An ice chest contains a mixture of ice and water at 32°F, a temperature which is constant during the melting of the ice. Th e walls of the chest are made of three materials sandwiched together as shown in Figure 2.11. Th e outside shell is made of stamped sheet metal—0.040 in. thick,

Question

An ice chest contains a mixture of ice and water at [latex]32°F[/latex], a temperature which is constant during the melting of the ice. The walls of the chest are made of three materials sandwiched together as shown in Figure 2.11. The outside shell is made of stamped sheet metal—[latex]0.040[/latex] in. thick, low-carbon [latex](1C)[/latex] steel. The insulating material in the wall is [latex]{3}/{4}[/latex] in. thick styrofoam [latex](k = 0.02 {BTU}/{hr · ft · °R})[/latex]. The inside material is a fiberglass liner [latex](k = 0.09 {BTU}/{hr · ft · °R})[/latex], [latex]{1}/{4}[/latex] in. thick. Outside the ice chest is air at [latex]90°F[/latex], and the convection coefficient between the air and the vertical steel wall is [latex]0.79 {BTU}/{(hr · ft 2 · °R)}[/latex]. Inside, the convection coefficient between the ice water and the fiberglass is [latex]150 {BTU}/{(hr · ft 2 · °R)}[/latex]. Determine the heat-transfer rate through the wall of the ice chest based on a per-square-foot area. Determine also the overall heat-transfer coefficient [latex]U[/latex] for the wall.

Accepted Answer

The thermal circuit and an expected temperature profile are shown in Figure 2.11.

Assumptions

1. One-dimensional steady heat flow exists.
2. Material properties are constant.
3. Convection coefficients are constant and apply to the appropriate wall.

The thermal conductivity for [latex]1C[/latex] steel is [latex]24.8 {BTU}/{(hr · ft · °R)}[/latex] from Appendix Table B.2.

We can first determine [latex]U[/latex] and then calculate the heat transferred. Equation 2.22 applies

[latex]U=\frac{1}{\frac{1}{\bar{h}_{c1}}+\frac{L}{K}+\frac{1}{\bar{h} _{c2}} } =\frac{1}{A(R_{c1}+R_{k}+R_{c3})}[/latex]

[latex]UA=1/\sum\limits_{i}{R_{i}} =1/(R_{c1}+R_{k1}+R_{k2}+R_{k3}+R_{c2})[/latex]

where

[latex]R_{c1}=\frac{1}{\bar{h} _{c1}A} =\frac{1}{0.79(1)} [/latex]

[latex]=1.266°F.{hr}/{BTU}[/latex] (air to sheet metal)

[latex]R_{k1}=\frac{\Delta x_{1}}{K_{1}A} =\frac{{0.04}/{12}}{24.8(1)} [/latex]

[latex]=0.0001°F⋅{hr}/{BTU}[/latex] (sheet metal)

[latex]R_{k2}=\frac{\Delta x_{2}}{K_{2}A} =\frac{{0.75}/{12}}{0.02(1)} [/latex]

[latex]=3.125°F.{hr}/{BTU}[/latex] (styrofoam)

[latex]R_{k3}=\frac{\Delta x_{3}}{K_{3}A} =\frac{{0.25}/{12}}{0.09(1)} [/latex]

[latex]=0.231°F⋅{hr}/{BTU}[/latex] (fiberglass)

[latex]R_{c2}=\frac{1}{\bar{h} _{c2}A} =\frac{1}{150(1)} [/latex]

[latex]=0.0067°F.{hr}/{BTU}[/latex] (ice water to fiberglass)

As can be seen, the sheet metal offers a negligible resistance to the heat flow. The overall coefficient thus becomes

[latex]UA=1/\sum\limits_{i}{R_{i}} A=1/(1.266+0.0001+3.125+0.231+0.0067)(1)[/latex]

[latex]U=0.216{BTU}/{hr⋅ft^2⋅°F}[/latex]

The heat transferred can be calculated with Equation 2.20

[latex]q=\frac{T_{\infty 1}-T_{\infty 2}}{\frac{1}{\bar{h} _{c1}A}+\frac{L}{KA}+\frac{1}{\bar{h} _{c2}A} } =\frac{T_{\infty 1}-T_{\infty 2}}{\sum\limits_{i}{R_{i}} }[/latex]

[latex]q=UA(T_{\infty 1}-T_{\infty 2})[/latex]

Substituting and solving yields

[latex]q=0.216(1)(90-32)[/latex]

[latex]q=12.5{BTU}/{hr}[/latex]

Question 2.4: An ice chest contains a mixture of ice and water at 32°F, a ...

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