Air at a temperature of T∞ = 25 °C flows up a vertical flat plate with a velocity of u∞ = 0.5 m/s. The plate is L = 0.15 m long and is maintained at a temperature of Tw = 55 °C. What is the total heat transfer from one side of the plate? Compare this with that due to (a) forced convection alone

Question

Air at a temperature of [latex]T_{\infty}=25^{\circ} C[/latex] flows up a vertical flat plate with a velocity of [latex]u_{\infty}=0.5 m / s[/latex]. The plate is L = 0.15 m long and is maintained at a temperature of [latex]T_{w}=55^{\circ} C[/latex]. What is the total heat transfer from one side of the plate? Compare this with that due to (a) forced convection alone and (b) natural convection alone.

Accepted Answer

Step 1 Air properties are taken at the mean of wall and free-stream temperatures, i.e., at [latex]T_{m}=\frac{T_{w}+T_{\infty}}{2}=\frac{55+25}{2}=40{ }^{\circ} C[/latex] . We thus have the following air properties:

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

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

Prandtl number: [latex]P r=0.7[/latex]

Isobaric expansion coefficient of air is taken as [latex]\beta=\frac{1}{T_{\infty}}=\frac{1}{273+25}=0.00336 K^{−1}[/latex] .

Step 2 The Reynolds, Grashof number and Richardson numbers are now calculated.

[latex]\operatorname{Re}_{L}=\frac{u_{\infty} L}{ u}=\frac{0.5 imes 0.15}{17.07 imes 10^{-6}}=4393.7[/latex]

[latex]G r_{L}=\frac{g \beta\left(T_{w}-T_{\infty} ight) L^{3}}{ u^{2}}=\frac{9.81 imes 0.00336(55-25) 0.15^{3}}{\left(17.07 imes 10^{-6} ight)^{2}}=1.144 imes 10^{7}[/latex]

Both Reynolds and Grashof numbers fall in the laminar range. The Richardson number is then given by

[latex]R i_{L}=\frac{G r_{L}}{R e_{L}^{2}}=\frac{1.144 imes 10^{-7}}{4393.7^{2}}=0.593[/latex]

Step 3 Using Eq. 17.107 we calculate the average mixed convectionNusselt number as

[latex]\overline{N u}_{L}=1.32 P r^{0.315} R i_{L}^{-0.141} G r_{L}^{0.23}[/latex] (17.107)

[latex]\overline{N u}_{L}=1.32 imes 0.7^{0.315} imes 0.593^{-0.141} imes\left(1.144 imes 10^{7} ight)^{0.23}=53.38[/latex]

The mean heat transfer coefficient from the plate to air is then given by

[latex]\bar{h}_{L}=\frac{\overline{N u}_{L} k}{L}=\frac{53.38 imes 0.0274}{0.15}=9.75 W / m ^{2}{ }^{\circ} C[/latex]

Step 4 Heat transfer from one side of the plate is calculated based on a unit width of plate. Area of plate is then numerically the same as the length of the plate. Heat transfer from the plate to air is given by

[latex]Q=\bar{h}_{L} L\left(T_{w}-T_{\infty} ight)=9.75 imes 0.15 imes(55-25)=43.88 W[/latex]

Step 5 If heat transfer were to be by forced convection alone, the average Nusselt number is based on the Blasius solution (Eq. 13.54). We have

[latex]h_{x}=\frac{N u_{x} k_{\infty}}{x}=\frac{0.33206 R e_{x}^{\frac{1}{2}} \operatorname{Pr}^{\frac{1}{3}} k_{\infty}}{x}=0.33206 \sqrt{\frac{U_{\infty}}{ u_{\infty}}} \operatorname{Pr}^{\frac{1}{3}} k_{\infty} x^{-\frac{1}{2}}[/latex] (13.54)

[latex]\overline{N u}_{L}=0.664 R e_{L}^{\frac{1}{2}} P r^{\frac{1}{3}}=0.664 imes 4393.7^{\frac{1}{2}} 0.7^{\frac{1}{3}}=39.06[/latex]

The mean heat transfer coefficient from the plate to air is then given by

[latex]\bar{h}_{L}=\frac{\overline{N u}_{L} k}{L}=\frac{39.06 imes 0.0274}{0.15}=7.14 W / m ^{2}{ }^{\circ} C[/latex]

Step 6 Heat transfer from one side of the plate is calculated as

[latex]Q=\bar{h}_{L} L\left(T_{w}-T_{\infty} ight)=7.14 imes 0.15 imes(55-25)=32.11 W[/latex]

Step 7 Comment: Mixed convection augments heat loss by 43.88 − 32.11 = 11.77 W or as a percentage by [latex]\frac{11.77 imes 100}{32.11}=36.65 \%[/latex].

Step 8 Heat transfer by natural convection alone may be computed using the Ostrach solution. Using the curve fit, C (Eq. 16.43) is given by

[latex]C(P r)=\left[\frac{0.4 P r}{1+2 \sqrt{\operatorname{Pr}}+2 \operatorname{Pr}} ight]^{\frac{1}{4}}[/latex] (16.43)

[latex]C=\left(\frac{0.4 \operatorname{Pr}}{1+2 \sqrt{\operatorname{Pr}}+2 \operatorname{Pr}} ight)^{\frac{1}{4}}=\left(\frac{0.4 imes 0.7}{1+2 \sqrt{0.7}+2 imes 0.7} ight)^{\frac{1}{4}}=0.512[/latex]

Mean Nusselt number is then given by

[latex]\overline{N u}_{L}=C R a_{L}^{\frac{1}{4}}=C\left(G r_{L} P r ight)^{\frac{1}{4}}=0.512 imes\left(1.144 imes 10^{7} imes 0.7 ight)^{\frac{1}{4}}=27.22[/latex]

The mean heat transfer coefficient from the plate to air is then given by

[latex]\bar{h}_{L}=\frac{\overline{N u}_{L} k}{L}=\frac{27.22 imes 0.0274}{0.15}=4.97 W / m ^{2}{ }^{\circ} C[/latex]

Step 9 Heat transfer from one side of the plate is calculated as

[latex]Q=\bar{h}_{L} L\left(T_{w}-T_{\infty} ight)=4.97 imes 0.15 imes(55-25)=22.38 W[/latex]

Step 10 Comment: Mixed convection augments heat loss by 43.88 − 22.38 = 21.50 W or as a percentage by [latex]\frac{21.50 imes 100}{22.38}=96.08 \%[/latex].

Step 11 General comment: If the heat loss by forced convection and natural convection are added together we get a total heat loss from plate to air of 32.11 + 22.38 = 54.49 W. This is more than the heat loss by mixed convection and hence the mixed convection heat transfer rate is less than the algebraic sum of forced and natural convection heat transfers occurring in isolation. Essentially, there is an interaction between the two modes of heat transfer.

Question 17.12: Air at a temperature of T∞ = 25 °C flows up a vertical flat ...

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