Consider a semi-infinite solid subjected to a constant surface temperature excess of 100^◦C for t 0. Obtain an expression for the surface heat flux as a function of time for two materials. (a) Concrete with properties as in Example 5.1 and (b) Aluminum with the following properties: density ρ = 2

Question

Consider a semi-infinite solid subjected to a constant surface temperature excess of 100°C for t > 0. Obtain an expression for the surface heat flux as a function of time for two materials. (a) Concrete with properties as in Example 5.1 and (b) Aluminum with the following properties: density [latex] ρ = 2701 kg/m^{3}[/latex] , specific heat c = 903 J/kg°C, and thermal conductivity k = 237 W/m°C. Make a plot of surface heat flux variation with time for the two materials. Compare the energy that has entered the unit area of the two materials in 1000 s.

Accepted Answer

The expression for heat flux is given by Eq. 5.30. The property values are substituted to get the surface heat flux in the two cases. Surface temperature excess is interpreted as [latex]T_{s}-T_{i}[/latex] . We have

[latex] q(t)\mid _{x=0} = -k\frac{dT}{dx} \mid _{x=0} =-kAt^{B} \frac{dT}{d\eta }\mid _{\eta =0}=-\frac{k}{s\sqrt{\alpha t} }\left[-\frac{2}{\sqrt{\pi } }T_{s} ight] = \frac{kT_{s}}{\sqrt{\pi \alpha t} }[/latex] (5.30)

[latex] q_{s} =\frac{k(T_{s}-T_{i})}{\sqrt{\pi \alpha t} }=(T_{s}-T_{i})\sqrt{\frac{ ho ck}{\pi t} }=\frac{k}{\sqrt{t} }[/latex]

where K is a constant given by [latex](T_{s}-T_{i})\sqrt{\frac{ ho ck}{\pi } }[/latex]

The property values for concrete are taken from Example 5.1. The constant [latex]K = K_{c}[/latex] for concrete is obtained as

[latex]k_{c} =100 imes \sqrt{\frac{2150 imes 950 imes 1.06}{\pi } } =83015.5Ws^{\frac{1}{2} } /m^{2}[/latex]

More suitable way of expressing the above will be as [latex]K = 83.02 kW s^{\frac{1}{2} } /m^{2}[/latex]. The property values for aluminum are given above in the Example statement. The constant [latex]K = K_{a}[/latex] for aluminum is obtained as

[latex]k_{a} =100 imes \sqrt{\frac{2702 imes 903 imes 237}{\pi } } = 1356706 Ws^{\frac{1}{2} } /m^{2}[/latex]

More suitableway of expressing the above will be as [latex]K = 1357 kW s^{\frac{1}{2} } /m^{2}[/latex]. Figure 5.7 shows the plot of surface heat flux history for the two materials. It is seen that the surface heat flux is much larger in the case of aluminum than concrete. Thermal properties of the material play a very important role in the transient response. Both the density and heat capacity are of comparable size for the two materials. However, thermal conductivity of aluminum is more than 200 times that of concrete and the surface heat flux mirrors this! It is interesting to compare the total energy that has entered the two materials in 1000 s for which the plot has been made. We make use of expression (5.31) for this purpose. The thermal diffusivity of concrete has been calculated in Example 5.1 as [latex]α_{c} = 5.19 × 10^{-7}m^{2}/s[/latex].

Heat that has entered concrete in 1000 s is calculated using Eq. 5.31 as

[latex]q_{s}=\int_{0}^{t}{q\mid _{x=0} } dt=\frac{kT_{s} }{\sqrt{ \pi \alpha }} \int_{0}^{t}{\frac{dt}{\sqrt{t} } } =2kT_{s}\sqrt{\frac{t}{\pi \alpha } } (5.31)[/latex] (5.31)

[latex]q_{s}(concrete)=2 imes 1.06 imes 100\sqrt{\frac{1000}{\pi imes 5.19 imes 10^{-7} } } =5.25 imes 10^{6} J/m^{2}[/latex]

[latex]=5.25MJ/m^{2}[/latex]

From the given data, the thermal diffusivity of aluminum is

[latex]\alpha _{a} =\frac{237}{2702 imes 903} =9.71 imes 10^{-5} m^{2}/s[/latex]

Hence the heat that has entered aluminum in 1000 s is calculated as

[latex]q_{s} (aluminum)=2 imes 237 imes 100\sqrt{\frac{1000}{\pi imes 9.71 imes 10^{-5} } } =85.82 imes10^{6}[/latex]

[latex]=85.52MJ/m^{2 }[/latex]

Question 5.2: Consider a semi-infinite solid subjected to a constant surfa...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions