(1) Steel tube with a thermal conductivity 50 W/(m-K), the outer diameter of 108 mm and the wall thickness of 5 mm, is covered with the three layer of insulation with a thickness of: 25 mm/ A = 0,038 W/(m-K) 35 mm/ A2 = 0,052 WI(m-K) 4 mm /Ag = 0,12 W/(m-K) The inner temperature of tube wall is twi=218°C and the outside surface of the second insulating layer twvs= 76°C. Calculate the all unknown temperature on the contact layers and on the outside surface of the insulation. %3D %3!

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
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Chapter1: Basic Modes Of Heat Transfer
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CONDUCTION – CYLINDRICAL WALLS
(1) Steel tube with a thermal conductivity 50 WI(m-K), the outer diameter of
108 mm and the wall thickness of 5 mm, is covered with the three layer of
insulation with a thickness of:
25 mm/ A = 0,038 W/(m-K)
35 mm/ = 0,052 W/(m-K)
4 mm / Ag = 0,12 W/(m-K)
The inner temperature of tube wall is twi-218°C and the outside surface of
the second insulating layer tws= 76°C.
Calculate the all unknown temperature on the contact layers and on the
outside surface of the insulation.
Transcribed Image Text:CONDUCTION – CYLINDRICAL WALLS (1) Steel tube with a thermal conductivity 50 WI(m-K), the outer diameter of 108 mm and the wall thickness of 5 mm, is covered with the three layer of insulation with a thickness of: 25 mm/ A = 0,038 W/(m-K) 35 mm/ = 0,052 W/(m-K) 4 mm / Ag = 0,12 W/(m-K) The inner temperature of tube wall is twi-218°C and the outside surface of the second insulating layer tws= 76°C. Calculate the all unknown temperature on the contact layers and on the outside surface of the insulation.
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