## Lumped Capacitance Model

Consider a 3 m long metal cube has a convection heat transfer coefficient of 100 W/m^2∙K, and a thermal conductivity of 1,000 W/m∙K. Is the lumped capacitance model valid for this part?

Hint
$$Bi=\frac{hV}{kA_s}$$$where $$Bi$$ is the Biot Number, $$h$$ is the convection heat transfer coefficient, $$V$$ is the volume, $$k$$ is the thermal conductivity, and $$A_s$$ is the surface area. Hint 2 The lumped capacitance model is valid if $$Bi< 0.1$$ . The lumped capacitance model is valid if $$Biot\:Number, Bi=\frac{hV}{kA_s}< 0.1$$$
where $$h$$ is the convection heat transfer coefficient, $$V$$ is the volume, $$k$$ is the thermal conductivity, and $$A_s$$ is the surface area.
$$Bi=\frac{100W(3m)^3(m\cdot K)}{(m^2\cdot K)(1,000W)(6\cdot (3m)^2)}=\frac{2,700}{54,000}=0.05$$\$
Since $$Bi<0.1$$ , the Lumped Capacitance Model is valid .
Yes