Question

A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at a velocity of 5.50 m/s. Determine the rate of heat transfer from the wire to the air per meter length when it is first exposed to the air. The properties of air at 1 atm and the film temperature of (Ts + T[infinity])/2 = (280 + 20)/2 = 150°C are k = 0.03443 W/m·°C, ν = 2.860 × 10−5 m2/s, and Pr = 0.70275.

Answer 1

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, $T_{s}=280^{o}C$

Temperature of air, $T_{\infinity}=20^{o}C$

Velocity of air flow $V=5.5m/s$

The film temperature is determined as:

$T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C$

from the table, properties of air at 1 atm pressure

At $T_{f}=150^{o}C$

Thermal conductivity, $K = 0.03443 W/m^oC$; kinematic viscosity $v=2.860 \times 10^{-5} m^2/s$; Prandtl number $Pr=0.70275$

The reynolds number for the flow is determined as:

$Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92$

sice the obtained reynolds number is less than $2\times10^5$, the flow is said to be laminar.

The nusselt number is determined from the relation given by:

$Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}$

$Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11$

The covective heat transfer coefficient is given by:

$Nu_{cyl}=\frac{hD}{k}$

Rewrite and solve for $h$

$h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K$

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

$Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m$

The rate of heat transfer from the wire to the air per meter length is $Q=340.42W/m$