Consider a cold aluminum canned drink that is initially at a uniform temperature of 3°C. The can is 12.5 cm high and has a diameter of 6 cm. If the combined convection/radiation heat transfer coefficient between the can and the surrounding air at 25°C is 10 W/m^2 · °C, determine how long it will

Question

Consider a cold aluminum canned drink that is initially at a uniform temperature of 3°C. The can is 12.5 cm high and has a diameter of 6 cm. If the combined convection/radiation heat transfer coefficient between the can and the surrounding air at 25°C is [latex]10 W / m ^{2} \cdot{ }^{\circ} C[/latex], determine how long it will take for the average temperature of the drink to rise to 10°C. In an effort to slow down the warming of the cold drink, a person puts the can in a perfectly fitting 1-cm-thick cylindrical rubber insulation ([latex]k = 0.13 W / m \cdot{ }^{\circ} C[/latex]). Now how long will it take for the average temperature of the drink to rise to 10°C? Assume the top of the can is not covered.

Accepted Answer

A cold aluminum canned drink that is initially at a uniform temperature of 3°C is brought into a room air at 25°C. The time it will take for the average temperature of the drink to rise to 10°C with and without rubber insulation is to be determined.

Assumptions 1 The drink is at a uniform temperature at all times. 2 The thermal resistance of the can and the internal convection resistance are negligible so that the can is at the same temperature as the drink inside. 3 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 4 Thermal properties are constant. 5 The thermal contact resistance at the interface is negligible.

Properties The thermal conductivity of rubber insulation is given to be [latex]k = 0.13 W / m \cdot{ }^{\circ} C[/latex]. For the drink, we use the properties of water at room temperature, [latex] ho = 1000 kg / m ^{3}[/latex] and [latex] C_{ p } =4180 J / kg \cdot{ }^{\circ} C[/latex]

Analysis This is a transient heat conduction, and the rate of heat transfer will decrease as the drink warms up and the temperature difference between the drink and the surroundings decreases. However, we can solve this problem approximately by assuming a constant average temperature of (3+10)/2 = 6.5°C during the process. Then the average rate of heat transfer into the drink is

[latex] A_{o}=\pi D_{o} L+2 \frac{\pi D^{2}}{4}=\pi(0.06 m )(0.125 m )+2 \frac{\pi(0.06 m )^{2}}{4}=0.0292 m ^{2}[/latex]

[latex] \dot{Q}_{ ext {bare ,ave }}=h_{o} A\left(T_{ ext {air }}-T_{ ext {can }, ext { ave }} ight) [/latex]

[latex] =\left(10 W / m ^{2} \cdot{ }^{\circ} C ight)\left(0.0292 m ^{2} ight)(25-6.5){ }^{\circ} C =5.40 W[/latex]

The amount of heat that must be supplied to the drink to raise its temperature to 10 °C is

[latex] m = ho V= ho \pi r^{2} L=\left(1000 kg / m ^{3} ight) \pi(0.03 m )^{2}(0.125 m ) = 0.353 kg[/latex]

[latex] Q = m C_{p} \Delta T = (0.353 kg )(4180 J / kg )(10 - 3)^{\circ} C = 10,329 J[/latex]

Then the time required for this much heat transfer to take place is

[latex] \Delta t=\frac{Q}{\dot{Q}}=\frac{10,329 J }{5.4 J / s }=1912 s = 3 1 . 9 min[/latex]

We now repeat calculations after wrapping the can with 1-cm thick rubber insulation, except the top surface. The rate of heat transfer from the top surface is

[latex] \dot{Q}_{ ext {top }, ext { ave }}=h_{o} A_{ ext {top }}\left(T_{ ext {air }}-T_{ ext {can, ave }} ight)[/latex]

[latex] =\left(10 W / m ^{2} \cdot{ }^{\circ} C ight)\left[\pi(0.03 m )^{2} ight](25-6.5)^{\circ} C =0.52 W[/latex]

Heat transfer through the insulated side surface is

[latex] A_{o}=\pi D_{o} L=\pi(0.08 m )(0.125 m )=0.03142 m ^{2}[/latex]

[latex] R_{o}=\frac{1}{h_{o} A_{o}}=\frac{1}{\left(10 W / m ^{2} \cdot{ }^{\circ} C ight)\left(0.03142 m ^{2} ight)}=3.183{ }^{\circ} C / W[/latex]

[latex] R_{ ext {insulation }, ext { side }}=\frac{\ln \left(r_{2} / r_{1} ight)}{2 \pi k L}=\frac{\ln (4 / 3)}{2 \pi\left(0.13 W / m ^{2} \cdot{ }^{\circ} C ight)(0.125 m )}=2.818^{\circ} C / W[/latex]

[latex] R_{ ext {total }}=R_{o}+R_{ ext {insulation }}=3.183+2.818=6.001^{\circ} C / W[/latex]

[latex] \dot{Q}_{ ext {side }}=\frac{T_{ ext {air }}-T_{ ext {can, ave }}}{R_{ ext {conv }, o }}=\frac{(25-6.5)^{\circ} C }{6.001^{\circ} C / W }=3.08 W[/latex]

The ratio of bottom to the side surface areas is [latex]\left(\pi r^{2} ight) /(2 \pi r L)=r /(2 L)=3 /(2 imes 12.5)=0.12 .[/latex]

Therefore, the effect of heat transfer through the bottom surface can be accounted for
approximately by increasing the heat transfer from the side surface by 12%. Then,

[latex] \dot{Q}_{ ext {insulated }}=\dot{Q}_{ ext {side+bottom }}+\dot{Q}_{ ext {top }}=1.12 imes 3.08+0.52=3.97 W[/latex]

Then the time of heating becomes

[latex] \Delta t=\frac{Q}{\dot{Q}}=\frac{10,329 J }{3.97 J / s } = 2602 s = 43.4 min[/latex]

Question 3.73: Consider a cold aluminum canned drink that is initially at a...

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