Consider a horizontal circular plate of diameter D = 0.2m losing heat by natural convection to surrounding ambient air at T∞ = 20 °C. The plate is maintained at a temperature of Tw = 60 °C. Compare the heat loss from this plate with a square plate of the same area, all other data remaining the same

Question

Consider a horizontal circular plate of diameter D = 0.2m losing heat by natural convection to surrounding ambient air at [latex]T_{\infty}=20^{\circ} C[/latex]. The plate is maintained at a temperature of [latex]T_{w}=20^{\circ} C[/latex]. Compare the heat loss from this plate with a square plate of the same area, all other data remaining the same.

Accepted Answer

Case (a): Circular plate

We use the thermo-physical properties of air at the mean temperature of [latex]T_{m}=\frac{T_{w}+T_{\infty}}{2}=\frac{60+20}{2}=40^{\circ} C[/latex]

Kinematic viscosity: [latex]v_{m}=17.07 imes 10^{-6} m ^{2} / s[/latex]

Thermal conductivity: [latex]k_{m}=0.0274 W / m ^{\circ} C[/latex]

Prandtl number: [latex]P r_{m}=0.699[/latex]

Isobaric expansion coefficient: [latex]\beta_{m}=\frac{1}{T_{m}}=\frac{1}{273+40}=0.0032 K ^{-1}[/latex]

The Rayleigh number may now be calculated with [latex]L_{c h}=D=0.2 m[/latex].

[latex]R a_{D}=\frac{g \beta_{m}\left(T_{w}-T_{\infty} ight) D^{3}}{v_{m}^{2}} P r_{m}[/latex] [latex]=\frac{9.81 imes 0.0032(60-20) 0.2^{3}}{\left(17.07 imes 10^{-6} ight)^{2}} 0.699=2.41 imes 10^{7}[/latex]

Using Eq. 16.55, we than have

[latex]\overline{N u}_{L_{c h}}=0.27 R a_{L_{c h}}^{\frac{1}{4}}[/latex] (16.55)

[latex]\overline{N u}_{D}=0.27 R a_{D}^{\frac{1}{4}}=0.27 imes\left(2.41 imes 10^{7} ight)^{\frac{1}{4}}=18.92[/latex]

Correspondingly the mean heat transfer coefficient is

[latex]\bar{h}=\frac{\overline{N u}_{D} k_{m}}{D}=\frac{18.92 imes 0.0274}{0.2}=2.59 W / m ^{2 \circ} C[/latex]

Heat loss from the plate to the ambient air is

[latex]Q=\bar{h} \frac{\pi D^{2}}{4}\left(T_{w}-T_{\infty} ight)=2.59 imes \frac{\pi imes 0.2^{2}}{4}(60-20)=3.26 W[/latex]

Case (b): Square plate of equal area

The only change in this case is that the characteristic dimension changes from D to[latex]\frac{W+L}{2}=a[/latex] where a is the side of the square plate. Since the square plate has the same area as that of the circular plate, we have

[latex]a=\sqrt{\frac{\pi D^{2}}{4}}=\sqrt{\frac{\pi imes 0.2^{2}}{4}}=0.177 m[/latex]

The Rayleigh number in this case is

[latex]R a_{a}=\frac{g \beta_{m}\left(T_{w}-T_{\infty} ight) a^{3}}{v^{2}} P r_{m}[/latex] [latex]=\frac{9.81 imes 0.0032(60-20) 0.177^{3}}{\left(17.07 imes 10^{-6} ight)^{2}} 0.699=1.67 imes10^{7}[/latex]

Using Eq. 16.55, we than have

[latex]\overline{N u}_{L_{c h}}=0.27 R a_{L_{c h}}^{\frac{1}{4}}[/latex] (16.55)

[latex]\overline{N u}_{a}=0.27 R a_{a}^{\frac{1}{4}}=0.27 imes\left(1.67 imes 10^{7} ight)^{\frac{1}{4}}=17.26[/latex]

Correspondingly the mean heat transfer coefficient is

[latex]\bar{h}=\frac{\overline{N u}_{a} k_{m}}{a}=\frac{17.26 imes 0.0274}{0.177}=2.672 W / m ^{2}{ }^{\circ} C[/latex]

Heat loss from the plate to the ambient air is

[latex]Q=\bar{h} a^{2}\left(T_{w}-T_{\infty} ight)=2.672 imes 0.177^{2}(60-20)=3.35 W[/latex]

The square plate loses marginally more heat than the circular plate.

Question 16.6: Consider a horizontal circular plate of diameter D = 0.2m lo...

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