Required informationA 16-m x 8.5-m section of wall of an industrial furnace burning natural gas is not insulated, and the temperature at theouter surface of this section is measured to be 80°C. The temperature of the furnace room is 30°C, and the combinedconvection and radiation heat transfer coefficient at the surface of the outer furnace is 10 W/mK. it is proposed toinsulate this section of the furnace wall with glass wool insuiation (k = 0:038 W/m K) in order to reduce the heat loss by 90percent.DedAssuming that the outer surface temperature of the metal section still remains at about 110°C, determine the thickness of the insulation thatneeds to be used.

Given data as per question Temperature of surface room=300cTemperature of outer surface of this…

Answered: Required information A 16-m x 8.5-m…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Required information
A 16-m x 8.5-m section of wall of an industrial furnace burning natural gas is not insulated, and the temperature at the
outer surface of this section is measured to be 80°C. The temperature of the furnace room is 30°C, and the combined
convection and radiation heat transfer coefficient at the surface of the outer furnace is 10 W/mK. it is proposed to
insulate this section of the furnace wall with glass wool insuiation (k = 0:038 W/m K) in order to reduce the heat loss by 90
percent.
Ded
Assuming that the outer surface temperature of the metal section still remains at about 110°C, determine the thickness of the insulation that
needs to be used.

Expert Solution

Step 1

Given data as per question

$\begin{array}{l} Temperature of surface room = 30^{0} c \\ Temperature of outer surface of this section = 80^{0} c \\ heat transfer cofficient of surface = 10 \frac{w}{m^{2} k} \\ thermal conductivity of glass wool = 0.038 \frac{w}{m^{2} k} \\ O n o n e s i d e o f m e t a l s e c t i o n t e m p e r t u r e i s 80^{0} c a n d o t h e r s i d e i s 110^{0} c . \\ A r e a o f w a l l = 16 \times 8.5 = 136 m^{2} \\ Considering heat transfer through wall \\ we know that \\ Q = h A (T_{s} - T_{\infty}) \\ = 10 \times 136 (80 - 30) \\ = 68000 W \\ In order to reduce heat loss by 90% ,the new heat transfer \\ rate and thermal resistance must be \\ Q = 68000 \times 0.10 \\ = 6800 w \end{array}$