The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity k i = 0.046 W/m ⋅ K and thickness L i = 50 mm and steel panels, each of thermal conductivity k p = 60 W/m ⋅ K and thickness L p = 3 mm . If the wall separates refrigerated air at T ∞ , i = 4 ° C from ambient air at T ∞ , ρ = 25 ° C, what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as h i = h o = 5 W/m 2 ⋅ K .
The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity k i = 0.046 W/m ⋅ K and thickness L i = 50 mm and steel panels, each of thermal conductivity k p = 60 W/m ⋅ K and thickness L p = 3 mm . If the wall separates refrigerated air at T ∞ , i = 4 ° C from ambient air at T ∞ , ρ = 25 ° C, what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as h i = h o = 5 W/m 2 ⋅ K .
Solution Summary: The author explains the heat gain per unit surface area, the thermal conductivity of the fiber glass insulation, and the interior air temperatures.
The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity
k
i
=
0.046
W/m
⋅
K
and thickness
L
i
=
50
mm
and steel panels, each of thermal conductivity
k
p
=
60
W/m
⋅
K
and thickness
L
p
=
3
mm
.
If the wall separates refrigerated air at
T
∞
,
i
=
4
°
C
from ambient air at
T
∞
,
ρ
=
25
°
C,
what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as
h
i
=
h
o
=
5
W/m
2
⋅
K
.
Q1.
Consider a plane wall (thermal conductivity, k = 0.8 W/mK, and thickness, fb1 = 100 mm) of a
house as shown in Fig. Q1(a). The outer surface of the wall is exposed to solar radiation and has
an absorptivity of a = 0.5 for solar energy, or=600 W/m². The temperature of the interior of
the house is maintained at T1 = 25 °C, while the ambient air temperature outside remains at
T2 = 5 °C. The sky, the ground and the surfaces of the surrounding structures at this location
can be modelled as a surface at an effective temperature of Tsky = 255 K for radiation exchange
on the outer surface. The radiation exchange inside the house is negligible. The convection heat
transfer coefficients on the inner and the outer surfaces of the wall are h₁ = 5 W/m²-K and
/1₂ = 20 W/m².K, respectively. The emissivity of the outer surface is = 0.9.
T1 = 25 °C
Ţ₁
Too1 = 25 °C
T₁
k
100 mm
Fig. Q1(a)
Assuming the heat transfer through the wall to be steady and one-dimensional:
(a) Solve the steady 1D heat…
Q1/ Consider a large plane wall of thickness L=0.03 m. The wall surface at x =0
is insulated, while the surface at x =L is maintained at a temperature of 30°C. The
thermal conductivity of the wall is k=25 W/m °C, and heat is generated in the
wall at a rate of g = 9oe0.5x/L W/m³ Where g, = 8 x 10 W /m². Assuming
steady one-dimensional heat transfer, (a) express the differential equation and the
boundary conditions for heat conduction through the wall, (b) obtain a relation for
the variation of temperature in the wall by solving the differential equation, and (c)
determine the temperature of the insulated surface of the wall.
=
Consider a large plane wall of thickness L=0.3 m, thermal conductivity k = 2.5 W/m.K,
and surface area A = 12 m². The left side of the wall at x=0 is subjected to a net heat
flux of ɖo = 700 W/m² while the temperature at that surface is measured to be T₁ =
80°C. Assuming constant thermal conductivity and no heat generation in the wall, (a)
express the differential equation and the boundary equations for steady one-
dimensional heat conduction through the wall, (b) obtain a relation for the variation of
the temperature in the wall by solving the differential equation, and (c) evaluate the
temperature of the right surface of the wall at x=L.
Ti
до
L
X
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