Respuesta :
Answer:
[tex]Nu = 30.311[/tex]
Explanation:
Let consider that pipe is a horizontal cylinder. The Nusselt number is equal to:
[tex]Nu = \left\{0.6+\frac{0.387\cdot Ra_{D}^{\frac{1}{6} }}{[1+\left(\frac{0.559}{Pr} \right)^{\frac{9}{16}} ]^{\frac{8}{27} }} \right\}^{2}[/tex], for [tex]Ra_{D} \le 10^{12}[/tex].
Where [tex]Ra_{D}[/tex] is the Rayleigh number associated with the cylinder.
The Rayleigh number is:
[tex]Ra_{D} = \frac{g\cdot \beta\cdot (T_{pipe}-T_{air})\cdot D^{3}}{\nu^{2}}\cdot Pr[/tex]
By assuming that air behaves ideally, the coefficient of volume expansion is:
[tex]\beta = \frac{1}{T}[/tex]
[tex]\beta = \frac{1}{283.15\,K}[/tex]
[tex]\beta = 3.532\times 10^{-3}\,\frac{1}{K}[/tex]
The cinematic and dynamic viscosities, thermal conductivity and isobaric specific heat of air at 10 °C and 1 atm are:
[tex]\nu = 1.426\times 10^{-5}\,\frac{m^{2}}{s}[/tex]
[tex]\mu = 1.778\times 10^{-5}\,\frac{kg}{m\cdot s}[/tex]
[tex]k = 0.02439\,\frac{W}{m\cdot ^{\textdegree}C}[/tex]
[tex]c_{p} = 1006\,\frac{J}{kg\cdot ^{\textdegree}C}[/tex]
The Prandtl number is:
[tex]Pr = \frac{\mu\cdot c_{p}}{k}[/tex]
[tex]Pr = \frac{(1.778\times 10^{-5}\,\frac{kg}{m\cdot s} )\cdot (1006\,\frac{J}{kg\cdot ^{\textdegree}C} )}{0.02439\,\frac{W}{m\cdot ^{\textdegree}C} }[/tex]
[tex]Pr = 0.733[/tex]
Likewise, the Rayleigh number is:
[tex]Ra_{D} = \frac{(9.807\,\frac{m}{s^{2}} )\cdot (3.532\times 10^{-3}\,\frac{1}{K} )\cdot (110^{\textdegree}C-10^{\textdegree}C)\cdot (0.1\,m)^{3}}{(1.426\times 10^{-5}\,\frac{m^{2}}{s})^{2} }\cdot (0.733)[/tex]
[tex]Ra_{D} = 12.486\times 10^{6}[/tex]
Finally, the Nusselt number is:
[tex]Nu = \left\{0.6+\frac{0.387\cdot (12.486\times 10^{6})^{\frac{1}{6} }}{\left[1 + \left(\frac{0.559}{0.733}\right)^{\frac{9}{16} }\right]^{\frac{8}{27} }} \right\}^{2}[/tex]
[tex]Nu = 30.311[/tex]
Answer:
The Nusselt number = 124
Explanation:
Our assumption is that air is an ideal gas and that the radiation effect is negligible
Surface temperature, [tex]T_{s} = 110^{0} C[/tex]
[tex]T_{\infty} = 10^{0} C[/tex]
Velocity, v = 8 m/s
The film temperature can be calculated as, [tex]T_{f} = \frac{T_{s} + T_{\infty} }{2}[/tex]
[tex]T_{f} = \frac{110 +10 }{2} \\T_{f} = 60^{0} C[/tex]
At the film temperature, [tex]T_{f} = 60^{0} C[/tex] and 1 atm pressure, air has the following properties:
[tex]K = 0.02808 W/m-k[/tex]
[tex]P_{r} = 0.7202[/tex]
Reynold number, [tex]Re = \frac{vD}{V}[/tex]
D = 10 cm = 0.1 m
V = 1.896 * 10⁻⁵ m²/s
[tex]Re = \frac{0.1 * 8}{1.896 * 10^{-5} }[/tex]
Re = 4.2194 * 10⁴
The Nusselt number will be calculated using the relation:
[tex]Nu = 0.3 + \frac{0.62 Re^{1/2} Pr^{1/3} }{[1 +( 0.41 Pr)^{2/3}] ^{1/4} } + [1 + (\frac{Re}{282000} )^{5/8} ]^{4/5}[/tex]
Substituting Re = 4.2194 * 10⁴ and [tex]P_{r} = 0.7202[/tex] into the equation above
the Nusselt number, Nu = 124
