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

`Nu_(D) = 49.047`,
`h = 24.621\,(W)/(m^(2)\cdot ^(\textdegree)C)`,
`\dot Q \approx 381.091\,W`

Step-by-step 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\,(kJ)/(kg\cdot ^(\textdegree)C)`

`k = 0.0251* 10^(-3)\,(kW)/(m\cdot ^(\textdegree)C)`

`\rho = 0.5978\,(kg)/(m^(3))`

`\mu = 1.227* 10^(-5)\,(kg)/(m\cdot s)`

The Reynolds and Prandtl numbers are, respectively:

`Re = (\rho\cdot v\cdot D)/(\mu)`

`Re = ((0.5978\,(kg)/(m^(3)) )\cdot (6\,(m)/(s) )\cdot (0.05\,m))/(1.227* 10^(-5)\,(kg)/(m\cdot s) )`

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

`Pr = (c_(p)\cdot \mu)/(k)`

`Pr = ((2.010\,(kJ)/(kg\cdot ^(\textdegree)C) )\cdot (1.227* 10^(-5)\,(kg)/(m\cdot s) ))/(0.0251* 10^(-3)\,(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) = (h\cdot D)/(k)`

The heat transfer coefficient is:

`h = (Nu_(D)\cdot k)/(D)`

`h = ((49.047)\cdot (0.0251* 10^(-3)\,(kW)/(m\cdot ^(\textdegree)C)))/(0.05\,m )`

`h = 24.621\,(W)/(m^(2)\cdot ^(\textdegree)C)`

The convection is the dominant heat transfer mechanism, then:

`\dot Q = (24.621\,(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`