Estimating Cooling Time of Steel Block at High Temperatures

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SUMMARY

The discussion focuses on estimating the cooling time of a steel block measuring 2.75m x 1.6m x 0.5m, initially at 1000°C, using heat transfer equations. The lumped capacitance approach is recommended for transient analysis, leading to the equation: t = (ρVc/hAs) ln((T - T∞)/(Ti - T∞)). Key parameters include specific heat capacity of steel (440 J/kg·K) and convective heat transfer coefficient for air (20 W/m²·K). The influence of radiation heat transfer at high temperatures is also emphasized, suggesting the need for numerical integration methods for accurate modeling.

PREREQUISITES
  • Understanding of heat transfer principles, specifically convective and radiative heat transfer.
  • Familiarity with the lumped capacitance method in transient heat transfer analysis.
  • Knowledge of thermodynamic properties such as specific heat capacity and thermal conductivity.
  • Experience with numerical methods for solving differential equations, particularly Newton-Raphson method.
NEXT STEPS
  • Research the Biot number and its significance in heat transfer analysis.
  • Learn about finite element (FE) and finite difference (FD) modeling techniques for heat transfer problems.
  • Explore numerical integration methods for solving nonlinear equations in thermal analysis.
  • Investigate the effects of radiation on heat transfer at high temperatures and how to incorporate it into models.
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Engineers, thermal analysts, and researchers involved in materials processing, particularly those dealing with high-temperature applications and heat transfer modeling.

1988ajk
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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\DeltaT to find the power in joules per second, then transpose to find seconds somehow?

Any help will be appreciated

Adam.
 
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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:

<br /> <br /> \rho = \text{material density} \\<br /> V = \text{volume} \\<br /> c = \text{specific heat capacity} \\<br /> h = \text{convective heat transfer coefficient} \\<br /> A_{s} = \text{surface area exposed to the air} \\<br /> T = \text{final temperature} \\<br /> T_{i} = \text{initial temperature} \\<br /> T_{\infty} = \text{temperature of the air} \\<br /> t = \text{time taken to cool from } T_{i} \text{ to } T \text{ in seconds}\\<br /> <br />

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:

<br /> <br /> c_{steel} = \text{440 } \dfrac{\text{J}}{\text{kg}\cdot{\text{K}}} \\<br /> h_{air} = \text{20} \dfrac{\text{W}}{\text{m}^{2}\cdot{\text{K}}} <br /> <br />


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.

<br /> <br /> Bi = \dfrac{hL_{c}}{k}\\<br /> <br /> \text{Where:} \\<br /> <br /> L_{c} = \dfrac{V}{A_{s}} = \text{characteristic length} \\<br /> k = \text{thermal conductivity of the material} \\<br /> <br />

For a good approximation, Bi should be much less than 1.

Hope that helps!
 
You have a bit more of a problem here. The block is at 1000 C. Therefore radiation is the dominant heat transfer mode. Natural convection is dwarfed by radiation at these temperatures. My advice is to create a FE for FD model of it and include the radiation boundary condition along with the natural comvection condition.

If you cannot do this you can approximate the effect of radiation and lump it into the convection coefficient. It's crude but better than ignoring the major heat loss.
 
If you got access to the ford automobile factory in detroit, you might be able to measure this in real life.

Last time I was there, I remember red hot ingots emerging from where they were made and them sitting (still red) before they were moved to where they would be worked.
 
Assuming the block to be lumped, you can equate convection loss + radiation loss to heat term at steady state.
 
Due to its nonlinearity you'll have to perform a numerical integration in order to solve it. Newton Raphson embedded in a fully implicit scheme would work well.
 

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