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Question

Consider a thin 16-cm-long and 20-cm-wide horizontal plate suspended in air at 20C.20^\circ C. The plate is equipped with electric resistance heating elements with a rating of 20 W. Now the heater is turned on and the plate temperature rises. Determine the temperature of the plate when steady operating conditions are reached. The plate has an emissivity of 0.90 and the surrounding surfaces are at 17C.17^\circ C. As an initial guess, assume a surface temperature of 50C.50^\circ C. Is this a good assumption?

Solution

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Answered 2 years ago
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We are given following data:

L=0.16 mL=0.16\text{ m}

t=0.20 mt=0.20\text{ m}

Q˙=20 W\dot Q=20\text{ W}

TC=20 C=293 KT_{C}=20\text{ C}=293\text{ K}

Tsur=17 C=290 KT_{sur}=17\text{ C}=290\text{ K}

TSC=50 C=323 KT_{SC}=50\text{ C}=323\text{ K}

ϵ=0.9\epsilon=0.9

Film temperature is equal to:

Tf=TSC+TC2=50+202=35 C=308 KT_f=\dfrac{T_{SC}+T_C}{2}=\dfrac{50+20}{2}=35\text{ C}=308\text{ K}

From properties of air table A-15 we can find values we need:

β=1308 K1=0.00325 K1ρ=1.145 kg m3k=0.02625 W m KPr=0.7268μ=1.895105 Pa s\begin{align*} \beta&=\dfrac{1}{308}\text{ K}^{-1}=0.00325\text{ K}^{-1}\\ \rho&=1.145\frac{\text{ kg}}{\text{ m}^3}\\ k&=0.02625\frac{\text{ W}}{\text{ m K}}\\ P_r&=0.7268\\ \mu&=1.895\cdot 10^{-5}\text{ Pa s}\\ \end{align*}

Before we can calculate Grashof number, we have to determine characteristic length:

Lc=Lt2(L+t)=0.160.22(˙0.2+0.16)=0.044 mL_c=\dfrac{L\cdot t}{2(L+t)}=\dfrac{0.16\cdot 0.2}{2\dot (0.2+0.16)}=0.044\text{ m}

Calculating Grashof number:

Gr=gβ(TSCTC)Lc3(μρ)2=9.810.00325(5020)0.0443(1.8951051.145)2=3105\begin{align*} G_r&=\dfrac{g\cdot \beta\cdot (T_{SC}-T_C)\cdot L_c^3}{\left(\dfrac{\mu}{\rho}\right)^2}=\dfrac{9.81\cdot 0.00325\cdot (50-20)\cdot 0.044^3}{\left(\dfrac{1.895\cdot 10^{-5}}{1.145}\right)^2}\\\\ &=3\cdot 10^{5} \end{align*}

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