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

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

Step-by-step 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 = (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) = (\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\,(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) = (\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) = (7)/(3000)\,m`

And by (Eq. 1):

`h = (k\cdot Nu)/(L_(c))`

`h = (\left(0.028\,(W)/(m\cdot ^(\circ)C) \right)\cdot (14.2))/((7)/(3000)\,m )`

`h = 170.4\,(W)/(m^(2)\cdot ^(\circ)C)`

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