We have a tube with a diameter of 5 inches that is 1 foot long. The tube then reduces the diameter to 3 inches. According to the
2 answers:
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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 = ...................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