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?
Expand 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
Time Analysis
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