(1) Steel tube with a thermal conductivity 50 WI(m-K), the outer diameter of108 mm and the wall thickness of 5 mm, is covered with the three layer ofinsulation with a thickness of:25 mm/ A, = 0,038 WI(m-K)35 mm/ = 0,052 WI(m-K)4 mm /Ag = 0,12 WI(m-K)The inner temperature of tube wall is twi=218°C and the outside surface ofthe second insulating layer tws = 76°C.Calculate the all unknown temperature on the contact layers and on theoutside surface of the insulation.

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) 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 WI(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 tws= 76°C. Calculate the all unknown temperature on the contact layers and on the

) 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 WI(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 tws= 76°C. Calculate the all unknown temperature on the contact layers and on the

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter4: Numerical Analysis Of Heat Conduction

Section: Chapter Questions

Problem 4.54P

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