Temperature Dynamics of a Heated Object Consider a steel sphere with a radius of r = 0.01 m submerged in a hot water bath with a heat transfer coefficient of h = 350 W/(m^2·°C). For steel, the density is ρ = 7850 kg/m3, the specific heat capacity is c = 440 J/(kg·°C), and the thermal conductivity

Question

Temperature Dynamics of a Heated Object

Consider a steel sphere with a radius of [latex]r=0.01 \mathrm{~m}[/latex] submerged in a hot water bath with a heat transfer coefficient of [latex]h=350 \mathrm{~W} /\left(\mathrm{m}^{2 .}{ }^{\circ} \mathrm{C}\right)[/latex]. For steel, the density is [latex]\rho=7850 \mathrm{~kg} / \mathrm{m}^{3}[/latex], the specific heat capacity is [latex]\mathrm{c}=440 \mathrm{~J} /\left(\mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)[/latex], and the thermal conductivity is [latex]k=43 \mathrm{~W} /\left(\mathrm{m} \cdot{ }^{\circ} \mathrm{C}\right)[/latex]. The temperature of the water [latex]T_{\mathrm{f}}[/latex] is [latex]100^{\circ} \mathrm{C}[/latex] and the initial temperature of the sphere [latex]T_{0}[/latex] is [latex]25^{\circ} \mathrm{C}[/latex].

a. Determine whether the sphere's temperature can be considered uniform.

b. Derive the differential equation relating the sphere's temperature [latex]T(t)[/latex] and the water's temperature [latex]T_{\mathrm{f}}[/latex].

c. Using the differential equation obtained in Part (b), construct a Simulink block diagram to find the sphere's temperature [latex]T(t)[/latex].

Accepted Answer

a. The characteristic length of the sphere is

[latex]L_{\mathrm{c}}=\frac{V_{ ext {body }}}{A_{ ext {surface }}}=\frac{4 / 3 \pi r^{3}}{4 \pi r^{2}}=\frac{1}{3} r=\frac{0.01}{3} .[/latex]

The Biot number of the steel sphere is

[latex]B i=\frac{h L_{c}}{k}=\frac{350(0.01)}{43(3)}=2.71 imes 10^{-2}<0.1.[/latex]

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,

[latex]\frac{\mathrm{d} U}{\mathrm{~d} t}=q_{\mathrm{hi}}-q_{\mathrm{ho}}.[/latex]

Note that [latex]U=m c T= ho V c T[/latex], and we have

[latex]\frac{\mathrm{d} U}{\mathrm{~d} t}=\frac{\mathrm{d}}{\mathrm{d} t}\left( ho V_{c} T ight)= ho V_{c} \frac{\mathrm{d} T}{\mathrm{~d} t}.[/latex]

The heat flow rate into the body is

[latex]q_{\mathrm{hi}}=\frac{T_{\mathrm{f}}-T}{R}[/latex]

and the heat flow rate out of the body is [latex]q_{\mathrm{ho}}=0[/latex]. Thus, the differential equation of the system is

[latex] ho V c(\mathrm{~d} T / \mathrm{d} t)=\left(T_{\mathrm{f}}-T ight) / R.[/latex]

Introducing the expression for the thermal capacitance [latex] ho V C=C[/latex], we find

[latex]R C \frac{\mathrm{d} T}{\mathrm{~d} t}+T=T_{\mathrm{f}}.[/latex]

The thermal capacitance of the sphere is

[latex]C= ho V_{C}=7850\left(\frac{4}{3} ight)(\pi)(0.01)^{3}(440)=14.47 \mathrm{~J} /{ }^{\circ} \mathrm{C},[/latex]

and the thermal resistance due to convection is

[latex]R=\frac{1}{h A}=\frac{1}{350(4)(\pi)(0.01)^{2}}=2.27^{\circ} \mathrm{Cs} / \mathrm{J}.[/latex]

Thus, the dynamic model of the sphere's temperature is

[latex]32.85 \frac{\mathrm{d} T}{\mathrm{~d} t}+T=T_{\mathrm{f}}.[/latex]

C. Given [latex]T_{\mathrm{f}}=100^{\circ} \mathrm{C}[/latex], solving for the highest derivative of the output [latex]T[/latex] gives

[latex]\frac{\mathrm{d} T}{\mathrm{~d} t}=\frac{1}{32.85}(100-T),[/latex]

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 [latex]25^{\circ} \mathrm{C}[/latex]. Run the simulation. The results can be plotted as shown in Figure 7.29.

Question 7.7: Temperature Dynamics of a Heated Object Consider a steel sph...

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