Respuesta :
Answer:
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 = [tex]\frac{L}{Ka*A}[/tex] ...................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 = [tex]\frac{0.008}{0.094*0.65*2.5}[/tex] = 0.0549 K/W ................2
and resistance for stud is
resistance Rs = [tex]\frac{0.100}{0.16*0.04*2.5}[/tex] = 6.25 K/W ....................3
and resistance for insulation
resistance Ri = [tex]\frac{0.100}{0.038*(0.65 - 0.04)*2.5}[/tex] = 1.7256 K/W .................4
and resistance for wall board
resistance Rg = [tex]\frac{0.012}{0.17*0.65*2.5}[/tex] = 0.4343 K/W .................5
so here stud and insulated are parallel
so resistance = [tex]( Rs^{-1} + Ri^{-1} )^{-1}[/tex]
we get resistance = [tex]( 6.25^{-1} + 1.7256^{-1} )^{-1}[/tex] = 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 = [tex]\frac{1.8414}{10}[/tex]
total resistance = 0.18414 K/W
