Question

If diameter and length of a cylinder are 0.01 m and 0.07 m respectively, the thermal conductivity of air is 0.028 W/mK and the Nusselt number is 14.2, what will be the convective heat transfer coefficient

Answer 1

Answer:

The convective heat transfer coefficient of the fluid is 170.4 watts per square meter-degree Celsius.

Explanation:

The Nusselt number ($Nu$) is a dimensionless factor which compares the sensitivity of a fluid due to convection with those due to conduction:

$Nu = \frac{h\cdot L_{c}}{k}$ (Eq. 1)

Where:

$h$ - Convective heat transfer coefficient, measured in watts per square meter-degree Celsius.

$k$ - Conductive heat transfer coefficient, measured in watts per meter-degree Celsius.

$L_{c}$ - Characteristic length, measured in meters.

In addition, the characteristic length of a cylinder is defined by the following formula:

$L_{c} = \frac{\pi\cdot r^{3}\cdot l}{2\pi\cdot r^{2}+2\pi\cdot r \cdot l}$ (Eq. 2)

Where:

$r$ - Radius of the cylinder, measured in meters.

$l$ - Length of the cylinder, measured in meters.

If we know that $Nu = 14.2$, $k = 0.028\,\frac{W}{m\cdot ^{\circ}C}$, $r = 0.005\,m$ and $l = 0.07\,m$, then the convective heat coefficient is:

From (Eq. 2):

$L_{c} = \frac{\pi\cdot (0.005\,m)^{2}\cdot (0.07\,m)}{2\pi\cdot (0.005\,m)^{2}+2\pi\cdot (0.005\,m)\cdot (0.07\,m)}$

$L_{c} = \frac{7}{3000}\,m$

And by (Eq. 1):

$h = \frac{k\cdot Nu}{L_{c}}$

$h = \frac{\left(0.028\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (14.2)}{\frac{7}{3000}\,m }$

$h = 170.4\,\frac{W}{m^{2}\cdot ^{\circ}C}$

The convective heat transfer coefficient of the fluid is 170.4 watts per square meter-degree Celsius.