A flat surface of height 0.2mat a uniform temperature of 65 °C is placed in still air at 27 °C. Calculate the maximum velocity within the natural convection boundary layer based on the Ostrach solution. Compare this with that estimated using the integral solution.

Question

A flat surface of height 0.2mat a uniform temperature of [latex]65^{\circ} C[/latex] is placed in still air at [latex]27^{\circ} C[/latex]. Calculate the maximum velocity within the natural convection boundary layer based on the Ostrach solution. Compare this with that estimated using the integral solution.

Accepted Answer

Maximum velocity based on Ostrach solution:
The Ostrach solution labeled [latex]P r=0.733[/latex] is used for the purpose of evaluating the desired maximum velocity at L = 0.2m.

Step 1 The air properties are evaluated at the ambient temperature of [latex]T_{\infty}=27^{\circ} C=300 K[/latex].

[latex]\beta=\frac{1}{T_{\infty}}=\frac{1}{300}=0.00333 K ^{-1}[/latex] , [latex]v=15.89 imes 10^{-6} m ^{2} / s[/latex]

Step 2 The wall temperature is [latex]T_{w}=65^{\circ} C[/latex] and hence the temperature difference is

[latex]\Delta T=T_{w}-T_{\infty}=65-27=38^{\circ} C[/latex]

The maximum velocity in the boundary layer will occur as the flow leaves at the top of the plate. Hence, we calculate the Grashof number based on plate length as

[latex]G r_{L}=\frac{g \beta \Delta T L^{3}}{v^{2}}=\frac{9.81 imes 0.000333 imes 38 imes 0.2^{3}}{\left(15.89 imes 10^{-6} ight)^{2}}=3.937 imes 10^{7}[/latex]

Step 3 We make use of Fig. 16.5 to evaluate the maximum velocity. We note that for [latex]P r=0.733[/latex] (typical value for air), the maximum velocity occurs at [latex]\eta=0.95[/latex]. The maximum value itself is read off the graph as

[latex]\frac{u_{\max } . L}{\frac{2 v}{L} \sqrt{G r_{L}}}=0.28[/latex] or [latex]u_{\max , L}=0.28 imes \frac{2 imes 15.89 imes 10^{-6}}{0.2} \sqrt{3.937 imes 10^{7}}=0.279 m / s[/latex]

Maximum velocity based on integral solution:

Step 4 The cubic velocity profile is given by Eq. 16.19 as [latex]\frac{u}{U}=\eta(1-\eta)^{2}[/latex], where [latex]\eta=\frac{y}{\delta}[/latex] . The maximum velocity occurs when

[latex]\frac{u}{U}=\eta-2 \eta^{2}+\eta^{3}=\eta(1-\eta)^{2}[/latex] (16.19)

[latex]\frac{d}{d \eta}\left(\frac{u}{U} ight)=(1-\eta)^{2}-2 \eta(1-\eta)=0[/latex]

This happens when [latex]\eta=\frac{1}{3}[/latex]. Thus the maximum velocity within the boundary layer is

[latex]u_{\max , L}=U_{L} \frac{1}{3}\left(1-\frac{1}{3} ight)^{2}=\frac{4}{27} U_{L}[/latex]

where [latex]U_{L}[/latex] is the velocity scale evaluated at x = L.

Step 5 Using Eq. 16.30, we have

[latex]U=A x^{m}=5.17\left[\frac{g \beta\left(T_{w}-T_{\infty} ight)}{\operatorname{Pr}+0.952} ight]^{\frac{1}{2}} \cdot x^{\frac{1}{2}}[/latex] (16.30)

[latex]U_{L}=5.17\left[\frac{9.81 imes 0.000333 imes 38}{0.733+0.952} ight]^{\frac{1}{2}} imes 0.2^{\frac{1}{2}}=1.986 m / s[/latex]

Step 6 Thus, the maximum velocity according to the integral solution is

[latex]u_{\max , L}=\frac{4}{27} imes 1.986=0.294 m / s[/latex]

The exact and the integral solution estimates are remarkably close to each other.

Question 16.4: A flat surface of height 0.2mat a uniform temperature of 65 ...

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