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Natural convection from a vertical plate

Short answer

Grashof number Gr = 6.14×10^8 (ratio of buoyancy to viscous force), Rayleigh number Ra = 4.42×10^8; using Nu = 0.59 Ra^1/4 gives Nu = 85.5, average h = 4.95 W/m^2·K, heat loss from one side = 119 W.

A vertical plate 0.5 m high and 0.8 m wide is maintained at 90 °C in still air at 30 °C. Using properties at the film temperature of 60 °C (nu = 18.97 x 10^-6 m^2/s, k = 0.02896 W/m K, Pr = 0.7202, beta = 1/Tf), calculate (a) the Grashof and Rayleigh numbers, (b) the average Nusselt number and heat transfer coefficient, and (c) the heat lost from one side of the plate. Use Nu = 0.59 Ra^(1/4) for 10^4 < Ra < 10^9. Also explain the physical meaning of the Grashof number.

Grashof number: physical meaning

Gr=gβ ΔT L3ν2Gr = \dfrac{g\beta\,\Delta T\,L^3}{\nu^2} is the ratio of buoyancy force to viscous force acting on the fluid. In natural convection it plays the role that the Reynolds number plays in forced convection. A higher GrGr means stronger buoyancy-driven motion and higher hh.

(a) Grashof and Rayleigh numbers

Tf=60T_f = 60 °C =333.15= 333.15 K, so β=1/333.15=3.0×10−3 K−1\beta = 1/333.15 = 3.0\times10^{-3}\ \text{K}^{-1}.

GrL=9.81×(1/333.15)×(90−30)×(0.5)3(18.97×10−6)2=6.14×108RaL=GrL Pr=6.14×108×0.7202=4.42×108\begin{aligned} Gr_L &= \frac{9.81\times(1/333.15)\times(90-30)\times(0.5)^3}{(18.97\times10^{-6})^2} = 6.14\times10^{8}\\ Ra_L &= Gr_L\,Pr = 6.14\times10^8\times0.7202 = 4.42\times10^{8} \end{aligned}

Ra<109Ra < 10^9, so the flow is laminar and the given correlation is valid.

(b) Nusselt number and hh

Nu=0.59 Ra1/4=0.59 (4.42×108)0.25=0.59×145.0=85.5Nu = 0.59\,Ra^{1/4} = 0.59\,(4.42\times10^8)^{0.25} = 0.59\times145.0 = 85.5 hˉ=Nu kL=85.5×0.028960.5=4.95 W/m2K\bar h = \frac{Nu\,k}{L} = \frac{85.5\times0.02896}{0.5} = 4.95\ \text{W/m}^2\text{K}

(c) Heat loss

Q=hˉA ΔT=4.95×(0.5×0.8)×60=119 WQ = \bar h A\,\Delta T = 4.95\times(0.5\times0.8)\times60 = 119\ \text{W}

Answer: Gr=6.14×108Gr = 6.14\times10^8, Ra=4.42×108Ra = 4.42\times10^8; Nu=85.5Nu = 85.5; hˉ=4.95 W/m2\bar h = 4.95\ \text{W/m}^2K; Q≈119Q \approx 119 W.

✓ Answer checked by Claude Sonnet 5.5

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