Question 4 of 6: Free versus Forced Convection from a Vertical Plate

Question B1: Free versus Forced Convection from a Vertical Plate (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A square vertical plate losing heat to still air; then the same plate in a forced stream whose velocity is twice the peak velocity attained inside the natural-convection boundary layer. Air properties are evaluated at the film temperature $T_f=\tfrac12(40+20)=30\ \mathrm{°C}=303.15\ \mathrm{K}$, interpolated linearly between the 300 K and 310 K rows of the supplied table ($\mu=18.53\to19.00\times10^{-6}$, $\rho=1.1769\to1.1389$, $k=26.14\to26.87\times10^{-3}$, $\alpha=22.07\to23.43\times10^{-6}$).

Find. The ratio of the free-convection heat loss to the forced-convection heat loss (at $U=2u_{max}$).

Approach. Compute the Rayleigh number and free-convection $\overline{\mathrm{Nu}}$; obtain the peak boundary-layer velocity from the similarity solution, double it, and evaluate the forced flat-plate $\overline{\mathrm{Nu}}$; take the ratio of the two heat rates.

1. Grashof and Rayleigh numbers. With $L=0.15$ m, $$\mathrm{Gr}_L=\frac{g\beta\Delta T L^3}{\nu^2}=\frac{9.81(1/303.15)(20)(0.15)^3}{(1.603\times10^{-5})^2}=8.50\times10^{6},\quad \mathrm{Ra}_L=\mathrm{Gr}_L\Pr=6.06\times10^{6}.$$ Laminar free convection ($\mathrm{Ra}<10^9$). (Check: the table’s own $g\beta/(\nu\alpha)$ column interpolates to $89.9\times10^6$, and $89.9\times10^6(20)(0.15)^3=6.07\times10^6$.)
2. Free-convection heat rate. Using $\overline{\mathrm{Nu}}=0.59\,\mathrm{Ra}_L^{1/4}=29.3$, $\bar h_{free}=\overline{\mathrm{Nu}}\,k/L=5.15\ \mathrm{W/m^2K}$, so over one face $A=0.15^2=0.0225\ \mathrm{m^2}$: $$Q_{free}=\bar h_{free}A\,\Delta T = 5.15(0.0225)(20)=2.32\ \mathrm{W}.$$ $Q_{free}=2.32\ \mathrm{W}$
3. Peak boundary-layer velocity. The similarity solution for a vertical plate gives $u=\dfrac{2\nu}{x}\mathrm{Gr}_x^{1/2}f'(\eta)=2\sqrt{g\beta\Delta T\,x}\;f'(\eta)$, which peaks inside the layer and grows along the plate, so the largest value is at the top edge $x=L$. For air ($\Pr\approx0.71$) the maximum of the dimensionless profile is $f'_{max}\approx0.275$, so $$u_{max}=2(0.275)\sqrt{g\beta\Delta T\,L}=0.55\sqrt{9.81(1/303.15)(20)(0.15)}=0.55(0.312)=0.171\ \mathrm{m/s}.$$ (Cross-check: the integral profile $u=u_x(y/\delta)(1-y/\delta)^2$ gives $u_{max}=\tfrac{4}{27}u_x=0.185\ \mathrm{m/s}$, within 8%.) The forced velocity is $U=2u_{max}=0.343\ \mathrm{m/s}$. $U=0.343\ \mathrm{m/s}$
4. Forced-convection heat rate. $\mathrm{Re}_L=UL/\nu=0.343(0.15)/1.603\times10^{-5}=3.21\times10^3$ (laminar), so the flat-plate average $\overline{\mathrm{Nu}}=0.664\mathrm{Re}_L^{1/2}\Pr^{1/3}=33.6$, $\bar h_{forced}=5.90\ \mathrm{W/m^2K}$: $$Q_{forced}=5.90(0.0225)(20)=2.66\ \mathrm{W}.$$ $Q_{forced}=2.66\ \mathrm{W}$
5. Compare. $$\frac{Q_{free}}{Q_{forced}}=\frac{2.32}{2.66}=0.87.$$ The forced stream removes about 15% more heat than free convection, but the two rates are the same order — a mixed-convection regime, as expected when the forced velocity is set by the buoyant velocity itself ($\mathrm{Gr}/\mathrm{Re}^2=8.5\times10^6/(3.21\times10^3)^2\approx0.8$). The integral-profile velocity gives 0.84, so the conclusion does not depend on the profile model. $Q_{free}/Q_{forced}\approx0.87$ (forced ≈15% higher)