# Cooling Time of Steel Block

by 1988ajk
Tags: block, cooling, steel, time
 P: 23 I have a steel block (2.75m x 1.6m x 0.5m) sat at 1000°C in ambient air (22°C) How long will it take for a surface to drop to say 800°C should i use the heat transfer eq: Q=KA$\Delta$T to find the power in joules per second, then transpose to find seconds somehow??? Any help will be appreciated Adam.
 P: 3 Hi Adam. Your block sitting in open air is undergoing convective heat transfer. For transient analysis of this type, you'll need to use the lumped capacitance approach. We ignore the temperature gradient inside the block and assume that the entire block is at the same temperature. Then, with an energy balance, we equate the rate of heat transfer out of the block (convection) to the rate of change in internal energy of the block. $-\dot{E}_{out} = \dot{E}_{st}$ Then: $-hA_{s}(T - T_{\infty}) = \rho Vc\dfrac{dT}{dt}$ With a bunch of exciting math, we can arrive at: $\dfrac{\rho Vc}{hA_{s}} \ln\dfrac{T - T_{\infty}}{T_{i} - T_{\infty}} = t$ Where: $\rho = \text{material density} \\ V = \text{volume} \\ c = \text{specific heat capacity} \\ h = \text{convective heat transfer coefficient} \\ A_{s} = \text{surface area exposed to the air} \\ T = \text{final temperature} \\ T_{i} = \text{initial temperature} \\ T_{\infty} = \text{temperature of the air} \\ t = \text{time taken to cool from } T_{i} \text{ to } T \text{ in seconds}\\$ At this point, consult some references to determine the specific heat capacity and the convective heat transfer coefficient, and then it's plug and chug. Watch out for units! According to my reference here: $c_{steel} = \text{440 } \dfrac{\text{J}}{\text{kg}\cdot{\text{K}}} \\ h_{air} = \text{20} \dfrac{\text{W}}{\text{m}^{2}\cdot{\text{K}}}$ Note ******************** Keep in mind this is an approximation because we are assuming the gradient in the block has no effect on the rate of cooling (although most likely good enough for your needs). A measure of the validity of the approximation can be found through calculating the Biot number. $Bi = \dfrac{hL_{c}}{k}\\ \text{Where:} \\ L_{c} = \dfrac{V}{A_{s}} = \text{characteristic length} \\ k = \text{thermal conductivity of the material} \\$ For a good approximation, Bi should be much less than 1. Hope that helps!