Answer:
Explanation:
Given:
Diameter of aluminum wire, D = 3mm
Temperature of aluminum wire, [tex]T_{s}=280^{o}C[/tex]
Temperature of air, [tex]T_{\infinity}=20^{o}C[/tex]
Velocity of air flow [tex]V=5.5m/s[/tex]
The film temperature is determined as:
[tex]T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C[/tex]
from the table, properties of air at 1 atm pressure
At [tex]T_{f}=150^{o}C[/tex]
Thermal conductivity, [tex]K = 0.03443 W/m^oC[/tex]; kinematic viscosity [tex]v=2.860 \times 10^{-5} m^2/s[/tex]; Prandtl number [tex]Pr=0.70275[/tex]
The reynolds number for the flow is determined as:
[tex]Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92[/tex]
sice the obtained reynolds number is less than [tex]2\times10^5[/tex], the flow is said to be laminar.
The nusselt number is determined from the relation given by:
[tex]Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}[/tex]
[tex]Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11[/tex]
The covective heat transfer coefficient is given by:
[tex]Nu_{cyl}=\frac{hD}{k}[/tex]
Rewrite and solve for [tex]h[/tex]
[tex]h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K[/tex]
The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:
[tex]Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m[/tex]
The rate of heat transfer from the wire to the air per meter length is [tex]Q=340.42W/m[/tex]
