A metal wall 3mm thick made of steel with a thermal conductivity of 45W/m°C is covered with a 10 cm layer of an insulating material of thermal conductivity 0.1W/m°C. The temperature of the inner surface of steel wall is measured to be 60 °C. The ambient is atmospheric air at 27 °C. The wall is 0.4m

Question

A metal wall 3mm thick made of steel with a thermal conductivity of [latex]45 W / m ^{\circ} C[/latex] is covered with a 10 cm layer of an insulating material of thermal conductivity [latex]0.1 W / m ^{\circ} C[/latex]. The temperature of the inner surface of steel wall is measured to be [latex]60^{\circ} C[/latex]. The ambient is atmospheric air at [latex]27^{\circ} C[/latex]. The wall is 0.4m wide and 1.5m tall. Determine the heat transfer across the composite wall. Refer to Fig. 16.9 for pictorial representation of the problem.

Accepted Answer

Figure 16.9 shows that there are two things that are not known viz. the temperature [latex]T_{0}[/latex] of the outer surface of the insulation and the heat transfer Q through the composite wall. An iterative solution is required. We expect a small temperature difference [latex]T_{0}-T_{\infty}[/latex] since the insulating layer is fairly thick and it has a very low thermal conductivity. Hence, the air properties are taken at the ambient temperature of [latex]T_{\infty}=27^{\circ} C[/latex].

Step 1 Air properties at [latex]27^{\circ} C[/latex] are

[latex]v=15.89 imes 10^{-6} m ^{2} / s[/latex] , [latex]k=0.0263 ; W / m ^{\circ} C[/latex] , [latex]\operatorname{Pr}=0.707[/latex]

Other data specified in the problem are listed below.

Temperature of inner surface of steel: [latex]T_{i}=60^{\circ} C[/latex]

Thermal conductivity of steel: [latex]k_{s}=45 W / m ^{\circ} C[/latex]

Thickness of steel: [latex]t_{s}[/latex] = 3mm = 0.003m

Thermal conductivity of insulation: [latex]k_{i}=0.1 W / m ^{\circ} C[/latex]

Thickness of insulation: [latex]t_{i}[/latex] = 10 cm = 0.1m

Step 2 The length scale for free convection from the outer surface of the insulation layer is the same as the height of the wall, i.e., H = 1.5m. The width of the wall is W = 0.4 m, and hence the heat transfer area is [latex]A = HW =1.5 × 0.4 = 0.6m^{2}[/latex].

The isobaric expansion coefficient is

[latex]\beta=\frac{1}{T_{\infty}}=\frac{1}{273+27}=0.0033 K ^{-1}[/latex]

Temperature [latex]T_{0}[/latex] is unknown.We start by assuming a suitable value for this and iterate to home in on the correct value.

Step 3 Let us start with [latex]T_{0}=T_{\infty}+10=27+10=37^{\circ} C[/latex] . The temperature on the absolute scale is [latex]T_{0}=273 + 37 = 310 K[/latex]. The Rayleigh number is calculated as

[latex]R a_{H}=\frac{g \beta\left(T_{0}-T_{\infty} ight) H^{3}}{v^{2}} imes \operatorname{Pr}[/latex] [latex]=\frac{9.81 imes 0.00033(310-300) imes 1.5^{3}}{\left(15.89 imes 10^{-6} ight)^{2}} imes 0.707=3.09 imes 10^{9}[/latex]

The regime is hence turbulent. According to McAdams recommended formula 16.52

[latex]\overline{N u}_{L}=0.13 R a_{L}^{\frac{1}{3}}[/latex] (16.52)

[latex]\overline{N u}_{H}=0.13\left(3.091 imes 10^{9} ight)^{\frac{1}{3}}=189.3[/latex]

The average heat transfer coefficient over the height of the plate is then obtained as

[latex]\bar{h}=\frac{\overline{N u}_{H} k}{H}=\frac{189.3 imes 0.0263}{1.5}=3.32 W / m ^{2}{ }^{\circ} C[/latex]

Step 4 Convective heat transfer from the outer surface of the insulation is calculated as

[latex]Q_{c}=\bar{h} A\left(T_{0}-T_{\infty} ight)=3.32 imes 0.6 imes 10=19.92 W[/latex]

Conductive heat transfer through the composite wall is

[latex]Q_{k}=A \frac{T_{i}-T_{0}}{\frac{t_{s}}{k_{s}}+\frac{t_{i}}{k_{i}}}[/latex] [latex]=0.6 \frac{60-37}{\frac{0.003}{45}+\frac{0.1}{0.1}}=13.79 W[/latex]

Since [latex]Q_{c}>Q_{k}[/latex] , it is clear that [latex]T_{0}[/latex] should be smaller than the value assumedabove. We may reduce it to a lower value and repeat the calculation.

Step 5 Instead we have systematically varied [latex]T_{0}[/latex] between 30 and [latex]37 ^{\circ} C[/latex] and plotted the conductive heat transfer across the composite wall against the convective heat transfer from the exposed surface of the insulation (Fig. 16.10). It is seen that the two curves cross when [latex]T_{0}=35.1 ^{\circ} C[/latex] and Qc = Qk = 15 W. These are the desired answers.

Question 16.5: A metal wall 3mm thick made of steel with a thermal conducti...

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