm{hi}}-q_{\mathrm{ho}}.
Note that U=m c T=\rho V c T, and we have
\frac{\mathrm{d} U}{\mathrm{~d} t}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\rho V_{c} T\right)=\rho V_{c} \frac{\mathrm{d} T}{\mathrm{~d} t}.
The heat flow rate into the body is
q_{\mathrm{hi}}=\frac{T_{\mathrm{f}}-T}{R}
and the heat flow rate out of the body is q_{\mathrm{ho}}=0. Thus, the differential equation of the system is
\rho V c(\mathrm{~d} T / \mathrm{d} t)=\left(T_{\mathrm{f}}-T\right) / R.
Introducing the expression for the thermal capacitance \rho V C=C, we find
R C \frac{\mathrm{d} T}{\mathrm{~d} t}+T=T_{\mathrm{f}}.
The thermal capacitance of the sphere is
C=\rho V_{C}=7850\left(\frac{4}{3}\right)(\pi)(0.01)^{3}(440)=14.47 \mathrm{~J} /{ }^{\circ} \mathrm{C},
and the thermal resistance due to convection is
R=\frac{1}{h A}=\frac{1}{350(4)(\pi)(0.01)^{2}}=2.27^{\circ} \mathrm{Cs} / \mathrm{J}.
Thus, the dynamic model of the sphere’s temperature is
32.85 \frac{\mathrm{d} T}{\mathrm{~d} t}+T=T_{\mathrm{f}}.
C. Given T_{\mathrm{f}}=100^{\circ} \mathrm{C}, solving for the highest derivative of the output T gives
\frac{\mathrm{d} T}{\mathrm{~d} t}=\frac{1}{32.85}(100-T),
which can be represented using the Simulink block diagram shown in Figure 7.28. Doubleclick on the Integrator block and define the initial temperature of the sphere to be 25^{\circ} \mathrm{C}. Run the simulation. The results can be plotted as shown in Figure 7.29.
