A storage unit with a height of 2.5 m is made of plastic pieces with thickness...
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A storage unit with a height of 2.5 m is made of plastic pieces with thickness 10 mm. It is 3 m by 4 m in size. The unit has 2 mm-thick glass in the sides, with total area of 3 m 2, and the top has an area of 15 m 2. The thermal conductivity of plastic is k = 0.1 W/mK and of glass is k = 0.26 W/m.K. Assume that the heat transfer coefficient from the inside air to the surfaces is h = 5 W/m2.K and from the outer surfaces to the outside air is h = 50 W/m2.K. The bottom conducts heat to the ground with an additional thermal resistance of R = 0.2 Km 2/.W. The inside is to be kept at 18 oC while the outside air temperature is -5 oC and the ground temperature +2 oC. Calculate: (1) The heat loss to the ground. (ii) The heat loss to the air through the top. (iii) The heat loss to the air through the plastic sides. (iv) The heat loss to the air through the glass. (v) Adding all contributions, give the total heat load. (vi) If you could insulate only one component, which one would you choose? (e) If a layer of insulating material of thickness 30 mm with conductivity k = 0.04 W/m.K is added to the sides, and an additional pane of glass with an air gap of 10 mm between the panes (conductivity of air is 0.026 W/m.K) added to the glass, calculate the new total heat load. A storage unit with a height of 2.5 m is made of plastic pieces with thickness 10 mm. It is 3 m by 4 m in size. The unit has 2 mm-thick glass in the sides, with total area of 3 m 2, and the top has an area of 15 m 2. The thermal conductivity of plastic is k = 0.1 W/mK and of glass is k = 0.26 W/m.K. Assume that the heat transfer coefficient from the inside air to the surfaces is h = 5 W/m2.K and from the outer surfaces to the outside air is h = 50 W/m2.K. The bottom conducts heat to the ground with an additional thermal resistance of R = 0.2 Km 2/.W. The inside is to be kept at 18 oC while the outside air temperature is -5 oC and the ground temperature +2 oC. Calculate: (1) The heat loss to the ground. (ii) The heat loss to the air through the top. (iii) The heat loss to the air through the plastic sides. (iv) The heat loss to the air through the glass. (v) Adding all contributions, give the total heat load. (vi) If you could insulate only one component, which one would you choose? (e) If a layer of insulating material of thickness 30 mm with conductivity k = 0.04 W/m.K is added to the sides, and an additional pane of glass with an air gap of 10 mm between the panes (conductivity of air is 0.026 W/m.K) added to the glass, calculate the new total heat load.
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Loss Calculations for the Storage Unit 1 Heat Loss to the Ground Area of bottom surface Abottom 3m 4... View the full answer
Related Book For
Income Tax Fundamentals 2013
ISBN: 9781285586618
31st Edition
Authors: Gerald E. Whittenburg, Martha Altus Buller, Steven L Gill
Posted Date:
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