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27.9 A Ugroove weld is used to butt weld 2 pieces of 7.0mmthick titanium plate. The Ugroove is
prepared using a milling cutter so the radius of the groove is 3.0 mm. During welding, the penetration of
the weld causes an additional 1.5 mm of material to be melted. The final crosssectional area of the weld
can be approximated by a semicircle with a radius of 4.5 mm. The length of the weld is 200 mm. The
melting factor of the setup is 0.57 and the heat transfer factor is 0.86. (a) What is the quantity of heat (in
Joules) required to melt the volume of metal in this weld (filler metal plus base metal)? Assume the
resulting top surface of the weld bead is flush with the top surface of the plates. (b) What is the required
heat generated at the welding source?
Solution:
(a) From Table 27.2,
T
m
for titanium is 2070°K
U
m
= 3.33 x 10
6
(2070)
2
= 14.29 J/mm
3
A
w
=
π
r
2
/2 =
π
(4.5)
2
/2 = 31.8 mm
2
V
=
A
w
L
= 31.8(200) = 6360 mm
3
H
w
=
U
m
V
= 14.29(6360) =
90,770 J
(b)
H
=
H
w
/(
f
1
f
2
) = 90,770/(0.86 x 0.57) =
185,200 J
27.13 Compute the unit melting energy for (a) aluminum and (b) steel as the sum of: (1) the heat
required to raise the temperature of the metal from room temperature to its melting point, which is the
volumetric specific heat multiplied by the temperature rise; and (2) the heat of fusion, so that this value
can be compared to the unit melting energy calculated by Eq. (27.2). Use either the SI units. Find the
values of the properties needed in these calculations either in this text or in other references. Are the
values close enough to validate Eq. (27.2)?
Solution
: (a) Aluminum properties (from standard sources):
heat of fusion
H
f
= 395,390 J/kg, melting temperature
T
m
= 660°C, density
ρ
= 2700 kg/m
3
, specific heat
C
= 900 J/kg°C.
U
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 Fall '11
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