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

Step-by-step 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)=(T_(s)-T_(\infinity))/(2)\\\\=(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 * 10^(-5) m^2/s`; Prandtl number
`Pr=0.70275`

The reynolds number for the flow is determined as:

`Re=(VD)/(v)\\\\=(5.5 *(3*10^(-3)))/(2.86*10^(-5))\\\\=576.92`

sice the obtained reynolds number is less than
`2*10^5`, 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^{(1)/(3)}}{[1+((0.4)/(Pr))^{(2)/(3)}]^{(1)/(4)}}[1+((Re)/(282000))^{(5)/(8)}]^{(4)/(5)}`

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

The covective heat transfer coefficient is given by:

`Nu_(cyl)=(hD)/(k)`

Rewrite and solve for
`h`

`h=(Nu_(cyl)*k)/(D)\\\\=(12.11*0.03443)/(3*10^(-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*(\pi*DL)*(T_(s)-T{\infinity})\\\\=138.92*(\pi*3*10^(-3)*1)*(280-20)\\\\=340.42W/m`

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