Convective heat transfer Problems
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Transcript of Convective heat transfer Problems
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8/12/2019 Convective heat transfer Problems
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Example:
A heavy lubricating oil ( = 0.8 N.s/m
2
, k =0.15 W/m. K) flows in the clearance between a
shaft and its bearing. If the bearing and the
shaft are kept at 10C and 30C respectively and
the clearance between them is 2mm,determine
the maximum temperature rise and the heat
flux to the plates for a velocity U = 6 m/s.
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Sol:
q(0) at the bearing = 2
2-
(TLT0)
= -. 6
22-
.5
2 (30 -10)
= -7200 -1500 =-8.7 kW/2
And q(L) at the shaft =
2
2 -
(TLT0)= 72001500 =5.7 kW/2
the maximum temperature in the oil will occur
where
= 0
2
2
[
] +
(TL T0)
= 0
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or ymax= [
(TLT0) +
] . L
= [.5
.6(3010 ) +] . L = 0.604 L
Substituting this value of ymax in the expression
for T
Tmax = T0+2
2
[.
( .
)2]+(TLT0)(.6
)
= 10+.
2.5*36[0.604=0.365]+(30-10)(0.604)
= 10+22.944+12.8
= 45.02C
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Example :
Air at 20 is flowing along a heated flat plate
at 134at a velocity of 3 m/s. the plate is 2m
long and 1.5m wide. Calculate the thickness of
the hydrodynamic boundary layer and the skin
friction coefficient at 40 cm from the leadingedge of the plate. The kinematic viscosity of air
at 20may be taken as 15.06 x 106m2/s.
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Sol:
At = 40 ; =
= ()(.)
.
= 7.9 X 10< 5 X 105So the boundary layer is laminar, its thickness is
calculated from
= 5
= (5)(.)(7. )
= 0.71 X 102 m
= 7.1 mm
The local skin friction coefficient is given by:
=.66
=
.66
(7. )= 2.36 X 10
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Sol:
The film temperature, Tf=
:2
2 = 77The physical properties of air at 77 are
= 0.998 /,= 1.009 kJ/kg ,
= 20.76 X 106 2
= 0.03 /, = 0.697, = 0.4
R
= .
= ()(.).
= 5.78 X10
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N=.
= 0.332 R/P
/
= (0.332)(5.78 X 10)1/2(0.697)1/3= 70.6
=(7.6)(.)
.= 5.3 W/m2K
The average value of heat transfer coefficient istwice this value of
=(2)(5.3) = 10.6 W/m2K
The heat flow is
= A( -)
= (10.6)(0.4)(1.5)(134-20) =725 WThe heat flow from the both sides of the plate
=(2)(725) = 1450 W
