Question

Lubricating oil at a temperature of ${60}^{o}C$ enter 1 cm diameter tube with velocity of 3m/s. the tube surface is maintained at ${40}^{o}C$ and the properties of oil are
$\rho =865Kg/m^3$, $\mu =0.0216N-s/m^2,$
$K=0.140W/m^{o}C,C_P=1780J/K{g}^{o}C$

The heat transfer coefficient for the lubricating oil is

• A. $7.32W/m^{2o}C$
• B. $3.66W/m^{2o}C$
• C. $51.24W/m^{2o}C$
• D. $102.5W/m^{2o}C$

Answer 1

The correct option is C $51.24W/m^{2o}C$


d = 1 cm = 0.01
V = 3 m/s
$T_s={40}^{o}C$
$T_{\infty }={60}^{o}C$
$\rho =865kg/m^3$
$\mu =0.0216Pa-s$
$k=0.140W/m-K$
$c_p=1780J/kgK$

Reynolds number (Re):

$\frac{\rho .v.d}{\mu }=\frac{865\times 3\times 0.01}{0.0216}$

$=1201.388<2000$

Re<2000, then flow is laminar for laminar flow

Nu = 3.66 $\to$ for constant wall temperature condition

$\frac{h.d}{k}=3.66$

$h=3.66\times \frac{k}{d}=\frac{3.66\times 0.14}{0.01}$

$h=51.24W/m^2-K$