Compare the heat loss (in Watts) from two different windows in a room. One windo

Question

Compare the heat loss (in Watts) from two different windows in a room. One window gas a single sheet of glass 10mm thick whereas the other has two sheets of glass, 5mm each but separated by an air gap of 5mm. The second window is called a thermopane. The windows sizes are 1mx1m. The air temperature in the room is 72 degree F and the outside temperature is 18 degree F. The thermal conductivity of the glass is 0.9 W/m.K and the average thermal conductivity at of air is 0.03 W/m.K. The convection coefficient inside the room is h = 10 W/m^2 K and outside due to a strong wind is h_o = 200 W/m K. Compare the heat losses between these two windows design, assuming that the air in the gap is stagnant, i.e. heat though the air gap is only transmitted by conduction Consider that in the air gap you also have some radiation and that the radiation coefficient is h_r = (T_1 + T_2)(T^2_1 + T^2_2), using the main equation used for radiation the validity of the expression to estimate h_r. Assuming that h_r, almostequalto 4sigmaT^4, estimate and compare how the resistance to heat transfer changes when radiation is also considered in addition to conduction alone. Therefore, you have to estimate the differences in the resistance to the heat transfer considering and without considering radiation in the air gap, and what is the effect in the heat flow?

Explanation / Answer

a)

With sigle sheet of glass 10 mm thick:

Equivalent thermal resistance R = [1 / (hi*A)] + [b / (k*A)] + [1 / (ho*A)]

= [(1 / 10) + (0.01 / 0.9) + (1 / 200)] / (1*1)

= 0.11611 K/W

72 deg F = 22.2 deg C

18 deg F = -7.8 deg C

Heat transfer Q = (T1 - T2) / R

= (22.2 - (-7.8)) / 0.1161

= 258.4 W

With thermopane:

Equivalent Thermal resistance R = [1 / (hi*A)] + 2*[b / (k*A)]glass + [b / (k*A)]air + [1 / (ho*A)]

= [(1 / 10) + 2*(0.005 / 0.9) + (0.005 / 0.03) + (1 / 200)] / (1*1)

= 0.28277 K/W

Heat transfer Q = (22.2 - (-7.8)) / 0.28277

= 106.1 W

b)

For radiation in air gap, heat transfer Q = sigma*A*(T14 - T24) = sigma*A*hr * (T1 - T2)

Hence, hr = (T14 - T24) / (T1 - T2)

= (T12 - T22) (T12 + T22) / (T1 - T2)

= (T1 - T2) (T1 + T2) (T12 + T22) / (T1 - T2)

= (T1 + T2) (T12 + T22)

Hence, expression is valid.

c)

Tavg = (22.2 + (-7.8)) /2 = 7.2 deg C = 280.2 K

hr = 4*(5.67*10-8)*280.24 = 1398 W

Thermal resistance = [(1 / 10) + 2*(0.005 / 0.9) + (0.005 / 0.03) + (1/1398) + (1 / 200)] / (1*1)

= 0.28349 K/W

Hence, resistance changes by only (0.28349 - 0.28277) / 0.28277 * 100 = 0.25 %

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.