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2
q  t (  t ), W
r 1 2
ln 2
r
1
For the given sample problem,

0
T 1 = 50 C
0
T 2 = 180 C
r 1 = 203 mm = 0.203 m

λ= 0.04 W/m·K

l = length of the cylinder

q = heat loss per unit length of pipe

q = 80 W/m

____________________

r 2 - ?

2  t (  t )
1
2
r q
r 
q  ln 2  2   (t  t 2 2 r e 
)
1
r 1
1
Hence, inserting the given numbers in the radial heat transfer rate equation from above,
80 = 2π × 0.04 × (180-50) ÷ ln(r 2/0.203)
ln(r 2/0.203) = 2π × 0.04 × (180-50) / 80 = 0.408

Hence, r 2= r 1 × e 0.408

r 2= 0.203 × 1.504 = 0.305 m

Hence, insulation thickness = r 2 – r 1

thickness = 305 – 203 = 102.5 mm

*Some margin should be taken on the insulation thickness because if the conductive heat transfer rate

happens to be higher than the convective heat transfer rate outside the insulation wall, the outer
0
insulation wall temperature will shoot up to higher values than 50 C. Hence conductive heat transfer
rate should be limited to lower values than estimates used in this sample problem. The purpose of this
sample problem is to demonstrate radial heat conduction calculations and practical calculations of

insulation thickness also require consideration of convective heat transfer on the outside of insulation

wall.

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