Question

A small metal bead (radius $0.5mm$), initially ${100}^{\circ }C$, When placed in a stream of fluid at ${20}^{\circ }C$ attains a temperature of ${28}^{\circ }C$ in$4.35second.$ The density and specific heat of the metal are $8500kg/m^3$ and $400J/kgK,$ respectively. If the bead is considered as lumped system, the convective heat transfer coefficient (in $W/m^{2}K$) between the metal bead and the fluid stream

• A.
  $149.9$
• B.
  $449.7$
• C.
  $283.3$
• D.
  $299.8$

Answer 1

The correct option is D

$299.8$
Given data:
$r=0.5mm:c_p=400J/kgK:\rho =8500kg/m^3;t=4.35sec{T}_{\circ }={100}^{\circ }C,T_{\infty }={20}^{\circ }CT={28}^{\circ }C,\tau =4.35seconds$

$\because \frac{V}{A}=\frac{\frac{4}{3}\pi R^3}{4\pi R^2}=\frac{R}{3}$

$\because \frac{\theta }{{\theta }_{\circ }}=\frac{T-T_{\infty }}{T_i-T_{\infty }}=e^{\frac{-hA\tau }{\rho C_{p}V}}\Rightarrow \frac{28-20}{100-20}=e^{-\left(\frac{h.3000}{0.5\times 8500\times 400}\right)\times 4.35}$

$\Rightarrow h=299.95W/m^{2}K$