Module 4: Activity No. 4: W MK Q 439.6 W, T W MK
Module 4: Activity No. 4: W MK Q 439.6 W, T W MK
Module 4: Activity No. 4: W MK Q 439.6 W, T W MK
4
Heat Transfer Exercises Problems
Q
r2
r3
r1
T2 T1
T3 K1
K2
SOLUTION:
r 3=r 2+ t insulation
r 3=0.127 m+ 0.0254 m
r 3=0.1524 m
T 2:
T 1−T 2
Q=
r 2−r 1
4 п K 1 r2 r1
426.7˚ C−T 2
439.6 W =
0.127 m−0.0762 m
W
4 п( 46.14 )(0.127 m)( 0.0762m)
mk
T 2=422.72 ˚ C
T 3:
T 2−T 3
Q=
r 3−r 2
4 п K 2 r3 r2
422.72˚ C−T 3
439.6 W =
0.1524 m−0.127 m
W
4 п( 0.2 )(0.1524 m)(0.127 m)
mk
T 3=193.18 ˚ C
Problem No. 2:
A hollow sphere with inner radius of 8 cm, outer radius of 13 cm and inside surface
W
temp of 430 deg C is made of material with thermal conductivity of 46.15 . If the
mK
outside surface temp is 420 deg C, find the heat loss from the sphere in W/m 2 based on
the inside surface area and again based on the outer surface area.
W
GIVEN:r 1=8 cm=0.08 m,r 2=13 cm=0.13 m,T 1=430 ˚ C , K=46.15 , T =420 ˚ C
mk 2
Solution:
Q
Ai
Ai=4 п r 21 =4 п ¿
T 1−T 2 A i
Q=
r 2−r 1 A i
4 п K 1 r2 r1
Q T 1−T 2
=
A i Ai (r 2−r 1 )
4 п K 1 r2 r1
Q 430 ˚ C−420˚ C
=
Ai 4п¿¿¿
Q W
=14998.75 2
Ai m
Q
:
A0
A0 =4 пr 22=4 п¿
T 1−T 2 A 0
Q=
r 2−r 1 A 0
4 п K 1 r2 r1
Q T 1−T 2
=
A o A 0(r 2−r 1 )
4 п K 1 r2 r1
Q 430 ˚ C−420 ˚ C
=
Ao 4 п ¿ ¿¿
Q W
=5680 2
Ao m