What was the solution to the last question in the morning session of the 2013 case study on heat transfer through double glass?
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This post was last edited by Pingdan Shi on 2015-6-16 at 11:31. 25. For a single-layer glass window with a glass thickness of 6 mm and a thermal conductivity of 1.4 W/(m·K), the temperatures of the inner and outer surfaces of the glass in winter are 15°C and -25°C respectively. To reduce heat loss, a double-layer glass structure is used; both layers have a thickness of 6 mm, and there is a layer of still air between them with a gap of 10 mm. The temperatures of the glass surfaces in contact with this still air layer are 10°C and -20°C respectively. The thermal conductivity of air is 0.024 W/(m·K). It is assumed that heat conduction in this glass window occurs stably. What is the value of heat loss (w) per square meter for single-layer and double-layer glass windows, among the following options? (A) 9333 ； 60（B）8167 ； 49（C）8167 ； 89（D）9333 ； I can calculate the heat loss for 72 single-glazed units at 9333, but I can’t do it for double-glazed units. Please help me out! Thank you! @zhanghp30 @Higee
For this question, it clearly states that it’s safe to use the temperatures on both sides of the air layer for calculation, and this way you can get an answer. If you try to use the air temperatures on both the inside and outside of the glass, you won’t get an answer. I don’t know either why this happens.
I’ve calculated it many times already, but I can’t get an answer
ZWP997 is correct; choose D. After changing it to double-glazed windows, the temperatures on both sides of the glass (the inner side of the inner layer and the outer side of the outer layer) are unknown, and they will not be the same as the temperatures on the inner and outer sides of single-glazed windows. Additionally, since the convective heat transfer coefficient of air is unknown, it’s also impossible to determine the temperature of the air. Therefore, for this problem, it’s necessary to use the method outlined in ZWP997. The temperatures given in the actual problem represent the temperatures on both sides of the air gap, which clearly indicates that the calculation method is already specified
The air temperatures on the inner and outer sides of the glass are unknown, so the temperature values for both sides applicable to a single-layer structure cannot be used
Single layer: Q/S = t1 – tf / b / lanbuda = (15 + 25) / 0.006 / 1.4 = 9333 W; Double layer: The temperatures of the glass surfaces in contact with the still air layers are 10°C and -20°C respectively, which represent the temperatures on both sides of the air layer. Since it is steady-state heat conduction, the heat flux through the total thickness is the same as that through any single layer. Therefore, Q/S = t1 – tf / b / lanbuda = (10 + 20) / 0.01 / 0.024 = 72 W. There’s an answer! Choose (D) – it’s an easy question to score points on! It is essential to deeply understand the concept of steady-state heat conduction! Hurry up and study hard!*