3-20 A double-pane window consists of two 3-mm thick layers of glass separated by an evacuated space.
For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner
surface temperature of the window are to be determined.
Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures
remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant
temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivities
of the glass and air are constant. 4 Heat transfer by radiation is negligible.
Properties The thermal conductivity of the glass is given to be kglass = 0.78 W/m⋅°C.
Analysis Heat cannot be conducted through an evacuated space since the thermal conductivity of vacuum
is zero (no medium to conduct heat) and thus its thermal resistance is zero. Therefore, if radiation is
disregarded, the heat transfer through the window will be zero. Then the answer of this problem is zero
since the problem states to disregard radiation.
Discussion In reality, heat will be transferred between the glasses
by radiation. We do not know the inner surface temperatures of
windows. In order to determine radiation heat resistance we
assume them to be 5°C and 15°C, respectively, and take the
Vacuum
emissivity to be 1. Then individual resistances are
A = (12
. m) × (2 m) = 2.4 m2
1
1
=
= 0.0417 °C/W
h1 A (10 W/m 2 .°C)(2.4 m 2 )
L
0.003 m
R1 = R3 = Rglass = 1 =
= 0.0016 °C/W
k1 A (0.78 W/m.°C)(2.4 m 2 )
Ri = R conv,1 =
R rad =
1
2
εσA(Ts + Tsurr 2 )(Ts + Tsurr )
=
Ri
T∞1
W/m 2 .K 4 )(2.4 m 2 )[288 2 + 278 2 ][288 + 278]K 3
R1
Rrad
R3
1
−8
1(5.67 × 10
= 0.0810 °C/W
Ro
T∞2
1
1
=
= 0.0167 °C/W
2
h2 A (25 W/m .°C)(2.4 m 2 )
= R conv,1 + 2 R1 + R rad + R conv, 2 = 0.0417 + 2(0.0016) + 0.0810 + 0.0167
R o = R conv, 2 =
Rtotal
= 0.1426 °C/W
The steady rate of heat transfer through window glass then becomes
T −T
[24 − ( −5)]° C
Q& = ∞1 ∞ 2 =
= 203 W
01426
Rtotal
.
°C / W
The inner surface temperature of the window glass can be determined from
T −T
&
Q& = ∞1 1 
→ T1 = T∞1 − QR
conv ,1 = 24° C − ( 203 W)(0.0417° C / W) = 15.5° C
Rconv ,1
Similarly, the inner surface temperatures of the glasses are calculated to be 15.2 and -1.2°C (we had
assumed them to be 15 and 5°C when determining the radiation resistance). We can improve the result
obtained by reevaluating the radiation resistance and repeating the calculations.