Assuming a constant flow, the equation governing the transfer of heat energy Q as a function...
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Assuming a constant flow, the equation governing the transfer of heat energy Q as a function of time t through unit area due to a gradient in temperature T over a distance x in a medium with coefficient of thermal conductivity k is given by dQ dT = k- dx dt Figure 1 below shows a sealed, double-glazed unit consisting of two sheets of glass of thickness L enclosing a layer of air also of thickness L. The temperature at the interface of the air and the 1st glass sheet is T1, at the interface of the air and the 2nd glass sheet it is T2, the air inside the room is at temperature Ti, and the air outside is at temperature To. The coefficients of thermal conductivity for glass and air are kg = 0.6 Wm-2K1 and ka = 0.016 Wm-2K1. T; T, T, T. inside k, k, k, outside Figure 1 (a) Assuming a constant flow of heat energy write down an expression of the form A = B = C for the energy flow across the 3 regions: inner sheet of glass; air; outer sheet of glass. Using your answer to part (a) derive two simultaneous equations for the (b) temperatures T, and T2 at the interfaces of the air with the glass. (c) Assuming Ti = 293 K and To = 263 K Solve the equations obtained in part (b) explicitly for T1 and T2. (Note: use the absolute temperature scale of Kelvin for all temperatures in your calculations.) Consider the values of T1 and T2 compared to Ti and To. What does this tell you about the effectiveness of double glazing? (d) Using your answer to part (c) or otherwise, show that the effective coefficient of thermal conductivity ke for this unit as a whole is approximately 3x ka. Assuming a constant flow, the equation governing the transfer of heat energy Q as a function of time t through unit area due to a gradient in temperature T over a distance x in a medium with coefficient of thermal conductivity k is given by dQ dT = k- dx dt Figure 1 below shows a sealed, double-glazed unit consisting of two sheets of glass of thickness L enclosing a layer of air also of thickness L. The temperature at the interface of the air and the 1st glass sheet is T1, at the interface of the air and the 2nd glass sheet it is T2, the air inside the room is at temperature Ti, and the air outside is at temperature To. The coefficients of thermal conductivity for glass and air are kg = 0.6 Wm-2K1 and ka = 0.016 Wm-2K1. T; T, T, T. inside k, k, k, outside Figure 1 (a) Assuming a constant flow of heat energy write down an expression of the form A = B = C for the energy flow across the 3 regions: inner sheet of glass; air; outer sheet of glass. Using your answer to part (a) derive two simultaneous equations for the (b) temperatures T, and T2 at the interfaces of the air with the glass. (c) Assuming Ti = 293 K and To = 263 K Solve the equations obtained in part (b) explicitly for T1 and T2. (Note: use the absolute temperature scale of Kelvin for all temperatures in your calculations.) Consider the values of T1 and T2 compared to Ti and To. What does this tell you about the effectiveness of double glazing? (d) Using your answer to part (c) or otherwise, show that the effective coefficient of thermal conductivity ke for this unit as a whole is approximately 3x ka.
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HIAA heat wrvent at To LL heat curvent I dQ kdT dt a Assuming lons hant flow of heat I Heat current ... View the full answer
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