Engineering, 29.10.2019 05:31 livirose1514

Ahollow aluminum sphere, with an electrical heater in the center, is used in tests to determine the thermal conductivity of insulating materials. the inner and outer radii of the sphere are 0.18 and 0.21 m, respectively, and testing is done under steady-state conditions with the inner surface of the aluminum maintained at 250°c. in a particular test, a spherical shell of insulation is cast on the outer surface of the sphere to a thickness of 0.15 m. the system is in a room for which the air temperature is 20°c and the convection coefficient at the outer surface of the insulation is 30 w/m2 ⋅ k. if 80 w are dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?

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Answer from: riiba002

$K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K$

Explanation:

Generally, thermal resistance for conduction heat transfer in a sphere.

$R_{cond} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi K}}$

Where $R_{cond}$ is the thermal resistance for conduction, K is the thermal conductivity of the material, $r_{i}$ is the inner radius of the sphere, and $r_{o}$ is the outer radius of the sphere.

The surface area of sphere, $A_{s}$ is given by

$A_{s}=4\pi {r^2}$

For aluminum sphere, the thermal resistance for conductive heat transfer is given by

Calculate the thermal resistance for conductive heat transfer through the aluminum sphere.

$R_{cond,s{\rm{ - 1}}} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}}$

Where $K_{Al}$ is aluminum’s thermal conductivity at $T_{s$ }

Thermal resistance for conductive heat transfer through the insulation.

$R_{cond,1{\rm{ - 2}}} = \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}}$

Thermal resistance for convection is given by

$R_{conv} = \frac{1}{{hA}}$

Where h is convective heat transfer coefficient, $R_{conv}$ is thermal resistance for convection and A is the cross-sectional area normal to the direction of flow of heat energy

Thermal resistance for convective heat transfer in-between the outer surface of the insulation and the ambient air.

$R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h{A_s}}}$

Where h represents convective heat transfer coefficient at the outer surface of the insulation. Since $A_{s}$ is already defined, substituting it into the above formula yields

$R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h\left( {4\pi {r^2}} \right)}}$

To obtain radial distance of the outer surface of the insulation from the center of the sphere.

$r = r_{o} + t$ where t is thickness of insulation

r=0.21+0.15=0.36m

Total thermal resistance

$R_{eq} = {R_{cond,s{\rm{ - 1}}}} + {R_{cond,1{\rm{ - 2}}}} +{R_{conv,2{\rm{ - }}\infty }}$

Where $R_{eq}$ is total thermal resistance

$R_{eq} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}$

Consider the thermal conductivity of aluminum at temperature $T_{s}$ as 234W/m.K

Rate of heat transfer for the given process

$\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{{R_{eq}}}}$

Where $\dot Q_{s - \infty }$ } is the steady state heat transfer rate in-between the inner surface of the sphere and the ambient air.

Substituting $\left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right)$ for $R_{eq}$ we obtain

$\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{\left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right)}}$

$\begin{array}{l}\\80{\rm{ W}} = \frac{{250{\rm{ }}^\circ {\rm{C}} - 20{\rm{ }}^\circ {\rm{C}}}}{{\left( {\frac{{\left( {\frac{1}{{0.18{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.21{\rm{ m}}}}} \right)}}{{4\pi \left( {234{\rm{ W/m}} \cdot {\rm{K}}} \right)}} + \frac{1}{{30{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}\left( {4\pi {{\left( {0.36{\rm{ m}}} \right)}^2}} \right)}}\frac{{\left( {\frac{1}{{0.21{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.36{\rm{ m}}}}} \right)}}{{4\pi {K_{ins}}}} + } \right)}}\\\\80{\rm{ W}}\left( {{\rm{0}}{\rm{0.020737 K/W}} + \frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}}} \right) = 230{\rm{ K}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = \frac{{230{\rm{ K}}}}{{80{\rm{ W}}}} - {\rm{0}}{\rm{0.020737 K/W}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = {\rm{2}}{\rm{.854263 K/W}}\\\end{array}$

$K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K$

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