Question

Steam at 1 atm and 100C is flowing across a 5-cm-OD tube at a velocity of 6 m/s. Estimatethe Nusselt number, the heat transfer coefficient, and the rateof heat transfer per meterlength of pipe if the pipe is at 200C.

Answer 1

Answer:

$Nu_{D} = 49.047$, $h = 24.621\,\frac{W}{m^{2}\cdot ^{\textdegree}C}$, $\dot Q \approx 381.091\,W$

Explanation:

Let assume that steam has a fully developed and turbulent flow and that tube is smooth and thin-walled. The steam is heated while flowing through the tube. The Nusselt number is given by the Dittus-Boelter equation:

$Nu_{D} = 0.023\cdot Re^{0.8}\cdot Pr^{0.4}$

Properties of steam at given pressure and temperature are:

$c_{p} = 2.010\,\frac{kJ}{kg\cdot ^{\textdegree}C}$

$k = 0.0251\times 10^{-3}\,\frac{kW}{m\cdot ^{\textdegree}C}$

$\rho = 0.5978\,\frac{kg}{m^{3}}$

$\mu = 1.227\times 10^{-5}\,\frac{kg}{m\cdot s}$

The Reynolds and Prandtl numbers are, respectively:

$Re = \frac{\rho\cdot v\cdot D}{\mu}$

$Re = \frac{(0.5978\,\frac{kg}{m^{3}} )\cdot (6\,\frac{m}{s} )\cdot (0.05\,m)}{1.227\times 10^{-5}\,\frac{kg}{m\cdot s} }$

$Re = 14616.136$ (which demonstrates the reasonability of the supposition of turbulent flow)

$Pr = \frac{c_{p}\cdot \mu}{k}$

$Pr = \frac{(2.010\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (1.227\times 10^{-5}\,\frac{kg}{m\cdot s} )}{0.0251\times 10^{-3}\,\frac{kW}{m\cdot ^{\textdegree}C} }$

$Pr = 0.983$

The Nusselt number is:

$Nu_{D} = 0.023\cdot (14616.136)^{0.8}\cdot (0.983)^{0.4}$

$Nu_{D} = 49.047$

The Nusselt number has the following definition:

$Nu_{D} = \frac{h\cdot D}{k}$

The heat transfer coefficient is:

$h = \frac{Nu_{D}\cdot k}{D}$

$h = \frac{(49.047)\cdot (0.0251\times 10^{-3}\,\frac{kW}{m\cdot ^{\textdegree}C})}{0.05\,m }$

$h = 24.621\,\frac{W}{m^{2}\cdot ^{\textdegree}C}$

The convection is the dominant heat transfer mechanism, then:

$\dot Q = (24.621\,\frac{W}{m^{2}\cdot ^{\textdegree}C} )\cdot [\pi\cdot (0.05\,m)\cdot (1\,m) ]\cdot (200^{\textdegree}C-100^{\textdegree}C)$

$\dot Q \approx 381.091\,W$