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Homework Statement
A short metal cylinder 145 mm in diameter and 145 mm high at 1045 K
is suddenly exposed (all sides exposed) to a room air temperature at
300 K with h=25 W/m².K. Assume that for the metal k=40W/m.K, den=7800 kg/m3
and Cp=c=600 J/kg.K. Estimate (a) the time required for the cylinder to
cool down to 600K at center and heat transferred.
Homework Equations
$$V = \frac {πD^2L} {4}$$
$$A_s = \frac {πD^2} {2} + πDL$$
$$Bi = \frac {hV}{kA}$$
$$\frac {T_o - T_∞}{T_i - T_∞} = e^{ \frac {htA_s} {ρVc}} $$
$$Q_conduction = -kA\frac{dT}{dr}$$
$$Q_convection = hA_exposed (T_s - T_∞)$$
The Attempt at a Solution
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$$ t = \frac {ρVc}{hA_s} [ln {\frac {T_o - T_∞}{T_i - T_∞}}]$$
t ≈ 4115s
Where T∞ = 300 K, To = 600 K, Ti = 1045 K
My main issue is determing the heat loss transferred. From my understanding, a short cylinder means that we can't assume the top and bottom areas are negligible in determining heat loss so they have to be factored in. But I can't figure out equation setup for total heat loss. I'm almost positive that heat loss is by convection AND conduction and when we combine the two, we multiply it by the time to determine the heat loss in Joules.