A very tall vertical channel consists of two parallel plates maintained at temperatures of T1 = 25 °C and T2 = 55 °C. The channel spacing is equal to 2b = 2.5 cm. Air at To = 40 °C enters the channel with a uniform vertical velocity of uo = 0.5 m/s. Would you expect flow reversal in this case

Question

A very tall vertical channel consists of two parallel plates maintained at temperatures of [latex]T_{1}=25^{\circ} C[/latex] and [latex]T_{2}=55^{\circ} C[/latex]. The channel spacing is equal to 2b = 2.5 cm. Air at [latex]T_{0}=40^{\circ} C[/latex] enters the channel with a uniform vertical velocity of [latex]u_{o}=0.5 m / s[/latex]. Would you expect flow reversal in this case because of natural convection? What is the axial pressure gradient in this mixed convection flow? What is the maximum velocity in this case? Consider the fully developed part of the flow far downstream of the entry.

Accepted Answer

Step 1 Air properties are taken at the mean temperature [latex]T_{m}=\frac{T_{1}+T_{2}}{2}=\frac{25+55}{2}=40^{\circ} C[/latex].

Density: [latex] ho_{m}=1.1169 kg / m ^{3}[/latex]

Dynamic viscosity: [latex]\mu_{m}=19.1 imes 10^{-6} kg / m s[/latex]

Kinematic viscosity: [latex] u_{m}=17.1 imes 10^{-6} m ^{2} / s[/latex]

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

Prandtl number: [latex]P r=0.699[/latex]

Isobaric coefficient of expansion: [latex]\beta_{o}=\frac{1}{T_{o}}=\frac{1}{273+40}=3.19 imes 10^{-3} K ^{-1}[/latex]

In addition, we take [latex]g=9.81 m / s ^{2}[/latex], semi-channel spacing b = 1.25 cm = 0.0125 m.

Step 2 Calculate the appropriate non-dimensional parameters:

The Reynolds number is calculated as

[latex]R e_{b}=\frac{u_{o} b}{ u_{m}}=\frac{0.5 imes 0.0125}{17.1 imes 10^{-6}}=366.14[/latex]

We note that the flow is laminar. The Grashof number is calculated as

[latex]G r_{b}=\frac{g \beta_{m}\left(T_{2}-T_{1} ight) b^{3}}{ u_{m}^{2}}=\frac{9.81 imes 3.19 imes 10^{-3}(55-25) imes 0.0125^{3}}{\left(17.1 imes 10^{-6} ight)^{2}}=6302.44[/latex]

Mixed convection parameter is then given by [latex]\frac{G r_{b}}{\operatorname{Re}_{b}}=\frac{6302.44}{366.14}=17.2[/latex]. Since the mixed convection parameter is greater than the critical value of 9, flow reversal is expected in the channel.

Step 3 Non-dimensional entry air temperature is calculated as [latex] heta_{o}=\frac{T_{m}-T_{o}}{T_{2}-T_{m}}=[/latex] [latex]\frac{40-40}{55-40}=0[/latex] . Hence, the non-dimensional pressure drop is given by [latex]\frac{ d \ddot{P}}{ d X}=[/latex] [latex]-3+\frac{G r_{b}}{R e_{b}} heta_{o}=-3[/latex] . Converting this to dimensional variables, we get

[latex]\frac{ d p}{ d x}=\frac{\mu_{m} u_{o}}{b^{2}} \frac{ d P}{ d X}=\frac{19.1 imes 10^{-6} imes 0.5}{0.0125^{2}}(-3)=-0.183 Pa / m[/latex]

Step 4 The maximum velocity occurs when the slope is zero. Differentiating Eq. 17.112 with respect to Y , we get [latex]\frac{ d U}{ d Y}=\frac{G r_{b}}{6 \operatorname{Re}_{h}}\left(1-3 Y^{2} ight)-3 Y[/latex] . Setting it to 0, we get a quadratic in Y which may be solved to get

[latex]U=\frac{G r_{b}}{6 R e_{b}}\left(Y-Y^{3} ight)+\frac{3\left(1-Y^{2} ight)}{2}[/latex] (17.112)

[latex]Y=-\frac{3 R e_{b}}{G r_{b}} \pm \sqrt{\left(\frac{3 R e_{b}}{G r_{b}} ight)^{2}+\frac{1}{3}}[/latex]

The root that is required for locating the maximum velocity is obtained by taking the plus sign for the second term. We then have

[latex]Y=-\frac{3}{17.2} \pm \sqrt{\left(\frac{3}{17.2} ight)^{2}+\frac{1}{3}}=0.429[/latex]

The maximum non-dimensional velocity is then given by the use of Eq. 17.112.

[latex]U_{\max }=U(y=0.429)=\frac{17.2}{6}\left(0.429-0.429^{3} ight)+1.4\left(1-0.429^{2} ight)=2.228[/latex]

The maximum velocity is then given by [latex]u_{\max }=U_{\max } u_{o}=2.228 imes 0.5=1.114 m/s[/latex]

Question 17.13: A very tall vertical channel consists of two parallel plates...

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