Question

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.

Answer 1

total resistance = 0.18414 K/W

Step-by-step 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 =
`(L)/(Ka*A)` ...................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 =
`(0.008)/(0.094*0.65*2.5)` = 0.0549 K/W ................2

and resistance for stud is

resistance Rs =
`(0.100)/(0.16*0.04*2.5)` = 6.25 K/W ....................3

and resistance for insulation

resistance Ri =
`(0.100)/(0.038*(0.65 - 0.04)*2.5)` = 1.7256 K/W .................4

and resistance for wall board

resistance Rg =
`(0.012)/(0.17*0.65*2.5)` = 0.4343 K/W .................5

so here stud and insulated are parallel

so resistance =
`( Rs^(-1) + Ri^(-1) )^(-1)`

we get resistance =
`( 6.25^(-1) + 1.7256^(-1) )^(-1)` = 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 =
`(1.8414)/(10)`

total resistance = 0.18414 K/W