a long aluminum wire of diameter 3 mm is extruded at a temperature of 370°c. the wire is subjected to cross air flow at 30°c at a velocity of 6 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.
answer:
Explanation:
Given:
Diameter of aluminum wire, D = 3mm
Temperature of aluminum wire, T_{s}=280^{o}CT
s
=280
o
C
Temperature of air, T_{infinity}=20^{o}CT
infinity
=20
o
C
Velocity of air flow V=5.5m/sV=5.5m/s
The film temperature is determined as:
begin{gathered}T_{f}=frac{T_{s}-T_{infinity}}{2}\\=frac{280-20}{2}\\=150^{o}Cend{gathered}
T
f
=
2
T
s
−T
infinity
=
2
280−20
=150
o
C
from the table, properties of air at 1 atm pressure
At T_{f}=150^{o}CT
f
=150
o
C
Thermal conductivity, K = 0.03443 W/m^oCK=0.03443W/m
o
C ; kinematic viscosity v=2.860 times 10^{-5} m^2/sv=2.860×10
−5
m
2
/s ; Prandtl number Pr=0.70275Pr=0.70275
The reynolds number for the flow is determined as:
begin{gathered}Re=frac{VD}{v}\\=frac{5.5 times(3times10^{-3})}{2.86times10^{-5}}\\=576.92end{gathered}
Re=
v
VD
=
2.86×10
−5
5.5×(3×10
−3
)
=576.92
sice the obtained reynolds number is less than 2times10^52×10
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^{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+
[1+(
Pr
0.4
)
3
2
]
4
1
0.62Re
0.5
Pr
3
1
[1+(
282000
Re
)
8
5
]
5
4
begin{gathered}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}}\\=12.11end{gathered}
Nu
cyl
=0.3+
[1+(
(0.70275)
0.4
)
3
2
]
4
1
0.62(576.92)
0.5
(0.70275)
3
1
[1+(
282000
576.92
)
8
5
]
5
4
=12.11
The covective heat transfer coefficient is given by:
Nu_{cyl}=frac{hD}{k}Nu
cyl
=
k
hD
Rewrite and solve for hh
begin{gathered}h=frac{Nu_{cyl}timesk}{D}\\=frac{12.11times0.03443}{3times10^{-3}}\\=138.98 W/m^{2}.Kend{gathered}
h=
D
Nu
cyl
timesk
=
3×10
−3
12.11×0.03443
=138.98W/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:
begin{gathered}Q=hA_{s}(T_{s}-T{infin})\\=htimes(pitimesDL)times(T_{s}-T{infinity})\\=138.92times(pitimes3times10^{-3}times1)times(280-20)\\=340.42W/mend{gathered}
Q=hA
s
(T
s
−T∞)
=h×(πtimesDL)×(T
s
−Tinfinity)
=138.92×(π×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/mQ=340.42W/m
Penjelasan:
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