Consider a composite wall that includes an 8-mm-thick hardwood siding, 40-mm by 100-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)? assume surfaces normal to the x-direction are isothermal.

total resistance = 0.18414 K/W

Explanation:

given data

length L = 8 mm

siding = 40 mm

siding = 100 mm

studs = 0.65-m

paper faced, 28 kg/m³

gypsum layer = 12-mm

to find out

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)

solution

we will apply here resistance formula that is

resistance =    ...................1

here L is length and Ka is thermal conductivity and A id area

thermal conductivity of hard wood siding = 0.94 W/m-K

and thermal conductivity of hard wood stud = 0.16 W/m-K

and thermal conductivity of glass fiber insulation = 0.038 W/m-K

and thermal conductivity of gypsum wall board  = 0.17 W/m-K

so resistance for wood is

resistance Rw = = 0.0549 K/W      ................2

and resistance for stud is

resistance Rs = = 6.25 K/W      ....................3

and resistance for insulation

resistance Ri = = 1.7256 K/W    .................4

and resistance for wall board

resistance Rg = = 0.4343 K/W    .................5

so here stud and insulated are parallel

so resistance =

we get resistance =   = 1.3522 K/W     ..........................6

so total resistance is

total resistance add equation 2 and equation 5 and 6

total resistance = 0.0549  +  0.4343 + 1.3522

total resistance = 1.8414 K/W

and

studs are 10

so total resistance will be

total resistance =

total resistance = 0.18414 K/W