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Consider a composite wall that includes an 8-mm-thick hardboard siding, 40-mm by 170-mm hardwood studs on 0.65-m centers with glass fiber insulation (paper faced, 28 kg/m3 ), and a 12-mm layer of gypsum (vermiculite) wall board. What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having 10 studs, each 2.5 m high)?

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Answer:

dimensions and materials information that are provided about composite wall (2.5mx6.5m 10 studs each 2.5m high)

to find:wall thermal resistance=??

properties:

Ka=0.094W/m.K

Kb=0.16W/m.K

Kc=0.17W/m.K

Kd=0.038W/m.K

using isothermal surface assumption:

(La/KaAa)=0.008/0.094(0.65x2.5)

=0.0524K/W

(LB/KbAb)=0.13/0.16(0.04x2.5)

=8.125K/W

(Ld/KdAd)=0.13/0.38(0.6x2.5)

=2.243K/W

core's equivalent resistance is:

Req=(1/Rb+1/Rd)⁻¹

=(1/8.125+1/2.243)⁻¹

=1.758K/W

For total resistance we will find sum of

Rtot=Ra+Req+Rc

=1.854K/W

Whereas 10 studs are used so:

Rtotal=(10x1/Rtot)⁻¹

=0.1854K/W

User Evgeniy Labunskiy
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