Large brass plate of 40 mm thickness having temperature of 20- C- - 8530 kg-m3- Cp - 380 K-kgK and K - 110 W-m K are heated in a furnace- maintained at 500- C- Taking the combined convection and radiation coefficient as 120 W-m2 K- the surface temperature of plate after 7 min with a Fourier number of 35-6 is

T_i= 20^∘ C, T_∞ = 500^∘ C, F_e = 35.6, h = 120W/m^2 k, k = 110W/m characteristic length (L) = 12(for plate) = 0.042 = 0.02 m Biot number = hLk = 120× 0.0 ...

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Lumped Heat Analysis

Large brass p...

Question

Large brass plate of 40 mm thickness having temperature of $20^{\circ} C, ρ = 8530 k g / m^{3}, C p = 380 K / k g K$ and $K = 110 W / m K$ are heated in a furnace, maintained at $500^{\circ} C .$ Taking the combined convection and radiation coefficient as $120 W / m^{2} K,$ the surface temperature of plate after 7 min with a Fourier number of 35.6 is

280^{\circ} C

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500^{\circ} C

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420^{\circ} C

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210^{\circ} C

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Solution

The correct option is A $280^{\circ} C$
$T_{i} = 20^{\circ} C, T_{\infty} = 500^{\circ} C, F_{e} = 35.6, h = 120 W / m^{2} - k, k = 110 W / m characteristic length (L) = \frac{1}{2} (f o r p l a t e) = \frac{0.04}{2} = 0.02 m Biot number = \frac{h L}{k} = \frac{120 \times 0.02}{110} = 0.0281 < 0.1 B i < 0.1; lumped capacity analysis can be applied. According to lumped capacity analysis \frac{T_{i} - T_{\infty}}{T - T_{\infty}} = e^{(B i \times F_{e})} \frac{- 480}{T - 500} = e^{0.0281 \times 35.6} \frac{- 480}{T - 500} = 2.1743 T - 500 = - 220.756 T = 279.24 T \approx 280^{\circ} C$

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