A house has a composite wall of wood, fiberglass insulation and plaster board, as indicated in Figure 2.1. The total wall surface area is 300 m²" h = 30 W/m*K Tni = 20°C | h. = 60 W/m³K |Teo --15°C Fiberglass L=10 cm k= 0.04 W/mK Plywood L-2 cm k=0.13 W/mK Plaster board L=1 cm k = 0.15 W/mK Figure 2.1 (i) Represent the composite wall by a thermal circuit, and label all the components in the circuit ii) Determine the inner surface temperature of the house.|

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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A house has a composite wall of wood, fiberglass insulation and plaster board,
as indicated in Figure 2.1. The total wall surface area is 300 m²"
h = 30 W/m°K
Tni = 20°C
| h. = 60 W/m²K
Teo =-15°C
Fiberglass
L=10 cm
Plaster board
k= 0.04 W/mK
L=1 cm
k = 0.15 W/mK
Plywood
L-2 cm
k= 0.13 W/mK
Figure 2.1
(i) Represent the composite wall by a thermal circuit, and label all the components in the circuit
ii) Determine the inner surface temperature of the house.
Transcribed Image Text:A house has a composite wall of wood, fiberglass insulation and plaster board, as indicated in Figure 2.1. The total wall surface area is 300 m²" h = 30 W/m°K Tni = 20°C | h. = 60 W/m²K Teo =-15°C Fiberglass L=10 cm Plaster board k= 0.04 W/mK L=1 cm k = 0.15 W/mK Plywood L-2 cm k= 0.13 W/mK Figure 2.1 (i) Represent the composite wall by a thermal circuit, and label all the components in the circuit ii) Determine the inner surface temperature of the house.
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